3.1119 \(\int \frac{(a+i a \tan (e+f x))^3}{\sqrt{c+d \tan (e+f x)}} \, dx\)
Optimal. Leaf size=126 \[ \frac{4 a^3 (-4 d+i c) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)}}{3 d f}-\frac{8 i a^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]
[Out]
((-8*I)*a^3*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (4*a^3*(I*c - 4*d)*Sqrt[c + d
*Tan[e + f*x]])/(3*d^2*f) - (2*(a^3 + I*a^3*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*d*f)
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Rubi [A] time = 0.299572, antiderivative size = 126, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used =
{3556, 3592, 3537, 63, 208} \[ \frac{4 a^3 (-4 d+i c) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)}}{3 d f}-\frac{8 i a^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f \sqrt{c-i d}} \]
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[e + f*x])^3/Sqrt[c + d*Tan[e + f*x]],x]
[Out]
((-8*I)*a^3*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (4*a^3*(I*c - 4*d)*Sqrt[c + d
*Tan[e + f*x]])/(3*d^2*f) - (2*(a^3 + I*a^3*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*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 3592
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] && !LeQ[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))^3}{\sqrt{c+d \tan (e+f x)}} \, dx &=-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{(2 a) \int \frac{(a+i a \tan (e+f x)) (a (i c+2 d)+a (c+4 i d) \tan (e+f x))}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d}\\ &=\frac{4 a^3 (i c-4 d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{(2 a) \int \frac{6 a^2 d+6 i a^2 d \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d}\\ &=\frac{4 a^3 (i c-4 d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)}}{3 d f}+\frac{\left (24 i a^5 d\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{i x}{6 a^2}} \left (-36 a^4 d^2+6 a^2 d x\right )} \, dx,x,6 i a^2 d \tan (e+f x)\right )}{f}\\ &=\frac{4 a^3 (i c-4 d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)}}{3 d f}-\frac{\left (288 a^7 d\right ) \operatorname{Subst}\left (\int \frac{1}{-36 i a^4 c d-36 a^4 d^2+36 i a^4 d x^2} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{f}\\ &=-\frac{8 i a^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d} f}+\frac{4 a^3 (i c-4 d) \sqrt{c+d \tan (e+f x)}}{3 d^2 f}-\frac{2 \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)}}{3 d f}\\ \end{align*}
Mathematica [A] time = 4.67758, size = 168, normalized size = 1.33 \[ \frac{a^3 (\cos (e+f x)+i \sin (e+f x))^3 \left (\frac{2 (\sin (3 e)+i \cos (3 e)) (2 c-d \tan (e+f x)+9 i d) \sqrt{c+d \tan (e+f x)}}{3 d^2}-\frac{8 i e^{-3 i e} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )}{\sqrt{c-i d}}\right )}{f (\cos (f x)+i \sin (f x))^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + I*a*Tan[e + f*x])^3/Sqrt[c + d*Tan[e + f*x]],x]
[Out]
(a^3*(Cos[e + f*x] + I*Sin[e + f*x])^3*(((-8*I)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I
)*(e + f*x)))]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*E^((3*I)*e)) + (2*(I*Cos[3*e] + Sin[3*e])*(2*c + (9*I)*d - d*Tan
[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*d^2)))/(f*(Cos[f*x] + I*Sin[f*x])^3)
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Maple [B] time = 0.079, size = 1748, normalized size = 13.9 \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))^3/(c+d*tan(f*x+e))^(1/2),x)
[Out]
4*I/f*a^3/(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))+4*I/f*a^3*d^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^
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-6/f*a^3/d*(c+d*ta
n(f*x+e))^(1/2)+4*I/f*a^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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^2-4*I/f*a^3/(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
/3*I/f*a^3/d^2*(c+d*tan(f*x+e))^(3/2)-4*I/f*a^3*d^2/((c^2+d^2)^(1/2)*c+c^2+d^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))-2/f*a^3*d/(2*(c
^2+d^2)^(1/2)+2*c)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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/f*a^3*d/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^
2+d^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^2-2/f*a^3*d^3
/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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))+2*I/f*a^3/d^2*c*(c+d*tan(f*x+e))^(1/2)-4/f*a^3*d/((c^2+d^
