3.1108 \(\int (a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=131 \[ -\frac{2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac{4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^2 (d+i c) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 i a^2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.320963, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac{4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}+\frac{4 a^2 (d+i c) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 i a^2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*a^2*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + (4*a^2*(I*c + d)*Sqrt[c + d*T
an[e + f*x]])/f + (((4*I)/3)*a^2*(c + d*Tan[e + f*x])^(3/2))/f - (2*a^2*(c + d*Tan[e + f*x])^(5/2))/(5*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))^{3/2} \, dx &=-\frac{2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\int \left (2 a^2+2 i a^2 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2} \, dx\\ &=\frac{4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\int \sqrt{c+d \tan (e+f x)} \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) \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\int \frac{2 a^2 (c-i d)^2+2 i a^2 (c-i d)^2 \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{4 a^2 (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac{\left (4 i a^4 (c-i d)^4\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-4 a^4 (c-i d)^4+2 a^2 (c-i d)^2 x\right ) \sqrt{c-\frac{i d x}{2 a^2 (c-i d)^2}}} \, dx,x,2 i a^2 (c-i d)^2 \tan (e+f x)\right )}{f}\\ &=\frac{4 a^2 (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}-\frac{\left (16 a^6 (c-i d)^6\right ) \operatorname{Subst}\left (\int \frac{1}{-4 a^4 (c-i d)^4-\frac{4 i a^4 c (c-i d)^4}{d}+\frac{4 i a^4 (c-i d)^4 x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}\\ &=-\frac{4 i a^2 (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}+\frac{4 a^2 (i c+d) \sqrt{c+d \tan (e+f x)}}{f}+\frac{4 i a^2 (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{2 a^2 (c+d \tan (e+f x))^{5/2}}{5 d f}\\ \end{align*}

Mathematica [A]  time = 4.05517, size = 221, normalized size = 1.69 \[ \frac{a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (-\frac{(\cos (2 e)-i \sin (2 e)) \sec ^2(e+f x) \sqrt{c+d \tan (e+f x)} \left (\left (3 c^2-40 i c d-33 d^2\right ) \cos (2 (e+f x))+3 c^2+2 d (3 c-5 i d) \sin (2 (e+f x))-40 i c d-27 d^2\right )}{15 d}-4 i e^{-2 i e} (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )\right )}{f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[e + f*x] + I*Sin[e + f*x])^2*(((-4*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]])/E^((2*I)*e) - (Sec[e + f*x]^2*(Cos[2*e] - I*Sin[2*e])*(3*c^2 -
(40*I)*c*d - 27*d^2 + (3*c^2 - (40*I)*c*d - 33*d^2)*Cos[2*(e + f*x)] + 2*(3*c - (5*I)*d)*d*Sin[2*(e + f*x)])*S
qrt[c + d*Tan[e + f*x]])/(15*d)))/(f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B]  time = 0.027, size = 2484, normalized size = 19. \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))^2*(c+d*tan(f*x+e))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.848, size = 1639, normalized size = 12.51 \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))^(3/2),x, algorithm="fricas")

[Out]

-1/60*(15*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*a^4*c^3 - 48*I*a^4*c^2*d - 48*
a^4*c*d^2 + 16*I*a^4*d^3)/f^2)*log(-1/2*(4*I*a^2*c^2 + 4*a^2*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(-(16*a^4*c^3 - 48*I*a^4*c^2*d - 48*a^4*c*d^2 +
16*I*a^4*d^3)/f^2) + (4*I*a^2*c^2 + 8*a^2*c*d - 4*I*a^2*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(-I*a^
2*c - a^2*d)) - 15*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(16*a^4*c^3 - 48*I*a^4*c^
2*d - 48*a^4*c*d^2 + 16*I*a^4*d^3)/f^2)*log(-1/2*(4*I*a^2*c^2 + 4*a^2*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(-(16*a^4*c^3 - 48*I*a^4*c^2*d - 48*a^
4*c*d^2 + 16*I*a^4*d^3)/f^2) + (4*I*a^2*c^2 + 8*a^2*c*d - 4*I*a^2*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*
e)/(-I*a^2*c - a^2*d)) + (24*a^2*c^2 - 272*I*a^2*c*d - 184*a^2*d^2 + (24*a^2*c^2 - 368*I*a^2*c*d - 344*a^2*d^2
)*e^(4*I*f*x + 4*I*e) + (48*a^2*c^2 - 640*I*a^2*c*d - 432*a^2*d^2)*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^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*
f)

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

[Out]

a**2*(Integral(c*sqrt(c + d*tan(e + f*x)), x) + Integral(-c*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Int
egral(d*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(-d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3, x) +
Integral(2*I*c*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(2*I*d*sqrt(c + d*tan(e + f*x))*tan(e + f*
x)**2, x))

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Giac [B]  time = 1.48394, size = 389, normalized size = 2.97 \begin{align*} \frac{2 \,{\left (8 i \, a^{2} c^{2} + 16 \, a^{2} c d - 8 i \, a^{2} 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{6 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a^{2} d^{4} f^{4} - 20 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} a^{2} d^{5} f^{4} - 60 i \, \sqrt{d \tan \left (f x + e\right ) + c} a^{2} c d^{5} f^{4} - 60 \, \sqrt{d \tan \left (f x + e\right ) + c} a^{2} d^{6} f^{4}}{15 \, d^{5} f^{5}} \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))^(3/2),x, algorithm="giac")

[Out]

2*(8*I*a^2*c^2 + 16*a^2*c*d - 8*I*a^2*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)) - 1/15*(6*(d*tan
(f*x + e) + c)^(5/2)*a^2*d^4*f^4 - 20*I*(d*tan(f*x + e) + c)^(3/2)*a^2*d^5*f^4 - 60*I*sqrt(d*tan(f*x + e) + c)
*a^2*c*d^5*f^4 - 60*sqrt(d*tan(f*x + e) + c)*a^2*d^6*f^4)/(d^5*f^5)