3.1150 \(\int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=378 \[ -\frac{\sqrt [4]{-1} a^{3/2} \left (45 c^2 d+5 i c^3-55 i c d^2-23 d^3\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{8 \sqrt{d} f}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{2 i \sqrt{2} a^{3/2} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}+\frac{a (7 d+5 i c) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a (c-3 i d) (3 d+5 i c) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 f} \]

[Out]

-((-1)^(1/4)*a^(3/2)*((5*I)*c^3 + 45*c^2*d - (55*I)*c*d^2 - 23*d^3)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*T
an[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/(8*Sqrt[d]*f) - ((2*I)*Sqrt[2]*a^(3/2)*(c - I*d)^(5/2)*ArcT
anh[(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*(c - (3*I)*
d)*((5*I)*c + 3*d)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(8*f) + (a*((5*I)*c + 7*d)*Sqrt[a + I*
a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(12*f) + (a^2*(c + I*d)*(c + d*Tan[e + f*x])^(5/2))/(3*d*f*Sqrt[a
+ I*a*Tan[e + f*x]]) - (a^2*(c + d*Tan[e + f*x])^(7/2))/(3*d*f*Sqrt[a + I*a*Tan[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.75567, antiderivative size = 378, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ -\frac{\sqrt [4]{-1} a^{3/2} \left (45 c^2 d+5 i c^3-55 i c d^2-23 d^3\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{8 \sqrt{d} f}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{2 i \sqrt{2} a^{3/2} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f}+\frac{a (7 d+5 i c) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a (c-3 i d) (3 d+5 i c) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

-((-1)^(1/4)*a^(3/2)*((5*I)*c^3 + 45*c^2*d - (55*I)*c*d^2 - 23*d^3)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*T
an[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/(8*Sqrt[d]*f) - ((2*I)*Sqrt[2]*a^(3/2)*(c - I*d)^(5/2)*ArcT
anh[(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*(c - (3*I)*
d)*((5*I)*c + 3*d)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(8*f) + (a*((5*I)*c + 7*d)*Sqrt[a + I*
a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(12*f) + (a^2*(c + I*d)*(c + d*Tan[e + f*x])^(5/2))/(3*d*f*Sqrt[a
+ I*a*Tan[e + f*x]]) - (a^2*(c + d*Tan[e + f*x])^(7/2))/(3*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))^{5/2} \, dx &=-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}+\frac{a \int \frac{\left (-\frac{1}{2} a (i c-13 d)-\frac{1}{2} a (c-11 i d) \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx}{3 d}\\ &=\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{\int \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \left (-\frac{1}{2} a^2 (7 c-5 i d) d-\frac{1}{2} a^2 d (5 i c+7 d) \tan (e+f x)\right ) \, dx}{3 a d}\\ &=\frac{a (5 i c+7 d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 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{3}{4} a^3 d \left (11 c^2-14 i c d-7 d^2\right )-\frac{3}{4} a^3 d \left (18 c d+i \left (5 c^2-9 d^2\right )\right ) \tan (e+f x)\right ) \, dx}{6 a^2 d}\\ &=\frac{a (c-3 i d) (5 i c+3 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 f}+\frac{a (5 i c+7 d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (-\frac{3}{8} a^4 (3 c-i d) d \left (9 c^2-14 i c d-9 d^2\right )-\frac{3}{8} a^4 d \left (5 i c^3+45 c^2 d-55 i c d^2-23 d^3\right ) \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{6 a^3 d}\\ &=\frac{a (c-3 i d) (5 i c+3 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 f}+\frac{a (5 i c+7 