3.1171 \(\int \frac{1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=354 \[ \frac{d (c-3 i d) \left (3 c^2+22 i c d+13 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 f (c-i d)^2 (c+i d)^4 \sqrt{c+d \tan (e+f x)}}+\frac{d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 f (c-i d) (c+i d)^3 (c+d \tan (e+f x))^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f (c-i d)^{5/2}}+\frac{-5 d+i c}{2 a f (c+i d)^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac{1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \]

[Out]

((-I/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[
2]*a^(3/2)*(c - I*d)^(5/2)*f) - 1/(3*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)) + (I
*c - 5*d)/(2*a*(c + I*d)^2*f*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)) + (d*(3*c^2 + (14*I)*c*d +
21*d^2)*Sqrt[a + I*a*Tan[e + f*x]])/(6*a^2*(c - I*d)*(c + I*d)^3*f*(c + d*Tan[e + f*x])^(3/2)) + ((c - (3*I)*
d)*d*(3*c^2 + (22*I)*c*d + 13*d^2)*Sqrt[a + I*a*Tan[e + f*x]])/(6*a^2*(c - I*d)^2*(c + I*d)^4*f*Sqrt[c + d*Tan
[e + f*x]])

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Rubi [A]  time = 1.23827, antiderivative size = 354, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3559, 3596, 3598, 12, 3544, 208} \[ \frac{d (c-3 i d) \left (3 c^2+22 i c d+13 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 f (c-i d)^2 (c+i d)^4 \sqrt{c+d \tan (e+f x)}}+\frac{d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 f (c-i d) (c+i d)^3 (c+d \tan (e+f x))^{3/2}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f (c-i d)^{5/2}}+\frac{-5 d+i c}{2 a f (c+i d)^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac{1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

((-I/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[
2]*a^(3/2)*(c - I*d)^(5/2)*f) - 1/(3*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)) + (I
*c - 5*d)/(2*a*(c + I*d)^2*f*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)) + (d*(3*c^2 + (14*I)*c*d +
21*d^2)*Sqrt[a + I*a*Tan[e + f*x]])/(6*a^2*(c - I*d)*(c + I*d)^3*f*(c + d*Tan[e + f*x])^(3/2)) + ((c - (3*I)*
d)*d*(3*c^2 + (22*I)*c*d + 13*d^2)*Sqrt[a + I*a*Tan[e + f*x]])/(6*a^2*(c - I*d)^2*(c + I*d)^4*f*Sqrt[c + d*Tan
[e + f*x]])

Rule 3559

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

Rule 3596

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

Rule 3598

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}-\frac{\int \frac{-\frac{3}{2} a (i c-3 d)-3 i a d \tan (e+f x)}{\sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx}{3 a^2 (i c-d)}\\ &=-\frac{1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac{i c-5 d}{2 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (-\frac{3}{4} a^2 \left (c^2+6 i c d-21 d^2\right )-3 a^2 (c+5 i d) d \tan (e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx}{3 a^4 (c+i d)^2}\\ &=-\frac{1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac{i c-5 d}{2 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac{d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c+i d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac{2 \int \frac{\sqrt{a+i a \tan (e+f x)} \left (-\frac{3}{8} a^3 \left (3 c^3+15 i c^2 d-37 c d^2+39 i d^3\right )-\frac{3}{4} a^3 d \left (3 c^2+14 i c d+21 d^2\right ) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{9 a^5 (c+i d)^2 \left (c^2+d^2\right )}\\ &=-\frac{1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac{i c-5 d}{2 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac{d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c+i d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{(c-3 i d) d \left (3 c^2+22 i c d+13 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c-i d)^2 (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}-\frac{4 \int -\frac{9 a^4 (c+i d)^4 \sqrt{a+i a \tan (e+f x)}}{16 \sqrt{c+d \tan (e+f x)}} \, dx}{9 a^6 (c+i d)^2 \left (c^2+d^2\right )^2}\\ &=-\frac{1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac{i c-5 d}{2 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac{d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c+i d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{(c-3 i d) d \left (3 c^2+22 i c d+13 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c-i d)^2 (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{4 a^2 (c-i d)^2}\\ &=-\frac{1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac{i c-5 d}{2 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac{d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c+i d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{(c-3 i d) d \left (3 c^2+22 i c d+13 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c-i d)^2 (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}-\frac{i \operatorname{Subst}\left (\int \frac{1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{a+i a \tan (e+f x)}}\right )}{2 (c-i d)^2 f}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} (c-i d)^{5/2} f}-\frac{1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac{i c-5 d}{2 a (c+i d)^2 f \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac{d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c+i d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{(c-3 i d) d \left (3 c^2+22 i c d+13 d^2\right ) \sqrt{a+i a \tan (e+f x)}}{6 a^2 (c-i d)^2 (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 9.10935, size = 803, normalized size = 2.27 \[ \frac{\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \sqrt{\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))} \left (\frac{i (5 c+21 i d) \cos (2 f x)}{12 (c+i d)^4}+\frac{\left (i \cos (e) c^4-3 d \cos (e) c^3+i d \sin (e) c^3+9 i d^2 \cos (e) c^2-3 d^2 \sin (e) c^2+29 d^3 \cos (e) c+9 i d^3 \sin (e) c-10 i d^4 \cos (e)+3 d^4 \sin (e)\right ) \left (\frac{1}{3} \cos (2 e)+\frac{1}{3} i \sin (2 e)\right )}{(c-i d)^2 (c+i d)^4 (c \cos (e)+d \sin (e))}+\frac{\cos (4 f x) \left (\frac{1}{12} i \cos (2 e)+\frac{1}{12} \sin (2 e)\right )}{(c+i d)^3}+\frac{(5 c+21 i d) \sin (2 f x)}{12 (c+i d)^4}+\frac{\left (\frac{1}{12} \cos (2 e)-\frac{1}{12} i \sin (2 e)\right ) \sin (4 f x)}{(c+i d)^3}-\frac{2 \left (\frac{5}{2} \cos (2 e-f x) d^5-\frac{5}{2} \cos (2 e+f x) d^5+\frac{5}{2} i \sin (2 e-f x) d^5-\frac{5}{2} i \sin (2 e+f x) d^5+\frac{13}{2} i c \cos (2 e-f x) d^4-\frac{13}{2} i c \cos (2 e+f x) d^4-\frac{13}{2} c \sin (2 e-f x) d^4+\frac{13}{2} c \sin (2 e+f x) d^4\right )}{3 (c-i d)^2 (c+i d)^4 (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac{\frac{2}{3} \cos (2 e) d^5+\frac{2}{3} i \sin (2 e) d^5}{(c-i d)^2 (c+i d)^4 (c \cos (e+f x)+d \sin (e+f x))^2}\right )}{f (i \tan (e+f x) a+a)^{3/2}}-\frac{i e^{2 i e} \sqrt{e^{i f x}} \log \left (2 \left (e^{i (e+f x)} \sqrt{c-i d}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right ) \sec ^{\frac{3}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{3/2}}{2 \sqrt{2} (c-i d)^{5/2} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} f (i \tan (e+f x) a+a)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

