3.1115 \(\int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.293557, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3528, 3537, 63, 208} \[ \frac{2 i a (c-i d)^2 \sqrt{c+d \tan (e+f x)}}{f}+\frac{2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac{2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac{2 i a (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 3.16956, size = 208, normalized size = 1.59 \[ \frac{\cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x)) \left (\frac{1}{15} (\sin (e)+i \cos (e)) \sec ^2(e+f x) \sqrt{c+d \tan (e+f x)} \left (\left (23 c^2-35 i c d-18 d^2\right ) \cos (2 (e+f x))+23 c^2+d (11 c-5 i d) \sin (2 (e+f x))-35 i c d-12 d^2\right )-2 i e^{-i e} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt{c-i d}}\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(Cos[f*x] - I*Sin[f*x])*(a + I*a*Tan[e + f*x])*(((-2*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^(I*e) + (Sec[e + f*x]^2*(I*Cos[e] + Sin
[e])*(23*c^2 - (35*I)*c*d - 12*d^2 + (23*c^2 - (35*I)*c*d - 18*d^2)*Cos[2*(e + f*x)] + (11*c - (5*I)*d)*d*Sin[
2*(e + f*x)])*Sqrt[c + d*Tan[e + f*x]])/15))/f

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Maple [B]  time = 0.026, size = 2785, 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))*(c+d*tan(f*x+e))^(5/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.21952, size = 1858, normalized size = 14.18 \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))*(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/60*(15*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(4*a^2*c^5 - 20*I*a^2*c^4*d - 40*a^2*c^3
*d^2 + 40*I*a^2*c^2*d^3 + 20*a^2*c*d^4 - 4*I*a^2*d^5)/f^2)*log((2*a*c^3 - 4*I*a*c^2*d - 2*a*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(-(4*a^2*c
^5 - 20*I*a^2*c^4*d - 40*a^2*c^3*d^2 + 40*I*a^2*c^2*d^3 + 20*a^2*c*d^4 - 4*I*a^2*d^5)/f^2) + (2*a*c^3 - 6*I*a*
c^2*d - 6*a*c*d^2 + 2*I*a*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*c^2 - 2*I*a*c*d - a*d^2)) - 15*(f*
e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(4*a^2*c^5 - 20*I*a^2*c^4*d - 40*a^2*c^3*d^2 + 40*I*a
^2*c^2*d^3 + 20*a^2*c*d^4 - 4*I*a^2*d^5)/f^2)*log((2*a*c^3 - 4*I*a*c^2*d - 2*a*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(-(4*a^2*c^5 - 20*I*a^
2*c^4*d - 40*a^2*c^3*d^2 + 40*I*a^2*c^2*d^3 + 20*a^2*c*d^4 - 4*I*a^2*d^5)/f^2) + (2*a*c^3 - 6*I*a*c^2*d - 6*a*
c*d^2 + 2*I*a*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*c^2 - 2*I*a*c*d - a*d^2)) - (184*I*a*c^2 + 192
*a*c*d - 104*I*a*d^2 + (184*I*a*c^2 + 368*a*c*d - 184*I*a*d^2)*e^(4*I*f*x + 4*I*e) + (368*I*a*c^2 + 560*a*c*d
- 192*I*a*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)))
/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + 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))*(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.52255, size = 414, normalized size = 3.16 \begin{align*} -\frac{2}{15} \, a{\left (\frac{30 \,{\left (-2 i \, c^{3} - 6 \, c^{2} d + 6 i \, c d^{2} + 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{-3 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} f^{4} - 5 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c f^{4} - 15 i \, \sqrt{d \tan \left (f x + e\right ) + c} c^{2} f^{4} - 5 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} d f^{4} - 30 \, \sqrt{d \tan \left (f x + e\right ) + c} c d f^{4} + 15 i \, \sqrt{d \tan \left (f x + e\right ) + c} d^{2} f^{4}}{f^{5}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*a*(30*(-2*I*c^3 - 6*c^2*d + 6*I*c*d^2 + 2*d^3)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sq
rt(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)) + (-3*
I*(d*tan(f*x + e) + c)^(5/2)*f^4 - 5*I*(d*tan(f*x + e) + c)^(3/2)*c*f^4 - 15*I*sqrt(d*tan(f*x + e) + c)*c^2*f^
4 - 5*(d*tan(f*x + e) + c)^(3/2)*d*f^4 - 30*sqrt(d*tan(f*x + e) + c)*c*d*f^4 + 15*I*sqrt(d*tan(f*x + e) + c)*d
^2*f^4)/f^5)