d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}+\left (2 a (c-i d)^3\right ) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx-\frac{1}{16} \left (5 c^3-45 i c^2 d-55 c d^2+23 i d^3\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 (c-3 i d) (5 i c+3 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 f}+\frac{a (5 i c+7 d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (4 a^3 (i c+d)^3\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 (5 c^3-45 i c^2 d-55 c d^2+23 i d^3\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{16 f}\\ &=-\frac{2 i \sqrt{2} a^{3/2} (c-i d)^{5/2} \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 (c-3 i d) (5 i c+3 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 f}+\frac{a (5 i c+7 d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (a \left (5 i c^3+45 c^2 d-55 i c d^2-23 d^3\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 )}{8 f}\\ &=-\frac{2 i \sqrt{2} a^{3/2} (c-i d)^{5/2} \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 (c-3 i d) (5 i c+3 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 f}+\frac{a (5 i c+7 d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (a \left (5 i c^3+45 c^2 d-55 i c d^2-23 d^3\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 )}{8 f}\\ &=-\frac{\sqrt [4]{-1} a^{3/2} \left (5 i c^3+45 c^2 d-55 i c d^2-23 d^3\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 )}{8 \sqrt{d} f}-\frac{2 i \sqrt{2} a^{3/2} (c-i d)^{5/2} \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 (c-3 i d) (5 i c+3 d) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 f}+\frac{a (5 i c+7 d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 f}+\frac{a^2 (c+i d) (c+d \tan (e+f x))^{5/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}-\frac{a^2 (c+d \tan (e+f x))^{7/2}}{3 d f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 8.08611, size = 645, normalized size = 1.71 \[ \frac{i (\cos (e)-i \sin (e)) \sec (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x))^{3/2} \left (\sqrt{c+d \tan (e+f x)} \left (\left (33 c^2-68 i c d-35 d^2\right ) \cos (2 (e+f x))+33 c^2+2 d (13 c-7 i d) \sin (2 (e+f x))-68 i c d-19 d^2\right )-\frac{(3-3 i) \cos ^3(e+f x) \left (\left (45 c^2 d+5 i c^3-55 i c d^2-23 d^3\right ) \left (\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}-i c e^{i (e+f x)}+c-d e^{i (e+f x)}+i d\right )}{\sqrt{d} \left (-45 c^2 d-5 i c^3+55 i c d^2+23 d^3\right ) \left (e^{i (e+f x)}+i\right )}\right )-\log \left (-\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt{d} \left (-45 c^2 d-5 i c^3+55 i c d^2+23 d^3\right ) \left (e^{i (e+f x)}-i\right )}\right )\right )+(32+32 i) \sqrt{d} (c-i d)^{5/2} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )\right )}{\sqrt{d} \sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{48 f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((I/48)*Sec[e + f*x]*(Cos[e] - I*Sin[e])*(Cos[f*x] - I*Sin[f*x])*(a + I*a*Tan[e + f*x])^(3/2)*(((-3 + 3*I)*Cos
[e + f*x]^3*(((5*I)*c^3 + 45*c^2*d - (55*I)*c*d^2 - 23*d^3)*(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]*((-5*I)*c^3 - 45*c^2*d + (55*I)*c*d^2 + 23*d^3)*(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]*Sq
rt[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]*((
-5*I)*c^3 - 45*c^2*d + (55*I)*c*d^2 + 23*d^3)*(-I + E^(I*(e + f*x))))]) + (32 + 32*I)*(c - I*d)^(5/2)*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)]]) + (33*c^2 - (68*I)
*c*d - 19*d^2 + (33*c^2 - (68*I)*c*d - 35*d^2)*Cos[2*(e + f*x)] + 2*(13*c - (7*I)*d)*d*Sin[2*(e + f*x)])*Sqrt[
c + d*Tan[e + f*x]]))/f

