3.1114 \(\int (a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=166 \[ -\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac{4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^2 (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 i a^2 (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

[Out]

((-4*I)*a^2*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((4*I)*a^2*(c - I*d)^2*Sqrt[c
+ d*Tan[e + f*x]])/f + (4*a^2*(I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (((4*I)/5)*a^2*(c + d*Tan[e + f*x
])^(5/2))/f - (2*a^2*(c + d*Tan[e + f*x])^(7/2))/(7*d*f)

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Rubi [A]  time = 0.441134, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3543, 3528, 3537, 63, 208} \[ -\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac{4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{4 a^2 (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 i a^2 (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*a^2*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((4*I)*a^2*(c - I*d)^2*Sqrt[c
+ d*Tan[e + f*x]])/f + (4*a^2*(I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (((4*I)/5)*a^2*(c + d*Tan[e + f*x
])^(5/2))/f - (2*a^2*(c + d*Tan[e + f*x])^(7/2))/(7*d*f)

Rule 3543

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

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))^2 (c+d \tan (e+f x))^{5/2} \, dx &=-\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\int \left (2 a^2+2 i a^2 \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2} \, dx\\ &=\frac{4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\int (c+d \tan (e+f x))^{3/2} \left (2 a^2 (c-i d)+2 a^2 (i c+d) \tan (e+f x)\right ) \, dx\\ &=\frac{4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\int \left (2 a^2 (c-i d)^2+2 i a^2 (c-i d)^2 \tan (e+f x)\right ) \sqrt{c+d \tan (e+f x)} \, dx\\ &=\frac{4 i a^2 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\int \frac{2 a^2 (c-i d)^3-2 a^2 (i c+d)^3 \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{4 i a^2 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}+\frac{\left (4 i a^4 (c-i d)^6\right ) \operatorname{Subst}\left (\int \frac{1}{\left (4 a^4 (i c+d)^6+2 a^2 (c-i d)^3 x\right ) \sqrt{c-\frac{d x}{2 a^2 (i c+d)^3}}} \, dx,x,-2 a^2 (i c+d)^3 \tan (e+f x)\right )}{f}\\ &=\frac{4 i a^2 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}-\frac{\left (16 a^6 (c-i d)^9\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{4 a^4 c (c-i d)^3 (i c+d)^3}{d}+4 a^4 (i c+d)^6-\frac{4 a^4 (c-i d)^3 (i c+d)^3 x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac{4 i a^2 (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{4 i a^2 (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 a^2 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 i a^2 (c+d \tan (e+f x))^{5/2}}{5 f}-\frac{2 a^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\\ \end{align*}

Mathematica [A]  time = 7.67805, size = 271, normalized size = 1.63 \[ \frac{a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (-\frac{(\cos (2 e)-i \sin (2 e)) \sec ^2(e+f x) \sqrt{c+d \tan (e+f x)} \left (d \left (45 c^2-154 i c d-55 d^2\right ) \tan (e+f x)+\cos (2 (e+f x)) \left (d \left (45 c^2-154 i c d-85 d^2\right ) \tan (e+f x)-322 i c^2 d+15 c^3-535 c d^2+252 i d^3\right )-322 i c^2 d+15 c^3-445 c d^2+168 i d^3\right )}{105 d}-4 i e^{-2 i e} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )\right )}{f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[e + f*x] + I*Sin[e + f*x])^2*(((-4*I)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x)
)))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c - I*d]])/E^((2*I)*e) - (Sec[e + f*x]^2*(Cos[2*e] - I*Sin[2*e])*Sqrt[c +
d*Tan[e + f*x]]*(15*c^3 - (322*I)*c^2*d - 445*c*d^2 + (168*I)*d^3 + d*(45*c^2 - (154*I)*c*d - 55*d^2)*Tan[e +
f*x] + Cos[2*(e + f*x)]*(15*c^3 - (322*I)*c^2*d - 535*c*d^2 + (252*I)*d^3 + d*(45*c^2 - (154*I)*c*d - 85*d^2)*
Tan[e + f*x])))/(105*d)))/(f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B]  time = 0.029, size = 2886, normalized size = 17.4 \begin{align*} \text{output 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))^2*(c+d*tan(f*x+e))^(5/2),x)

[Out]

4/5*I*a^2*(c+d*tan(f*x+e))^(5/2)/f-2/7*a^2*(c+d*tan(f*x+e))^(7/2)/d/f+4/3/f*a^2*d*(c+d*tan(f*x+e))^(3/2)+2/f*a
^2*d^3/(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))-4*I/f*a^2*d^2*(c+d*tan(f*x+e))^(1/2)+4*I/f*a^2*c^2*(c+d*tan(f*x+e))^(1/2)+4/3*I/f*a^2*(c
+d*tan(f*x+e))^(3/2)*c+8/f*a^2*d*(c+d*tan(f*x+e))^(1/2)*c+1/f*a^2*d^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))-1/f*a^2*d^3/(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/f*a^2*d^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))+I/f*a^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^3-6/f*a^2*d/(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+6/f*a^2*d/(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/f*a^2*d/(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+3/f*a^2*d/(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-I/f*a^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^3-2*I/f*a^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*I/f*a^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^3-3*I/f
*a^2*d^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*I/f*a^2*d^4/(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))-I/f*a^2*d^4/(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))+3*I/f*a^2*
d^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/f*a^2*d^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+I/f*a^2/(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^4-6*I/f*a^2*d^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^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*I/f*a^2*d^4/(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))+4/f*a^2*d^3/(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-4/f*a^2*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))*c^3-4/f*a^2*d^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+
I/f*a^2*d^4/(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^2/(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^4+2*I/f*a^2/(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^4-I/f*a^2/(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^4-2/f*a^2*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))*c^3+2/f*a^2*d/(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+4/f*a^2*d/(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+2/f*a^2*d^3/(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

