3.598 \(\int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=397 \[ \frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}+\frac {b^2}{a d \left (a^2+b^2\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac {2 a^2+5 b^2}{3 a^2 d \left (a^2+b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}+\frac {b^{7/2} \left (9 a^2+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{7/2} d \left (a^2+b^2\right )^2}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 d \left (a^2+b^2\right ) \sqrt {\tan (c+d x)}} \]

[Out]

b^(7/2)*(9*a^2+5*b^2)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/a^(7/2)/(a^2+b^2)^2/d-1/2*(a^2-2*a*b-b^2)*arcta
n(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)-1/2*(a^2-2*a*b-b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a
^2+b^2)^2/d*2^(1/2)+1/4*(a^2+2*a*b-b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^2/d*2^(1/2)-1/4*(a
^2+2*a*b-b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^2/d*2^(1/2)+b*(4*a^2+5*b^2)/a^3/(a^2+b^2)/d/
tan(d*x+c)^(1/2)+1/3*(-2*a^2-5*b^2)/a^2/(a^2+b^2)/d/tan(d*x+c)^(3/2)+b^2/a/(a^2+b^2)/d/tan(d*x+c)^(3/2)/(a+b*t
an(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 1.00, antiderivative size = 397, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3569, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac {b^{7/2} \left (9 a^2+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{7/2} d \left (a^2+b^2\right )^2}+\frac {b^2}{a d \left (a^2+b^2\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}-\frac {2 a^2+5 b^2}{3 a^2 d \left (a^2+b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 d \left (a^2+b^2\right ) \sqrt {\tan (c+d x)}}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^2),x]

[Out]

((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*d) - ((a^2 - 2*a*b - b^2)*
ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*d) + (b^(7/2)*(9*a^2 + 5*b^2)*ArcTan[(Sqrt[b]*S
qrt[Tan[c + d*x]])/Sqrt[a]])/(a^(7/2)*(a^2 + b^2)^2*d) + ((a^2 + 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x
]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) - ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Ta
n[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^2*d) - (2*a^2 + 5*b^2)/(3*a^2*(a^2 + b^2)*d*Tan[c + d*x]^(3/2)) + (b*(4*a^
2 + 5*b^2))/(a^3*(a^2 + b^2)*d*Sqrt[Tan[c + d*x]]) + b^2/(a*(a^2 + b^2)*d*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*
x]))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2} \, dx &=\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac {\int \frac {\frac {1}{2} \left (2 a^2+5 b^2\right )-a b \tan (c+d x)+\frac {5}{2} b^2 \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac {2 a^2+5 b^2}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac {2 \int \frac {\frac {3}{4} b \left (4 a^2+5 b^2\right )+\frac {3}{2} a^3 \tan (c+d x)+\frac {3}{4} b \left (2 a^2+5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx}{3 a^2 \left (a^2+b^2\right )}\\ &=-\frac {2 a^2+5 b^2}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 \left (a^2+b^2\right ) d \sqrt {\tan (c+d x)}}+\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac {4 \int \frac {-\frac {3}{8} \left (2 a^4-4 a^2 b^2-5 b^4\right )+\frac {3}{4} a^3 b \tan (c+d x)+\frac {3}{8} b^2 \left (4 a^2+5 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{3 a^3 \left (a^2+b^2\right )}\\ &=-\frac {2 a^2+5 b^2}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 \left (a^2+b^2\right ) d \sqrt {\tan (c+d x)}}+\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac {4 \int \frac {-\frac {3}{4} a^3 \left (a^2-b^2\right )+\frac {3}{2} a^4 b \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{3 a^3 \left (a^2+b^2\right )^2}+\frac {\left (b^4 \left (9 a^2+5 b^2\right )\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a^3 \left (a^2+b^2\right )^2}\\ &=-\frac {2 a^2+5 b^2}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 \left (a^2+b^2\right ) d \sqrt {\tan (c+d x)}}+\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}+\frac {8 \operatorname {Subst}\left (\int \frac {-\frac {3}{4} a^3 \left (a^2-b^2\right )+\frac {3}{2} a^4 b x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{3 a^3 \left (a^2+b^2\right )^2 d}+\frac {\left (b^4 \left (9 a^2+5 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{2 a^3 \left (a^2+b^2\right )^2 d}\\ &=-\frac {2 a^2+5 b^2}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 \left (a^2+b^2\right ) d \sqrt {\tan (c+d x)}}+\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac {\left (a^2-2 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac {\left (b^4 \left (9 a^2+5 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 \left (a^2+b^2\right )^2 d}\\ &=\frac {b^{7/2} \left (9 a^2+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{7/2} \left (a^2+b^2\right )^2 d}-\frac {2 a^2+5 b^2}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 \left (a^2+b^2\right ) d \sqrt {\tan (c+d x)}}+\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac {\left (a^2-2 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}-\frac {\left (a^2-2 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}\\ &=\frac {b^{7/2} \left (9 a^2+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{7/2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {2 a^2+5 b^2}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 \left (a^2+b^2\right ) d \sqrt {\tan (c+d x)}}+\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}-\frac {\left (a^2-2 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}\\ &=\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {b^{7/2} \left (9 a^2+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{7/2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {2 a^2+5 b^2}{3 a^2 \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \left (4 a^2+5 b^2\right )}{a^3 \left (a^2+b^2\right ) d \sqrt {\tan (c+d x)}}+\frac {b^2}{a \left (a^2+b^2\right ) d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 4.41, size = 230, normalized size = 0.58 \[ \frac {-\frac {2 a^2+5 b^2}{a \tan ^{\frac {3}{2}}(c+d x)}+\frac {3 b \left (4 a^2+5 b^2\right )}{a^2 \sqrt {\tan (c+d x)}}+\frac {3 \left (\sqrt [4]{-1} a^{7/2} (a+i b)^2 \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt [4]{-1} a^{7/2} (a-i b)^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+b^{7/2} \left (9 a^2+5 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )\right )}{a^{5/2} \left (a^2+b^2\right )}+\frac {3 b^2}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}}{3 a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^2),x]

