3.1132 \(\int \frac{(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=127 \[ \frac{4 i a^2}{f (c-i d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 a^2 (-d+i c)}{3 d f (d+i c) (c+d \tan (e+f x))^{3/2}}-\frac{4 i a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.328827, antiderivative size = 127, 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 = {3542, 3529, 3537, 63, 208} \[ \frac{4 i a^2}{f (c-i d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 a^2 (-d+i c)}{3 d f (d+i c) (c+d \tan (e+f x))^{3/2}}-\frac{4 i a^2 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3542

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

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

Mathematica [A]  time = 5.48462, size = 218, normalized size = 1.72 \[ \frac{a^2 (\cos (e+f x)+i \sin (e+f x))^2 \left (\frac{2 (\cos (2 e)-i \sin (2 e)) \cos (e+f x) \sqrt{c+d \tan (e+f x)} \left (\left (c^2+6 i c d+d^2\right ) \cos (e+f x)+6 i d^2 \sin (e+f x)\right )}{3 d (c-i d)^2 (c \cos (e+f x)+d \sin (e+f x))^2}-\frac{4 i e^{-2 i e} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2}}\right )}{f (\cos (f x)+i \sin (f x))^2} \]

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)*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^((2*I)*e)) + (2*Cos[e + f*x]*(Cos[2*e] - I*Sin[2*e])*((c^2 +
(6*I)*c*d + d^2)*Cos[e + f*x] + (6*I)*d^2*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(3*(c - I*d)^2*d*(c*Cos[e +
f*x] + d*Sin[e + f*x])^2)))/(f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B]  time = 0.035, size = 2699, normalized size = 21.3 \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))^(5/2),x)

[Out]

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

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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 = 3.28653, size = 2268, normalized size = 17.86 \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]

(((3*c^4*d - 12*I*c^3*d^2 - 18*c^2*d^3 + 12*I*c*d^4 + 3*d^5)*f*e^(4*I*f*x + 4*I*e) + (6*c^4*d - 12*I*c^3*d^2 -
12*I*c*d^4 - 6*d^5)*f*e^(2*I*f*x + 2*I*e) + 3*(c^4*d + 2*c^2*d^3 + d^5)*f)*sqrt(-16*I*a^4/((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))*log(1/2*(4*a^2*c + ((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(-16*I*a^4/((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))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))
+ (4*a^2*c - 4*I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) - ((3*c^4*d - 12*I*c^3*d^2 - 18*c^2*d^
3 + 12*I*c*d^4 + 3*d^5)*f*e^(4*I*f*x + 4*I*e) + (6*c^4*d - 12*I*c^3*d^2 - 12*I*c*d^4 - 6*d^5)*f*e^(2*I*f*x + 2
*I*e) + 3*(c^4*d + 2*c^2*d^3 + d^5)*f)*sqrt(-16*I*a^4/((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))*log(1/2*(4*a^2*c + ((-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(-16*I*a^4/((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)
)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) + (4*a^2*c - 4*I*a^2*d)*e^(2*I*f*x
+ 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) + (8*a^2*c^2 + 48*I*a^2*c*d - 40*a^2*d^2 + (8*a^2*c^2 + 48*I*a^2*c*d + 56
*a^2*d^2)*e^(4*I*f*x + 4*I*e) + (16*a^2*c^2 + 96*I*a^2*c*d + 16*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)))/((12*c^4*d - 48*I*c^3*d^2 - 72*c^2*d^3 + 48*I*c*d^4
+ 12*d^5)*f*e^(4*I*f*x + 4*I*e) + (24*c^4*d - 48*I*c^3*d^2 - 48*I*c*d^4 - 24*d^5)*f*e^(2*I*f*x + 2*I*e) + 12*
(c^4*d + 2*c^2*d^3 + d^5)*f)

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

[Out]

a**2*(Integral(-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(2*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) + 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))

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Giac [B]  time = 1.55753, size = 336, normalized size = 2.65 \begin{align*} \frac{16 \, a^{2} \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{2 i \, a^{2} c^{2} - 12 \,{\left (d \tan \left (f x + e\right ) + c\right )} a^{2} d + 2 i \, a^{2} d^{2}}{{\left (3 i \, c^{2} d f + 6 \, c d^{2} f - 3 i \, d^{3} 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))^2/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

16*a^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))))/((-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)) + (2*I*a^2*c^2 - 12*(d*
tan(f*x + e) + c)*a^2*d + 2*I*a^2*d^2)/((3*I*c^2*d*f + 6*c*d^2*f - 3*I*d^3*f)*(d*tan(f*x + e) + c)^(3/2))