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

[Out]

Timed out

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Fricas [B]  time = 4.91557, size = 2323, normalized size = 13.99 \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))^2*(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/420*(105*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*
a^4*c^5 - 80*I*a^4*c^4*d - 160*a^4*c^3*d^2 + 160*I*a^4*c^2*d^3 + 80*a^4*c*d^4 - 16*I*a^4*d^5)/f^2)*log((4*a^2*
c^3 - 8*I*a^2*c^2*d - 4*a^2*c*d^2 - (I*f*e^(2*I*f*x + 2*I*e) + I*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(-(16*a^4*c^5 - 80*I*a^4*c^4*d - 160*a^4*c^3*d^2 + 160*I*a^4*c^2*d^3 + 80*
a^4*c*d^4 - 16*I*a^4*d^5)/f^2) + (4*a^2*c^3 - 12*I*a^2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)*e^(2*I*f*x + 2*I*e)
)*e^(-2*I*f*x - 2*I*e)/(2*a^2*c^2 - 4*I*a^2*c*d - 2*a^2*d^2)) - 105*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*
x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*a^4*c^5 - 80*I*a^4*c^4*d - 160*a^4*c^3*d^2 + 160*I*a^4
*c^2*d^3 + 80*a^4*c*d^4 - 16*I*a^4*d^5)/f^2)*log((4*a^2*c^3 - 8*I*a^2*c^2*d - 4*a^2*c*d^2 - (-I*f*e^(2*I*f*x +
2*I*e) - I*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(-(16*a^4*c^5 - 8
0*I*a^4*c^4*d - 160*a^4*c^3*d^2 + 160*I*a^4*c^2*d^3 + 80*a^4*c*d^4 - 16*I*a^4*d^5)/f^2) + (4*a^2*c^3 - 12*I*a^
2*c^2*d - 12*a^2*c*d^2 + 4*I*a^2*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(2*a^2*c^2 - 4*I*a^2*c*d - 2*a
^2*d^2)) + (120*a^2*c^3 - 2216*I*a^2*c^2*d - 3048*a^2*c*d^2 + 1336*I*a^2*d^3 + (120*a^2*c^3 - 2936*I*a^2*c^2*d
- 5512*a^2*c*d^2 + 2696*I*a^2*d^3)*e^(6*I*f*x + 6*I*e) + (360*a^2*c^3 - 8088*I*a^2*c^2*d - 12632*a^2*c*d^2 +
4904*I*a^2*d^3)*e^(4*I*f*x + 4*I*e) + (360*a^2*c^3 - 7368*I*a^2*c^2*d - 10168*a^2*c*d^2 + 4504*I*a^2*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)))/(d*f*e^(6*I*f*x +
6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + 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))**2*(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.59728, size = 506, normalized size = 3.05 \begin{align*} \frac{2 \,{\left (8 i \, a^{2} c^{3} + 24 \, a^{2} c^{2} d - 24 i \, a^{2} c d^{2} - 8 \, a^{2} d^{3}\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{30 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{7}{2}} a^{2} d^{6} f^{6} - 84 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{2} d^{7} f^{6} - 140 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{2} c d^{7} f^{6} - 420 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{2} c^{2} d^{7} f^{6} - 140 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{2} d^{8} f^{6} - 840 \, \sqrt{d \tan \left (f x + e\right ) + c} a^{2} c d^{8} f^{6} + 420 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{2} d^{9} f^{6}}{105 \, d^{7} f^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(8*I*a^2*c^3 + 24*a^2*c^2*d - 24*I*a^2*c*d^2 - 8*a^2*d^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))))/(sqrt(-8*c + 8*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1))
- 1/105*(30*(d*tan(f*x + e) + c)^(7/2)*a^2*d^6*f^6 - 84*I*(d*tan(f*x + e) + c)^(5/2)*a^2*d^7*f^6 - 140*I*(d*t
an(f*x + e) + c)^(3/2)*a^2*c*d^7*f^6 - 420*I*sqrt(d*tan(f*x + e) + c)*a^2*c^2*d^7*f^6 - 140*(d*tan(f*x + e) +
c)^(3/2)*a^2*d^8*f^6 - 840*sqrt(d*tan(f*x + e) + c)*a^2*c*d^8*f^6 + 420*I*sqrt(d*tan(f*x + e) + c)*a^2*d^9*f^6
)/(d^7*f^7)