2)^(1/2)*c+c^2+d^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-4/f*a^3*d/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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-
4/f*a^3*d^3/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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))-2*I/f*a^3/(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+4*I
/f*a^3/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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+2/f*a^3*d/(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))+2*I/f*a^3*d^2
/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/((c^2+d^2)^(1/2)*c+c^2+d^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))-2*I/f*a^3/(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))+4/f*a^3*d/(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))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt{d \tan \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((I*a*tan(f*x + e) + a)^3/sqrt(d*tan(f*x + e) + c), x)
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Fricas [B] time = 1.99316, size = 1106, normalized size = 8.78 \begin{align*} \frac{3 \,{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt{-\frac{64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac{{\left (8 \, a^{3} c + \sqrt{-\frac{64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}}{\left ({\left (i \, c + d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (i \, c + d\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}} +{\left (8 \, a^{3} c - 8 i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 3 \,{\left (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt{-\frac{64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac{{\left (8 \, a^{3} c + \sqrt{-\frac{64 i \, a^{6}}{{\left (i \, c + d\right )} f^{2}}}{\left ({\left (-i \, c - d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-i \, c - d\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}} +{\left (8 \, a^{3} c - 8 i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - 16 \,{\left (-i \, a^{3} c + 4 \, a^{3} d +{\left (-i \, a^{3} c + 5 \, a^{3} 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 (d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
1/12*(3*(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-64*I*a^6/((I*c + d)*f^2))*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/
((I*c + d)*f^2))*((I*c + d)*f*e^(2*I*f*x + 2*I*e) + (I*c + d)*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)) + (8*a^3*c - 8*I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - 3*(d^2*f
*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-64*I*a^6/((I*c + d)*f^2))*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c + d)*f^2
))*((-I*c - d)*f*e^(2*I*f*x + 2*I*e) + (-I*c - d)*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)) + (8*a^3*c - 8*I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(-I*a^3*c + 4*a^3
*d + (-I*a^3*c + 5*a^3*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)))/(d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int - \frac{3 \tan ^{2}{\left (e + f x \right )}}{\sqrt{c + d \tan{\left (e + f x \right )}}}\, dx + \int \frac{3 i \tan{\left (e + f x \right )}}{\sqrt{c + d \tan{\left (e + f x \right )}}}\, dx + \int - \frac{i \tan ^{3}{\left (e + f x \right )}}{\sqrt{c + d \tan{\left (e + f x \right )}}}\, dx + \int \frac{1}{\sqrt{c + d \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))**3/(c+d*tan(f*x+e))**(1/2),x)
[Out]
a**3*(Integral(-3*tan(e + f*x)**2/sqrt(c + d*tan(e + f*x)), x) + Integral(3*I*tan(e + f*x)/sqrt(c + d*tan(e +
f*x)), x) + Integral(-I*tan(e + f*x)**3/sqrt(c + d*tan(e + f*x)), x) + Integral(1/sqrt(c + d*tan(e + f*x)), x)
)
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Giac [B] time = 1.56497, size = 328, normalized size = 2.6 \begin{align*} \frac{32 i \, a^{3} \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{2 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{3} d^{4} f^{2} - 6 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{3} c d^{4} f^{2} + 18 \, \sqrt{d \tan \left (f x + e\right ) + c} a^{3} d^{5} f^{2}}{3 \, d^{6} f^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")
[Out]
32*I*a^3*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*sqr
t(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)) - 1/3*(2*I*(d*tan(f*x + e) + c)^(3/2)*a^3*d^4*f^2
- 6*I*sqrt(d*tan(f*x + e) + c)*a^3*c*d^4*f^2 + 18*sqrt(d*tan(f*x + e) + c)*a^3*d^5*f^2)/(d^6*f^3)