3.12.44 \(\int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx\) [1144]

Optimal. Leaf size=315 \[ -\frac {\sqrt [4]{-1} a^{3/2} \left (3 i c^2+18 c d-11 i d^2\right ) \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{4 \sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}+\frac {a (3 i c+5 d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

-2*I*a^(3/2)*(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-1/4*(-1)^(1/4)*a^(3/2)*(3*I*c^2+18*c*d-11*I*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))/f/d^(1/2)+1/4*a*(3*I*c+5*d)*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/
2)/f+1/2*a^2*(c+I*d)*(c+d*tan(f*x+e))^(3/2)/d/f/(a+I*a*tan(f*x+e))^(1/2)-1/2*a^2*(c+d*tan(f*x+e))^(5/2)/d/f/(a
+I*a*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.87, antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3637, 3676, 3678, 3682, 3625, 214, 3680, 65, 223, 212} \begin {gather*} -\frac {\sqrt [4]{-1} a^{3/2} \left (3 i c^2+18 c d-11 i 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 )}{4 \sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} (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 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}+\frac {a (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((-1)^(1/4)*a^(3/2)*((3*I)*c^2 + 18*c*d - (11*I)*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]])])/(Sqrt[d]*f) - ((2*I)*Sqrt[2]*a^(3/2)*(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*((3*I)*c + 5*d)*Sqrt[a
+ I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/(4*f) + (a^2*(c + I*d)*(c + d*Tan[e + f*x])^(3/2))/(2*d*f*Sqrt[
a + I*a*Tan[e + f*x]]) - (a^2*(c + d*Tan[e + f*x])^(5/2))/(2*d*f*Sqrt[a + I*a*Tan[e + f*x]])

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 223

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 3625

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 3637

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 3676

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 3678

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 3680

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 3682

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))^{3/2} \, dx &=-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}+\frac {a \int \frac {\left (-\frac {1}{2} a (i c-9 d)-\frac {1}{2} a (c-7 i d) \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx}{2 d}\\ &=\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 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 {1}{2} a^2 (5 c-3 i d) d-\frac {1}{2} a^2 d (3 i c+5 d) \tan (e+f x)\right ) \, dx}{2 a d}\\ &=\frac {a (3 i c+5 d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {\int \frac {\sqrt {a+i a \tan (e+f x)} \left (-\frac {1}{4} a^3 d \left (13 c^2-14 i c d-5 d^2\right )-\frac {1}{4} a^3 d \left (18 c d+i \left (3 c^2-11 d^2\right )\right ) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 a^2 d}\\ &=\frac {a (3 i c+5 d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}+\left (2 a (c-i d)^2\right ) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx-\frac {1}{8} \left (3 c^2-18 i c d-11 d^2\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 (3 i c+5 d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {\left (4 i a^3 (c-i d)^2\right ) \text {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 (3 c^2-18 i c d-11 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {2 i \sqrt {2} a^{3/2} (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 (3 i c+5 d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}+\frac {\left (a \left (3 i c^2+18 c d-11 i d^2\right )\right ) \text {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 )}{4 f}\\ &=-\frac {2 i \sqrt {2} a^{3/2} (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 (3 i c+5 d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}+\frac {\left (a \left (3 i c^2+18 c d-11 i d^2\right )\right ) \text {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 )}{4 f}\\ &=-\frac {\sqrt [4]{-1} a^{3/2} \left (3 i c^2+18 c d-11 i 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 )}{4 \sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} (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 (3 i c+5 d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 7.20, size = 574, normalized size = 1.82 \begin {gather*} \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \cos (e+f x) (\cos (e)-i \sin (e)) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x))^{3/2} \left (-\frac {\cos (e+f x) \left (\left (3 i c^2+18 c d-11 i d^2\right ) \left (\log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left (d+i d e^{i (e+f x)}-c \left (i+e^{i (e+f x)}\right )+(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 (-3 c^2+18 i c d+11 d^2\right ) \left (i+e^{i (e+f x)}\right )}\right )-\log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left (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-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{\sqrt {d} \left (3 i c^2+18 c d-11 i d^2\right ) \left (-i+e^{i (e+f x)}\right )}\right )\right )+(16+16 i) (c-i d)^{3/2} \sqrt {d} \log \left (2 \left (\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)}\right )\right )\right )}{\sqrt {d} \sqrt {1+\cos (2 (e+f x))+i \sin (2 (e+f x))}}+(1+i) \sqrt {c+d \tan (e+f x)} (5 c-5 i d+2 d \tan (e+f x))\right )}{f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/8 + I/8)*Cos[e + f*x]*(Cos[e] - I*Sin[e])*(Cos[f*x] - I*Sin[f*x])*(a + I*a*Tan[e + f*x])^(3/2)*(-((Cos[e +
f*x]*(((3*I)*c^2 + 18*c*d - (11*I)*d^2)*(Log[((2 + 2*I)*E^((I/2)*e)*(d + I*d*E^(I*(e + f*x)) - 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]*(-3*c^2 + (18*I)*c*d + 11*d^2)*(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]*((3*I)*c^2 + 18*c*d - (11*I)*d^2)*(-I +
E^(I*(e + f*x))))]) + (16 + 16*I)*(c - I*d)^(3/2)*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)]])) + (1 + I)*Sqrt[c + d*Tan[e + f*x]]*(5*c - (5*I)*d + 2*d*Tan[e + f*x]
)))/f

