3.1131 \(\int \frac{(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=158 \[ \frac{4 a^3 (-d+i c) (c-4 i d)}{3 d^2 f (c-i d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (c-i d) (c+d \tan (e+f x))^{3/2}}-\frac{8 i a^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]

[Out]

((-8*I)*a^3*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*f) + (2*(c + I*d)*(a^3 + I*a^3*T
an[e + f*x]))/(3*(c - I*d)*d*f*(c + d*Tan[e + f*x])^(3/2)) + (4*a^3*(I*c - d)*(c - (4*I)*d))/(3*(c - I*d)^2*d^
2*f*Sqrt[c + d*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.420889, antiderivative size = 158, 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 = {3553, 3591, 3537, 63, 208} \[ \frac{4 a^3 (-d+i c) (c-4 i d)}{3 d^2 f (c-i d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (c-i d) (c+d \tan (e+f x))^{3/2}}-\frac{8 i a^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*a^3*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*f) + (2*(c + I*d)*(a^3 + I*a^3*T
an[e + f*x]))/(3*(c - I*d)*d*f*(c + d*Tan[e + f*x])^(3/2)) + (4*a^3*(I*c - d)*(c - (4*I)*d))/(3*(c - I*d)^2*d^
2*f*Sqrt[c + d*Tan[e + f*x]])

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{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] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1))/(b*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*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 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 \frac{(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx &=\frac{2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}-\frac{2 \int \frac{(a+i a \tan (e+f x)) \left (-a^2 (c+4 i d)+a^2 (i c+2 d) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d (i c+d)}\\ &=\frac{2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac{4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt{c+d \tan (e+f x)}}-\frac{2 \int \frac{-6 a^3 (i c-d) d+6 a^3 (c+i d) d \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d (i c+d) \left (c^2+d^2\right )}\\ &=\frac{2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac{4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\left (24 a^6 (c+i d) d\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{x}{6 a^3 (c+i d)}} \left (36 a^6 (c+i d)^2 d^2-6 a^3 (i c-d) d x\right )} \, dx,x,6 a^3 (c+i d) d \tan (e+f x)\right )}{(c-i d)^2 f}\\ &=\frac{2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac{4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\left (288 a^9 (c+i d)^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{36 a^6 c (i c-d) (c+i d) d+36 a^6 (c+i d)^2 d^2-36 a^6 (i c-d) (c+i d) d x^2} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c-i d)^2 f}\\ &=-\frac{8 i a^3 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2} f}+\frac{2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac{4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 8.19149, size = 232, normalized size = 1.47 \[ \frac{a^3 (\cos (e+f x)+i \sin (e+f x))^3 \left (\frac{2 (c+i d) (\sin (3 e)+i \cos (3 e)) \cos (e+f x) \sqrt{c+d \tan (e+f x)} \left (\left (2 c^2-9 i c d-d^2\right ) \cos (e+f x)+3 d (c-3 i d) \sin (e+f x)\right )}{3 d^2 (c-i d)^2 (c \cos (e+f x)+d \sin (e+f x))^2}-\frac{8 i e^{-3 i e} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2}}\right )}{f (\cos (f x)+i \sin (f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(5/2),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]])/((c - I*d)^(5/2)*E^((3*I)*e)) + (2*(c + I*d)*Cos[e + f*x]*(I*Cos[3*e] + Sin[3*e
])*((2*c^2 - (9*I)*c*d - d^2)*Cos[e + f*x] + 3*(c - (3*I)*d)*d*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*(c -
I*d)^2*d^2*(c*Cos[e + f*x] + d*Sin[e + f*x])^2)))/(f*(Cos[f*x] + I*Sin[f*x])^3)

________________________________________________________________________________________

Maple [B]  time = 0.045, size = 2772, normalized size = 17.5 \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))^3/(c+d*tan(f*x+e))^(5/2),x)

[Out]

-12/f*a^3*d/(c^2+d^2)^(5/2)/(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+12/f*a^3*d/(c^2+d^2)^(5/2)/(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^2-8/f*a^3*d/(c^2+d^
2)^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+8/f*a^3*d/(c^2+d^2)^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)^2/(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+4/f*a^3*d/(c^2+d^
2)^2/(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-6/f*a^3*d/(c^2+d^2)^(5/2)/(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+6/f*a^3*d/(c^2+d^2)^(5/2)/(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))*c^2-4*I/f*a^3/(c^2+d^2)
^(5/2)/(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^3-2*I/f*a^3/(c^2+d^2)^(5/2)/(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^3-4*I/f*a^3/(c^2+d^2)^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+2*I/f
*a^3/(c^2+d^2)^(5/2)/(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*t
an(f*x+e)-c-(c^2+d^2)^(1/2))*c^3+4*I/f*a^3/(c^2+d^2)^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+2*I/f*a^3/(c^2+d^2)^2/(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^2-2*
I/f*a^3/(c^2+d^2)^2/(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+4*I/f*a^3/(c^2+d^2)^(5/2)/(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^3-2/3*I/f*a^3/d^2/(c^2+d^2)/(c+d*ta
n(f*x+e))^(3/2)*c^3+2*I/f*a^3/d^2/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c^4-2/3/f*a^3*d/(c^2+d^2)/(c+d*tan(f*x+e)
)^(3/2)-12*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(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+6*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(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+12*I/f*a^3*d^2
/(c^2+d^2)^(5/2)/(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-6*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(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-4/f*a^3*d^3/(c^2+d^2)^(5/2)/(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))+2/f*a^3/d/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*c^2-16/f*a^3*d/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c-2/f*a^3*
d^3/(c^2+d^2)^(5/2)/(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))+2/f*a^3*d^3/(c^2+d^2)^(5/2)/(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))+4/f*a^3*d^3/(c^2+d^2)^(5/2)/(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))+12*
I/f*a^3/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c^2+2*I/f*a^3/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*c-6*I/f*a^3*d^2/(c^2
+d^2)^2/(c+d*tan(f*x+e))^(1/2)+4*I/f*a^3*d^2/(c^2+d^2)^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*d^2/(c^2+d^2)^2/(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))+2*I/
f*a^3*d^2/(c^2+d^2)^2/(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*I/f*a^3*d^2/(c^2+d^2)^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))

