3.1127 \(\int \frac{a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac{2 a}{f (d+i c) \sqrt{c+d \tan (e+f x)}}-\frac{2 i a \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.161195, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3529, 3537, 63, 208} \[ -\frac{2 a}{f (d+i c) \sqrt{c+d \tan (e+f x)}}-\frac{2 i a \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{3/2}} \]

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

Rule 3529

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

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

Mathematica [B]  time = 2.4667, size = 158, normalized size = 2.08 \[ \frac{2 i a e^{-i e} (\cos (e)+i \sin (e)) \left (\sqrt{c-i d} \cos (e+f x) \sqrt{c+d \tan (e+f x)}-\tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right ) (c \cos (e+f x)+d \sin (e+f x))\right )}{f (c-i d)^{3/2} (c \cos (e+f x)+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.031, size = 2446, normalized size = 32.2 \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]

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

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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 = 1.95708, size = 1342, normalized size = 17.66 \begin{align*} \frac{{\left ({\left (c^{2} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (c^{2} + d^{2}\right )} f\right )} \sqrt{\frac{4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac{{\left (2 \, a c +{\left ({\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, c^{2} + 2 \, c d - i \, d^{2}\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 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} +{\left (2 \, a c - 2 i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) -{\left ({\left (c^{2} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (c^{2} + d^{2}\right )} f\right )} \sqrt{\frac{4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac{{\left (2 \, a c +{\left ({\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\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 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} +{\left (2 \, a c - 2 i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) -{\left (-8 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a\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}}}{4 \,{\left ({\left (c^{2} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (c^{2} + d^{2}\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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{i \tan{\left (e + f x \right )}}{c \sqrt{c + d \tan{\left (e + f x \right )}} + d \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}}\, dx + \int \frac{1}{c \sqrt{c + d \tan{\left (e + f x \right )}} + d \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\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(I*tan(e + f*x)/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x) + Integr
al(1/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x))

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Giac [B]  time = 1.4431, size = 259, normalized size = 3.41 \begin{align*} -2 \, a{\left (\frac{1}{{\left (i \, c f + d f\right )} \sqrt{d \tan \left (f x + e\right ) + c}} - \frac{4 i \, \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 )}{{\left (c f - i \, d f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}}\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*a*(1/((I*c*f + d*f)*sqrt(d*tan(f*x + e) + c)) - 4*I*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))))/((c*f - I*d*f)*sqrt(-8*c + 8*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)
) + 1)))