3.1143 \(\int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=329 \[ \frac{a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d f}-\frac{\sqrt [4]{-1} a^{5/2} (c-3 i d) \left (c^2+18 i c d+15 d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{8 d^{3/2} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{4 i \sqrt{2} a^{5/2} (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f} \]

[Out]

-((-1)^(1/4)*a^(5/2)*(c - (3*I)*d)*(c^2 + (18*I)*c*d + 15*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]])])/(8*d^(3/2)*f) - ((4*I)*Sqrt[2]*a^(5/2)*(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^2*(c^2 + (14*I)*
c*d + 19*d^2)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(8*d*f) + (a^2*(c + (13*I)*d)*Sqrt[a + I*a*
Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(12*d*f) - (a^2*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2
))/(3*d*f)

________________________________________________________________________________________

Rubi [A]  time = 1.35906, antiderivative size = 329, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.281, Rules used = {3556, 3597, 3601, 3544, 208, 3599, 63, 217, 206} \[ \frac{a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d f}-\frac{\sqrt [4]{-1} a^{5/2} (c-3 i d) \left (c^2+18 i c d+15 d^2\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{8 d^{3/2} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{4 i \sqrt{2} a^{5/2} (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-1)^(1/4)*a^(5/2)*(c - (3*I)*d)*(c^2 + (18*I)*c*d + 15*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]])])/(8*d^(3/2)*f) - ((4*I)*Sqrt[2]*a^(5/2)*(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^2*(c^2 + (14*I)*
c*d + 19*d^2)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(8*d*f) + (a^2*(c + (13*I)*d)*Sqrt[a + I*a*
Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(12*d*f) - (a^2*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2
))/(3*d*f)

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 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))^{5/2} (c+d \tan (e+f x))^{3/2} \, dx &=-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{a \int \sqrt{a+i a \tan (e+f x)} \left (\frac{1}{2} a (i c+11 d)+\frac{1}{2} a (c+13 i d) \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2} \, dx}{3 d}\\ &=\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\int \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (\frac{3}{4} a^2 \left (18 c d+i \left (c^2-13 d^2\right )\right )+\frac{3}{4} a^2 \left (c^2+14 i c d+19 d^2\right ) \tan (e+f x)\right ) \, dx}{6 d}\\ &=\frac{a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d f}+\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (\frac{3}{8} a^3 \left (i c^3+49 c^2 d-59 i c d^2-19 d^3\right )+\frac{3}{8} a^3 (c-3 i d) \left (c^2+18 i c d+15 d^2\right ) \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{6 a d}\\ &=\frac{a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d f}+\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\left (4 a^2 (c-i d)^2\right ) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{\left (a (i c+3 d) \left (c^2+18 i c d+15 d^2\right )\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}{16 d}\\ &=\frac{a^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d f}+\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}-\frac{\left (8 i a^4 (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^3 (i c+3 d) \left (c^2+18 i c d+15 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 )}{16 d f}\\ &=-\frac{4 i \sqrt{2} a^{5/2} (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^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d f}+\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\left (a^2 (c-3 i d) \left (c^2+18 i c d+15 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 )}{8 d f}\\ &=-\frac{4 i \sqrt{2} a^{5/2} (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^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d f}+\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}+\frac{\left (a^2 (c-3 i d) \left (c^2+18 i c d+15 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 )}{8 d f}\\ &=-\frac{\sqrt [4]{-1} a^{5/2} (c-3 i d) \left (c^2+18 i c d+15 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 )}{8 d^{3/2} f}-\frac{4 i \sqrt{2} a^{5/2} (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^2 \left (c^2+14 i c d+19 d^2\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{8 d f}+\frac{a^2 (c+13 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{12 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{3 d f}\\ \end{align*}

Mathematica [B]  time = 9.0701, size = 686, normalized size = 2.09 \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) \cos ^2(e+f x) (a+i a \tan (e+f x))^{5/2} \left (\frac{\left (\frac{1}{6}+\frac{i}{6}\right ) (\sin (2 e)+i \cos (2 e)) \sec ^2(e+f x) \sqrt{c+d \tan (e+f x)} \left (\left (3 c^2-68 i c d-65 d^2\right ) \cos (2 (e+f x))+3 c^2+2 d (7 c-13 i d) \sin (2 (e+f x))-68 i c d-49 d^2\right )}{d}-\frac{i (\cos (2 e)-i \sin (2 e)) \cos (e+f x) \left (\left (15 c^2 d-i c^3-69 i c d^2-45 d^3\right ) \left (\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left (-(1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{\sqrt{d} \left (-15 c^2 d+i c^3+69 i c d^2+45 d^3\right ) \left (e^{i (e+f x)}+i\right )}\right )-\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt{d} \left (-15 c^2 d+i c^3+69 i c d^2+45 d^3\right ) \left (e^{i (e+f x)}-i\right )}\right )\right )+(64-64 i) d^{3/2} (c-i d)^{3/2} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )\right )}{d^{3/2} \sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{f (\cos (f x)+i \sin (f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/16 + I/16)*Cos[e + f*x]^2*(a + I*a*Tan[e + f*x])^(5/2)*(((-I)*Cos[e + f*x]*(((-I)*c^3 + 15*c^2*d - (69*I)*
c*d^2 - 45*d^3)*(Log[((2 + 2*I)*E^((I/2)*e)*((-I)*d + d*E^(I*(e + f*x)) + I*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]*(I*c^3 - 15*c^2*d + (69*I)*c*d^2 + 45*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]*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]*(I*c^3 - 15*c^2*d + (69*I)*c*d^2 + 45*d^3)*(-I
+ E^(I*(e + f*x))))]) + (64 - 64*I)*(c - I*d)^(3/2)*d^(3/2)*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]])])*(Cos[2*e] - I*Sin
[2*e]))/(d^(3/2)*Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]]) + ((1/6 + I/6)*Sec[e + f*x]^2*(I*Cos[2*e] +
Sin[2*e])*(3*c^2 - (68*I)*c*d - 49*d^2 + (3*c^2 - (68*I)*c*d - 65*d^2)*Cos[2*(e + f*x)] + 2*(7*c - (13*I)*d)*d
*Sin[2*(e + f*x)])*Sqrt[c + d*Tan[e + f*x]])/d))/(f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B]  time = 0.083, size = 1518, normalized size = 4.6 \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))^(5/2)*(c+d*tan(f*x+e))^(3/2),x)

