∫ tan5 _2 (c+dx)__ (a+b tan(c+dx))3 dx

3.602 \(\int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=390 \[ \frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\sqrt {a} \left (a^4+18 a^2 b^2-15 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{3/2} d \left (a^2+b^2\right )^3} \]

[Out]

-1/2*(a-b)*(a^2+4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)-1/2*(a-b)*(a^2+4*a*b+b^2)

*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)-1/4*(a+b)*(a^2-4*a*b+b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1

/2)+tan(d*x+c))/(a^2+b^2)^3/d*2^(1/2)+1/4*(a+b)*(a^2-4*a*b+b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2

+b^2)^3/d*2^(1/2)+1/4*(a^4+18*a^2*b^2-15*b^4)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))*a^(1/2)/b^(3/2)/(a^2+b^

2)^3/d-1/2*a^2*tan(d*x+c)^(1/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^2+1/4*a*(a^2+9*b^2)*tan(d*x+c)^(1/2)/b/(a^2+b^2

)^2/d/(a+b*tan(d*x+c))

________________________________________________________________________________________

Rubi [A] time = 0.87, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3565, 3649, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac {\sqrt {a} \left (18 a^2 b^2+a^4-15 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{3/2} d \left (a^2+b^2\right )^3}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Tan[c + d*x]^(5/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

((a - b)*(a^2 + 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - ((a - b)*(a^2

+ 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (Sqrt[a]*(a^4 + 18*a^2*b^2

- 15*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(4*b^(3/2)*(a^2 + b^2)^3*d) - ((a + b)*(a^2 - 4*a*b +

b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) + ((a + b)*(a^2 - 4*a*b

+ b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - (a^2*Sqrt[Tan[c + d*x

]])/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (a*(a^2 + 9*b^2)*Sqrt[Tan[c + d*x]])/(4*b*(a^2 + b^2)^2*d*(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 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si

mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D

ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(

m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3

*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x

], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt

Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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, -

Rubi steps

\begin {align*} \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {\frac {a^2}{2}-2 a b \tan (c+d x)+\frac {1}{2} \left (a^2+4 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {\frac {1}{4} a^2 \left (a^2-7 b^2\right )-2 a b \left (a^2-b^2\right ) \tan (c+d x)+\frac {1}{4} a^2 \left (a^2+9 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a b \left (a^2+b^2\right )^2}\\ &=-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {-2 a b^2 \left (3 a^2-b^2\right )-2 a^2 b \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a b \left (a^2+b^2\right )^3}+\frac {\left (a \left (a^4+18 a^2 b^2-15 b^4\right )\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 b \left (a^2+b^2\right )^3}\\ &=-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\operatorname {Subst}\left (\int \frac {-2 a b^2 \left (3 a^2-b^2\right )-2 a^2 b \left (a^2-3 b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a b \left (a^2+b^2\right )^3 d}+\frac {\left (a \left (a^4+18 a^2 b^2-15 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 b \left (a^2+b^2\right )^3 d}\\ &=-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\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 )^3 d}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\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 )^3 d}+\frac {\left (a \left (a^4+18 a^2 b^2-15 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 b \left (a^2+b^2\right )^3 d}\\ &=\frac {\sqrt {a} \left (a^4+18 a^2 b^2-15 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{3/2} \left (a^2+b^2\right )^3 d}-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\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 )^3 d}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\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 )^3 d}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\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 )^3 d}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\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 )^3 d}\\ &=\frac {\sqrt {a} \left (a^4+18 a^2 b^2-15 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{3/2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 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 )^3 d}+\frac {(a+b) \left (a^2-4 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 )^3 d}-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\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 )^3 d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\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 )^3 d}\\ &=\frac {(a-b) \left (a^2+4 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 )^3 d}-\frac {(a-b) \left (a^2+4 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 )^3 d}+\frac {\sqrt {a} \left (a^4+18 a^2 b^2-15 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{3/2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 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 )^3 d}+\frac {(a+b) \left (a^2-4 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 )^3 d}-\frac {a^2 \sqrt {\tan (c+d x)}}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (a^2+9 b^2\right ) \sqrt {\tan (c+d x)}}{4 b \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] time = 4.22, size = 325, normalized size = 0.83 \[ \frac {-\frac {a (a+b \tan (c+d x)) \left (2 a \sqrt {b} \left (a^2+b^2\right )^2 \sqrt {\tan (c+d x)}-a \sqrt {b} \left (a^2+9 b^2\right ) \left (a^2+b^2\right ) \sqrt {\tan (c+d x)}-2 b^{3/2} \left (a^2+b^2\right )^2 \tan ^{\frac {3}{2}}(c+d x)-(a+b \tan (c+d x)) \left (\sqrt {a} \left (a^4+18 a^2 b^2-15 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )-4 (-1)^{3/4} b^{3/2} (a+i b)^3 \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-4 \sqrt [4]{-1} b^{3/2} (b+i a)^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right )\right )}{\left (a^2+b^2\right )^2}-2 b^{5/2} \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))+2 b^{7/2} \tan ^{\frac {7}{2}}(c+d x)}{4 a b^{3/2} d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Tan[c + d*x]^(5/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

