3.809 \(\int \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{\sqrt{c-i c \tan (e+f x)}} \, dx\)
Optimal. Leaf size=227 \[ \frac{3 a^{5/2} (3 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{c} f}-\frac{3 a^2 (3 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{a (3 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}} \]
[Out]
(3*a^(5/2)*((2*I)*A + 3*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/
(Sqrt[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (3*a^2*((2*I)*A + 3*B)
*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*f) - (a*((2*I)*A + 3*B)*(a + I*a*Tan[e + f*x])^(3
/2)*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*f)
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Rubi [A] time = 0.304716, antiderivative size = 227, 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} \[ \frac{3 a^{5/2} (3 B+2 i A) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{\sqrt{c} f}-\frac{3 a^2 (3 B+2 i A) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{a (3 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]
[Out]
(3*a^(5/2)*((2*I)*A + 3*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/
(Sqrt[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (3*a^2*((2*I)*A + 3*B)
*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*f) - (a*((2*I)*A + 3*B)*(a + I*a*Tan[e + f*x])^(3
/2)*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*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 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+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{\sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{(a (2 A-3 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{\left (3 a^2 (2 A-3 i B)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{3 a^2 (2 i A+3 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{\left (3 a^3 (2 A-3 i B)\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{(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{3 a^2 (2 i A+3 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}+\frac{\left (3 a^2 (2 i A+3 B)\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 )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{3 a^2 (2 i A+3 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}+\frac{\left (3 a^2 (2 i A+3 B)\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 )}{f}\\ &=\frac{3 a^{5/2} (2 i A+3 B) \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{c} f}-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt{c-i c \tan (e+f x)}}-\frac{3 a^2 (2 i A+3 B) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c f}-\frac{a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c f}\\ \end{align*}
Mathematica [A] time = 10.1899, size = 239, normalized size = 1.05 \[ \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \left (\frac{3 (3 B+2 i A) e^{-3 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt{\frac{c}{1+e^{2 i (e+f x)}}}}-\frac{(\tan (e+f x)+i) \sqrt{\sec (e+f x)} \sqrt{c-i c \tan (e+f x)} ((-5 B-2 i A) \sin (2 (e+f x))+(10 A-13 i B) \cos (2 (e+f x))+5 (2 A-3 i B))}{4 c}\right )}{f \sec ^{\frac{7}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]
[Out]
((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x])*((3*((2*I)*A + 3*B)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e +
f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^((3*I)*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) - (Sqrt[Sec[e + f*x]
]*(5*(2*A - (3*I)*B) + (10*A - (13*I)*B)*Cos[2*(e + f*x)] + ((-2*I)*A - 5*B)*Sin[2*(e + f*x)])*(I + Tan[e + f*
x])*Sqrt[c - I*c*Tan[e + f*x]])/(4*c)))/(f*Sec[e + f*x]^(7/2)*(A*Cos[e + f*x] + B*Sin[e + f*x]))
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Maple [B] time = 0.183, size = 565, 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+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x)
[Out]
1/2*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c*(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 = 2.81444, size = 1342, normalized size = 5.91 \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))^(1/2),x, algorithm="maxima")
[Out]
-((32*A - 112*I*B)*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*(2*I*A + 7*B)*a^2*sin(3/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - ((48*A - 72*I*B)*a^2*cos(4*f*x + 4*e) + (96*A - 144*I*B)*a^2*cos(2
*f*x + 2*e) - 24*(-2*I*A - 3*B)*a^2*sin(4*f*x + 4*e) - 48*(-2*I*A - 3*B)*a^2*sin(2*f*x + 2*e) + (48*A - 72*I*B
)*a^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) - ((48*A - 72*I*B)*a^2*cos(4*f*x + 4*e) + (96*A - 144*I*B)*a^2*cos(2*f*x + 2*e) - 24*(-2*I*A