[Out]

1/96/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^2*(-16*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)-28*2^(1/2)*tan(f*x+e)*c*d*(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)+207*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^2+136*I*(a*(c+d*t
an(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+96*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^2-96*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+96*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+52*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
)*d^2-45*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+135*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+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*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^3-96*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^2+96*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-96*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^2-6*(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^2+114*(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^2-96*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))*(I*a*d)^(1/2)*a*c*d+96*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))*(I*a*d)^(1/2)*a*d^2)*2^(1/2)/d/(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))^(5/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.09736, size = 3725, normalized size = 11.32 \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))^(5/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/48*(2*sqrt(2)*(3*a^2*c^2 - 54*I*a^2*c*d - 39*a^2*d^2 + (3*a^2*c^2 - 82*I*a^2*c*d - 91*a^2*d^2)*e^(4*I*f*x +
4*I*e) + (6*a^2*c^2 - 136*I*a^2*c*d - 98*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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - 3*(d*f*e^(4*I*f*x + 4*
I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt((I*a^5*c^6 - 30*a^5*c^5*d - 87*I*a^5*c^4*d^2 - 1980*a^5*c^3*d^3 +
6111*I*a^5*c^2*d^4 + 6210*a^5*c*d^5 - 2025*I*a^5*d^6)/(d^3*f^2))*log((2*d^2*f*sqrt((I*a^5*c^6 - 30*a^5*c^5*d
- 87*I*a^5*c^4*d^2 - 1980*a^5*c^3*d^3 + 6111*I*a^5*c^2*d^4 + 6210*a^5*c*d^5 - 2025*I*a^5*d^6)/(d^3*f^2))*e^(2*
I*f*x + 2*I*e) + sqrt(2)*(I*a^2*c^3 - 15*a^2*c^2*d + 69*I*a^2*c*d^2 + 45*a^2*d^3 + (I*a^2*c^3 - 15*a^2*c^2*d +
69*I*a^2*c*d^2 + 45*a^2*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)/(I*a^2*c^3 - 15*a^2*c^2
*d + 69*I*a^2*c*d^2 + 45*a^2*d^3)) + 3*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt((I*a^5
*c^6 - 30*a^5*c^5*d - 87*I*a^5*c^4*d^2 - 1980*a^5*c^3*d^3 + 6111*I*a^5*c^2*d^4 + 6210*a^5*c*d^5 - 2025*I*a^5*d
^6)/(d^3*f^2))*log(-(2*d^2*f*sqrt((I*a^5*c^6 - 30*a^5*c^5*d - 87*I*a^5*c^4*d^2 - 1980*a^5*c^3*d^3 + 6111*I*a^5
*c^2*d^4 + 6210*a^5*c*d^5 - 2025*I*a^5*d^6)/(d^3*f^2))*e^(2*I*f*x + 2*I*e) - sqrt(2)*(I*a^2*c^3 - 15*a^2*c^2*d
+ 69*I*a^2*c*d^2 + 45*a^2*d^3 + (I*a^2*c^3 - 15*a^2*c^2*d + 69*I*a^2*c*d^2 + 45*a^2*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)/(I*a^2*c^3 - 15*a^2*c^2*d + 69*I*a^2*c*d^2 + 45*a^2*d^3)) - 24*(d*f*e^(4*
I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(32*a^5*c^3 - 96*I*a^5*c^2*d - 96*a^5*c*d^2 + 32*I*a^5
*d^3)/f^2)*log(1/4*(4*sqrt(2)*(-I*a^2*c - a^2*d + (-I*a^2*c - a^2*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*sq
rt(-(32*a^5*c^3 - 96*I*a^5*c^2*d - 96*a^5*c*d^2 + 32*I*a^5*d^3)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)
/(-I*a^2*c - a^2*d)) + 24*(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt(-(32*a^5*c^3 - 96*I
*a^5*c^2*d - 96*a^5*c*d^2 + 32*I*a^5*d^3)/f^2)*log(1/4*(4*sqrt(2)*(-I*a^2*c - a^2*d + (-I*a^2*c - a^2*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(-(32*a^5*c^3 - 96*I*a^5*c^2*d - 96*a^5*c*d^2 + 32*I*a^5*d^3)/f^2)*e^(2*
I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(-I*a^2*c - a^2*d)))/(d*f*e^(4*I*f*x + 4*I*e) + 2*d*f*e^(2*I*f*x + 2*I*e)
+ 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))**(5/2)*(c+d*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 5.44467, size = 338, normalized size = 1.03 \begin{align*} \frac{{\left (2 i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} a c + 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} d - 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} 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}{\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 )}{2 \,{\left ({\left (i \, a \tan \left (f x + e\right ) + a\right )} a - 2 \, a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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