((-I/2)*E^((2*I)*e)*Sqrt[E^(I*f*x)]*Log[2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[
c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])]*Sec[e + f*x]^(3/2)*(Cos[f*x] + I*Sin[f*x])^(
3/2))/(Sqrt[2]*(c - I*d)^(5/2)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]*f
*(a + I*a*Tan[e + f*x])^(3/2)) + (Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] +
d*Sin[e + f*x])]*(((I/12)*(5*c + (21*I)*d)*Cos[2*f*x])/(c + I*d)^4 + ((I*c^4*Cos[e] - 3*c^3*d*Cos[e] + (9*I)*
c^2*d^2*Cos[e] + 29*c*d^3*Cos[e] - (10*I)*d^4*Cos[e] + I*c^3*d*Sin[e] - 3*c^2*d^2*Sin[e] + (9*I)*c*d^3*Sin[e]
+ 3*d^4*Sin[e])*(Cos[2*e]/3 + (I/3)*Sin[2*e]))/((c - I*d)^2*(c + I*d)^4*(c*Cos[e] + d*Sin[e])) + (Cos[4*f*x]*(
(I/12)*Cos[2*e] + Sin[2*e]/12))/(c + I*d)^3 + ((5*c + (21*I)*d)*Sin[2*f*x])/(12*(c + I*d)^4) + ((Cos[2*e]/12 -
(I/12)*Sin[2*e])*Sin[4*f*x])/(c + I*d)^3 + ((2*d^5*Cos[2*e])/3 + ((2*I)/3)*d^5*Sin[2*e])/((c - I*d)^2*(c + I*
d)^4*(c*Cos[e + f*x] + d*Sin[e + f*x])^2) - (2*(((13*I)/2)*c*d^4*Cos[2*e - f*x] + (5*d^5*Cos[2*e - f*x])/2 - (
(13*I)/2)*c*d^4*Cos[2*e + f*x] - (5*d^5*Cos[2*e + f*x])/2 - (13*c*d^4*Sin[2*e - f*x])/2 + ((5*I)/2)*d^5*Sin[2*
e - f*x] + (13*c*d^4*Sin[2*e + f*x])/2 - ((5*I)/2)*d^5*Sin[2*e + f*x]))/(3*(c - I*d)^2*(c + I*d)^4*(c*Cos[e] +
d*Sin[e])*(c*Cos[e + f*x] + d*Sin[e + f*x]))))/(f*(a + I*a*Tan[e + f*x])^(3/2))

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Maple [B]  time = 0.098, size = 7061, normalized size = 20. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 3.16052, size = 3780, normalized size = 10.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