[Out]

((3*((-1)^(1/4)*a^(7/2)*(a + I*b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + b^(7/2)*(9*a^2 + 5*b^2)*ArcTan[(Sq
rt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] + (-1)^(1/4)*a^(7/2)*(a - I*b)^2*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]))/(
a^(5/2)*(a^2 + b^2)) - (2*a^2 + 5*b^2)/(a*Tan[c + d*x]^(3/2)) + (3*b*(4*a^2 + 5*b^2))/(a^2*Sqrt[Tan[c + d*x]])
+ (3*b^2)/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])))/(3*a*(a^2 + b^2)*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} \tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((b*tan(d*x + c) + a)^2*tan(d*x + c)^(5/2)), x)

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maple [A]  time = 0.23, size = 595, normalized size = 1.50 \[ \frac {b^{4} \left (\sqrt {\tan }\left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{2} a \left (a +b \tan \left (d x +c \right )\right )}+\frac {b^{6} \left (\sqrt {\tan }\left (d x +c \right )\right )}{d \,a^{3} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {9 b^{4} \arctan \left (\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) b}{\sqrt {a b}}\right )}{d \left (a^{2}+b^{2}\right )^{2} a \sqrt {a b}}+\frac {5 b^{6} \arctan \left (\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) b}{\sqrt {a b}}\right )}{d \,a^{3} \left (a^{2}+b^{2}\right )^{2} \sqrt {a b}}-\frac {2}{3 d \,a^{2} \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {4 b}{d \,a^{3} \sqrt {\tan \left (d x +c \right )}}-\frac {\sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right ) a^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right ) b^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right ) a^{2}}{4 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right ) b^{2}}{4 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right ) a^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right ) b^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {a b \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {a b \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d \left (a^{2}+b^{2}\right )^{2}}+\frac {a b \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{2 d \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x)

