3.1144 \(\int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=315 \[ -\frac{\sqrt [4]{-1} a^{3/2} \left (3 i c^2+18 c d-11 i d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{4 \sqrt{d} f}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{2 i \sqrt{2} a^{3/2} (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}+\frac{a (5 d+3 i c) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f} \]

[Out]

-((-1)^(1/4)*a^(3/2)*((3*I)*c^2 + 18*c*d - (11*I)*d^2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])
/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/(4*Sqrt[d]*f) - ((2*I)*Sqrt[2]*a^(3/2)*(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 + (a*((3*I)*c + 5*d)*Sqrt[a +
I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*f) + (a^2*(c + I*d)*(c + d*Tan[e + f*x])^(3/2))/(2*d*f*Sqrt[a
+ I*a*Tan[e + f*x]]) - (a^2*(c + d*Tan[e + f*x])^(5/2))/(2*d*f*Sqrt[a + I*a*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.33338, antiderivative size = 315, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3556, 3595, 3597, 3601, 3544, 208, 3599, 63, 217, 206} \[ -\frac{\sqrt [4]{-1} a^{3/2} \left (3 i c^2+18 c d-11 i d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{4 \sqrt{d} f}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{2 i \sqrt{2} a^{3/2} (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}+\frac{a (5 d+3 i c) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-1)^(1/4)*a^(3/2)*((3*I)*c^2 + 18*c*d - (11*I)*d^2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])
/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/(4*Sqrt[d]*f) - ((2*I)*Sqrt[2]*a^(3/2)*(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 + (a*((3*I)*c + 5*d)*Sqrt[a +
I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*f) + (a^2*(c + I*d)*(c + d*Tan[e + f*x])^(3/2))/(2*d*f*Sqrt[a
+ I*a*Tan[e + f*x]]) - (a^2*(c + d*Tan[e + f*x])^(5/2))/(2*d*f*Sqrt[a + I*a*Tan[e + f*x]])

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3595

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

Rule 3597

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

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 (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx &=-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}+\frac{a \int \frac{\left (-\frac{1}{2} a (i c-9 d)-\frac{1}{2} a (c-7 i d) \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx}{2 d}\\ &=\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{\int \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (-\frac{1}{2} a^2 (5 c-3 i d) d-\frac{1}{2} a^2 d (3 i c+5 d) \tan (e+f x)\right ) \, dx}{2 a d}\\ &=\frac{a (3 i c+5 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (-\frac{1}{4} a^3 d \left (13 c^2-14 i c d-5 d^2\right )-\frac{1}{4} a^3 d \left (18 c d+i \left (3 c^2-11 d^2\right )\right ) \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 a^2 d}\\ &=\frac{a (3 i c+5 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}+\left (2 a (c-i d)^2\right ) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx-\frac{1}{8} \left (3 c^2-18 i c d-11 d^2\right ) \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\\ &=\frac{a (3 i c+5 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (4 i a^3 (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{\left (a^2 \left (3 c^2-18 i c d-11 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{2 i \sqrt{2} a^{3/2} (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{a (3 i c+5 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (a \left (3 i c^2+18 c d-11 i d^2\right )\right ) \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 )}{4 f}\\ &=-\frac{2 i \sqrt{2} a^{3/2} (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{a (3 i c+5 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (a \left (3 i c^2+18 c d-11 i d^2\right )\right ) \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 )}{4 f}\\ &=-\frac{\sqrt [4]{-1} a^{3/2} \left (3 i c^2+18 c d-11 i d^2\right ) \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 )}{4 \sqrt{d} f}-\frac{2 i \sqrt{2} a^{3/2} (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{a (3 i c+5 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 6.10813, size = 574, normalized size = 1.82 \[ \frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (\cos (e)-i \sin (e)) \cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x))^{3/2} \left ((1+i) \sqrt{c+d \tan (e+f x)} (5 c+2 d \tan (e+f x)-5 i d)-\frac{\cos (e+f x) \left (\left (3 i c^2+18 c d-11 i d^2\right ) \left (\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1-i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}-c \left (e^{i (e+f x)}+i\right )+i d e^{i (e+f x)}+d\right )}{\sqrt{d} \left (-3 c^2+18 i c d+11 d^2\right ) \left (e^{i (e+f x)}+i\right )}\right )-\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt{d} \left (3 i c^2+18 c d-11 i d^2\right ) \left (e^{i (e+f x)}-i\right )}\right )\right )+(16+16 i) \sqrt{d} (c-i d)^{3/2} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )\right )}{\sqrt{d} \sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/8 + I/8)*Cos[e + f*x]*(Cos[e] - I*Sin[e])*(Cos[f*x] - I*Sin[f*x])*(a + I*a*Tan[e + f*x])^(3/2)*(-((Cos[e +
f*x]*(((3*I)*c^2 + 18*c*d - (11*I)*d^2)*(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)))]))/(Sqrt[d]*(-3*c^2 + (18*I)*c*d + 11*d^2)*(I + E^(I*(e + f*x))))] - 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)))]))/(Sqrt[d]*((3*I)*c^2 + 18*c*d - (11*I)*d^2)*(-I +
E^(I*(e + f*x))))]) + (16 + 16*I)*(c - I*d)^(3/2)*Sqrt[d]*Log[2*(Sqrt[c - I*d]*Cos[e + f*x] + I*Sqrt[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[d]*Sqrt[1 +
Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]])) + (1 + I)*Sqrt[c + d*Tan[e + f*x]]*(5*c - (5*I)*d + 2*d*Tan[e + f*x]
)))/f

