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]

-(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)

________________________________________________________________________________________

Rubi [A]  time = 0.304214, antiderivative size = 222, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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 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])

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 = 11.9281, 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]

((-I/4)*(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*Se
c[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*Sin
[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*Cos[
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]))

________________________________________________________________________________________

Maple [A]  time = 0.098, size = 350, normalized size = 1.6 \begin{align*}{\frac{{a}^{2}c}{24\,f}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 6\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+8\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-3\,iB\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) ac+3\,iB\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) +8\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+8\,iA\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }+12\,A\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) ac+12\,A\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\tan \left ( fx+e \right ) +8\,B\sqrt{ac}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) } \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

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

________________________________________________________________________________________

Maxima [B]  time = 6.78353, size = 1845, normalized size = 8.31 \begin{align*} \text{result too large to display} \end{align*}

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]

-((4608*A - 1152*I*B)*a^2*c*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (7680*A - 20352*I*B)*a^2*c*
cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (16896*A - 4224*I*B)*a^2*c*cos(3/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) - (4608*A - 1152*I*B)*a^2*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 11
52*(-4*I*A - B)*a^2*c*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 384*(20*I*A + 53*B)*a^2*c*sin(5/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4224*(4*I*A + B)*a^2*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) - 1152*(4*I*A + B)*a^2*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((2304*A - 576*I
*B)*a^2*c*cos(8*f*x + 8*e) + (9216*A - 2304*I*B)*a^2*c*cos(6*f*x + 6*e) + (13824*A - 3456*I*B)*a^2*c*cos(4*f*x
+ 4*e) + (9216*A - 2304*I*B)*a^2*c*cos(2*f*x + 2*e) - 576*(-4*I*A - B)*a^2*c*sin(8*f*x + 8*e) - 2304*(-4*I*A
- B)*a^2*c*sin(6*f*x + 6*e) - 3456*(-4*I*A - B)*a^2*c*sin(4*f*x + 4*e) - 2304*(-4*I*A - B)*a^2*c*sin(2*f*x + 2
*e) + (2304*A - 576*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) + ((2304*A - 576*I*B)*a^2*c*cos(8*f*x + 8*e) + (9216*A - 2304*I*B)*a
^2*c*cos(6*f*x + 6*e) + (13824*A - 3456*I*B)*a^2*c*cos(4*f*x + 4*e) + (9216*A - 2304*I*B)*a^2*c*cos(2*f*x + 2*
e) - 576*(-4*I*A - B)*a^2*c*sin(8*f*x + 8*e) - 2304*(-4*I*A - B)*a^2*c*sin(6*f*x + 6*e) - 3456*(-4*I*A - B)*a^
2*c*sin(4*f*x + 4*e) - 2304*(-4*I*A - B)*a^2*c*sin(2*f*x + 2*e) + (2304*A - 576*I*B)*a^2*c)*arctan2(cos(1/2*ar
ctan2(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) - (288*
(-4*I*A - B)*a^2*c*cos(8*f*x + 8*e) + 1152*(-4*I*A - B)*a^2*c*cos(6*f*x + 6*e) + 1728*(-4*I*A - B)*a^2*c*cos(4
*f*x + 4*e) + 1152*(-4*I*A - B)*a^2*c*cos(2*f*x + 2*e) + (1152*A - 288*I*B)*a^2*c*sin(8*f*x + 8*e) + (4608*A -
1152*I*B)*a^2*c*sin(6*f*x + 6*e) + (6912*A - 1728*I*B)*a^2*c*sin(4*f*x + 4*e) + (4608*A - 1152*I*B)*a^2*c*sin
(2*f*x + 2*e) + 288*(-4*I*A - 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*a
rctan2(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) -
(288*(4*I*A + B)*a^2*c*cos(8*f*x + 8*e) + 1152*(4*I*A + B)*a^2*c*cos(6*f*x + 6*e) + 1728*(4*I*A + B)*a^2*c*cos
(4*f*x + 4*e) + 1152*(4*I*A + B)*a^2*c*cos(2*f*x + 2*e) - (1152*A - 288*I*B)*a^2*c*sin(8*f*x + 8*e) - (4608*A
- 1152*I*B)*a^2*c*sin(6*f*x + 6*e) - (6912*A - 1728*I*B)*a^2*c*sin(4*f*x + 4*e) - (4608*A - 1152*I*B)*a^2*c*si
n(2*f*x + 2*e) + 288*(4*I*A + 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*a
rctan2(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))*s
qrt(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*co
s(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))

________________________________________________________________________________________

Fricas [B]  time = 1.64753, size = 1581, normalized size = 7.12 \begin{align*} \text{result too large to display} \end{align*}

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \end{align*}

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]

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