________________________________________________________________________________________

Maxima [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))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 3.33166, size = 2376, normalized size = 15.04 \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))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

(sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((3*c^4*d^2 - 12*I*c^3*
d^3 - 18*c^2*d^4 + 12*I*c*d^5 + 3*d^6)*f*e^(4*I*f*x + 4*I*e) + (6*c^4*d^2 - 12*I*c^3*d^3 - 12*I*c*d^5 - 6*d^6)
*f*e^(2*I*f*x + 2*I*e) + 3*(c^4*d^2 + 2*c^2*d^4 + d^6)*f)*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d
- 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e
) + (I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*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) - sqrt(-64*I*a^6/((I*c^5 + 5*c^4
*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((3*c^4*d^2 - 12*I*c^3*d^3 - 18*c^2*d^4 + 12*I*c*d^5 +
3*d^6)*f*e^(4*I*f*x + 4*I*e) + (6*c^4*d^2 - 12*I*c^3*d^3 - 12*I*c*d^5 - 6*d^6)*f*e^(2*I*f*x + 2*I*e) + 3*(c^4
*d^2 + 2*c^2*d^4 + d^6)*f)*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5
*I*c*d^4 + d^5)*f^2))*((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*e^(2*I*f*x + 2*I*e) + (-I*c^3 - 3*c^2*d + 3*I*c*
d^2 + d^3)*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^3 + 32*a^3*c^2*d + 112*I*a^3*c*d^2 - 64*a^3*d^3
+ (16*I*a^3*c^3 + 80*a^3*c^2*d + 16*I*a^3*c*d^2 + 80*a^3*d^3)*e^(4*I*f*x + 4*I*e) + (32*I*a^3*c^3 + 112*a^3*c
^2*d + 128*I*a^3*c*d^2 + 16*a^3*d^3)*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)))/((12*c^4*d^2 - 48*I*c^3*d^3 - 72*c^2*d^4 + 48*I*c*d^5 + 12*d^6)*f*e^(4*I*f*x + 4*I*e) +
(24*c^4*d^2 - 48*I*c^3*d^3 - 48*I*c*d^5 - 24*d^6)*f*e^(2*I*f*x + 2*I*e) + 12*(c^4*d^2 + 2*c^2*d^4 + d^6)*f)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int - \frac{3 \tan ^{2}{\left (e + f x \right )}}{c^{2} \sqrt{c + d \tan{\left (e + f x \right )}} + 2 c d \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )} + d^{2} \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\, dx + \int \frac{3 i \tan{\left (e + f x \right )}}{c^{2} \sqrt{c + d \tan{\left (e + f x \right )}} + 2 c d \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )} + d^{2} \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\, dx + \int - \frac{i \tan ^{3}{\left (e + f x \right )}}{c^{2} \sqrt{c + d \tan{\left (e + f x \right )}} + 2 c d \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )} + d^{2} \sqrt{c + d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\, dx + \int \frac{1}{c^{2} \sqrt{c + d \tan{\left (e + f x \right )}} + 2 c d \sqrt{c + d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )} + d^{2} \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))**3/(c+d*tan(f*x+e))**(5/2),x)

[Out]

a**3*(Integral(-3*tan(e + f*x)**2/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)
+ d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x) + Integral(3*I*tan(e + f*x)/(c**2*sqrt(c + d*tan(e + f*x
)) + 2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x) + Integr
al(-I*tan(e + f*x)**3/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) + d**2*sqrt
(c + d*tan(e + f*x))*tan(e + f*x)**2), x) + Integral(1/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d*tan(e
+ f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x))

________________________________________________________________________________________

Giac [B]  time = 1.58571, size = 421, normalized size = 2.66 \begin{align*} \frac{32 \, 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 )}{{\left (-i \, c^{2} f - 2 \, c d f + i \, d^{2} f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} + \frac{{\left (6 i \, d \tan \left (f x + e\right ) + 6 i \, c\right )} a^{3} c^{2} - 2 i \, a^{3} c^{3} + 12 \,{\left (d \tan \left (f x + e\right ) + c\right )} a^{3} c d + 2 \, a^{3} c^{2} d +{\left (18 i \, d \tan \left (f x + e\right ) + 18 i \, c\right )} a^{3} d^{2} - 2 i \, a^{3} c d^{2} + 2 \, a^{3} d^{3}}{{\left (3 \, c^{2} d^{2} f - 6 i \, c d^{3} f - 3 \, d^{4} f\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}} \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))^(5/2),x, algorithm="giac")

[Out]

32*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*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))))/((-I*c^2*f
- 2*c*d*f + I*d^2*f)*sqrt(-8*c + 8*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + ((6*I*d*tan(f*x + e)
+ 6*I*c)*a^3*c^2 - 2*I*a^3*c^3 + 12*(d*tan(f*x + e) + c)*a^3*c*d + 2*a^3*c^2*d + (18*I*d*tan(f*x + e) + 18*I*c
)*a^3*d^2 - 2*I*a^3*c*d^2 + 2*a^3*d^3)/((3*c^2*d^2*f - 6*I*c*d^3*f - 3*d^4*f)*(d*tan(f*x + e) + c)^(3/2))