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Maple [B]  time = 0.045, size = 1234, normalized size = 3.9 \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))^(3/2)*(c+d*tan(f*x+e))^(3/2),x)

[Out]

1/16/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a*(-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)*a*c^2-8*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*c+8*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*d+11*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^2+10*I*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c+8*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-8*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+18*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*c*d+10*(I*a
*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d-8*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))*a*d*2^(1/2)*(-a*(I*d
-c))^(1/2)+8*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+4*I*(I*a*d)^(1/2)*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)*d-8*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*ta
n(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+8*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*(I*a*d)^(1/2))/(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)

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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))^(3/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.99069, size = 2908, normalized size = 9.23 \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))^(3/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/8*(2*sqrt(2)*(5*I*a*c + 3*a*d + (5*I*a*c + 7*a*d)*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - (f*e^(2*I*f*x + 2*I*e
) + f)*sqrt((-9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d^3 - 121*I*a^3*d^4)/(d*f^2))*log((2
*d*f*sqrt((-9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d^3 - 121*I*a^3*d^4)/(d*f^2))*e^(2*I*f
*x + 2*I*e) + sqrt(2)*(3*a*c^2 - 18*I*a*c*d - 11*a*d^2 + (3*a*c^2 - 18*I*a*c*d - 11*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*
e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/(3*a*c^2 - 18*I*a*c*d - 11*a*d^2)) + (f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-
9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d^3 - 121*I*a^3*d^4)/(d*f^2))*log(-(2*d*f*sqrt((-9
*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d^3 - 121*I*a^3*d^4)/(d*f^2))*e^(2*I*f*x + 2*I*e) -
sqrt(2)*(3*a*c^2 - 18*I*a*c*d - 11*a*d^2 + (3*a*c^2 - 18*I*a*c*d - 11*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*
e))*e^(-2*I*f*x - 2*I*e)/(3*a*c^2 - 18*I*a*c*d - 11*a*d^2)) + 4*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(8*a^3*c^3 -
24*I*a^3*c^2*d - 24*a^3*c*d^2 + 8*I*a^3*d^3)/f^2)*log(1/2*(2*sqrt(2)*(-I*a*c - a*d + (-I*a*c - a*d)*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))*sqrt(a/(e^(2*I*f*x + 2*I
*e) + 1))*e^(I*f*x + I*e) + f*sqrt(-(8*a^3*c^3 - 24*I*a^3*c^2*d - 24*a^3*c*d^2 + 8*I*a^3*d^3)/f^2)*e^(2*I*f*x
+ 2*I*e))*e^(-2*I*f*x - 2*I*e)/(-I*a*c - a*d)) - 4*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(8*a^3*c^3 - 24*I*a^3*c^2
*d - 24*a^3*c*d^2 + 8*I*a^3*d^3)/f^2)*log(1/2*(2*sqrt(2)*(-I*a*c - a*d + (-I*a*c - a*d)*e^(2*I*f*x + 2*I*e))*s
qrt(((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(-(8*a^3*c^3 - 24*I*a^3*c^2*d - 24*a^3*c*d^2 + 8*I*a^3*d^3)/f^2)*e^(2*I*f*x + 2*I*e))*e^(
-2*I*f*x - 2*I*e)/(-I*a*c - a*d)))/(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))**(3/2)*(c+d*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 4.95281, size = 344, normalized size = 1.09 \begin{align*} -\frac{i \,{\left ({\left (a \tan \left (f x + e\right ) - i \, a\right )} a c -{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} d +{\left (i \, a \tan \left (f x + e\right ) + a\right )} a d\right )} \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} d^{2}{\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^{2} - 2 i \, a d^{2}\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*((a*tan(f*x + e) - I*a)*a*c - (I*a*tan(f*x + e) + a)^2*d + (I*a*tan(f*x + e) + a)*a*d)*sqrt(2*a^2*c + 2*sqr
t(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)*d^2*((-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^2 - 2*I*a*d^2)*a)