[Out]

1/d*b^4/(a^2+b^2)^2/a*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))+1/d*b^6/a^3/(a^2+b^2)^2*tan(d*x+c)^(1/2)/(a+b*tan(d*x+
c))+9/d*b^4/(a^2+b^2)^2/a/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))+5/d*b^6/a^3/(a^2+b^2)^2/(a*b)^(1/
2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))-2/3/d/a^2/tan(d*x+c)^(3/2)+4/d/a^3*b/tan(d*x+c)^(1/2)-1/2/d/(a^2+b^2
)^2*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^2+1/2/d/(a^2+b^2)^2*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/
2))*b^2-1/4/d/(a^2+b^2)^2*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d
*x+c)))*a^2+1/4/d/(a^2+b^2)^2*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+t
an(d*x+c)))*b^2-1/2/d/(a^2+b^2)^2*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2+1/2/d/(a^2+b^2)^2*2^(1/2)*arc
tan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^2+1/d/(a^2+b^2)^2*a*b*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))+1/d/(a^2+b
^2)^2*a*b*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+1/2/d/(a^2+b^2)^2*a*b*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1
/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))

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maxima [A]  time = 0.62, size = 342, normalized size = 0.86 \[ \frac {\frac {12 \, {\left (9 \, a^{2} b^{4} + 5 \, b^{6}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a b}} - \frac {4 \, {\left (2 \, a^{4} + 2 \, a^{2} b^{2} - 3 \, {\left (4 \, a^{2} b^{2} + 5 \, b^{4}\right )} \tan \left (d x + c\right )^{2} - 10 \, {\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )\right )}}{{\left (a^{5} b + a^{3} b^{3}\right )} \tan \left (d x + c\right )^{\frac {5}{2}} + {\left (a^{6} + a^{4} b^{2}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}} - \frac {3 \, {\left (2 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(12*(9*a^2*b^4 + 5*b^6)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^7 + 2*a^5*b^2 + a^3*b^4)*sqrt(a*b)) -
4*(2*a^4 + 2*a^2*b^2 - 3*(4*a^2*b^2 + 5*b^4)*tan(d*x + c)^2 - 10*(a^3*b + a*b^3)*tan(d*x + c))/((a^5*b + a^3*b
^3)*tan(d*x + c)^(5/2) + (a^6 + a^4*b^2)*tan(d*x + c)^(3/2)) - 3*(2*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sqr
t(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(t
an(d*x + c)))) + sqrt(2)*(a^2 + 2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*(a^2
+ 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^4 + 2*a^2*b^2 + b^4))/d

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mupad [B]  time = 9.64, size = 6886, normalized size = 17.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(c + d*x)^(5/2)*(a + b*tan(c + d*x))^2),x)

[Out]