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1233 vs. \(2 (251 ) = 502\).
time = 0.54, size = 1234, normalized size = 3.92

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1234\)
default	\(\text {Expression too large to display}\)	\(1234\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/16/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a*(4*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^
(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*d*tan(f*x+e)+18*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)
+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))*a*c*d+10*I*(a*(c+d*tan(
f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*c-8*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1
/2*(2*I*a*d*tan(f*x+e)+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))*a
*c-8*I*ln(1/2*(2*I*a*d*tan(f*x+e)+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))*a*d*2^(1/2)*(-a*(I*d-c))^(1/2)-3*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+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))*a*c^2+11*2^(1/2)*(-a*(I*d-c))^(1/2)*l
n(1/2*(2*I*a*d*tan(f*x+e)+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)
)*a*d^2+8*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+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+8*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x
+e)+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+10*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*d+8*2^(1/2)*(-a*(I*d
-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+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))*a*c-8*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+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))*a*d-8*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*t
an(f*x+e)+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+8*ln((
3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)
))^(1/2))/(tan(f*x+e)+I))*a*d*(I*a*d)^(1/2))/(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)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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))^(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)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1142 vs. \(2 (249) = 498\).
time = 0.91, size = 1142, normalized size = 3.63 \begin {gather*} \frac {8 \, \sqrt {2} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {a^{3} c^{3} - 3 i \, a^{3} c^{2} d - 3 \, a^{3} c d^{2} + i \, a^{3} d^{3}}{f^{2}}} \log \left (\frac {{\left (\sqrt {2} f \sqrt {-\frac {a^{3} c^{3} - 3 i \, a^{3} c^{2} d - 3 \, a^{3} c d^{2} + i \, a^{3} d^{3}}{f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (-i \, a c - a d + {\left (-i \, a c - a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{-i \, a c - a d}\right ) - 8 \, \sqrt {2} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {a^{3} c^{3} - 3 i \, a^{3} c^{2} d - 3 \, a^{3} c d^{2} + i \, a^{3} d^{3}}{f^{2}}} \log \left (-\frac {{\left (\sqrt {2} f \sqrt {-\frac {a^{3} c^{3} - 3 i \, a^{3} c^{2} d - 3 \, a^{3} c d^{2} + i \, a^{3} d^{3}}{f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (-i \, a c - a d + {\left (-i \, a c - a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{-i \, a c - a d}\right ) + 2 \, \sqrt {2} {\left ({\left (5 i \, a c + 7 \, a d\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (5 i \, a c + 3 \, a d\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-9 i \, a^{3} c^{4} - 108 \, a^{3} c^{3} d + 390 i \, a^{3} c^{2} d^{2} + 396 \, a^{3} c d^{3} - 121 i \, a^{3} d^{4}}{d f^{2}}} \log \left (\frac {{\left (2 \, d f \sqrt {\frac {-9 i \, a^{3} c^{4} - 108 \, a^{3} c^{3} d + 390 i \, a^{3} c^{2} d^{2} + 396 \, a^{3} c d^{3} - 121 i \, a^{3} d^{4}}{d f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (3 \, a c^{2} - 18 i \, a c d - 11 \, a d^{2} + {\left (3 \, a c^{2} - 18 i \, a c d - 11 \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{3 \, a c^{2} - 18 i \, a c d - 11 \, a d^{2}}\right ) + {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {\frac {-9 i \, a^{3} c^{4} - 108 \, a^{3} c^{3} d + 390 i \, a^{3} c^{2} d^{2} + 396 \, a^{3} c d^{3} - 121 i \, a^{3} d^{4}}{d f^{2}}} \log \left (-\frac {{\left (2 \, d f \sqrt {\frac {-9 i \, a^{3} c^{4} - 108 \, a^{3} c^{3} d + 390 i \, a^{3} c^{2} d^{2} + 396 \, a^{3} c d^{3} - 121 i \, a^{3} d^{4}}{d f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (3 \, a c^{2} - 18 i \, a c d - 11 \, a d^{2} + {\left (3 \, a c^{2} - 18 i \, a c d - 11 \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{3 \, a c^{2} - 18 i \, a c d - 11 \, a d^{2}}\right )}{8 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

