3.830 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx\)
Optimal. Leaf size=228 \[ \frac{3 c^{5/2} (-3 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{3 c^2 (-3 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a f}+\frac{c (-3 B+2 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a f}+\frac{(-B+i A) (c-i c \tan (e+f x))^{5/2}}{f \sqrt{a+i a \tan (e+f x)}} \]
[Out]
(3*((2*I)*A - 3*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]])])/
(Sqrt[a]*f) + (3*((2*I)*A - 3*B)*c^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*a*f) + (((2*I)*
A - 3*B)*c*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2))/(2*a*f) + ((I*A - B)*(c - I*c*Tan[e + f*x]
)^(5/2))/(f*Sqrt[a + I*a*Tan[e + f*x]])
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Rubi [A] time = 0.296598, antiderivative size = 228, 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, 78, 50, 63, 217, 203} \[ \frac{3 c^{5/2} (-3 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{3 c^2 (-3 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a f}+\frac{c (-3 B+2 i A) \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a f}+\frac{(-B+i A) (c-i c \tan (e+f x))^{5/2}}{f \sqrt{a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2))/Sqrt[a + I*a*Tan[e + f*x]],x]
[Out]
(3*((2*I)*A - 3*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]])])/
(Sqrt[a]*f) + (3*((2*I)*A - 3*B)*c^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*a*f) + (((2*I)*
A - 3*B)*c*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2))/(2*a*f) + ((I*A - B)*(c - I*c*Tan[e + f*x]
)^(5/2))/(f*Sqrt[a + I*a*Tan[e + f*x]])
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 78
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ
[p, n]))))
Rule 50
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[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
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 \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{3/2}}{(a+i a x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{((2 A+3 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(2 i A-3 B) c \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (3 (2 A+3 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{\sqrt{a+i a x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{3 (2 i A-3 B) c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a f}+\frac{(2 i A-3 B) c \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (3 (2 A+3 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 )}{2 f}\\ &=\frac{3 (2 i A-3 B) c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a f}+\frac{(2 i A-3 B) c \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (3 (2 i A-3 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 )}{a f}\\ &=\frac{3 (2 i A-3 B) c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a f}+\frac{(2 i A-3 B) c \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{\left (3 (2 i A-3 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 )}{a f}\\ &=\frac{3 (2 i A-3 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 )}{\sqrt{a} f}+\frac{3 (2 i A-3 B) c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 a f}+\frac{(2 i A-3 B) c \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a f}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 8.249, size = 185, normalized size = 0.81 \[ -\frac{c^3 \sec (e+f x) (\cos (f x)+i \sin (f x)) \left (6 (3 B-2 i A) (\cos (f x)-i \sin (f x)) \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))+\frac{1}{2} \sec ^2(e+f x) (\cos (e+2 f x)-i \sin (e+2 f x)) ((2 A+5 i B) \sin (2 (e+f x))+(13 B-10 i A) \cos (2 (e+f x))+5 (3 B-2 i A))\right )}{2 f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2))/Sqrt[a + I*a*Tan[e + f*x]],x]
[Out]
-(c^3*Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*(6*((-2*I)*A + 3*B)*ArcTan[Cos[e + f*x] + I*Sin[e + f*x]]*(Cos[f*x]
- I*Sin[f*x]) + (Sec[e + f*x]^2*(5*((-2*I)*A + 3*B) + ((-10*I)*A + 13*B)*Cos[2*(e + f*x)] + (2*A + (5*I)*B)*S
in[2*(e + f*x)])*(Cos[e + 2*f*x] - I*Sin[e + 2*f*x]))/2))/(2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e +
f*x]])
________________________________________________________________________________________
Maple [B] time = 0.196, size = 566, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x)
[Out]
1/2*I/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a*(6*I*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))*tan(f*x+e)^2*a*c+18*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))*tan(f*x+e)*a*c+4*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-9*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))*tan(f*x+e)^2*a*c+B*(a*c*(1+tan(f*x+e)^
2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^3-6*I*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*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)+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))*tan(f*x+e)*a*c+2*A*tan(f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-
