3.1155 \(\int \frac{(a+i a \tan (e+f x))^{5/2}}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=200 \[ -\frac{\sqrt [4]{-1} a^{5/2} (c+5 i d) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}-\frac{4 i \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f \sqrt{c-i d}} \]

[Out]

-(((-1)^(1/4)*a^(5/2)*(c + (5*I)*d)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c +
d*Tan[e + f*x]])])/(d^(3/2)*f)) - ((4*I)*Sqrt[2]*a^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(S
qrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*f) - (a^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[
e + f*x]])/(d*f)

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Rubi [A]  time = 0.667043, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3556, 3601, 3544, 208, 3599, 63, 217, 206} \[ -\frac{\sqrt [4]{-1} a^{5/2} (c+5 i d) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}-\frac{4 i \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((-1)^(1/4)*a^(5/2)*(c + (5*I)*d)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c +
d*Tan[e + f*x]])])/(d^(3/2)*f)) - ((4*I)*Sqrt[2]*a^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(S
qrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*f) - (a^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[
e + f*x]])/(d*f)

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3601

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

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]

Rule 3599

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^{5/2}}{\sqrt{c+d \tan (e+f x)}} \, dx &=-\frac{a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{a \int \frac{\sqrt{a+i a \tan (e+f x)} \left (\frac{1}{2} a (i c+3 d)+\frac{1}{2} a (c+5 i d) \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{d}\\ &=-\frac{a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\left (4 a^2\right ) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{(a (i c-5 d)) \int \frac{(a-i a \tan (e+f x)) \sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 d}\\ &=-\frac{a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}-\frac{\left (8 i a^4\right ) \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 )}{f}+\frac{\left (a^3 (i c-5 d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 d f}\\ &=-\frac{4 i \sqrt{2} a^{5/2} \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 )}{\sqrt{c-i d} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\left (a^2 (c+5 i d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+i d-\frac{i d x^2}{a}}} \, dx,x,\sqrt{a+i a \tan (e+f x)}\right )}{d f}\\ &=-\frac{4 i \sqrt{2} a^{5/2} \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 )}{\sqrt{c-i d} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}+\frac{\left (a^2 (c+5 i d)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{i d x^2}{a}} \, dx,x,\frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{d f}\\ &=-\frac{\sqrt [4]{-1} a^{5/2} (c+5 i d) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac{4 i \sqrt{2} a^{5/2} \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 )}{\sqrt{c-i d} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{d f}\\ \end{align*}

Mathematica [B]  time = 6.60638, size = 602, normalized size = 3.01 \[ \frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \cos ^2(e+f x) (a+i a \tan (e+f x))^{5/2} \left (\frac{(-\cos (2 e)+i \sin (2 e)) \cos (e+f x) \left ((8+8 i) d^{3/2} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )+\sqrt{c-i d} (c+5 i d) \log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left (-(1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{\sqrt{d} (5 d-i c) \left (e^{i (e+f x)}+i\right )}\right )-\sqrt{c-i d} (c+5 i d) \log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt{d} (5 d-i c) \left (e^{i (e+f x)}-i\right )}\right )\right )}{\sqrt{c-i d} \sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}+(1+i) \sqrt{d} \sin (2 e) \sqrt{c+d \tan (e+f x)}+(-1+i) \sqrt{d} \cos (2 e) \sqrt{c+d \tan (e+f x)}\right )}{d^{3/2} f (\cos (f x)+i \sin (f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/2 + I/2)*Cos[e + f*x]^2*(a + I*a*Tan[e + f*x])^(5/2)*((Cos[e + f*x]*(Sqrt[c - I*d]*(c + (5*I)*d)*Log[((2 +
2*I)*E^((I/2)*e)*((-I)*d + d*E^(I*(e + f*x)) + I*c*(I + E^(I*(e + f*x))) - (1 + I)*Sqrt[d]*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)))]))/(Sqrt[d]*((-I)*c + 5*d)*(I
+ E^(I*(e + f*x))))] - Sqrt[c - I*d]*(c + (5*I)*d)*Log[((2 + 2*I)*E^((I/2)*e)*(c + I*d + I*c*E^(I*(e + f*x)) +
d*E^(I*(e + f*x)) + (1 + I)*Sqrt[d]*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)))]))/(Sqrt[d]*((-I)*c + 5*d)*(-I + E^(I*(e + f*x))))] + (8 + 8*I)*d^(3/2)*Log[2*(Sqrt[c
- I*d]*Cos[e + f*x] + I*Sqrt[c - I*d]*Sin[e + f*x] + Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]]*Sqrt[c +
d*Tan[e + f*x]])])*(-Cos[2*e] + I*Sin[2*e]))/(Sqrt[c - I*d]*Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]])
- (1 - I)*Sqrt[d]*Cos[2*e]*Sqrt[c + d*Tan[e + f*x]] + (1 + I)*Sqrt[d]*Sin[2*e]*Sqrt[c + d*Tan[e + f*x]]))/(d^(
3/2)*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B]  time = 0.138, size = 1295, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^2*(I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*tan(f*
x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3+I*ln(1/2*(2
*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2
)*(-a*(I*d-c))^(1/2)*a*c*d^2+4*I*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2
)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d-4*I*d^2*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+
2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*2^(1/2)*(-a*(I*d-c))^(1/2)-5
*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/
2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2*d-5*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e
)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^3-4*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*t
an(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(
tan(f*x+e)+I))*a*c*d+4*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c)
)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*a*d^2+4*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+
2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c
*d+4*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)
^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^2-2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(
-a*(I*d-c))^(1/2)*c^2-2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d
^2-4*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*t
an(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*c*d-4*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*tan(f*x+e)*d+2*2
^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*a*d^2)*2^
(1/2)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(c^2+d^2)/d/(I*a*d)^(1/2)/(-a*(I*d-c))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.90891, size = 2095, normalized size = 10.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*sqrt(2)*a^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^(I*f*x + I*e) + d*f*sqrt((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*log((2*d^2*f*s
qrt((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(2)*(-I*a^2*c + 5*a^2*d + (-I
*a^2*c + 5*a^2*d)*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))*e^(-2*I*f*x - 2*I*e)/(-I*a^2*c + 5*a^2*d)) - d*f*sqrt((I
*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d^3*f^2))*log(-(2*d^2*f*sqrt((I*a^5*c^2 - 10*a^5*c*d - 25*I*a^5*d^2)/(d
^3*f^2))*e^(2*I*f*x + 2*I*e) - sqrt(2)*(-I*a^2*c + 5*a^2*d + (-I*a^2*c + 5*a^2*d)*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))*e^(-2*I*f*x - 2*I*e)/(-I*a^2*c + 5*a^2*d)) - sqrt(-32*I*a^5/((I*c + d)*f^2))*d*f*log(1/4*(sqrt(-32*I*a
^5/((I*c + d)*f^2))*(I*c + d)*f*e^(2*I*f*x + 2*I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^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^(I*f*x + I*e))
*e^(-2*I*f*x - 2*I*e)/a^2) + sqrt(-32*I*a^5/((I*c + d)*f^2))*d*f*log(1/4*(sqrt(-32*I*a^5/((I*c + d)*f^2))*(-I*
c - d)*f*e^(2*I*f*x + 2*I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^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^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/a
^2))/(d*f)

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

[Out]

Timed out