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

Optimal. Leaf size=226 \[ -\frac {a^{3/2} c^{5/2} (-B+4 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a c^2 (4 A+i B) \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {c (-B+4 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 226, 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} \[ -\frac {a^{3/2} c^{5/2} (-B+4 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a c^2 (4 A+i B) \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}-\frac {c (-B+4 i A) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 10.83, size = 241, normalized size = 1.07 \[ \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \left (\frac {c^3 (B-4 i A) e^{-2 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}+\frac {1}{12} c^2 (1-i \tan (e+f x)) \sec ^{\frac {3}{2}}(e+f x) \sqrt {c-i c \tan (e+f x)} (3 \tan (e+f x) ((4 A+i B) \cos (2 (e+f x))+4 A-3 i B)+16 (B-i A))\right )}{4 f \sec ^{\frac {5}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x])*((((-4*I)*A + B)*c^3*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e
+ f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^((2*I)*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) + (c^2*Sec[e + f*x]
^(3/2)*(1 - I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]*(16*((-I)*A + B) + 3*(4*A - (3*I)*B + (4*A + I*B)*Cos[2
*(e + f*x)])*Tan[e + f*x]))/12))/(4*f*Sec[e + f*x]^(5/2)*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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fricas [B]  time = 0.80, size = 611, normalized size = 2.70 \[ -\frac {3 \, \sqrt {\frac {{\left (16 \, A^{2} + 8 i \, A B - B^{2}\right )} a^{3} c^{5}}{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 c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-16 i \, A + 4 \, B\right )} a c^{2} 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^{3} c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-4 i \, A + B\right )} a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-4 i \, A + B\right )} a c^{2}}\right ) - 3 \, \sqrt {\frac {{\left (16 \, A^{2} + 8 i \, A B - B^{2}\right )} a^{3} c^{5}}{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 c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-16 i \, A + 4 \, B\right )} a c^{2} 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^{3} c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-4 i \, A + B\right )} a c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-4 i \, A + B\right )} a c^{2}}\right ) - 4 \, {\left ({\left (-12 i \, A + 3 \, B\right )} a c^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-44 i \, A + 11 \, B\right )} a c^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-20 i \, A + 53 \, B\right )} a c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (12 i \, A - 3 \, B\right )} a c^{2} 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))^(3/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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maple [A]  time = 0.53, size = 350, normalized size = 1.55 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, c^{2} a \left (6 i B \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+8 i A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-3 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +3 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-8 B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+8 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-12 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -12 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-8 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{24 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c^2*a*(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)

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maxima [B]  time = 2.55, size = 1386, normalized size = 6.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1152*(4*A + I*B)*a*c^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4224*(4*A + I*B)*a*c^2*cos(5/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 384*(20*A + 53*I*B)*a*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), co
s(2*f*x + 2*e))) - 1152*(4*A + I*B)*a*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-4608*I*A +
1152*B)*a*c^2*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-16896*I*A + 4224*B)*a*c^2*sin(5/2*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-7680*I*A + 20352*B)*a*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*
f*x + 2*e))) - (4608*I*A - 1152*B)*a*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (576*(4*A + I*
B)*a*c^2*cos(8*f*x + 8*e) + 2304*(4*A + I*B)*a*c^2*cos(6*f*x + 6*e) + 3456*(4*A + I*B)*a*c^2*cos(4*f*x + 4*e)
+ 2304*(4*A + I*B)*a*c^2*cos(2*f*x + 2*e) - (-2304*I*A + 576*B)*a*c^2*sin(8*f*x + 8*e) - (-9216*I*A + 2304*B)*
a*c^2*sin(6*f*x + 6*e) - (-13824*I*A + 3456*B)*a*c^2*sin(4*f*x + 4*e) - (-9216*I*A + 2304*B)*a*c^2*sin(2*f*x +
2*e) + 576*(4*A + I*B)*a*c^2)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (576*(4*A + I*B)*a*c^2*cos(8*f*x + 8*e) + 2304*(4*A + I*B)*a*c^2*co
s(6*f*x + 6*e) + 3456*(4*A + I*B)*a*c^2*cos(4*f*x + 4*e) + 2304*(4*A + I*B)*a*c^2*cos(2*f*x + 2*e) - (-2304*I*
A + 576*B)*a*c^2*sin(8*f*x + 8*e) - (-9216*I*A + 2304*B)*a*c^2*sin(6*f*x + 6*e) - (-13824*I*A + 3456*B)*a*c^2*
sin(4*f*x + 4*e) - (-9216*I*A + 2304*B)*a*c^2*sin(2*f*x + 2*e) + 576*(4*A + I*B)*a*c^2)*arctan2(cos(1/2*arctan
2(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*c^2*cos(8*f*x + 8*e) + (-4608*I*A + 1152*B)*a*c^2*cos(6*f*x + 6*e) + (-6912*I*A + 1728*B)*a*c^2*
cos(4*f*x + 4*e) + (-4608*I*A + 1152*B)*a*c^2*cos(2*f*x + 2*e) + 288*(4*A + I*B)*a*c^2*sin(8*f*x + 8*e) + 1152
*(4*A + I*B)*a*c^2*sin(6*f*x + 6*e) + 1728*(4*A + I*B)*a*c^2*sin(4*f*x + 4*e) + 1152*(4*A + I*B)*a*c^2*sin(2*f
*x + 2*e) + (-1152*I*A + 288*B)*a*c^2)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*ar
ctan2(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*c^2*cos(8*f*x + 8*e) + (4608*I*A - 1152*B)*a*c^2*cos(6*f*x + 6*e) + (6912*I*A - 1728*B)*a
*c^2*cos(4*f*x + 4*e) + (4608*I*A - 1152*B)*a*c^2*cos(2*f*x + 2*e) - 288*(4*A + I*B)*a*c^2*sin(8*f*x + 8*e) -
1152*(4*A + I*B)*a*c^2*sin(6*f*x + 6*e) - 1728*(4*A + I*B)*a*c^2*sin(4*f*x + 4*e) - 1152*(4*A + I*B)*a*c^2*sin
(2*f*x + 2*e) + (1152*I*A - 288*B)*a*c^2)*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))

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

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

[Out]

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

[Out]

Timed out

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