3.806 \(\int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx\)
Optimal. Leaf size=213 \[ -\frac{3 i a^{5/2} A c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{4 f}+\frac{3 a^2 A c^2 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}+\frac{a A c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{5 f} \]
[Out]
(((-3*I)/4)*a^(5/2)*A*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])
])/f + (3*a^2*A*c^2*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f) + (a*A*c*Tan[e +
f*x]*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(4*f) + (B*(a + I*a*Tan[e + f*x])^(5/2)*(c -
I*c*Tan[e + f*x])^(5/2))/(5*f)
________________________________________________________________________________________
Rubi [A] time = 0.274512, antiderivative size = 213, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used =
{3588, 80, 38, 63, 217, 203} \[ -\frac{3 i a^{5/2} A c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{4 f}+\frac{3 a^2 A c^2 \tan (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{8 f}+\frac{a A c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{5 f} \]
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])^(5/2),x]
[Out]
(((-3*I)/4)*a^(5/2)*A*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])
])/f + (3*a^2*A*c^2*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f) + (a*A*c*Tan[e +
f*x]*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(4*f) + (B*(a + I*a*Tan[e + f*x])^(5/2)*(c -
I*c*Tan[e + f*x])^(5/2))/(5*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 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))^{5/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int (a+i a x)^{3/2} (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))^{5/2} (c-i c \tan (e+f x))^{5/2}}{5 f}+\frac{(a A c) \operatorname{Subst}\left (\int (a+i a x)^{3/2} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a A c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{5 f}+\frac{\left (3 a^2 A 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{3 a^2 A 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{a A c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{5 f}+\frac{\left (3 a^3 A 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{3 a^2 A 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{a A c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac{\left (3 i a^2 A 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{3 a^2 A 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{a A c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac{\left (3 i a^2 A 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{3 i a^{5/2} A 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{3 a^2 A 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{a A c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac{B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}{5 f}\\ \end{align*}
Mathematica [B] time = 13.9205, size = 459, normalized size = 2.15 \[ \frac{\cos ^3(e+f x) (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left (A c^2 \sec (e) \left (\frac{1}{4} \cos (2 e)-\frac{1}{4} i \sin (2 e)\right ) \sin (f x) \sec ^3(e+f x)+A c^2 \sec (e) \left (\frac{3}{8} \cos (2 e)-\frac{3}{8} i \sin (2 e)\right ) \sin (f x) \sec (e+f x)+\tan (e) \sec ^2(e+f x) \left (\frac{1}{4} A c^2 \cos (2 e)-\frac{1}{4} i A c^2 \sin (2 e)\right )+\tan (e) \left (\frac{3}{8} A c^2 \cos (2 e)-\frac{3}{8} i A c^2 \sin (2 e)\right )+\sec ^4(e+f x) \left (\frac{1}{5} B c^2 \cos (2 e)-\frac{1}{5} i B c^2 \sin (2 e)\right )\right )}{f (\cos (f x)+i \sin (f x))^2 (A \cos (e+f x)+B \sin (e+f x))}-\frac{3 i A c^3 \sqrt{e^{i f x}} e^{-i (3 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{4 f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{7}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{5/2} (A \cos (e+f x)+B \sin (e+f x))} \]
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])^(5/2),x]
[Out]
(((-3*I)/4)*A*c^3*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 + f*x]^4*((B*c^2*Cos[2*e])/5 - (I/5)*B*c^2*Sin[2*e]) + A*c^2*Sec
[e]*Sec[e + f*x]^3*(Cos[2*e]/4 - (I/4)*Sin[2*e])*Sin[f*x] + A*c^2*Sec[e]*Sec[e + f*x]*((3*Cos[2*e])/8 - ((3*I)
/8)*Sin[2*e])*Sin[f*x] + Sec[e + f*x]^2*((A*c^2*Cos[2*e])/4 - (I/4)*A*c^2*Sin[2*e])*Tan[e] + ((3*A*c^2*Cos[2*e
])/8 - ((3*I)/8)*A*c^2*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*S
in[f*x])^2*(A*Cos[e + f*x] + B*Sin[e + f*x]))
________________________________________________________________________________________
Maple [A] time = 0.095, size = 252, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}{c}^{2}}{40\,f}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) } \left ( 8\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+10\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+16\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+15\,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+25\,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))^(5/2),x)
