3.807 \(\int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=222 \[ -\frac {a^{5/2} c^{3/2} (B+4 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a^2 c (4 A-i B) \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a (B+4 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f} \]

[Out]

-1/4*a^(5/2)*(4*I*A+B)*c^(3/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/f+1/8
*a^2*(4*A-I*B)*c*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)*tan(f*x+e)/f+1/12*a*(4*I*A+B)*(a+I*a*tan(f*
x+e))^(3/2)*(c-I*c*tan(f*x+e))^(3/2)/f+1/4*B*(a+I*a*tan(f*x+e))^(5/2)*(c-I*c*tan(f*x+e))^(3/2)/f

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3588, 80, 49, 38, 63, 217, 203} \[ -\frac {a^{5/2} c^{3/2} (B+4 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a^2 c (4 A-i B) \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a (B+4 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^(5/2)*((4*I)*A + B)*c^(3/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]
])])/(4*f) + (a^2*(4*A - I*B)*c*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f) + (a
*((4*I)*A + B)*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(12*f) + (B*(a + I*a*Tan[e + f*x])^(
5/2)*(c - I*c*Tan[e + f*x])^(3/2))/(4*f)

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 49

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*(m
+ n + 1)), x] + Dist[(2*c*n)/(m + n + 1), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 203

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

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 3588

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*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int (a+i a x)^{3/2} (A+B x) \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac {(a (4 A-i B) c) \operatorname {Subst}\left (\int (a+i a x)^{3/2} \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {a (4 i A+B) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac {\left (a^2 (4 A-i B) c\right ) \operatorname {Subst}\left (\int \sqrt {a+i a x} \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {a^2 (4 A-i B) c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a (4 i A+B) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac {\left (a^3 (4 A-i B) c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {a^2 (4 A-i B) c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a (4 i A+B) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f}-\frac {\left (a^2 (4 i A+B) c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{4 f}\\ &=\frac {a^2 (4 A-i B) c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a (4 i A+B) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f}-\frac {\left (a^2 (4 i A+B) c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{4 f}\\ &=-\frac {a^{5/2} (4 i A+B) c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a^2 (4 A-i B) c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a (4 i A+B) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 13.33, size = 460, normalized size = 2.07 \[ \frac {\cos ^3(e+f x) (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left (\sec (e) \left (\frac {1}{12} c \cos (2 e)-\frac {1}{12} i c \sin (2 e)\right ) \sec ^2(e+f x) (4 i A \cos (e)+3 i B \sin (e)+4 B \cos (e))+\sec (e) \left (\frac {1}{8} \cos (2 e)-\frac {1}{8} i \sin (2 e)\right ) \sec (e+f x) (4 A c \sin (f x)-i B c \sin (f x))+(4 A-i B) \tan (e) \left (\frac {1}{8} c \cos (2 e)-\frac {1}{8} i c \sin (2 e)\right )+i B c \sec (e) \left (\frac {1}{4} \cos (2 e)-\frac {1}{4} i \sin (2 e)\right ) \sin (f x) \sec ^3(e+f x)\right )}{f (\cos (f x)+i \sin (f x))^2 (A \cos (e+f x)+B \sin (e+f x))}-\frac {i c^2 (4 A-i B) \sqrt {e^{i f x}} e^{-i (3 e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{4 f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \sec ^{\frac {7}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{5/2} (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/4*I)*(4*A - I*B)*c^2*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x
))]*(a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(E^(I*(3*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*
Sec[e + f*x]^(7/2)*(Cos[f*x] + I*Sin[f*x])^(5/2)*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (Cos[e + f*x]^3*Sqrt[Sec
[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(Sec[e]*Sec[e + f*x]^2*((4*I)*A*Cos[e] + 4*B*Cos[e] + (3*I)*B*S
in[e])*((c*Cos[2*e])/12 - (I/12)*c*Sin[2*e]) + I*B*c*Sec[e]*Sec[e + f*x]^3*(Cos[2*e]/4 - (I/4)*Sin[2*e])*Sin[f
*x] + Sec[e]*Sec[e + f*x]*(Cos[2*e]/8 - (I/8)*Sin[2*e])*(4*A*c*Sin[f*x] - I*B*c*Sin[f*x]) + (4*A - I*B)*((c*Co
s[2*e])/8 - (I/8)*c*Sin[2*e])*Tan[e])*(a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[
f*x])^2*(A*Cos[e + f*x] + B*Sin[e + f*x]))

