3.1149 \(\int (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=415 \[ \frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 \left (95 i c^2 d+5 c^3+273 c d^2-149 i d^3\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{64 d f}-\frac{\sqrt [4]{-1} a^{5/2} \left (690 c^2 d^2+100 i c^3 d+5 c^4-900 i c d^3-363 d^4\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{64 d^{3/2} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{4 i \sqrt{2} a^{5/2} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f} \]

[Out]

-((-1)^(1/4)*a^(5/2)*(5*c^4 + (100*I)*c^3*d + 690*c^2*d^2 - (900*I)*c*d^3 - 363*d^4)*ArcTanh[((-1)^(3/4)*Sqrt[
d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/(64*d^(3/2)*f) - ((4*I)*Sqrt[2]*a^(5/2)*(c
- I*d)^(5/2)*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*(5*c^3 + (95*I)*c^2*d + 273*c*d^2 - (149*I)*d^3)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])
/(64*d*f) + (a^2*(5*c^2 + (90*I)*c*d + 107*d^2)*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(96*d*f
) + (a^2*(c + (17*I)*d)*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(24*d*f) - (a^2*Sqrt[a + I*a*Ta
n[e + f*x]]*(c + d*Tan[e + f*x])^(7/2))/(4*d*f)

________________________________________________________________________________________

Rubi [A]  time = 1.72307, antiderivative size = 415, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ \frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 \left (95 i c^2 d+5 c^3+273 c d^2-149 i d^3\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{64 d f}-\frac{\sqrt [4]{-1} a^{5/2} \left (690 c^2 d^2+100 i c^3 d+5 c^4-900 i c d^3-363 d^4\right ) \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{64 d^{3/2} f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{4 i \sqrt{2} a^{5/2} (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-1)^(1/4)*a^(5/2)*(5*c^4 + (100*I)*c^3*d + 690*c^2*d^2 - (900*I)*c*d^3 - 363*d^4)*ArcTanh[((-1)^(3/4)*Sqrt[
d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/(64*d^(3/2)*f) - ((4*I)*Sqrt[2]*a^(5/2)*(c
- I*d)^(5/2)*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*(5*c^3 + (95*I)*c^2*d + 273*c*d^2 - (149*I)*d^3)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])
/(64*d*f) + (a^2*(5*c^2 + (90*I)*c*d + 107*d^2)*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(96*d*f
) + (a^2*(c + (17*I)*d)*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(5/2))/(24*d*f) - (a^2*Sqrt[a + I*a*Ta
n[e + f*x]]*(c + d*Tan[e + f*x])^(7/2))/(4*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))^{5/2} \, dx &=-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{a \int \sqrt{a+i a \tan (e+f x)} \left (\frac{1}{2} a (i c+15 d)+\frac{1}{2} a (c+17 i d) \tan (e+f x)\right ) (c+d \tan (e+f x))^{5/2} \, dx}{4 d}\\ &=\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{\int \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \left (\frac{1}{4} a^2 \left (5 i c^2+102 c d-85 i d^2\right )+\frac{1}{4} a^2 \left (5 c^2+90 i c d+107 d^2\right ) \tan (e+f x)\right ) \, dx}{12 d}\\ &=\frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{\int \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)} \left (\frac{3}{8} a^3 \left (5 i c^3+161 c^2 d-239 i c d^2-107 d^3\right )+\frac{3}{8} a^3 \left (5 c^3+95 i c^2 d+273 c d^2-149 i d^3\right ) \tan (e+f x)\right ) \, dx}{24 a d}\\ &=\frac{a^2 \left (5 c^3+95 i c^2 d+273 c d^2-149 i d^3\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{64 d f}+\frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (\frac{3}{16} a^4 \left (5 i c^4+412 c^3 d-846 i c^2 d^2-636 c d^3+149 i d^4\right )+\frac{3}{16} a^4 \left (5 c^4+100 i c^3 d+690 c^2 d^2-900 i c d^3-363 d^4\right ) \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{24 a^2 d}\\ &=\frac{a^2 \left (5 c^3+95 i c^2 d+273 c d^2-149 i d^3\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{64 d f}+\frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\left (4 a^2 (c-i d)^3\right ) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{\left (a \left (5 i c^4-100 c^3 d+690 i c^2 d^2+900 c d^3-363 i d^4\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}{128 d}\\ &=\frac{a^2 \left (5 c^3+95 i c^2 d+273 c d^2-149 i d^3\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{64 d f}+\frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{\left (8 a^4 (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^3 \left (5 i c^4-100 c^3 d+690 i c^2 d^2+900 c d^3-363 i d^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{128 d f}\\ &=-\frac{4 i \sqrt{2} a^{5/2} (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^2 \left (5 c^3+95 i c^2 d+273 c d^2-149 i d^3\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{64 d f}+\frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{\left (a^2 \left (5 c^4+100 i c^3 d+690 c^2 d^2-900 i c d^3-363 d^4\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 )}{64 d f}\\ &=-\frac{4 i \sqrt{2} a^{5/2} (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^2 \left (5 c^3+95 i c^2 d+273 c d^2-149 i d^3\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{64 d f}+\frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}+\frac{\left (a^2 \left (5 c^4+100 i c^3 d+690 c^2 d^2-900 i c d^3-363 d^4\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 )}{64 d f}\\ &=-\frac{\sqrt [4]{-1} a^{5/2} \left (5 c^4+100 i c^3 d+690 c^2 d^2-900 i c d^3-363 d^4\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 )}{64 d^{3/2} f}-\frac{4 i \sqrt{2} a^{5/2} (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^2 \left (5 c^3+95 i c^2 d+273 c d^2-149 i d^3\right ) \sqrt{a+i a \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{64 d f}+\frac{a^2 \left (5 c^2+90 i c d+107 d^2\right ) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{96 d f}+\frac{a^2 (c+17 i d) \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}}{24 d f}-\frac{a^2 \sqrt{a+i a \tan (e+f x)} (c+d \tan (e+f x))^{7/2}}{4 d f}\\ \end{align*}