(2*b^(7/2)*Tan[c + d*x]^(7/2) - 2*b^(5/2)*Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x]) - (a*(a + b*Tan[c + d*x])*(2

*a*Sqrt[b]*(a^2 + b^2)^2*Sqrt[Tan[c + d*x]] - a*Sqrt[b]*(a^2 + b^2)*(a^2 + 9*b^2)*Sqrt[Tan[c + d*x]] - 2*b^(3/

2)*(a^2 + b^2)^2*Tan[c + d*x]^(3/2) - (-4*(-1)^(3/4)*(a + I*b)^3*b^(3/2)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]

+ Sqrt[a]*(a^4 + 18*a^2*b^2 - 15*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] - 4*(-1)^(1/4)*b^(3/2)*(I*

a + b)^3*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])*(a + b*Tan[c + d*x])))/(a^2 + b^2)^2)/(4*a*b^(3/2)*(a^2 + b^2

)*d*(a + b*Tan[c + d*x])^2)

________________________________________________________________________________________

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(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [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(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [B] time = 0.31, size = 900, normalized size = 2.31 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^3,x)

[Out]

1/4/d*a^5/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)+5/2/d*a^3*b^2/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x

+c)^(3/2)+9/4/d*a/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(3/2)*b^4-1/4/d*a^6/b/(a^2+b^2)^3/(a+b*tan(d*x+c))

^2*tan(d*x+c)^(1/2)+3/2/d*a^4*b/(a^2+b^2)^3/(a+b*tan(d*x+c))^2*tan(d*x+c)^(1/2)+7/4/d*a^2/(a^2+b^2)^3/(a+b*tan

(d*x+c))^2*b^3*tan(d*x+c)^(1/2)+1/4/d*a^5/b/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))+9/2

/d*a^3*b/(a^2+b^2)^3/(a*b)^(1/2)*arctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))-15/4/d*a/(a^2+b^2)^3*b^3/(a*b)^(1/2)*a

rctan(tan(d*x+c)^(1/2)*b/(a*b)^(1/2))-3/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b+1/2/d

/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3-3/2/d/(a^2+b^2)^3*2^(1/2)*arctan(-1+2^(1/2)*tan(d*

x+c)^(1/2))*a^2*b+1/2/d/(a^2+b^2)^3*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^3-3/4/d/(a^2+b^2)^3*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*b+1/4/d/(a^2+b^2)^3*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)))*b^3-1/4/d/(a^2+b^2)^

3*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^3+3/4/d/(a^2+b

^2)^3*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*b^2-1/2/d/

(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+3/2/d/(a^2+b^2)^3*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+

c)^(1/2))*a*b^2-1/2/d/(a^2+b^2)^3*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+3/2/d/(a^2+b^2)^3*2^(1/2)*ar

ctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2

________________________________________________________________________________________

maxima [A] time = 0.51, size = 408, normalized size = 1.05 \[ \frac {\frac {{\left (a^{5} + 18 \, a^{3} b^{2} - 15 \, a b^{4}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (a^{3} b + 9 \, a b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} - {\left (a^{4} - 7 \, a^{2} b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5} + {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )}}{4 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((a^5 + 18*a^3*b^2 - 15*a*b^4)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^

7)*sqrt(a*b)) - (2*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))

) + 2*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*

(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(a^3 - 3*a^2*b -

3*a*b^2 + b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + ((a^

3*b + 9*a*b^3)*tan(d*x + c)^(3/2) - (a^4 - 7*a^2*b^2)*sqrt(tan(d*x + c)))/(a^6*b + 2*a^4*b^3 + a^2*b^5 + (a^4*

b^3 + 2*a^2*b^5 + b^7)*tan(d*x + c)^2 + 2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*tan(d*x + c)))/d

