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

[Out]

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

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Rubi [A]  time = 1.36, antiderivative size = 329, normalized size of antiderivative = 1.00, 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 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 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])

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 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 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 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 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 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]

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*}

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Mathematica [B]  time = 10.01, size = 686, normalized size = 2.09 \[ \frac {\left (\frac {1}{16}+\frac {i}{16}\right ) \cos ^2(e+f x) (a+i a \tan (e+f x))^{5/2} \left (\frac {\left (\frac {1}{6}+\frac {i}{6}\right ) (\sin (2 e)+i \cos (2 e)) \sec ^2(e+f x) \sqrt {c+d \tan (e+f x)} \left (\left (3 c^2-68 i c d-65 d^2\right ) \cos (2 (e+f x))+3 c^2+2 d (7 c-13 i d) \sin (2 (e+f x))-68 i c d-49 d^2\right )}{d}-\frac {i (\cos (2 e)-i \sin (2 e)) \cos (e+f x) \left (\left (-i c^3+15 c^2 d-69 i c d^2-45 d^3\right ) \left (\log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left (-(1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{\sqrt {d} \left (i c^3-15 c^2 d+69 i c d^2+45 d^3\right ) \left (e^{i (e+f x)}+i\right )}\right )-\log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left ((1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt {d} \left (i c^3-15 c^2 d+69 i c d^2+45 d^3\right ) \left (e^{i (e+f x)}-i\right )}\right )\right )+(64-64 i) d^{3/2} (c-i d)^{3/2} \log \left (2 \left (i \sqrt {c-i d} \sin (e+f x)+\sqrt {c-i d} \cos (e+f x)+\sqrt {i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt {c+d \tan (e+f x)}\right )\right )\right )}{d^{3/2} \sqrt {i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{f (\cos (f x)+i \sin (f x))^2} \]

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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fricas [B]  time = 0.61, size = 1467, normalized size = 4.46 \[ \text {result too large to display} \]

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 - 82*I*a^2*c*d - 91*a^2*d^2)*e^(5*I*f*x + 5*I*e) + 2*(3*a^2*c^2 - 68*I*a^2*c*d -
49*a^2*d^2)*e^(3*I*f*x + 3*I*e) + 3*(a^2*c^2 - 18*I*a^2*c*d - 13*a^2*d^2)*e^(I*f*x + 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)) - 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^
(I*f*x + 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)/(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^(I*f*x + 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)/(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)) + 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^(I*f*x + I*e))*e^(-I*f*x - 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)) - 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^(I*f*x + I*e))*e^(-I*f*x - 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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giac [A]  time = 4.25, size = 242, normalized size = 0.74 \[ -\frac {{\left (2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{3} a^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{2} a^{2} c - 2 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{2} a^{2} d\right )} \sqrt {2 \, a d^{2} + 2 \, \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} c + c^{2} + d^{2}} a d} {\left (\frac {i \, {\left (d \tan \left (f x + e\right ) + c\right )} a d - i \, a c d}{a d^{2} + \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} a^{2} d^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{2} c d^{2} + a^{2} c^{2} d^{2} + a^{2} d^{4}}} + 1\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{4 \, {\left ({\left (d \tan \left (f x + e\right ) + c\right )} d^{2} - c d^{2} + i \, d^{3}\right )}} \]

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]

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

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maple [B]  time = 0.40, size = 1518, normalized size = 4.61 \[ \text {result too large to display} \]

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*(-a*(I*d-c))^(1/2)*
(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*d*a)^(1/2)-28*2^(1/2)*tan(f*x+e)*c*d*(-a*(I*d-c))^(1/2)*(a*(c+d
*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*d*a)^(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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d^2+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*d*a)^(1/2)+d*a)/(I*d*a)^(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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^2+96*I*(I*d*a)^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*
c-I*d*a+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*(I*d*a)^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+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*d*a)^(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*d*a)^(1/2)+d*a)/(I*d*a)^(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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2
^(1/2)*(-a*(I*d-c))^(1/2)*a*d^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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c*d+136*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*
x+e)))^(1/2)*(I*d*a)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d+96*2^(1/2)*(-a*(I*d-c))^(1/2)*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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*a*c*d-96*2^(1/2)*(-
a*(I*d-c))^(1/2)*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*d*a)^(1/2)+
d*a)/(I*d*a)^(1/2))*a*d^2-6*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*d*a)^(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*d*a)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^2-96*(I*d*a)
^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+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+96*(I*d*a)^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+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)*2^(1/2)/d/(a*
(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(I*d*a)^(1/2)/(-a*(I*d-c))^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(3*d-c>0)', see `assume?` for m
ore details)Is 3*d-c positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^(5/2)*(c + d*tan(e + f*x))^(3/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(5/2)*(c + d*tan(e + f*x))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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