Mathematica [B]  time = 12.2592, size = 849, normalized size = 2.05 \[ \frac{\cos ^2(e+f x) \sqrt{\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))} \left (\left (-\frac{1}{4} i \cos (3 e+f x) d^2-\frac{1}{4} \sin (3 e+f x) d^2\right ) \sec ^3(e+f x)+(17 i c+23 d) \left (\frac{1}{24} i d \cos (2 e)+\frac{1}{24} d \sin (2 e)\right ) \sec ^2(e+f x)+\left (59 c^2-226 i d c-131 d^2\right ) \left (-\frac{1}{96} i \cos (3 e+f x)-\frac{1}{96} \sin (3 e+f x)\right ) \sec (e+f x)+\left (-15 c^3+719 i d c^2+1621 d^2 c-845 i d^3\right ) \left (\frac{\cos (2 e)}{192 d}-\frac{i \sin (2 e)}{192 d}\right )\right ) (i \tan (e+f x) a+a)^{5/2}}{f (\cos (f x)+i \sin (f x))^2}-\frac{\left (\frac{1}{128}+\frac{i}{128}\right ) \cos ^3(e+f x) \left ((512+512 i) d^{3/2} \log \left (2 \left (\sqrt{c-i d} \cos (e+f x)+i \sqrt{c-i d} \sin (e+f x)+\sqrt{\cos (2 e+2 f x)+i \sin (2 e+2 f x)+1} \sqrt{c+d \tan (e+f x)}\right )\right ) (c-i d)^{5/2}+\left (5 c^4+100 i d c^3+690 d^2 c^2-900 i d^3 c-363 d^4\right ) \left (\log \left (\frac{(2+2 i) e^{\frac{i e}{2}} \left (-i e^{i (e+f x)} c+c-d e^{i (e+f x)}+i d+(1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{\sqrt{d} \left (-5 c^4-100 i d c^3-690 d^2 c^2+900 i d^3 c+363 d^4\right ) \left (i+e^{i (e+f x)}\right )}\right )-\log \left (-\frac{(2+2 i) e^{\frac{i e}{2}} \left (i e^{i (e+f x)} c+c+d e^{i (e+f x)}+i d+(1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{\sqrt{d} \left (-5 c^4-100 i d c^3-690 d^2 c^2+900 i d^3 c+363 d^4\right ) \left (-i+e^{i (e+f x)}\right )}\right )\right )\right ) (\cos (2 e)-i \sin (2 e)) (i \tan (e+f x) a+a)^{5/2}}{d^{3/2} f (\cos (f x)+i \sin (f x))^2 \sqrt{\cos (2 (e+f x))+i \sin (2 (e+f x))+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/128 - I/128)*Cos[e + f*x]^3*((5*c^4 + (100*I)*c^3*d + 690*c^2*d^2 - (900*I)*c*d^3 - 363*d^4)*(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*c^4 - (100*I)*c^3*d -
690*c^2*d^2 + (900*I)*c*d^3 + 363*d^4)*(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]*(-5*c^4 - (100*I)*c^3*d - 690*c^2*d^2 + (900*I)*c*d^3 + 3
63*d^4)*(-I + E^(I*(e + f*x))))]) + (512 + 512*I)*(c - I*d)^(5/2)*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 + 2*f*x] + I*Sin[2*e + 2*f*x]]*Sqrt[c + d*Tan[e + f*x]])])*(Co
s[2*e] - I*Sin[2*e])*(a + I*a*Tan[e + f*x])^(5/2))/(d^(3/2)*f*(Cos[f*x] + I*Sin[f*x])^2*Sqrt[1 + Cos[2*(e + f*
x)] + I*Sin[2*(e + f*x)]]) + (Cos[e + f*x]^2*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] + d*Sin[e + f*x])]*((-15*c^3 +
(719*I)*c^2*d + 1621*c*d^2 - (845*I)*d^3)*(Cos[2*e]/(192*d) - ((I/192)*Sin[2*e])/d) + ((17*I)*c + 23*d)*Sec[e
+ f*x]^2*((I/24)*d*Cos[2*e] + (d*Sin[2*e])/24) + (59*c^2 - (226*I)*c*d - 131*d^2)*Sec[e + f*x]*((-I/96)*Cos[3*
e + f*x] - Sin[3*e + f*x]/96) + Sec[e + f*x]^3*((-I/4)*d^2*Cos[3*e + f*x] - (d^2*Sin[3*e + f*x])/4))*(a + I*a*
Tan[e + f*x])^(5/2))/(f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [B]  time = 0.084, size = 1851, normalized size = 4.5 \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))^(5/2),x)