14*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+9*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+19*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-10*A*(a*c*(1+tan(f*x+e)^2))^(1/2)
*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(-tan(f*x+e)+I)^2/(a*c)^(1/2)
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Maxima [B] time = 4.10663, size = 1779, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
((96*A + 144*I*B)*c^2*cos(4*f*x + 4*e) + (160*A + 240*I*B)*c^2*cos(2*f*x + 2*e) + 48*(2*I*A - 3*B)*c^2*sin(4*f
*x + 4*e) + 80*(2*I*A - 3*B)*c^2*sin(2*f*x + 2*e) + (64*A + 64*I*B)*c^2 + ((48*A + 72*I*B)*c^2*cos(5/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (96*A + 144*I*B)*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
))) + (48*A + 72*I*B)*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 24*(2*I*A - 3*B)*c^2*sin(5/2*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 48*(2*I*A - 3*B)*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 24*(2*I*A - 3*B)*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*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) + ((48*A + 7
2*I*B)*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (96*A + 144*I*B)*c^2*cos(3/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e))) + (48*A + 72*I*B)*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 24*
(2*I*A - 3*B)*c^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 48*(2*I*A - 3*B)*c^2*sin(3/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 24*(2*I*A - 3*B)*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)))*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) + (12*(2*I*A - 3*B)*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 24*(2*I*A - 3*B)
*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*(2*I*A - 3*B)*c^2*cos(1/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) - (24*A + 36*I*B)*c^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (48*A + 7
2*I*B)*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (24*A + 36*I*B)*c^2*sin(1/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e))))*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) + (12*(-2*I*
A + 3*B)*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 24*(-2*I*A + 3*B)*c^2*cos(3/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*(-2*I*A + 3*B)*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ (24*A + 36*I*B)*c^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (48*A + 72*I*B)*c^2*sin(3/2*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (24*A + 36*I*B)*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))))*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)/((-16*I*a*cos(5/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 32*I*a*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 16
*I*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*a*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x +
2*e))) + 32*a*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*a*sin(1/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))))*f)
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Fricas [B] time = 1.74717, size = 1593, normalized size = 6.99 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
1/4*(2*((-10*I*A + 14*B)*c^2*e^(5*I*f*x + 5*I*e) + (12*I*A - 18*B)*c^2*e^(4*I*f*x + 4*I*e) + (-20*I*A + 28*B)*
c^2*e^(3*I*f*x + 3*I*e) + (20*I*A - 30*B)*c^2*e^(2*I*f*x + 2*I*e) + (-10*I*A + 14*B)*c^2*e^(I*f*x + I*e) + (8*
I*A - 8*B)*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((36
*A^2 + 108*I*A*B - 81*B^2)*c^5/(a*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log((2*((6*I*A - 9
*B)*c^2*e^(2*I*f*x + 2*I*e) + (6*I*A - 9*B)*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((36*A^2 + 108*I*A*B - 81*B^2)*c^5/(a*f^2))*(a*f*e^(2*I*f*x + 2*I*e) - a*f))/((-1
2*I*A + 18*B)*c^2*e^(2*I*f*x + 2*I*e) + (-12*I*A + 18*B)*c^2)) + sqrt((36*A^2 + 108*I*A*B - 81*B^2)*c^5/(a*f^2
))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log((2*((6*I*A - 9*B)*c^2*e^(2*I*f*x + 2*I*e) + (6*I*A
- 9*B)*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((36*A^2
+ 108*I*A*B - 81*B^2)*c^5/(a*f^2))*(a*f*e^(2*I*f*x + 2*I*e) - a*f))/((-12*I*A + 18*B)*c^2*e^(2*I*f*x + 2*I*e)
+ (-12*I*A + 18*B)*c^2)))/(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))
________________________________________________________________________________________
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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(1/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{\sqrt{i \, a \tan \left (f x + e\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(5/2)/sqrt(I*a*tan(f*x + e) + a), x)