1/8*(8*sqrt(2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(a^3*c^3 - 3*I*a^3*c^2*d - 3*a^3*c*d^2 + I*a^3*d^3)/f^2)*log(
(sqrt(2)*f*sqrt(-(a^3*c^3 - 3*I*a^3*c^2*d - 3*a^3*c*d^2 + I*a^3*d^3)/f^2)*e^(I*f*x + I*e) + sqrt(2)*(-I*a*c -
a*d + (-I*a*c - a*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)/(-I*a*c - a*d)) - 8*sqrt(2)*(f*e^(2*I*f*x + 2*I*e) +
f)*sqrt(-(a^3*c^3 - 3*I*a^3*c^2*d - 3*a^3*c*d^2 + I*a^3*d^3)/f^2)*log(-(sqrt(2)*f*sqrt(-(a^3*c^3 - 3*I*a^3*c^
2*d - 3*a^3*c*d^2 + I*a^3*d^3)/f^2)*e^(I*f*x + I*e) - sqrt(2)*(-I*a*c - a*d + (-I*a*c - a*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)/(-I*a*c - a*d)) + 2*sqrt(2)*((5*I*a*c + 7*a*d)*e^(3*I*f*x + 3*I*e) + (5*I*a*c + 3*a*d)*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)) - (f*e^(2*I*f*x + 2*I*e) + f)*sqrt((-9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d
^3 - 121*I*a^3*d^4)/(d*f^2))*log((2*d*f*sqrt((-9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d^3
- 121*I*a^3*d^4)/(d*f^2))*e^(I*f*x + I*e) + sqrt(2)*(3*a*c^2 - 18*I*a*c*d - 11*a*d^2 + (3*a*c^2 - 18*I*a*c*d
- 11*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))*sqr
t(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(3*a*c^2 - 18*I*a*c*d - 11*a*d^2)) + (f*e^(2*I*f*x + 2*I*e) +
f)*sqrt((-9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d^3 - 121*I*a^3*d^4)/(d*f^2))*log(-(2*d
*f*sqrt((-9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d^3 - 121*I*a^3*d^4)/(d*f^2))*e^(I*f*x +
I*e) - sqrt(2)*(3*a*c^2 - 18*I*a*c*d - 11*a*d^2 + (3*a*c^2 - 18*I*a*c*d - 11*a*d^2)*e^(2*I*f*x + 2*I*e))*sqrt
(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I
*f*x - I*e)/(3*a*c^2 - 18*I*a*c*d - 11*a*d^2)))/(f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

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))**(3/2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 0.61index.cc index_m operator + Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^{3/2} \,d x \end {gather*}

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))^(3/2),x)

[Out]

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

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