3.1166 \(\int \frac{1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=349 \[ \frac{d \left (65 i c^2 d+15 c^3-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 f (c-i d) (c+i d)^4 \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 f (-d+i c)^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f (c-i d)^{3/2}}+\frac{-17 d+5 i c}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \]

[Out]

((-I/4)*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^(5/2)*(c - I*d)^(3/2)*f) - 1/(5*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]) + ((5*
I)*c - 17*d)/(30*a*(c + I*d)^2*f*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]) + (15*c^2 + (70*I)*c*d
- 151*d^2)/(60*a^2*(I*c - d)^3*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) + (d*(15*c^3 + (65*I)*c
^2*d - 117*c*d^2 + (317*I)*d^3)*Sqrt[a + I*a*Tan[e + f*x]])/(60*a^3*(c - I*d)*(c + I*d)^4*f*Sqrt[c + d*Tan[e +
f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.24015, antiderivative size = 349, 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} \[ \frac{d \left (65 i c^2 d+15 c^3-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 f (c-i d) (c+i d)^4 \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 f (-d+i c)^3 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f (c-i d)^{3/2}}+\frac{-17 d+5 i c}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/4)*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^(5/2)*(c - I*d)^(3/2)*f) - 1/(5*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]) + ((5*
I)*c - 17*d)/(30*a*(c + I*d)^2*f*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]]) + (15*c^2 + (70*I)*c*d
- 151*d^2)/(60*a^2*(I*c - d)^3*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) + (d*(15*c^3 + (65*I)*c
^2*d - 117*c*d^2 + (317*I)*d^3)*Sqrt[a + I*a*Tan[e + f*x]])/(60*a^3*(c - I*d)*(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))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{-\frac{1}{2} a (5 i c-11 d)-3 i a d \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx}{5 a^2 (i c-d)}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{-\frac{1}{4} a^2 \left (15 c^2+50 i c d-83 d^2\right )-a^2 (5 c+17 i d) d \tan (e+f x)}{\sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx}{15 a^4 (c+i d)^2}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (\frac{1}{8} a^3 \left (15 i c^3-75 c^2 d-185 i c d^2+317 d^3\right )+\frac{1}{4} a^3 d \left (15 i c^2-70 c d-151 i d^2\right ) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{15 a^6 (i c-d)^3}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}-\frac{2 \int \frac{15 i a^4 (c+i d)^4 \sqrt{a+i a \tan (e+f x)}}{16 \sqrt{c+d \tan (e+f x)}} \, dx}{15 a^7 (i c-d)^3 \left (c^2+d^2\right )}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 (c-i d) (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}{8 a^3 (c-i d)}\\ &=-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}+\frac{\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 )}{4 a (i c+d) 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 )}{4 \sqrt{2} a^{5/2} (c-i d)^{3/2} f}-\frac{1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)}}+\frac{5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c+d \tan (e+f x)}}+\frac{15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}+\frac{d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt{a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 8.9528, size = 788, normalized size = 2.26 \[ \frac{\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac{\left (17 c^2+77 i c d-126 d^2\right ) \left (-\frac{\sin (e)}{60}+\frac{1}{60} i \cos (e)\right ) \cos (2 f x)}{(c+i d)^4}+\frac{\left (17 c^2+77 i c d-126 d^2\right ) \left (\frac{\cos (e)}{60}+\frac{1}{60} i \sin (e)\right ) \sin (2 f x)}{(c+i d)^4}+\frac{\left (\frac{1}{120} \cos (3 e)+\frac{1}{120} i \sin (3 e)\right ) \left (91 i c^2 d^2 \sin (e)-109 c^2 d^2 \cos (e)+23 c^3 d \sin (e)+91 i c^3 d \cos (e)+23 c^4 \cos (e)-109 c d^3 \sin (e)+223 i c d^3 \cos (e)+223 i d^4 \sin (e)+240 d^4 \cos (e)\right )}{(c-i d) (c+i d)^4 (-i c \cos (e)-i d \sin (e))}+\frac{2 \left (\frac{1}{2} i d^5 \sin (3 e-f x)-\frac{1}{2} i d^5 \sin (3 e+f x)+\frac{1}{2} d^5 \cos (3 e-f x)-\frac{1}{2} d^5 \cos (3 e+f x)\right )}{(c-i d) (c+i d)^4 (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac{(7 c+16 i d) \left (\frac{\sin (e)}{60}+\frac{1}{60} i \cos (e)\right ) \cos (4 f x)}{(c+i d)^3}+\frac{\left (\frac{1}{40} \sin (3 e)+\frac{1}{40} i \cos (3 e)\right ) \cos (6 f x)}{(c+i d)^2}+\frac{(7 c+16 i d) \left (\frac{\cos (e)}{60}-\frac{1}{60} i \sin (e)\right ) \sin (4 f x)}{(c+i d)^3}+\frac{\left (\frac{1}{40} \cos (3 e)-\frac{1}{40} i \sin (3 e)\right ) \sin (6 f x)}{(c+i d)^2}\right ) \sqrt{\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))}}{f (a+i a \tan (e+f x))^{5/2}}-\frac{i e^{3 i e} \sqrt{e^{i f x}} \sec ^{\frac{5}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{5/2} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{4 \sqrt{2} f (c-i d)^{3/2} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} (a+i a \tan (e+f x))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I/4)*E^((3*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]^(5/2)*(Cos[f*x] + I*Sin[f*x])^(
5/2))/(Sqrt[2]*(c - I*d)^(3/2)*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])^(5/2)) + (Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] +
d*Sin[e + f*x])]*(((17*c^2 + (77*I)*c*d - 126*d^2)*Cos[2*f*x]*((I/60)*Cos[e] - Sin[e]/60))/(c + I*d)^4 + ((7*
c + (16*I)*d)*Cos[4*f*x]*((I/60)*Cos[e] + Sin[e]/60))/(c + I*d)^3 + ((23*c^4*Cos[e] + (91*I)*c^3*d*Cos[e] - 10
9*c^2*d^2*Cos[e] + (223*I)*c*d^3*Cos[e] + 240*d^4*Cos[e] + 23*c^3*d*Sin[e] + (91*I)*c^2*d^2*Sin[e] - 109*c*d^3
*Sin[e] + (223*I)*d^4*Sin[e])*(Cos[3*e]/120 + (I/120)*Sin[3*e]))/((c - I*d)*(c + I*d)^4*((-I)*c*Cos[e] - I*d*S
in[e])) + (Cos[6*f*x]*((I/40)*Cos[3*e] + Sin[3*e]/40))/(c + I*d)^2 + ((17*c^2 + (77*I)*c*d - 126*d^2)*(Cos[e]/
60 + (I/60)*Sin[e])*Sin[2*f*x])/(c + I*d)^4 + ((7*c + (16*I)*d)*(Cos[e]/60 - (I/60)*Sin[e])*Sin[4*f*x])/(c + I
*d)^3 + ((Cos[3*e]/40 - (I/40)*Sin[3*e])*Sin[6*f*x])/(c + I*d)^2 + (2*((d^5*Cos[3*e - f*x])/2 - (d^5*Cos[3*e +
f*x])/2 + (I/2)*d^5*Sin[3*e - f*x] - (I/2)*d^5*Sin[3*e + f*x]))/((c - I*d)*(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])^(5/2))

