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

[Out]

-2*I*a^(3/2)*(c-I*d)^(5/2)*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*(-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)*d^(1/2)*(a+I*a*ta
n(f*x+e))^(1/2)/a^(1/2)/(c+d*tan(f*x+e))^(1/2))/f/d^(1/2)+1/8*a*(c-3*I*d)*(5*I*c+3*d)*(a+I*a*tan(f*x+e))^(1/2)
*(c+d*tan(f*x+e))^(1/2)/f+1/12*a*(5*I*c+7*d)*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)/f+1/3*a^2*(c+I*d)
*(c+d*tan(f*x+e))^(5/2)/d/f/(a+I*a*tan(f*x+e))^(1/2)-1/3*a^2*(c+d*tan(f*x+e))^(7/2)/d/f/(a+I*a*tan(f*x+e))^(1/
2)

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Rubi [A]  time = 1.76, antiderivative size = 378, normalized size of antiderivative = 1.00, 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 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 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 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))^{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*}

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Mathematica [A]  time = 8.57, size = 645, normalized size = 1.71 \[ \frac {i (\cos (e)-i \sin (e)) \sec (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x))^{3/2} \left (\sqrt {c+d \tan (e+f x)} \left (\left (33 c^2-68 i c d-35 d^2\right ) \cos (2 (e+f x))+33 c^2+2 d (13 c-7 i d) \sin (2 (e+f x))-68 i c d-19 d^2\right )-\frac {(3-3 i) \cos ^3(e+f x) \left (\left (5 i c^3+45 c^2 d-55 i c d^2-23 d^3\right ) \left (\log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left ((1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}-i c e^{i (e+f x)}+c-d e^{i (e+f x)}+i d\right )}{\sqrt {d} \left (-5 i c^3-45 c^2 d+55 i c d^2+23 d^3\right ) \left (e^{i (e+f x)}+i\right )}\right )-\log \left (-\frac {(2+2 i) e^{\frac {i e}{2}} \left ((1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt {d} \left (-5 i c^3-45 c^2 d+55 i c d^2+23 d^3\right ) \left (e^{i (e+f x)}-i\right )}\right )\right )+(32+32 i) \sqrt {d} (c-i d)^{5/2} \log \left (2 \left (i \sqrt {c-i d} \sin (e+f x)+\sqrt {c-i d} \cos (e+f x)+\sqrt {i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt {c+d \tan (e+f x)}\right )\right )\right )}{\sqrt {d} \sqrt {i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{48 f} \]

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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fricas [B]  time = 0.67, size = 1487, normalized size = 3.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*sqrt(2)*((33*I*a*c^2 + 94*a*c*d - 49*I*a*d^2)*e^(5*I*f*x + 5*I*e) + (66*I*a*c^2 + 136*a*c*d - 38*I*a*d
^2)*e^(3*I*f*x + 3*I*e) + (33*I*a*c^2 + 42*a*c*d - 21*I*a*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*(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 + 257
5*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^(I*f*x +
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)/(-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 + 2575*I*a^3*c^4*d^2 + 518
0*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^(I*f*x + 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)/(-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(1/2*(2*sqrt(2)*(a*c^2 - 2*I*a*c*d - a*d
^2 + (a*c^2 - 2*I*a*c*d - 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)) + 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^(I*f*x + I*e))*e^(-I*f*x - I*e)/(a*c^2 - 2*I*a*c*d - 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(1/2*(2*sqrt(2)*(a*c^2 - 2*I*a*c*d - a*d^2 + (a
*c^2 - 2*I*a*c*d - 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)) - 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^(I*f*x + I*e))*e^(-I*f*x - I*e)/(a*c^2 - 2*I*a*c*d - 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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giac [A]  time = 1.97, size = 196, normalized size = 0.52 \[ \frac {\sqrt {2 \, a d^{2} + 2 \, \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} c + c^{2} + d^{2}} a d} {\left (d \tan \left (f x + e\right ) + c\right )}^{3} a {\left (\frac {i \, {\left (d \tan \left (f x + e\right ) + c\right )} a d - i \, a c d}{a d^{2} + \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} a^{2} d^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{2} c d^{2} + a^{2} c^{2} d^{2} + a^{2} d^{4}}} + 1\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{2 \, {\left ({\left (-i \, d \tan \left (f x + e\right ) - i \, c\right )} d + i \, c d + d^{2}\right )}} \]

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]

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

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maple [B]  time = 0.38, size = 1503, normalized size = 3.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(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*(48*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-54*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)*d^2-69*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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/
2)*(-a*(I*d-c))^(1/2)*a*d^3+48*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-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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/2)*(-a*(I*d-c))^
(1/2)*a*d+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*d*a)^(1/2)+d
*a)/(I*d*a)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2*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*d*a)^(1/2)+d*a)/(I*d*a)^(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*d*a)^(1/2)+d*a)/(I*d*a)^(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*d*a)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/
2)*tan(f*x+e)*d^2+16*I*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)+66*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^2+1
36*(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+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)*c*d-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*d*a)^(1/2)+d*a)/(I*d*a)^(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*d*a)^(1/2
)+d*a)/(I*d*a)^(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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d-48*(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+48*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))*d*a*(I*d*a)^(1/2))*2^(1/2)/(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))^(3/2)*(c+d*tan(f*x+e))^(5/2),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 \[ \int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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