________________________________________________________________________________________

mupad [B] time = 20.84, size = 12531, normalized size = 32.13 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(5/2)/(a + b*tan(c + d*x))^3,x)

[Out]

atan(((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2

*d^2)))^(1/2)*((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 1

5*a^4*b^2*d^2)))^(1/2)*((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^

2*20i + 15*a^4*b^2*d^2)))^(1/2)*(((832*a*b^22*d^4 + 5952*a^3*b^20*d^4 + 17664*a^5*b^18*d^4 + 26880*a^7*b^16*d^

4 + 18816*a^9*b^14*d^4 - 2688*a^11*b^12*d^4 - 16128*a^13*b^10*d^4 - 13056*a^15*b^8*d^4 - 4800*a^17*b^6*d^4 - 7

04*a^19*b^4*d^4)/(b^17*d^5 + a^16*b*d^5 + 8*a^2*b^15*d^5 + 28*a^4*b^13*d^5 + 56*a^6*b^11*d^5 + 70*a^8*b^9*d^5

+ 56*a^10*b^7*d^5 + 28*a^12*b^5*d^5 + 8*a^14*b^3*d^5) - (tan(c + d*x)^(1/2)*(-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5

*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*(512*b^26*d^4 + 4608*a^2*b

^24*d^4 + 17920*a^4*b^22*d^4 + 38400*a^6*b^20*d^4 + 46080*a^8*b^18*d^4 + 21504*a^10*b^16*d^4 - 21504*a^12*b^14

*d^4 - 46080*a^14*b^12*d^4 - 38400*a^16*b^10*d^4 - 17920*a^18*b^8*d^4 - 4608*a^20*b^6*d^4 - 512*a^22*b^4*d^4))

/(b^17*d^4 + a^16*b*d^4 + 8*a^2*b^15*d^4 + 28*a^4*b^13*d^4 + 56*a^6*b^11*d^4 + 70*a^8*b^9*d^4 + 56*a^10*b^7*d^

4 + 28*a^12*b^5*d^4 + 8*a^14*b^3*d^4))*(-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d

^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2) + (tan(c + d*x)^(1/2)*(8*a^19*b*d^2 - 1472*a*b^19*d^2 + 776*a^3

*b^17*d^2 + 11328*a^5*b^15*d^2 + 10208*a^7*b^13*d^2 - 5056*a^9*b^11*d^2 - 5328*a^11*b^9*d^2 + 4032*a^13*b^7*d^

2 + 3552*a^15*b^5*d^2 + 384*a^17*b^3*d^2))/(b^17*d^4 + a^16*b*d^4 + 8*a^2*b^15*d^4 + 28*a^4*b^13*d^4 + 56*a^6*

b^11*d^4 + 70*a^8*b^9*d^4 + 56*a^10*b^7*d^4 + 28*a^12*b^5*d^4 + 8*a^14*b^3*d^4)) - (10*a^16*b*d^2 - 2398*a^2*b

^15*d^2 + 5238*a^4*b^13*d^2 + 7386*a^6*b^11*d^2 - 8322*a^8*b^9*d^2 - 5498*a^10*b^7*d^2 + 2946*a^12*b^5*d^2 + 3

82*a^14*b^3*d^2)/(b^17*d^5 + a^16*b*d^5 + 8*a^2*b^15*d^5 + 28*a^4*b^13*d^5 + 56*a^6*b^11*d^5 + 70*a^8*b^9*d^5

+ 56*a^10*b^7*d^5 + 28*a^12*b^5*d^5 + 8*a^14*b^3*d^5)) + (tan(c + d*x)^(1/2)*(a^14 - 32*b^14 + 97*a^2*b^12 - 2

082*a^4*b^10 + 3631*a^6*b^8 - 2300*a^8*b^6 + 79*a^10*b^4 + 30*a^12*b^2))/(b^17*d^4 + a^16*b*d^4 + 8*a^2*b^15*d

^4 + 28*a^4*b^13*d^4 + 56*a^6*b^11*d^4 + 70*a^8*b^9*d^4 + 56*a^10*b^7*d^4 + 28*a^12*b^5*d^4 + 8*a^14*b^3*d^4))