(sqrt(2)*(I*c^5 - c^4*d + 2*I*c^3*d^2 - 2*c^2*d^3 + I*c*d^4 - d^5 + (4*I*c^5 - 4*c^4*d + 56*I*c^3*d^2 + 200*c^
2*d^3 - 204*I*c*d^4 - 52*d^5)*e^(8*I*f*x + 8*I*e) + (13*I*c^5 - 25*c^4*d + 134*I*c^3*d^2 + 314*c^2*d^3 - 87*I*
c*d^4 + 35*d^5)*e^(6*I*f*x + 6*I*e) + (15*I*c^5 - 39*c^4*d + 90*I*c^3*d^2 + 78*c^2*d^3 + 123*I*c*d^4 + 69*d^5)
*e^(4*I*f*x + 4*I*e) + (7*I*c^5 - 19*c^4*d + 14*I*c^3*d^2 - 38*c^2*d^3 + 7*I*c*d^4 - 19*d^5)*e^(2*I*f*x + 2*I*
e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)
)*e^(I*f*x + I*e) + (3*(a^2*c^8 + 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 + 4*a^2*c^2*d^6 + a^2*d^8)*f*e^(8*I*f*x + 8*I*
e) + (6*a^2*c^8 + 12*I*a^2*c^7*d + 12*a^2*c^6*d^2 + 36*I*a^2*c^5*d^3 + 36*I*a^2*c^3*d^5 - 12*a^2*c^2*d^6 + 12*
I*a^2*c*d^7 - 6*a^2*d^8)*f*e^(6*I*f*x + 6*I*e) + (3*a^2*c^8 + 12*I*a^2*c^7*d - 12*a^2*c^6*d^2 + 12*I*a^2*c^5*d
^3 - 30*a^2*c^4*d^4 - 12*I*a^2*c^3*d^5 - 12*a^2*c^2*d^6 - 12*I*a^2*c*d^7 + 3*a^2*d^8)*f*e^(4*I*f*x + 4*I*e))*s
qrt(-I/((2*I*a^3*c^5 + 10*a^3*c^4*d - 20*I*a^3*c^3*d^2 - 20*a^3*c^2*d^3 + 10*I*a^3*c*d^4 + 2*a^3*d^5)*f^2))*lo
g(((2*I*a^2*c^3 + 6*a^2*c^2*d - 6*I*a^2*c*d^2 - 2*a^2*d^3)*f*sqrt(-I/((2*I*a^3*c^5 + 10*a^3*c^4*d - 20*I*a^3*c
^3*d^2 - 20*a^3*c^2*d^3 + 10*I*a^3*c*d^4 + 2*a^3*d^5)*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(2)*sqrt(((c - I*d)*e^(2
*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) +
1)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)) - (3*(a^2*c^8 + 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 + 4*a^2*c^2*d^6 + a^2*d^8
)*f*e^(8*I*f*x + 8*I*e) + (6*a^2*c^8 + 12*I*a^2*c^7*d + 12*a^2*c^6*d^2 + 36*I*a^2*c^5*d^3 + 36*I*a^2*c^3*d^5 -
12*a^2*c^2*d^6 + 12*I*a^2*c*d^7 - 6*a^2*d^8)*f*e^(6*I*f*x + 6*I*e) + (3*a^2*c^8 + 12*I*a^2*c^7*d - 12*a^2*c^6
*d^2 + 12*I*a^2*c^5*d^3 - 30*a^2*c^4*d^4 - 12*I*a^2*c^3*d^5 - 12*a^2*c^2*d^6 - 12*I*a^2*c*d^7 + 3*a^2*d^8)*f*e
^(4*I*f*x + 4*I*e))*sqrt(-I/((2*I*a^3*c^5 + 10*a^3*c^4*d - 20*I*a^3*c^3*d^2 - 20*a^3*c^2*d^3 + 10*I*a^3*c*d^4
+ 2*a^3*d^5)*f^2))*log(((-2*I*a^2*c^3 - 6*a^2*c^2*d + 6*I*a^2*c*d^2 + 2*a^2*d^3)*f*sqrt(-I/((2*I*a^3*c^5 + 10*
a^3*c^4*d - 20*I*a^3*c^3*d^2 - 20*a^3*c^2*d^3 + 10*I*a^3*c*d^4 + 2*a^3*d^5)*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(2
)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*
(e^(2*I*f*x + 2*I*e) + 1)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)))/(12*(a^2*c^8 + 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 + 4
*a^2*c^2*d^6 + a^2*d^8)*f*e^(8*I*f*x + 8*I*e) + (24*a^2*c^8 + 48*I*a^2*c^7*d + 48*a^2*c^6*d^2 + 144*I*a^2*c^5*
d^3 + 144*I*a^2*c^3*d^5 - 48*a^2*c^2*d^6 + 48*I*a^2*c*d^7 - 24*a^2*d^8)*f*e^(6*I*f*x + 6*I*e) + (12*a^2*c^8 +
48*I*a^2*c^7*d - 48*a^2*c^6*d^2 + 48*I*a^2*c^5*d^3 - 120*a^2*c^4*d^4 - 48*I*a^2*c^3*d^5 - 48*a^2*c^2*d^6 - 48*
I*a^2*c*d^7 + 12*a^2*d^8)*f*e^(4*I*f*x + 4*I*e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))**(3/2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out