((10*b*tan(c + d*x))/(3*a^2) - 2/(3*a) + (tan(c + d*x)^2*(5*b^4 + 4*a^2*b^2))/(a^3*(a^2 + b^2)))/(a*d*tan(c +
d*x)^(3/2) + b*d*tan(c + d*x)^(5/2)) - atan(((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*
b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(
a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(tan(c + d*x)^(1/2)*(3200*a^22*b^28*d
^7 + 33920*a^24*b^26*d^7 + 158208*a^26*b^24*d^7 + 425536*a^28*b^22*d^7 + 727296*a^30*b^20*d^7 + 820672*a^32*b^
18*d^7 + 615936*a^34*b^16*d^7 + 304256*a^36*b^14*d^7 + 98432*a^38*b^12*d^7 + 22016*a^40*b^10*d^7 + 3072*a^42*b
^8*d^7 - 704*a^44*b^6*d^7 - 512*a^46*b^4*d^7 - 64*a^48*b^2*d^7) - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i -
a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(1280*a^24*b^28*d^8 + 13824*a^26*b^26*d^8 + 66944*a^28*b^24*d^8 + 190848
*a^30*b^22*d^8 + 352640*a^32*b^20*d^8 + 435840*a^34*b^18*d^8 + 354048*a^36*b^16*d^8 + 169728*a^38*b^14*d^8 + 2
4576*a^40*b^12*d^8 - 21760*a^42*b^10*d^8 - 13440*a^44*b^8*d^8 - 2176*a^46*b^6*d^8 + 384*a^48*b^4*d^8 + 128*a^5
0*b^2*d^8 + tan(c + d*x)^(1/2)*(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/
2)*(512*a^27*b^27*d^9 + 5120*a^29*b^25*d^9 + 22528*a^31*b^23*d^9 + 56320*a^33*b^21*d^9 + 84480*a^35*b^19*d^9 +
67584*a^37*b^17*d^9 - 67584*a^41*b^13*d^9 - 84480*a^43*b^11*d^9 - 56320*a^45*b^9*d^9 - 22528*a^47*b^7*d^9 - 5
120*a^49*b^5*d^9 - 512*a^51*b^3*d^9))) - 800*a^21*b^27*d^6 - 2080*a^23*b^25*d^6 + 12928*a^25*b^23*d^6 + 78464*
a^27*b^21*d^6 + 183616*a^29*b^19*d^6 + 238400*a^31*b^17*d^6 + 184960*a^33*b^15*d^6 + 84608*a^35*b^13*d^6 + 207
04*a^37*b^11*d^6 + 2016*a^39*b^9*d^6) - tan(c + d*x)^(1/2)*(9472*a^31*b^15*d^5 - 3040*a^23*b^23*d^5 - 9056*a^2
5*b^21*d^5 - 12352*a^27*b^19*d^5 - 4256*a^29*b^17*d^5 - 400*a^21*b^25*d^5 + 13760*a^33*b^13*d^5 + 7744*a^35*b^
11*d^5 + 1968*a^37*b^9*d^5 + 224*a^39*b^7*d^5 + 32*a^41*b^5*d^5))*1i - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*
4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b
^2*d^2)))^(1/2)*(12928*a^25*b^23*d^6 - 800*a^21*b^27*d^6 - 2080*a^23*b^25*d^6 - (-1i/(4*(a^4*d^2 + b^4*d^2 + a
*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(tan(c + d*x)^(1/2)*(3200*a^22*b^28*d^7 + 33920*a^24*b^26*
d^7 + 158208*a^26*b^24*d^7 + 425536*a^28*b^22*d^7 + 727296*a^30*b^20*d^7 + 820672*a^32*b^18*d^7 + 615936*a^34*
b^16*d^7 + 304256*a^36*b^14*d^7 + 98432*a^38*b^12*d^7 + 22016*a^40*b^10*d^7 + 3072*a^42*b^8*d^7 - 704*a^44*b^6
*d^7 - 512*a^46*b^4*d^7 - 64*a^48*b^2*d^7) + (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*
