3.1109 \(\int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=98 \[ \frac{2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 a (d+i c) \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 i a (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.225311, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3528, 3537, 63, 208} \[ \frac{2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{2 a (d+i c) \sqrt{c+d \tan (e+f x)}}{f}-\frac{2 i a (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3528

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

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 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 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 (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx &=\frac{2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\int \sqrt{c+d \tan (e+f x)} (a (c-i d)+a (i c+d) \tan (e+f x)) \, dx\\ &=\frac{2 a (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\int \frac{a (c-i d)^2+i a (c-i d)^2 \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{2 a (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 i a (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{\left (i a^2 (c-i d)^4\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-a^2 (c-i d)^4+a (c-i d)^2 x\right ) \sqrt{c-\frac{i d x}{a (c-i d)^2}}} \, dx,x,i a (c-i d)^2 \tan (e+f x)\right )}{f}\\ &=\frac{2 a (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 i a (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{\left (2 a^3 (c-i d)^6\right ) \operatorname{Subst}\left (\int \frac{1}{-a^2 (c-i d)^4-\frac{i a^2 c (c-i d)^4}{d}+\frac{i a^2 (c-i d)^4 x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac{2 i a (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{2 a (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 i a (c+d \tan (e+f x))^{3/2}}{3 f}\\ \end{align*}

Mathematica [A]  time = 1.87725, size = 111, normalized size = 1.13 \[ \frac{2 a \left ((4 i c+i d \tan (e+f x)+3 d) \sqrt{c+d \tan (e+f x)}-3 i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.023, size = 2398, normalized size = 24.5 \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))*(c+d*tan(f*x+e))^(3/2),x)

[Out]

2/3*I*a*(c+d*tan(f*x+e))^(3/2)/f-1/f*a/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e)
)^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2*d+1/2/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1
/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^
2*d-1/2*I/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d
^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^3-I/f*a/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*t
an(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3-1/2*I/f*a/(2*(c^2+d^2)^(1/2
)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^2-1/f*a
/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))
/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^3-1/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/
2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c*d+1/2/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln
((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*d^3-2/f*a/(2*(c^2+d^2)^(
1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))
*c*d+1/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-
c-(c^2+d^2)^(1/2))*c*d+2/f*a*(c+d*tan(f*x+e))^(1/2)*d+I/f*a/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arct
an(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c*d^2-I/f*a/(c^2+d^
2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2
+d^2)^(1/2)-2*c)^(1/2))*c*d^2+2*I/f*a*c*(c+d*tan(f*x+e))^(1/2)-1/2*I/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^
2)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c*d^2+1/2*I/f
*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan
(f*x+e)-c-(c^2+d^2)^(1/2))*c*d^2-1/2/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*
tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*d^3+1/2*I/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*l
n(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*d^2+1/2*I/f*a/(2*(c^2+d
^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c
^2+1/f*a/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e)
)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^3+I/f*a/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c
)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^2+I/f*a/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arcta
n((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+2/f*a/(2*(c^2+d^
2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1
/2))*c*d-I/f*a/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(
2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2-I/f*a/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2
+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^2-1/2*I/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan
(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*d^2+1/2*I/f*a/(2*(c^2+d^2)^(1/2)+
2*c)^(1/2)/(c^2+d^2)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1
/2))*c^3+I/f*a/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(
f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3-1/2/f*a/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*ln(d*t
an(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^2*d+1/f*a/(c^2+d^2)^(1/2)/
(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/
2)-2*c)^(1/2))*c^2*d

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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))*(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31266, size = 1288, normalized size = 13.14 \begin{align*} -\frac{3 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt{-\frac{4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}}{f^{2}}} \log \left (\frac{{\left (2 i \, a c^{2} + 2 \, a c d +{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}}{f^{2}}} +{\left (2 i \, a c^{2} + 4 \, a c d - 2 i \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{i \, a c + a d}\right ) - 3 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt{-\frac{4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}}{f^{2}}} \log \left (\frac{{\left (2 i \, a c^{2} + 2 \, a c d -{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{4 \, a^{2} c^{3} - 12 i \, a^{2} c^{2} d - 12 \, a^{2} c d^{2} + 4 i \, a^{2} d^{3}}{f^{2}}} +{\left (2 i \, a c^{2} + 4 \, a c d - 2 i \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{i \, a c + a d}\right ) + 16 \,{\left (-2 i \, a c - a d + 2 \,{\left (-i \, a c - a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(4*a^2*c^3 - 12*I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)/f^2)*log(
(2*I*a*c^2 + 2*a*c*d + (f*e^(2*I*f*x + 2*I*e) + f)*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(-(4*a^2*c^3 - 12*I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)/f^2) + (2*I*a*c^2 + 4*a*c*d - 2
*I*a*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*a*c + a*d)) - 3*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(4*a^
2*c^3 - 12*I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)/f^2)*log((2*I*a*c^2 + 2*a*c*d - (f*e^(2*I*f*x + 2*I*e) +
f)*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(-(4*a^2*c^3 - 12*I*a^2*c^2*d
- 12*a^2*c*d^2 + 4*I*a^2*d^3)/f^2) + (2*I*a*c^2 + 4*a*c*d - 2*I*a*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I
*e)/(I*a*c + a*d)) + 16*(-2*I*a*c - a*d + 2*(-I*a*c - a*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)))/(f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int c \sqrt{c + d \tan{\left (e + f x \right )}}\, dx + \int d \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx + \int i c \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx + \int i d \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(c*sqrt(c + d*tan(e + f*x)), x) + Integral(d*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(I
*c*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(I*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x))

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Giac [B]  time = 1.44779, size = 321, normalized size = 3.28 \begin{align*} \frac{2}{3} \, a{\left (\frac{6 \,{\left (2 i \, c^{2} + 4 \, c d - 2 i \, d^{2}\right )} \arctan \left (\frac{4 \,{\left (\sqrt{d \tan \left (f x + e\right ) + c} c - \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} - i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{\sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} f{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} - \frac{-i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} f^{2} - 3 i \, \sqrt{d \tan \left (f x + e\right ) + c} c f^{2} - 3 \, \sqrt{d \tan \left (f x + e\right ) + c} d f^{2}}{f^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a*(6*(2*I*c^2 + 4*c*d - 2*I*d^2)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e
) + c))/(c*sqrt(-8*c + 8*sqrt(c^2 + d^2)) - I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-8*c + 8
*sqrt(c^2 + d^2))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - (-I*(d*tan(f*x + e)
+ c)^(3/2)*f^2 - 3*I*sqrt(d*tan(f*x + e) + c)*c*f^2 - 3*sqrt(d*tan(f*x + e) + c)*d*f^2)/f^3)