3.1145 \(\int \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=196 \[ \frac{d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}-\frac{i \sqrt{2} \sqrt{a} (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}-\frac{\sqrt [4]{-1} \sqrt{a} \sqrt{d} (3 c-i d) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{f} \]

[Out]

-(((-1)^(1/4)*Sqrt[a]*(3*c - I*d)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqr
t[c + d*Tan[e + f*x]])])/f) - (I*Sqrt[2]*Sqrt[a]*(c - I*d)^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f
*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/f + (d*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])
/f

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Rubi [A]  time = 0.736266, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3560, 3601, 3544, 208, 3599, 63, 217, 206} \[ \frac{d \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{f}-\frac{i \sqrt{2} \sqrt{a} (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}-\frac{\sqrt [4]{-1} \sqrt{a} \sqrt{d} (3 c-i d) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((-1)^(1/4)*Sqrt[a]*(3*c - I*d)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqr
t[c + d*Tan[e + f*x]])])/f) - (I*Sqrt[2]*Sqrt[a]*(c - I*d)^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f
*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/f + (d*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])
/f

Rule 3560

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

Rule 3601

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{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]

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]

Rule 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 6.7868, size = 507, normalized size = 2.59 \[ \frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{a+i a \tan (e+f x)} \left ((1-i) d \sqrt{c+d \tan (e+f x)}-\frac{\cos (e+f x) \left (\sqrt{d} (3 c-i d) \log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1-i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}-c \left (e^{i (e+f x)}+i\right )+i d e^{i (e+f x)}+d\right )}{d^{3/2} (d+3 i c) \left (e^{i (e+f x)}+i\right )}\right )+i \sqrt{d} (d+3 i c) \log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{d^{3/2} (3 c-i d) \left (e^{i (e+f x)}-i\right )}\right )+(2+2 i) (c-i d)^{3/2} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )\right )}{\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/2 + I/2)*Sqrt[a + I*a*Tan[e + f*x]]*(-((Cos[e + f*x]*((3*c - I*d)*Sqrt[d]*Log[((2 + 2*I)*E^((I/2)*e)*(d +
I*d*E^(I*(e + f*x)) - c*(I + E^(I*(e + f*x))) + (1 - I)*Sqrt[d]*Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-
1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]))/(d^(3/2)*((3*I)*c + d)*(I + E^(I*(e + f*x))))] + I*Sqrt
[d]*((3*I)*c + d)*Log[((2 + 2*I)*E^((I/2)*e)*(c + I*d + I*c*E^(I*(e + f*x)) + d*E^(I*(e + f*x)) + (1 + I)*Sqrt
[d]*Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]))/((3*c
- I*d)*d^(3/2)*(-I + E^(I*(e + f*x))))] + (2 + 2*I)*(c - I*d)^(3/2)*Log[2*(Sqrt[c - I*d]*Cos[e + f*x] + I*Sqr
t[c - I*d]*Sin[e + f*x] + Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]]*Sqrt[c + d*Tan[e + f*x]])]))/Sqrt[1
+ Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]]) + (1 - I)*d*Sqrt[c + d*Tan[e + f*x]]))/f

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Maple [B]  time = 0.108, size = 1701, normalized size = 8.7 \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))^(1/2)*(c+d*tan(f*x+e))^(3/2),x)

[Out]

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

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

[Out]

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

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Fricas [B]  time = 1.68343, size = 2080, normalized size = 10.61 \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))^(1/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/2*(2*sqrt(2)*d*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(a/(e^(2*I*f*x
+ 2*I*e) + 1))*e^(I*f*x + I*e) + f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*log((sqrt(2)*((3*I*c + d)*e^(
2*I*f*x + 2*I*e) + 3*I*c + d)*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(a
/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + 2*f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*e^(2*I*f*x + 2
*I*e))*e^(-2*I*f*x - 2*I*e)/(3*I*c + d)) - f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*log((sqrt(2)*((3*I*
c + d)*e^(2*I*f*x + 2*I*e) + 3*I*c + d)*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(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - 2*f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*e^(2
*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(3*I*c + d)) - f*sqrt(-(2*a*c^3 - 6*I*a*c^2*d - 6*a*c*d^2 + 2*I*a*d^3)/f
^2)*log((sqrt(2)*((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c + d)*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(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + f*sqrt(-(2*a*c^3 - 6*I*a*c^2*d - 6*
a*c*d^2 + 2*I*a*d^3)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*c + d)) + f*sqrt(-(2*a*c^3 - 6*I*a*c^2*
d - 6*a*c*d^2 + 2*I*a*d^3)/f^2)*log((sqrt(2)*((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c + d)*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(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - f*sqrt(
-(2*a*c^3 - 6*I*a*c^2*d - 6*a*c*d^2 + 2*I*a*d^3)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*c + 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))**(1/2)*(c+d*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 5.19253, size = 309, normalized size = 1.58 \begin{align*} -\frac{i \, \sqrt{2 \, a^{2} c + 2 \, \sqrt{a^{2} c^{2} +{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} d^{2} - 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a d^{2} + a^{2} d^{2}} a}{\left (i \, a c +{\left (i \, a \tan \left (f x + e\right ) + a\right )} d - a d\right )} d^{3}{\left (\frac{-i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a d + i \, a^{2} d}{a^{2} c + \sqrt{a^{4} c^{2} +{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} a^{2} d^{2} - 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a^{3} d^{2} + a^{4} d^{2}}} + 1\right )} \log \left (\sqrt{i \, a \tan \left (f x + e\right ) + a}\right )}{{\left (-i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} d^{3} + 2 i \, a d^{3}\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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