[Out]
1/40/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2*c^2*(8*B*tan(f*x+e)^4*(a*c*(1+tan(f*x+e)^2)
)^(1/2)*(a*c)^(1/2)+10*A*tan(f*x+e)^3*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+16*B*tan(f*x+e)^2*(a*c*(1+tan(f
*x+e)^2))^(1/2)*(a*c)^(1/2)+15*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
+25*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 = 5.69319, size = 1955, normalized size = 9.18 \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))^(5/2),x, algorithm="maxima")
[Out]
-(60*A*a^2*c^2*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 280*A*a^2*c^2*cos(7/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) + 512*I*B*a^2*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 280*A*a^2*
c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 60*A*a^2*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(
2*f*x + 2*e))) + 60*I*A*a^2*c^2*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 280*I*A*a^2*c^2*sin(7/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 512*B*a^2*c^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) - 280*I*A*a^2*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 60*I*A*a^2*c^2*sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + (30*A*a^2*c^2*cos(10*f*x + 10*e) + 150*A*a^2*c^2*cos(8*f*x + 8*e) + 300*
A*a^2*c^2*cos(6*f*x + 6*e) + 300*A*a^2*c^2*cos(4*f*x + 4*e) + 150*A*a^2*c^2*cos(2*f*x + 2*e) + 30*I*A*a^2*c^2*
sin(10*f*x + 10*e) + 150*I*A*a^2*c^2*sin(8*f*x + 8*e) + 300*I*A*a^2*c^2*sin(6*f*x + 6*e) + 300*I*A*a^2*c^2*sin
(4*f*x + 4*e) + 150*I*A*a^2*c^2*sin(2*f*x + 2*e) + 30*A*a^2*c^2)*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) + (30*A*a^2*c^2*cos(10*f*x + 10*e)
+ 150*A*a^2*c^2*cos(8*f*x + 8*e) + 300*A*a^2*c^2*cos(6*f*x + 6*e) + 300*A*a^2*c^2*cos(4*f*x + 4*e) + 150*A*a^2
*c^2*cos(2*f*x + 2*e) + 30*I*A*a^2*c^2*sin(10*f*x + 10*e) + 150*I*A*a^2*c^2*sin(8*f*x + 8*e) + 300*I*A*a^2*c^2
*sin(6*f*x + 6*e) + 300*I*A*a^2*c^2*sin(4*f*x + 4*e) + 150*I*A*a^2*c^2*sin(2*f*x + 2*e) + 30*A*a^2*c^2)*arctan
2(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) - (-15*I*A*a^2*c^2*cos(10*f*x + 10*e) - 75*I*A*a^2*c^2*cos(8*f*x + 8*e) - 150*I*A*a^2*c^2*cos(6*f*x + 6*e
) - 150*I*A*a^2*c^2*cos(4*f*x + 4*e) - 75*I*A*a^2*c^2*cos(2*f*x + 2*e) + 15*A*a^2*c^2*sin(10*f*x + 10*e) + 75*
A*a^2*c^2*sin(8*f*x + 8*e) + 150*A*a^2*c^2*sin(6*f*x + 6*e) + 150*A*a^2*c^2*sin(4*f*x + 4*e) + 75*A*a^2*c^2*si
n(2*f*x + 2*e) - 15*I*A*a^2*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) - (15*I*A
*a^2*c^2*cos(10*f*x + 10*e) + 75*I*A*a^2*c^2*cos(8*f*x + 8*e) + 150*I*A*a^2*c^2*cos(6*f*x + 6*e) + 150*I*A*a^2
*c^2*cos(4*f*x + 4*e) + 75*I*A*a^2*c^2*cos(2*f*x + 2*e) - 15*A*a^2*c^2*sin(10*f*x + 10*e) - 75*A*a^2*c^2*sin(8
*f*x + 8*e) - 150*A*a^2*c^2*sin(6*f*x + 6*e) - 150*A*a^2*c^2*sin(4*f*x + 4*e) - 75*A*a^2*c^2*sin(2*f*x + 2*e)
+ 15*I*A*a^2*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*(-80
*I*cos(10*f*x + 10*e) - 400*I*cos(8*f*x + 8*e) - 800*I*cos(6*f*x + 6*e) - 800*I*cos(4*f*x + 4*e) - 400*I*cos(2
*f*x + 2*e) + 80*sin(10*f*x + 10*e) + 400*sin(8*f*x + 8*e) + 800*sin(6*f*x + 6*e) + 800*sin(4*f*x + 4*e) + 400
*sin(2*f*x + 2*e) - 80*I))
________________________________________________________________________________________
Fricas [B] time = 1.60908, size = 1523, normalized size = 7.15 \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))^(5/2),x, algorithm="fricas")
[Out]
1/80*(4*(-15*I*A*a^2*c^2*e^(8*I*f*x + 8*I*e) - 70*I*A*a^2*c^2*e^(6*I*f*x + 6*I*e) + 128*B*a^2*c^2*e^(4*I*f*x +
4*I*e) + 70*I*A*a^2*c^2*e^(2*I*f*x + 2*I*e) + 15*I*A*a^2*c^2)*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) - 15*sqrt(A^2*a^5*c^5/f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*
e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(1/4*(32*(A*a^2*c^2*e^(2*I*f*x + 2*I*e) + A*a^2
*c^2)*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) + sqrt(A^2*a^5*c^5/f
^2)*(16*I*f*e^(2*I*f*x + 2*I*e) - 16*I*f))/(A*a^2*c^2*e^(2*I*f*x + 2*I*e) + A*a^2*c^2)) + 15*sqrt(A^2*a^5*c^5/
f^2)*(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)
*log(1/4*(32*(A*a^2*c^2*e^(2*I*f*x + 2*I*e) + A*a^2*c^2)*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) + sqrt(A^2*a^5*c^5/f^2)*(-16*I*f*e^(2*I*f*x + 2*I*e) + 16*I*f))/(A*a^2*c^2*e^(2
*I*f*x + 2*I*e) + A*a^2*c^2)))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*
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))**(5/2),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{5}{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))^(5/2),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)^(5/2), x)