b^2*d^2)))^(1/2)*(1280*a^24*b^28*d^8 + 13824*a^26*b^26*d^8 + 66944*a^28*b^24*d^8 + 190848*a^30*b^22*d^8 + 3526
40*a^32*b^20*d^8 + 435840*a^34*b^18*d^8 + 354048*a^36*b^16*d^8 + 169728*a^38*b^14*d^8 + 24576*a^40*b^12*d^8 -
21760*a^42*b^10*d^8 - 13440*a^44*b^8*d^8 - 2176*a^46*b^6*d^8 + 384*a^48*b^4*d^8 + 128*a^50*b^2*d^8 - tan(c + d
*x)^(1/2)*(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(512*a^27*b^27*d^9
+ 5120*a^29*b^25*d^9 + 22528*a^31*b^23*d^9 + 56320*a^33*b^21*d^9 + 84480*a^35*b^19*d^9 + 67584*a^37*b^17*d^9
- 67584*a^41*b^13*d^9 - 84480*a^43*b^11*d^9 - 56320*a^45*b^9*d^9 - 22528*a^47*b^7*d^9 - 5120*a^49*b^5*d^9 - 51
2*a^51*b^3*d^9))) + 78464*a^27*b^21*d^6 + 183616*a^29*b^19*d^6 + 238400*a^31*b^17*d^6 + 184960*a^33*b^15*d^6 +
84608*a^35*b^13*d^6 + 20704*a^37*b^11*d^6 + 2016*a^39*b^9*d^6) + tan(c + d*x)^(1/2)*(9472*a^31*b^15*d^5 - 304
0*a^23*b^23*d^5 - 9056*a^25*b^21*d^5 - 12352*a^27*b^19*d^5 - 4256*a^29*b^17*d^5 - 400*a^21*b^25*d^5 + 13760*a^
33*b^13*d^5 + 7744*a^35*b^11*d^5 + 1968*a^37*b^9*d^5 + 224*a^39*b^7*d^5 + 32*a^41*b^5*d^5))*1i)/(160*a^24*b^20
*d^4 - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b
^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(12928*a^25*b^23*d^6 - 800*a^21*b^27*d^6 - 2080*
a^23*b^25*d^6 - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(tan(c + d*x
)^(1/2)*(3200*a^22*b^28*d^7 + 33920*a^24*b^26*d^7 + 158208*a^26*b^24*d^7 + 425536*a^28*b^22*d^7 + 727296*a^30*
b^20*d^7 + 820672*a^32*b^18*d^7 + 615936*a^34*b^16*d^7 + 304256*a^36*b^14*d^7 + 98432*a^38*b^12*d^7 + 22016*a^
40*b^10*d^7 + 3072*a^42*b^8*d^7 - 704*a^44*b^6*d^7 - 512*a^46*b^4*d^7 - 64*a^48*b^2*d^7) + (-1i/(4*(a^4*d^2 +
b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(1280*a^24*b^28*d^8 + 13824*a^26*b^26*d^8 + 669
44*a^28*b^24*d^8 + 190848*a^30*b^22*d^8 + 352640*a^32*b^20*d^8 + 435840*a^34*b^18*d^8 + 354048*a^36*b^16*d^8 +
169728*a^38*b^14*d^8 + 24576*a^40*b^12*d^8 - 21760*a^42*b^10*d^8 - 13440*a^44*b^8*d^8 - 2176*a^46*b^6*d^8 + 3
84*a^48*b^4*d^8 + 128*a^50*b^2*d^8 - tan(c + d*x)^(1/2)*(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*
4i - 6*a^2*b^2*d^2)))^(1/2)*(512*a^27*b^27*d^9 + 5120*a^29*b^25*d^9 + 22528*a^31*b^23*d^9 + 56320*a^33*b^21*d^
9 + 84480*a^35*b^19*d^9 + 67584*a^37*b^17*d^9 - 67584*a^41*b^13*d^9 - 84480*a^43*b^11*d^9 - 56320*a^45*b^9*d^9
- 22528*a^47*b^7*d^9 - 5120*a^49*b^5*d^9 - 512*a^51*b^3*d^9))) + 78464*a^27*b^21*d^6 + 183616*a^29*b^19*d^6 +