________________________________________________________________________________________

fricas [B]  time = 0.97, size = 611, normalized size = 2.75 \[ \frac {3 \, \sqrt {\frac {{\left (16 \, A^{2} - 8 i \, A B - B^{2}\right )} a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (16 i \, A + 4 \, B\right )} a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (16 i \, A + 4 \, B\right )} a^{2} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 2 \, \sqrt {\frac {{\left (16 \, A^{2} - 8 i \, A B - B^{2}\right )} a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (4 i \, A + B\right )} a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (4 i \, A + B\right )} a^{2} c}\right ) - 3 \, \sqrt {\frac {{\left (16 \, A^{2} - 8 i \, A B - B^{2}\right )} a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (16 i \, A + 4 \, B\right )} a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (16 i \, A + 4 \, B\right )} a^{2} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - 2 \, \sqrt {\frac {{\left (16 \, A^{2} - 8 i \, A B - B^{2}\right )} a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (4 i \, A + B\right )} a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (4 i \, A + B\right )} a^{2} c}\right ) + 4 \, {\left ({\left (-12 i \, A - 3 \, B\right )} a^{2} c e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (20 i \, A + 53 \, B\right )} a^{2} c e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (44 i \, A + 11 \, B\right )} a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (12 i \, A + 3 \, B\right )} a^{2} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{48 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*sqrt((16*A^2 - 8*I*A*B - B^2)*a^5*c^3/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2
*I*f*x + 2*I*e) + f)*log(2*(((16*I*A + 4*B)*a^2*c*e^(3*I*f*x + 3*I*e) + (16*I*A + 4*B)*a^2*c*e^(I*f*x + I*e))*
sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + 2*sqrt((16*A^2 - 8*I*A*B - B^2)*a^5*c^3/
f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((4*I*A + B)*a^2*c*e^(2*I*f*x + 2*I*e) + (4*I*A + B)*a^2*c)) - 3*sqrt((16*A^
2 - 8*I*A*B - B^2)*a^5*c^3/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f
)*log(2*(((16*I*A + 4*B)*a^2*c*e^(3*I*f*x + 3*I*e) + (16*I*A + 4*B)*a^2*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x
+ 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - 2*sqrt((16*A^2 - 8*I*A*B - B^2)*a^5*c^3/f^2)*(f*e^(2*I*f*x
+ 2*I*e) - f))/((4*I*A + B)*a^2*c*e^(2*I*f*x + 2*I*e) + (4*I*A + B)*a^2*c)) + 4*((-12*I*A - 3*B)*a^2*c*e^(7*I*
f*x + 7*I*e) + (20*I*A + 53*B)*a^2*c*e^(5*I*f*x + 5*I*e) + (44*I*A + 11*B)*a^2*c*e^(3*I*f*x + 3*I*e) + (12*I*A
+ 3*B)*a^2*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(f*e^(6*I*
f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)

________________________________________________________________________________________

giac [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)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.60, size = 350, normalized size = 1.58 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{2} c \left (6 i B \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+8 i A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-3 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +3 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+8 B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+8 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+12 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +12 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+8 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{24 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

1/24/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2*c*(6*I*B*tan(f*x+e)^3*(c*a*(1+tan(f*x+e)^2)
)^(1/2)*(c*a)^(1/2)+8*I*A*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-3*I*B*ln((c*a*tan(f*x+e)+(c*a*
(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*a*c+3*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e
)+8*B*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+8*I*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+12*
A*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*a*c+12*A*(c*a*(1+tan(f*x+e)^2))^(1
/2)*(c*a)^(1/2)*tan(f*x+e)+8*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a*(1+tan(f*x+e)^2))^(1/2)/(c*a)^(1
/2)

