3.594 \(\int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\)

Optimal. Leaf size=325 \[ \frac {(b (A-B)-a (A+B)) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}-\frac {(b (A-B)-a (A+B)) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}-\frac {(a (A-B)+b (A+B)) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )}+\frac {(a (A-B)+b (A+B)) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )}+\frac {2 a^{5/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d \left (a^2+b^2\right )}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)} \]

[Out]

2*a^(5/2)*(A*b-B*a)*arctan(a^(1/2)*cot(d*x+c)^(1/2)/b^(1/2))/b^(5/2)/(a^2+b^2)/d+2/3*B/b/d/cot(d*x+c)^(3/2)+1/
2*(a*(A-B)+b*(A+B))*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)+1/2*(a*(A-B)+b*(A+B))*arctan(1+2^(
1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)+1/4*(b*(A-B)-a*(A+B))*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(a^
2+b^2)/d*2^(1/2)-1/4*(b*(A-B)-a*(A+B))*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)+2*(A*b-B*
a)/b^2/d/cot(d*x+c)^(1/2)

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Rubi [A]  time = 1.09, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3581, 3609, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac {(b (A-B)-a (A+B)) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}-\frac {(b (A-B)-a (A+B)) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}+\frac {2 a^{5/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} d \left (a^2+b^2\right )}-\frac {(a (A-B)+b (A+B)) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )}+\frac {(a (A-B)+b (A+B)) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a*(A - B) + b*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d)) + ((a*(A - B) + b*
(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) + (2*a^(5/2)*(A*b - a*B)*ArcTan[(Sqrt
[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(b^(5/2)*(a^2 + b^2)*d) + (2*B)/(3*b*d*Cot[c + d*x]^(3/2)) + (2*(A*b - a*B))
/(b^2*d*Sqrt[Cot[c + d*x]]) + ((b*(A - B) - a*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*
Sqrt[2]*(a^2 + b^2)*d) - ((b*(A - B) - a*(A + B))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[
2]*(a^2 + b^2)*d)

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 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*(a + b*Tan[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[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(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 {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx &=\int \frac {B+A \cot (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (b+a \cot (c+d x))} \, dx\\ &=\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\frac {3}{2} (A b-a B)-\frac {3}{2} b B \cot (c+d x)-\frac {3}{2} a B \cot ^2(c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (b+a \cot (c+d x))} \, dx}{3 b}\\ &=\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {4 \int \frac {-\frac {3}{4} \left (a A b-a^2 B+b^2 B\right )-\frac {3}{4} A b^2 \cot (c+d x)-\frac {3}{4} a (A b-a B) \cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{3 b^2}\\ &=\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {4 \int \frac {-\frac {3}{4} b^2 (a A+b B)-\frac {3}{4} b^2 (A b-a B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{3 b^2 \left (a^2+b^2\right )}-\frac {\left (a^3 (A b-a B)\right ) \int \frac {1+\cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{b^2 \left (a^2+b^2\right )}\\ &=\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {8 \operatorname {Subst}\left (\int \frac {\frac {3}{4} b^2 (a A+b B)+\frac {3}{4} b^2 (A b-a B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{3 b^2 \left (a^2+b^2\right ) d}-\frac {\left (a^3 (A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{b^2 \left (a^2+b^2\right ) d}\\ &=\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {\left (2 a^3 (A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{b^2 \left (a^2+b^2\right ) d}-\frac {(b (A-B)-a (A+B)) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac {2 a^{5/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} \left (a^2+b^2\right ) d}+\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {(b (A-B)-a (A+B)) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(b (A-B)-a (A+B)) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {2 a^{5/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} \left (a^2+b^2\right ) d}+\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {(b (A-B)-a (A+B)) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {(b (A-B)-a (A+B)) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a (A-B)+b (A+B)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}\\ &=-\frac {(a (A-B)+b (A+B)) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 a^{5/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{5/2} \left (a^2+b^2\right ) d}+\frac {2 B}{3 b d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b-a B)}{b^2 d \sqrt {\cot (c+d x)}}+\frac {(b (A-B)-a (A+B)) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {(b (A-B)-a (A+B)) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 1.12, size = 272, normalized size = 0.84 \[ \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {6 \sqrt {2} (a (A-B)+b (A+B)) \left (\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )}{a^2+b^2}-\frac {3 \sqrt {2} (a (A+B)+b (B-A)) \left (\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )}{a^2+b^2}+\frac {24 a^{5/2} (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )}+\frac {24 (A b-a B) \sqrt {\tan (c+d x)}}{b^2}+\frac {8 B \tan ^{\frac {3}{2}}(c+d x)}{b}\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((6*Sqrt[2]*(a*(A - B) + b*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*
x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]))/(a^2 + b^2) + (24*a^(5/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[
Tan[c + d*x]])/Sqrt[a]])/(b^(5/2)*(a^2 + b^2)) - (3*Sqrt[2]*(b*(-A + B) + a*(A + B))*(Log[1 - Sqrt[2]*Sqrt[Tan
[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]))/(a^2 + b^2) + (24*(A*b - a*B
)*Sqrt[Tan[c + d*x]])/b^2 + (8*B*Tan[c + d*x]^(3/2))/b))/(12*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((A+B*tan(d*x+c))/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 3.69, size = 12107, normalized size = 37.25 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [A]  time = 0.64, size = 262, normalized size = 0.81 \[ -\frac {\frac {24 \, {\left (B a^{4} - A a^{3} b\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{2} b^{2} + b^{4}\right )} \sqrt {a b}} - \frac {8 \, {\left (B b - \frac {3 \, {\left (B a - A b\right )}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{b^{2}} - \frac {3 \, {\left (2 \, \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )}}{a^{2} + b^{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(24*(B*a^4 - A*a^3*b)*arctan(a/(sqrt(a*b)*sqrt(tan(d*x + c))))/((a^2*b^2 + b^4)*sqrt(a*b)) - 8*(B*b - 3*
(B*a - A*b)/tan(d*x + c))*tan(d*x + c)^(3/2)/b^2 - 3*(2*sqrt(2)*((A - B)*a + (A + B)*b)*arctan(1/2*sqrt(2)*(sq
rt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*((A - B)*a + (A + B)*b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d
*x + c)))) + sqrt(2)*((A + B)*a - (A - B)*b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - sqrt(2)*((
A + B)*a - (A - B)*b)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/(a^2 + b^2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)**(5/2)/(a+b*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))/((a + b*tan(c + d*x))*cot(c + d*x)**(5/2)), x)

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