238400*a^31*b^17*d^6 + 184960*a^33*b^15*d^6 + 84608*a^35*b^13*d^6 + 20704*a^37*b^11*d^6 + 2016*a^39*b^9*d^6)
+ tan(c + d*x)^(1/2)*(9472*a^31*b^15*d^5 - 3040*a^23*b^23*d^5 - 9056*a^25*b^21*d^5 - 12352*a^27*b^19*d^5 - 425
6*a^29*b^17*d^5 - 400*a^21*b^25*d^5 + 13760*a^33*b^13*d^5 + 7744*a^35*b^11*d^5 + 1968*a^37*b^9*d^5 + 224*a^39*
b^7*d^5 + 32*a^41*b^5*d^5)) - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2
)*((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d
^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(tan(c + d*x)^(1/2)*(3200*a^22*b^28*d^7 + 33920*a^24
*b^26*d^7 + 158208*a^26*b^24*d^7 + 425536*a^28*b^22*d^7 + 727296*a^30*b^20*d^7 + 820672*a^32*b^18*d^7 + 615936
*a^34*b^16*d^7 + 304256*a^36*b^14*d^7 + 98432*a^38*b^12*d^7 + 22016*a^40*b^10*d^7 + 3072*a^42*b^8*d^7 - 704*a^
44*b^6*d^7 - 512*a^46*b^4*d^7 - 64*a^48*b^2*d^7) - (-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i -
6*a^2*b^2*d^2)))^(1/2)*(1280*a^24*b^28*d^8 + 13824*a^26*b^26*d^8 + 66944*a^28*b^24*d^8 + 190848*a^30*b^22*d^8
+ 352640*a^32*b^20*d^8 + 435840*a^34*b^18*d^8 + 354048*a^36*b^16*d^8 + 169728*a^38*b^14*d^8 + 24576*a^40*b^12*
d^8 - 21760*a^42*b^10*d^8 - 13440*a^44*b^8*d^8 - 2176*a^46*b^6*d^8 + 384*a^48*b^4*d^8 + 128*a^50*b^2*d^8 + tan
(c + d*x)^(1/2)*(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*(512*a^27*b^
27*d^9 + 5120*a^29*b^25*d^9 + 22528*a^31*b^23*d^9 + 56320*a^33*b^21*d^9 + 84480*a^35*b^19*d^9 + 67584*a^37*b^1
7*d^9 - 67584*a^41*b^13*d^9 - 84480*a^43*b^11*d^9 - 56320*a^45*b^9*d^9 - 22528*a^47*b^7*d^9 - 5120*a^49*b^5*d^
9 - 512*a^51*b^3*d^9))) - 800*a^21*b^27*d^6 - 2080*a^23*b^25*d^6 + 12928*a^25*b^23*d^6 + 78464*a^27*b^21*d^6 +
183616*a^29*b^19*d^6 + 238400*a^31*b^17*d^6 + 184960*a^33*b^15*d^6 + 84608*a^35*b^13*d^6 + 20704*a^37*b^11*d^
6 + 2016*a^39*b^9*d^6) - tan(c + d*x)^(1/2)*(9472*a^31*b^15*d^5 - 3040*a^23*b^23*d^5 - 9056*a^25*b^21*d^5 - 12
352*a^27*b^19*d^5 - 4256*a^29*b^17*d^5 - 400*a^21*b^25*d^5 + 13760*a^33*b^13*d^5 + 7744*a^35*b^11*d^5 + 1968*a
^37*b^9*d^5 + 224*a^39*b^7*d^5 + 32*a^41*b^5*d^5)) + 1088*a^26*b^18*d^4 + 3040*a^28*b^16*d^4 + 4480*a^30*b^14*
d^4 + 3680*a^32*b^12*d^4 + 1600*a^34*b^10*d^4 + 288*a^36*b^8*d^4))*(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i -
a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*2i + (log(80*a^24*b^20*d^4 - ((((-1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*
d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*(12928*a^25*b^23*d^6 - 800*a^21*b^27*d^6 - 2080*a^23*b^25*d^6 - ((-