________________________________________________________________________________________

maxima [B]  time = 2.31, size = 1372, normalized size = 6.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-(1152*(4*A - I*B)*a^2*c*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 384*(20*A - 53*I*B)*a^2*c*cos(
5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4224*(4*A - I*B)*a^2*c*cos(3/2*arctan2(sin(2*f*x + 2*e), co
s(2*f*x + 2*e))) - 1152*(4*A - I*B)*a^2*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (4608*I*A + 1
152*B)*a^2*c*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (-7680*I*A - 20352*B)*a^2*c*sin(5/2*arctan
2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (-16896*I*A - 4224*B)*a^2*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))) + (-4608*I*A - 1152*B)*a^2*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (576*(4*A - I*
B)*a^2*c*cos(8*f*x + 8*e) + 2304*(4*A - I*B)*a^2*c*cos(6*f*x + 6*e) + 3456*(4*A - I*B)*a^2*c*cos(4*f*x + 4*e)
+ 2304*(4*A - I*B)*a^2*c*cos(2*f*x + 2*e) + (2304*I*A + 576*B)*a^2*c*sin(8*f*x + 8*e) + (9216*I*A + 2304*B)*a^
2*c*sin(6*f*x + 6*e) + (13824*I*A + 3456*B)*a^2*c*sin(4*f*x + 4*e) + (9216*I*A + 2304*B)*a^2*c*sin(2*f*x + 2*e
) + 576*(4*A - I*B)*a^2*c)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (576*(4*A - I*B)*a^2*c*cos(8*f*x + 8*e) + 2304*(4*A - I*B)*a^2*c*cos(6*
f*x + 6*e) + 3456*(4*A - I*B)*a^2*c*cos(4*f*x + 4*e) + 2304*(4*A - I*B)*a^2*c*cos(2*f*x + 2*e) + (2304*I*A + 5
76*B)*a^2*c*sin(8*f*x + 8*e) + (9216*I*A + 2304*B)*a^2*c*sin(6*f*x + 6*e) + (13824*I*A + 3456*B)*a^2*c*sin(4*f
*x + 4*e) + (9216*I*A + 2304*B)*a^2*c*sin(2*f*x + 2*e) + 576*(4*A - I*B)*a^2*c)*arctan2(cos(1/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((1152*I*A + 288*
B)*a^2*c*cos(8*f*x + 8*e) + (4608*I*A + 1152*B)*a^2*c*cos(6*f*x + 6*e) + (6912*I*A + 1728*B)*a^2*c*cos(4*f*x +
4*e) + (4608*I*A + 1152*B)*a^2*c*cos(2*f*x + 2*e) - 288*(4*A - I*B)*a^2*c*sin(8*f*x + 8*e) - 1152*(4*A - I*B)
*a^2*c*sin(6*f*x + 6*e) - 1728*(4*A - I*B)*a^2*c*sin(4*f*x + 4*e) - 1152*(4*A - I*B)*a^2*c*sin(2*f*x + 2*e) +
(1152*I*A + 288*B)*a^2*c)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((-1152*I*A -
288*B)*a^2*c*cos(8*f*x + 8*e) + (-4608*I*A - 1152*B)*a^2*c*cos(6*f*x + 6*e) + (-6912*I*A - 1728*B)*a^2*c*cos(4
*f*x + 4*e) + (-4608*I*A - 1152*B)*a^2*c*cos(2*f*x + 2*e) + 288*(4*A - I*B)*a^2*c*sin(8*f*x + 8*e) + 1152*(4*A
- I*B)*a^2*c*sin(6*f*x + 6*e) + 1728*(4*A - I*B)*a^2*c*sin(4*f*x + 4*e) + 1152*(4*A - I*B)*a^2*c*sin(2*f*x +
2*e) + (-1152*I*A - 288*B)*a^2*c)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)
*sqrt(c)/(f*(-4608*I*cos(8*f*x + 8*e) - 18432*I*cos(6*f*x + 6*e) - 27648*I*cos(4*f*x + 4*e) - 18432*I*cos(2*f*
x + 2*e) + 4608*sin(8*f*x + 8*e) + 18432*sin(6*f*x + 6*e) + 27648*sin(4*f*x + 4*e) + 18432*sin(2*f*x + 2*e) -
4608*I))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________