*1i - (-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2

*d^2)))^(1/2)*((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 1

5*a^4*b^2*d^2)))^(1/2)*((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^

2*20i + 15*a^4*b^2*d^2)))^(1/2)*(((832*a*b^22*d^4 + 5952*a^3*b^20*d^4 + 17664*a^5*b^18*d^4 + 26880*a^7*b^16*d^

4 + 18816*a^9*b^14*d^4 - 2688*a^11*b^12*d^4 - 16128*a^13*b^10*d^4 - 13056*a^15*b^8*d^4 - 4800*a^17*b^6*d^4 - 7

04*a^19*b^4*d^4)/(b^17*d^5 + a^16*b*d^5 + 8*a^2*b^15*d^5 + 28*a^4*b^13*d^5 + 56*a^6*b^11*d^5 + 70*a^8*b^9*d^5

+ 56*a^10*b^7*d^5 + 28*a^12*b^5*d^5 + 8*a^14*b^3*d^5) + (tan(c + d*x)^(1/2)*(-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5

*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2)*(512*b^26*d^4 + 4608*a^2*b

^24*d^4 + 17920*a^4*b^22*d^4 + 38400*a^6*b^20*d^4 + 46080*a^8*b^18*d^4 + 21504*a^10*b^16*d^4 - 21504*a^12*b^14

*d^4 - 46080*a^14*b^12*d^4 - 38400*a^16*b^10*d^4 - 17920*a^18*b^8*d^4 - 4608*a^20*b^6*d^4 - 512*a^22*b^4*d^4))

/(b^17*d^4 + a^16*b*d^4 + 8*a^2*b^15*d^4 + 28*a^4*b^13*d^4 + 56*a^6*b^11*d^4 + 70*a^8*b^9*d^4 + 56*a^10*b^7*d^

4 + 28*a^12*b^5*d^4 + 8*a^14*b^3*d^4))*(-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d

^2 - a^3*b^3*d^2*20i + 15*a^4*b^2*d^2)))^(1/2) - (tan(c + d*x)^(1/2)*(8*a^19*b*d^2 - 1472*a*b^19*d^2 + 776*a^3

*b^17*d^2 + 11328*a^5*b^15*d^2 + 10208*a^7*b^13*d^2 - 5056*a^9*b^11*d^2 - 5328*a^11*b^9*d^2 + 4032*a^13*b^7*d^

2 + 3552*a^15*b^5*d^2 + 384*a^17*b^3*d^2))/(b^17*d^4 + a^16*b*d^4 + 8*a^2*b^15*d^4 + 28*a^4*b^13*d^4 + 56*a^6*

b^11*d^4 + 70*a^8*b^9*d^4 + 56*a^10*b^7*d^4 + 28*a^12*b^5*d^4 + 8*a^14*b^3*d^4)) - (10*a^16*b*d^2 - 2398*a^2*b

^15*d^2 + 5238*a^4*b^13*d^2 + 7386*a^6*b^11*d^2 - 8322*a^8*b^9*d^2 - 5498*a^10*b^7*d^2 + 2946*a^12*b^5*d^2 + 3

82*a^14*b^3*d^2)/(b^17*d^5 + a^16*b*d^5 + 8*a^2*b^15*d^5 + 28*a^4*b^13*d^5 + 56*a^6*b^11*d^5 + 70*a^8*b^9*d^5

+ 56*a^10*b^7*d^5 + 28*a^12*b^5*d^5 + 8*a^14*b^3*d^5)) - (tan(c + d*x)^(1/2)*(a^14 - 32*b^14 + 97*a^2*b^12 - 2

082*a^4*b^10 + 3631*a^6*b^8 - 2300*a^8*b^6 + 79*a^10*b^4 + 30*a^12*b^2))/(b^17*d^4 + a^16*b*d^4 + 8*a^2*b^15*d

^4 + 28*a^4*b^13*d^4 + 56*a^6*b^11*d^4 + 70*a^8*b^9*d^4 + 56*a^10*b^7*d^4 + 28*a^12*b^5*d^4 + 8*a^14*b^3*d^4))

*1i)/((-1i/(4*(b^6*d^2 - a^6*d^2 + a*b^5*d^2*6i + a^5*b*d^2*6i - 15*a^2*b^4*d^2 - a^3*b^3*d^2*20i + 15*a^4*b^2