[Out]

1/768/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^2*(-96*2^(1/2)*tan(f*x+e)^3*d^3*(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)-272*2^(1/2)*tan(f*x+e)^2*c*d^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)-236*2^(1/2)*tan(f*x+e)*c^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)+904*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+768*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^2-768
*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+768*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-1089*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^4+272*I*2^(1/2)*tan(f*x+e)^2*d^3*(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)+15*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^4-300*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*d+2
700*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^3+428*(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^3+2070*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^2+1202*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)*c^2*d-30*(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^3+2066*(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^2-768*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^2-894*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^3+768*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-768*
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^2-768*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+768*ln((3*a*c+I*a*ta
n(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))/(t
an(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)

________________________________________________________________________________________

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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.73679, size = 5017, normalized size = 12.09 \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))^(5/2),x, algorithm="fricas")

[Out]

-1/384*(2*sqrt(2)*(15*a^2*c^3 - 483*I*a^2*c^2*d - 717*a^2*c*d^2 + 321*I*a^2*d^3 + (15*a^2*c^3 - 719*I*a^2*c^2*
d - 1621*a^2*c*d^2 + 845*I*a^2*d^3)*e^(6*I*f*x + 6*I*e) + (45*a^2*c^3 - 1921*I*a^2*c^2*d - 3415*a^2*c*d^2 + 12
75*I*a^2*d^3)*e^(4*I*f*x + 4*I*e) + (45*a^2*c^3 - 1685*I*a^2*c^2*d - 2511*a^2*c*d^2 + 1135*I*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) - 3*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e
) + d*f)*sqrt((25*I*a^5*c^8 - 1000*a^5*c^7*d - 3100*I*a^5*c^6*d^2 - 129000*a^5*c^5*d^3 + 652470*I*a^5*c^4*d^4
+ 1314600*a^5*c^3*d^5 - 1310940*I*a^5*c^2*d^6 - 653400*a^5*c*d^7 + 131769*I*a^5*d^8)/(d^3*f^2))*log(-(2*I*d^2*
f*sqrt((25*I*a^5*c^8 - 1000*a^5*c^7*d - 3100*I*a^5*c^6*d^2 - 129000*a^5*c^5*d^3 + 652470*I*a^5*c^4*d^4 + 13146
00*a^5*c^3*d^5 - 1310940*I*a^5*c^2*d^6 - 653400*a^5*c*d^7 + 131769*I*a^5*d^8)/(d^3*f^2))*e^(2*I*f*x + 2*I*e) -
sqrt(2)*(5*a^2*c^4 + 100*I*a^2*c^3*d + 690*a^2*c^2*d^2 - 900*I*a^2*c*d^3 - 363*a^2*d^4 + (5*a^2*c^4 + 100*I*a