- 3*B)*a^2*sin(4*f*x + 4*e) - 48*(-2*I*A - 3*B)*a^2*sin(2*f*x + 2*e) + (48*A - 72*I*B)*a^2)*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) + ((64*
A - 64*I*B)*a^2*cos(4*f*x + 4*e) + (128*A - 128*I*B)*a^2*cos(2*f*x + 2*e) + 64*(I*A + B)*a^2*sin(4*f*x + 4*e)
+ 128*(I*A + B)*a^2*sin(2*f*x + 2*e) + (96*A - 144*I*B)*a^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
))) + (12*(-2*I*A - 3*B)*a^2*cos(4*f*x + 4*e) + 24*(-2*I*A - 3*B)*a^2*cos(2*f*x + 2*e) + (24*A - 36*I*B)*a^2*s
in(4*f*x + 4*e) + (48*A - 72*I*B)*a^2*sin(2*f*x + 2*e) + 12*(-2*I*A - 3*B)*a^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) + (12*(2*I*A + 3*B)*a^2*cos(4*f*x + 4*e) + 24*(2*I*A + 3*B)*a^2*cos(2*f*
x + 2*e) - (24*A - 36*I*B)*a^2*sin(4*f*x + 4*e) - (48*A - 72*I*B)*a^2*sin(2*f*x + 2*e) + 12*(2*I*A + 3*B)*a^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) + (64*(I*A + B)*a^2*cos(4*f*x + 4*e) + 12
8*(I*A + B)*a^2*cos(2*f*x + 2*e) - (64*A - 64*I*B)*a^2*sin(4*f*x + 4*e) - (128*A - 128*I*B)*a^2*sin(2*f*x + 2*
e) + 48*(2*I*A + 3*B)*a^2)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-16*I*c*cos
(4*f*x + 4*e) - 32*I*c*cos(2*f*x + 2*e) + 16*c*sin(4*f*x + 4*e) + 32*c*sin(2*f*x + 2*e) - 16*I*c)*f)
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Fricas [B] time = 1.61783, size = 1347, normalized size = 5.93 \begin{align*} \frac{2 \,{\left ({\left (-8 i \, A - 8 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-20 i \, A - 30 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (-12 i \, A - 18 \, B\right )} a^{2}\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}} e^{\left (i \, f x + i \, e\right )} - \sqrt{\frac{{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}}{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac{2 \,{\left ({\left ({\left (24 i \, A + 36 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (24 i \, A + 36 \, B\right )} a^{2}\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}} e^{\left (i \, f x + i \, e\right )} + 2 \, \sqrt{\frac{{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}}{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )}\right )}}{{\left (6 i \, A + 9 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (6 i \, A + 9 \, B\right )} a^{2}}\right ) + \sqrt{\frac{{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}}{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac{2 \,{\left ({\left ({\left (24 i \, A + 36 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (24 i \, A + 36 \, B\right )} a^{2}\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}} e^{\left (i \, f x + i \, e\right )} - 2 \, \sqrt{\frac{{\left (36 \, A^{2} - 108 i \, A B - 81 \, B^{2}\right )} a^{5}}{c f^{2}}}{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )}\right )}}{{\left (6 i \, A + 9 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (6 i \, A + 9 \, B\right )} a^{2}}\right )}{4 \,{\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \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))^(1/2),x, algorithm="fricas")
[Out]
1/4*(2*((-8*I*A - 8*B)*a^2*e^(4*I*f*x + 4*I*e) + (-20*I*A - 30*B)*a^2*e^(2*I*f*x + 2*I*e) + (-12*I*A - 18*B)*a
^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)*a^5/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*f)*log(2*(((24*I*A + 36*B)*a^2*e^(2*I*f*x + 2*I*e) +
(24*I*A + 36*B)*a^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) + 2*s
qrt((36*A^2 - 108*I*A*B - 81*B^2)*a^5/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) - c*f))/((6*I*A + 9*B)*a^2*e^(2*I*f*x
+ 2*I*e) + (6*I*A + 9*B)*a^2)) + sqrt((36*A^2 - 108*I*A*B - 81*B^2)*a^5/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*
f)*log(2*(((24*I*A + 36*B)*a^2*e^(2*I*f*x + 2*I*e) + (24*I*A + 36*B)*a^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sq
rt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - 2*sqrt((36*A^2 - 108*I*A*B - 81*B^2)*a^5/(c*f^2))*(c*f*e^(2*
I*f*x + 2*I*e) - c*f))/((6*I*A + 9*B)*a^2*e^(2*I*f*x + 2*I*e) + (6*I*A + 9*B)*a^2)))/(c*f*e^(2*I*f*x + 2*I*e)
+ c*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))**(1/2),x)
[Out]
Timed out
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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 \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{\sqrt{-i \, c \tan \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="giac")
[Out]
integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(5/2)/sqrt(-I*c*tan(f*x + e) + c), x)