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Maple [B]  time = 0.046, size = 1503, normalized size = 4. \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))^(5/2),x)

[Out]

1/96/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a*(16*I*2^(1/2)*tan(f*x+e)^2*d^2*(I*a*d)^(1/2)*(-a*(I
*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+52*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)*c*d+48*I*(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*d+48*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-69*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^3-48*I*ln(1/2*(2*I*a*ta
n(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-15*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^3+165*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+28*(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+66*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^2-48*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+136*(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-54*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)*d^2+135*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^2*d+48*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-48*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-48*(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+48*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))*2^(1/2)/(a*(c+d*tan(f*x+e))
*(1+I*tan(f*x+e)))^(1/2)/(I*a*d)^(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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.14836, size = 3930, normalized size = 10.4 \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))^(5/2),x, algorithm="fricas")

[Out]

1/48*(2*sqrt(2)*(33*I*a*c^2 + 42*a*c*d - 21*I*a*d^2 + (33*I*a*c^2 + 94*a*c*d - 49*I*a*d^2)*e^(4*I*f*x + 4*I*e)
+ (66*I*a*c^2 + 136*a*c*d - 38*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + 3*(f*e^(4*I*f*x + 4*I*e) + 2*f*e
^(2*I*f*x + 2*I*e) + f)*sqrt((-25*I*a^3*c^6 - 450*a^3*c^5*d + 2575*I*a^3*c^4*d^2 + 5180*a^3*c^3*d^3 - 5095*I*a
^3*c^2*d^4 - 2530*a^3*c*d^5 + 529*I*a^3*d^6)/(d*f^2))*log((2*I*d*f*sqrt((-25*I*a^3*c^6 - 450*a^3*c^5*d + 2575*
I*a^3*c^4*d^2 + 5180*a^3*c^3*d^3 - 5095*I*a^3*c^2*d^4 - 2530*a^3*c*d^5 + 529*I*a^3*d^6)/(d*f^2))*e^(2*I*f*x +
2*I*e) + sqrt(2)*(-5*I*a*c^3 - 45*a*c^2*d + 55*I*a*c*d^2 + 23*a*d^3 + (-5*I*a*c^3 - 45*a*c^2*d + 55*I*a*c*d^2
+ 23*a*d^3)*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))*sqr
t(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/(-5*I*a*c^3 - 45*a*c^2*d + 55*I*a*c*d^2 +
23*a*d^3)) - 3*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-25*I*a^3*c^6 - 450*a^3*c^5*d + 25
75*I*a^3*c^4*d^2 + 5180*a^3*c^3*d^3 - 5095*I*a^3*c^2*d^4 - 2530*a^3*c*d^5 + 529*I*a^3*d^6)/(d*f^2))*log((-2*I*
d*f*sqrt((-25*I*a^3*c^6 - 450*a^3*c^5*d + 2575*I*a^3*c^4*d^2 + 5180*a^3*c^3*d^3 - 5095*I*a^3*c^2*d^4 - 2530*a^
3*c*d^5 + 529*I*a^3*d^6)/(d*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(2)*(-5*I*a*c^3 - 45*a*c^2*d + 55*I*a*c*d^2 + 23*a
*d^3 + (-5*I*a*c^3 - 45*a*c^2*d + 55*I*a*c*d^2 + 23*a*d^3)*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)/(-5*I*a*c^3 - 45*a*c^2*d + 55*I*a*c*d^2 + 23*a*d^3)) + 24*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*
e) + f)*sqrt(-(8*a^3*c^5 - 40*I*a^3*c^4*d - 80*a^3*c^3*d^2 + 80*I*a^3*c^2*d^3 + 40*a^3*c*d^4 - 8*I*a^3*d^5)/f^
2)*log((sqrt(2)*(2*a*c^2 - 4*I*a*c*d - 2*a*d^2 + (2*a*c^2 - 4*I*a*c*d - 2*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) + I*f*sqrt(-(8*a^3*c^5 - 40*I*a^3*c^4*d - 80*a^3*c^3*d^2 + 80*I*a^3*c^2*d^3 + 40*a^3*c*d^4 - 8*I*a^3*d^5
)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(2*a*c^2 - 4*I*a*c*d - 2*a*d^2)) - 24*(f*e^(4*I*f*x + 4*I*e)
+ 2*f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(8*a^3*c^5 - 40*I*a^3*c^4*d - 80*a^3*c^3*d^2 + 80*I*a^3*c^2*d^3 + 40*a^3*
c*d^4 - 8*I*a^3*d^5)/f^2)*log((sqrt(2)*(2*a*c^2 - 4*I*a*c*d - 2*a*d^2 + (2*a*c^2 - 4*I*a*c*d - 2*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) - I*f*sqrt(-(8*a^3*c^5 - 40*I*a^3*c^4*d - 80*a^3*c^3*d^2 + 80*I*a^3*c^2*d^3 + 40*
a^3*c*d^4 - 8*I*a^3*d^5)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(2*a*c^2 - 4*I*a*c*d - 2*a*d^2)))/(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))**(3/2)*(c+d*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 5.48738, size = 431, normalized size = 1.14 \begin{align*} -\frac{{\left ({\left (a \tan \left (f x + e\right ) - i \, a\right )} a^{2} c^{2} - 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} a c d + 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a^{2} c d + i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} d^{2} - 2 i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} a d^{2} + i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )} a^{2} d^{2}\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{\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 \, a \tan \left (f x + e\right ) + a\right )} a^{2} d - 2 \, a^{3} d} \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))^(5/2),x, algorithm="giac")

[Out]

-((a*tan(f*x + e) - I*a)*a^2*c^2 - 2*(I*a*tan(f*x + e) + a)^2*a*c*d + 2*(I*a*tan(f*x + e) + a)*a^2*c*d + I*(I*
a*tan(f*x + e) + a)^3*d^2 - 2*I*(I*a*tan(f*x + e) + a)^2*a*d^2 + I*(I*a*tan(f*x + e) + a)*a^2*d^2)*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)*d*((-I*(I*a*t
an(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*a*tan(f*x + e) + a)*a^2*d - 2*a^3*d)