^2*c^3*d + 690*a^2*c^2*d^2 - 900*I*a^2*c*d^3 - 363*a^2*d^4)*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*a^2*c^4 + 100*I*a^2*c^3*d + 690*a^2*c^2*d^2 - 900*I*a^2*c*d^3 - 363*a^2*d^4)) + 3*(d*f*e^(6*I*f*x +
6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f)*sqrt((25*I*a^5*c^8 - 1000*a^5*c^7*d - 31
00*I*a^5*c^6*d^2 - 129000*a^5*c^5*d^3 + 652470*I*a^5*c^4*d^4 + 1314600*a^5*c^3*d^5 - 1310940*I*a^5*c^2*d^6 - 6
53400*a^5*c*d^7 + 131769*I*a^5*d^8)/(d^3*f^2))*log(-(-2*I*d^2*f*sqrt((25*I*a^5*c^8 - 1000*a^5*c^7*d - 3100*I*a
^5*c^6*d^2 - 129000*a^5*c^5*d^3 + 652470*I*a^5*c^4*d^4 + 1314600*a^5*c^3*d^5 - 1310940*I*a^5*c^2*d^6 - 653400*
a^5*c*d^7 + 131769*I*a^5*d^8)/(d^3*f^2))*e^(2*I*f*x + 2*I*e) - sqrt(2)*(5*a^2*c^4 + 100*I*a^2*c^3*d + 690*a^2*
c^2*d^2 - 900*I*a^2*c*d^3 - 363*a^2*d^4 + (5*a^2*c^4 + 100*I*a^2*c^3*d + 690*a^2*c^2*d^2 - 900*I*a^2*c*d^3 - 3
63*a^2*d^4)*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*a^2*c^4 + 100*I*a^2*c^3*d + 690*a^2*c^
2*d^2 - 900*I*a^2*c*d^3 - 363*a^2*d^4)) - 192*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(
2*I*f*x + 2*I*e) + d*f)*sqrt(-(32*a^5*c^5 - 160*I*a^5*c^4*d - 320*a^5*c^3*d^2 + 320*I*a^5*c^2*d^3 + 160*a^5*c*
d^4 - 32*I*a^5*d^5)/f^2)*log((sqrt(2)*(4*a^2*c^2 - 8*I*a^2*c*d - 4*a^2*d^2 + (4*a^2*c^2 - 8*I*a^2*c*d - 4*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) + I*f*sqrt(-(32*a^5*c^5 - 160*I*a^5*c^4*d - 320*a^5*c^3*d^2 + 320*I*a^
5*c^2*d^3 + 160*a^5*c*d^4 - 32*I*a^5*d^5)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(4*a^2*c^2 - 8*I*a^2*
c*d - 4*a^2*d^2)) + 192*(d*f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*d*f*e^(2*I*f*x + 2*I*e) + d*f
)*sqrt(-(32*a^5*c^5 - 160*I*a^5*c^4*d - 320*a^5*c^3*d^2 + 320*I*a^5*c^2*d^3 + 160*a^5*c*d^4 - 32*I*a^5*d^5)/f^
2)*log((sqrt(2)*(4*a^2*c^2 - 8*I*a^2*c*d - 4*a^2*d^2 + (4*a^2*c^2 - 8*I*a^2*c*d - 4*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) - I*f*sqrt(-(32*a^5*c^5 - 160*I*a^5*c^4*d - 320*a^5*c^3*d^2 + 320*I*a^5*c^2*d^3 + 160*a^5*c*
d^4 - 32*I*a^5*d^5)/f^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(4*a^2*c^2 - 8*I*a^2*c*d - 4*a^2*d^2)))/(d*
f*e^(6*I*f*x + 6*I*e) + 3*d*f*e^(4*I*f*x + 4*I*e) + 3*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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 5.9837, size = 436, normalized size = 1.05 \begin{align*} \frac{{\left (2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} a^{2} c^{2} - 4 i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} a c d + 4 i \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} a^{2} c d - 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{4} d^{2} + 4 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} a d^{2} - 2 \,{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} 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}{\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 (a \tan \left (f x + e\right ) - i \, a\right )} a^{2} + 2 i \, a^{3}\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))^(5/2),x, algorithm="giac")

[Out]

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