1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*(tan(c + d*x)^(1/2)*(3200*a^22
*b^28*d^7 + 33920*a^24*b^26*d^7 + 158208*a^26*b^24*d^7 + 425536*a^28*b^22*d^7 + 727296*a^30*b^20*d^7 + 820672*
a^32*b^18*d^7 + 615936*a^34*b^16*d^7 + 304256*a^36*b^14*d^7 + 98432*a^38*b^12*d^7 + 22016*a^40*b^10*d^7 + 3072
*a^42*b^8*d^7 - 704*a^44*b^6*d^7 - 512*a^46*b^4*d^7 - 64*a^48*b^2*d^7) + ((-1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b
^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2)*(1280*a^24*b^28*d^8 + 13824*a^26*b^26*d^8 + 66944*a^28*b^24*d^8
+ 190848*a^30*b^22*d^8 + 352640*a^32*b^20*d^8 + 435840*a^34*b^18*d^8 + 354048*a^36*b^16*d^8 + 169728*a^38*b^14
*d^8 + 24576*a^40*b^12*d^8 - 21760*a^42*b^10*d^8 - 13440*a^44*b^8*d^8 - 2176*a^46*b^6*d^8 + 384*a^48*b^4*d^8 +
128*a^50*b^2*d^8 - (tan(c + d*x)^(1/2)*(-1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2
*6i))^(1/2)*(512*a^27*b^27*d^9 + 5120*a^29*b^25*d^9 + 22528*a^31*b^23*d^9 + 56320*a^33*b^21*d^9 + 84480*a^35*b
^19*d^9 + 67584*a^37*b^17*d^9 - 67584*a^41*b^13*d^9 - 84480*a^43*b^11*d^9 - 56320*a^45*b^9*d^9 - 22528*a^47*b^
7*d^9 - 5120*a^49*b^5*d^9 - 512*a^51*b^3*d^9))/2))/2))/2 + 78464*a^27*b^21*d^6 + 183616*a^29*b^19*d^6 + 238400
*a^31*b^17*d^6 + 184960*a^33*b^15*d^6 + 84608*a^35*b^13*d^6 + 20704*a^37*b^11*d^6 + 2016*a^39*b^9*d^6))/2 + ta
n(c + d*x)^(1/2)*(9472*a^31*b^15*d^5 - 3040*a^23*b^23*d^5 - 9056*a^25*b^21*d^5 - 12352*a^27*b^19*d^5 - 4256*a^
29*b^17*d^5 - 400*a^21*b^25*d^5 + 13760*a^33*b^13*d^5 + 7744*a^35*b^11*d^5 + 1968*a^37*b^9*d^5 + 224*a^39*b^7*
d^5 + 32*a^41*b^5*d^5))*(-1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2))/2 +
544*a^26*b^18*d^4 + 1520*a^28*b^16*d^4 + 2240*a^30*b^14*d^4 + 1840*a^32*b^12*d^4 + 800*a^34*b^10*d^4 + 144*a^
36*b^8*d^4)*(-1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2))/2 - log(80*a^24
*b^20*d^4 - (-1/(4*(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i)))^(1/2)*((-1/(4*(a^4
*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i)))^(1/2)*((-1/(4*(a^4*d^2*1i + b^4*d^2*1i +
4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i)))^(1/2)*(tan(c + d*x)^(1/2)*(3200*a^22*b^28*d^7 + 33920*a^24*b^26*
d^7 + 158208*a^26*b^24*d^7 + 425536*a^28*b^22*d^7 + 727296*a^30*b^20*d^7 + 820672*a^32*b^18*d^7 + 615936*a^34*
b^16*d^7 + 304256*a^36*b^14*d^7 + 98432*a^38*b^12*d^7 + 22016*a^40*b^10*d^7 + 3072*a^42*b^8*d^7 - 704*a^44*b^6
*d^7 - 512*a^46*b^4*d^7 - 64*a^48*b^2*d^7) - (-1/(4*(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2