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Maple [B]  time = 0.107, size = 7870, normalized size = 22.6 \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))^(5/2)/(c+d*tan(f*x+e))^(3/2),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))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.2659, size = 2824, normalized size = 8.09 \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))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-(sqrt(2)*(-3*I*c^4 + 6*c^3*d + 6*c*d^3 + 3*I*d^4 + (-23*I*c^4 + 68*c^3*d + 18*I*c^2*d^2 + 332*c*d^3 - 463*I*d
^4)*e^(8*I*f*x + 8*I*e) + (-57*I*c^4 + 200*c^3*d + 178*I*c^2*d^2 + 464*c*d^3 - 269*I*d^4)*e^(6*I*f*x + 6*I*e)
+ (-48*I*c^4 + 172*c^3*d + 172*I*c^2*d^2 + 172*c*d^3 + 220*I*d^4)*e^(4*I*f*x + 4*I*e) + (-17*I*c^4 + 46*c^3*d
+ 12*I*c^2*d^2 + 46*c*d^3 + 29*I*d^4)*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) - ((30*a^3*c^6 + 60*I*a^3*c^5*d + 30*
a^3*c^4*d^2 + 120*I*a^3*c^3*d^3 - 30*a^3*c^2*d^4 + 60*I*a^3*c*d^5 - 30*a^3*d^6)*f*e^(8*I*f*x + 8*I*e) + (30*a^
3*c^6 + 120*I*a^3*c^5*d - 150*a^3*c^4*d^2 - 150*a^3*c^2*d^4 - 120*I*a^3*c*d^5 + 30*a^3*d^6)*f*e^(6*I*f*x + 6*I
*e))*sqrt(I/((-8*I*a^5*c^3 - 24*a^5*c^2*d + 24*I*a^5*c*d^2 + 8*a^5*d^3)*f^2))*log(((4*I*a^3*c^2 + 8*a^3*c*d -
4*I*a^3*d^2)*f*sqrt(I/((-8*I*a^5*c^3 - 24*a^5*c^2*d + 24*I*a^5*c*d^2 + 8*a^5*d^3)*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)) + ((30*a^3*c^6 + 60*I*a^3*c^5*d + 30*a^3*c^
4*d^2 + 120*I*a^3*c^3*d^3 - 30*a^3*c^2*d^4 + 60*I*a^3*c*d^5 - 30*a^3*d^6)*f*e^(8*I*f*x + 8*I*e) + (30*a^3*c^6
+ 120*I*a^3*c^5*d - 150*a^3*c^4*d^2 - 150*a^3*c^2*d^4 - 120*I*a^3*c*d^5 + 30*a^3*d^6)*f*e^(6*I*f*x + 6*I*e))*s
qrt(I/((-8*I*a^5*c^3 - 24*a^5*c^2*d + 24*I*a^5*c*d^2 + 8*a^5*d^3)*f^2))*log(((-4*I*a^3*c^2 - 8*a^3*c*d + 4*I*a
^3*d^2)*f*sqrt(I/((-8*I*a^5*c^3 - 24*a^5*c^2*d + 24*I*a^5*c*d^2 + 8*a^5*d^3)*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)))/((120*a^3*c^6 + 240*I*a^3*c^5*d + 120*a^3*c^4*d
^2 + 480*I*a^3*c^3*d^3 - 120*a^3*c^2*d^4 + 240*I*a^3*c*d^5 - 120*a^3*d^6)*f*e^(8*I*f*x + 8*I*e) + (120*a^3*c^6
+ 480*I*a^3*c^5*d - 600*a^3*c^4*d^2 - 600*a^3*c^2*d^4 - 480*I*a^3*c*d^5 + 120*a^3*d^6)*f*e^(6*I*f*x + 6*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))**(5/2)/(c+d*tan(f*x+e))**(3/2),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))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out