*b^2*d^2*6i)))^(1/2)*(1280*a^24*b^28*d^8 + 13824*a^26*b^26*d^8 + 66944*a^28*b^24*d^8 + 190848*a^30*b^22*d^8 +
352640*a^32*b^20*d^8 + 435840*a^34*b^18*d^8 + 354048*a^36*b^16*d^8 + 169728*a^38*b^14*d^8 + 24576*a^40*b^12*d^
8 - 21760*a^42*b^10*d^8 - 13440*a^44*b^8*d^8 - 2176*a^46*b^6*d^8 + 384*a^48*b^4*d^8 + 128*a^50*b^2*d^8 + tan(c
+ d*x)^(1/2)*(-1/(4*(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i)))^(1/2)*(512*a^27*
b^27*d^9 + 5120*a^29*b^25*d^9 + 22528*a^31*b^23*d^9 + 56320*a^33*b^21*d^9 + 84480*a^35*b^19*d^9 + 67584*a^37*b
^17*d^9 - 67584*a^41*b^13*d^9 - 84480*a^43*b^11*d^9 - 56320*a^45*b^9*d^9 - 22528*a^47*b^7*d^9 - 5120*a^49*b^5*
d^9 - 512*a^51*b^3*d^9))) - 800*a^21*b^27*d^6 - 2080*a^23*b^25*d^6 + 12928*a^25*b^23*d^6 + 78464*a^27*b^21*d^6
+ 183616*a^29*b^19*d^6 + 238400*a^31*b^17*d^6 + 184960*a^33*b^15*d^6 + 84608*a^35*b^13*d^6 + 20704*a^37*b^11*
d^6 + 2016*a^39*b^9*d^6) - tan(c + d*x)^(1/2)*(9472*a^31*b^15*d^5 - 3040*a^23*b^23*d^5 - 9056*a^25*b^21*d^5 -
12352*a^27*b^19*d^5 - 4256*a^29*b^17*d^5 - 400*a^21*b^25*d^5 + 13760*a^33*b^13*d^5 + 7744*a^35*b^11*d^5 + 1968
*a^37*b^9*d^5 + 224*a^39*b^7*d^5 + 32*a^41*b^5*d^5)) + 544*a^26*b^18*d^4 + 1520*a^28*b^16*d^4 + 2240*a^30*b^14
*d^4 + 1840*a^32*b^12*d^4 + 800*a^34*b^10*d^4 + 144*a^36*b^8*d^4)*(-1/(4*(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^
2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i)))^(1/2) + (atan((b^11*tan(c + d*x)^(1/2)*(-a^7*b^7)^(5/2)*1250i - a^25*tan(c
+ d*x)^(1/2)*(-a^7*b^7)^(3/2)*12i - a^10*b*tan(c + d*x)^(1/2)*(-a^7*b^7)^(5/2)*1040i + a^2*b^9*tan(c + d*x)^(
1/2)*(-a^7*b^7)^(5/2)*9000i + a^4*b^7*tan(c + d*x)^(1/2)*(-a^7*b^7)^(5/2)*24300i + a^6*b^5*tan(c + d*x)^(1/2)*
(-a^7*b^7)^(5/2)*29160i + a^8*b^3*tan(c + d*x)^(1/2)*(-a^7*b^7)^(5/2)*12722i - a^19*b^6*tan(c + d*x)^(1/2)*(-a
^7*b^7)^(3/2)*144i - a^21*b^4*tan(c + d*x)^(1/2)*(-a^7*b^7)^(3/2)*1298i - a^23*b^2*tan(c + d*x)^(1/2)*(-a^7*b^
7)^(3/2)*8i + a^34*b^5*tan(c + d*x)^(1/2)*(-a^7*b^7)^(1/2)*8i + a^36*b^3*tan(c + d*x)^(1/2)*(-a^7*b^7)^(1/2)*2
i)/(1250*a^18*b^28 + 9000*a^20*b^26 + 24300*a^22*b^24 + 29160*a^24*b^22 + 12722*a^26*b^20 - 1040*a^28*b^18 + 1
44*a^30*b^16 + 1298*a^32*b^14 + 8*a^34*b^12 + 12*a^36*b^10 + 8*a^38*b^8 + 2*a^40*b^6))*(9*a^2 + 5*b^2)*(-a^7*b
^7)^(1/2)*1i)/(a^11*d + a^7*b^4*d + 2*a^9*b^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{2} \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(5/2)/(a+b*tan(d*x+c))**2,x)

[Out]

Integral(1/((a + b*tan(c + d*x))**2*tan(c + d*x)**(5/2)), x)

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