3.1033 \(\int \frac{(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=304 \[ \frac{63 i a^{11/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac{63 i a^5 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

((63*I)*a^(11/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(c^(5/2)*f
) - (((2*I)/5)*a*(a + I*a*Tan[e + f*x])^(9/2))/(f*(c - I*c*Tan[e + f*x])^(5/2)) + (((6*I)/5)*a^2*(a + I*a*Tan[
e + f*x])^(7/2))/(c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (((42*I)/5)*a^3*(a + I*a*Tan[e + f*x])^(5/2))/(c^2*f*Sqr
t[c - I*c*Tan[e + f*x]]) - (((63*I)/2)*a^5*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*f) - ((
(21*I)/2)*a^4*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*f)

________________________________________________________________________________________

Rubi [A]  time = 0.240694, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3523, 47, 50, 63, 217, 203} \[ \frac{63 i a^{11/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac{63 i a^5 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(11/2)/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((63*I)*a^(11/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(c^(5/2)*f
) - (((2*I)/5)*a*(a + I*a*Tan[e + f*x])^(9/2))/(f*(c - I*c*Tan[e + f*x])^(5/2)) + (((6*I)/5)*a^2*(a + I*a*Tan[
e + f*x])^(7/2))/(c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (((42*I)/5)*a^3*(a + I*a*Tan[e + f*x])^(5/2))/(c^2*f*Sqr
t[c - I*c*Tan[e + f*x]]) - (((63*I)/2)*a^5*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*f) - ((
(21*I)/2)*a^4*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*f)

Rule 3523

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((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), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{9/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{\left (9 a^2\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{7/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}+\frac{\left (21 a^3\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{5 c f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac{42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{\left (21 a^4\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{c^2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac{42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{\left (63 a^5\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 c^2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac{42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{63 i a^5 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{\left (63 a^6\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 c^2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac{42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{63 i a^5 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}+\frac{\left (63 i a^5\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 )}{c^2 f}\\ &=-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac{42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{63 i a^5 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}+\frac{\left (63 i a^5\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 )}{c^2 f}\\ &=\frac{63 i a^{11/2} \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 )}{c^{5/2} f}-\frac{2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac{42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{63 i a^5 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}-\frac{21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 c^3 f}\\ \end{align*}

Mathematica [A]  time = 17.3035, size = 459, normalized size = 1.51 \[ \frac{\cos ^5(e+f x) (a+i a \tan (e+f x))^{11/2} \left (\cos (6 f x) \left (\frac{4 \sin (e)}{5 c^3}-\frac{4 i \cos (e)}{5 c^3}\right )+\cos (4 f x) \left (\frac{16 \sin (e)}{5 c^3}+\frac{16 i \cos (e)}{5 c^3}\right )+\cos (2 f x) \left (-\frac{20 \sin (3 e)}{c^3}-\frac{20 i \cos (3 e)}{c^3}\right )+\sin (2 f x) \left (\frac{20 \cos (3 e)}{c^3}-\frac{20 i \sin (3 e)}{c^3}\right )+\sin (4 f x) \left (-\frac{16 \cos (e)}{5 c^3}+\frac{16 i \sin (e)}{5 c^3}\right )+\sin (6 f x) \left (\frac{4 \cos (e)}{5 c^3}+\frac{4 i \sin (e)}{5 c^3}\right )+\sec (e) \sin (f x) \left (\frac{\cos (5 e)}{2 c^3}-\frac{i \sin (5 e)}{2 c^3}\right ) \sec (e+f x)+\sec (e) (64 \cos (e)+i \sin (e)) \left (-\frac{\sin (5 e)}{2 c^3}-\frac{i \cos (5 e)}{2 c^3}\right )\right ) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))}}{f (\cos (f x)+i \sin (f x))^5}+\frac{63 i \sqrt{e^{i f x}} e^{-i (6 e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{11/2}}{c^2 f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \sec ^{\frac{11}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{11/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(11/2)/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((63*I)*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])^(11/2))/(c^2*E^(I*(6*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e + f*x]^(11/2)*(Cos[f*x] + I*
Sin[f*x])^(11/2)) + (Cos[e + f*x]^5*(Cos[6*f*x]*((((-4*I)/5)*Cos[e])/c^3 + (4*Sin[e])/(5*c^3)) + Cos[4*f*x]*((
((16*I)/5)*Cos[e])/c^3 + (16*Sin[e])/(5*c^3)) + Cos[2*f*x]*(((-20*I)*Cos[3*e])/c^3 - (20*Sin[3*e])/c^3) + Sec[
e]*(64*Cos[e] + I*Sin[e])*(((-I/2)*Cos[5*e])/c^3 - Sin[5*e]/(2*c^3)) + Sec[e]*Sec[e + f*x]*(Cos[5*e]/(2*c^3) -
((I/2)*Sin[5*e])/c^3)*Sin[f*x] + ((20*Cos[3*e])/c^3 - ((20*I)*Sin[3*e])/c^3)*Sin[2*f*x] + ((-16*Cos[e])/(5*c^
3) + (((16*I)/5)*Sin[e])/c^3)*Sin[4*f*x] + ((4*Cos[e])/(5*c^3) + (((4*I)/5)*Sin[e])/c^3)*Sin[6*f*x])*Sqrt[Sec[
e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^(11/2))/(f*(Cos[f*x] + I*Sin[f*x])^5)

________________________________________________________________________________________

Maple [B]  time = 0.046, size = 490, normalized size = 1.6 \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))^(11/2)/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

-1/10/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^5/c^3*(1260*I*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)^3*a*c+60*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan
(f*x+e)^4+315*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)^4*a*c-5*(a*
c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^5-1260*I*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-1964*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-1890*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-866*tan(f*x+e)^3*(a*c)^(1/
2)*(a*c*(1+tan(f*x+e)^2))^(1/2)+496*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+315*a*c*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))+1659*tan(f*x+e)*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(
a*c*(1+tan(f*x+e)^2))^(1/2)/(a*c)^(1/2)/(tan(f*x+e)+I)^4

________________________________________________________________________________________

Maxima [B]  time = 2.12403, size = 1472, normalized size = 4.84 \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))^(11/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

((6300*a^5*cos(4*f*x + 4*e) + 12600*a^5*cos(2*f*x + 2*e) + 6300*I*a^5*sin(4*f*x + 4*e) + 12600*I*a^5*sin(2*f*x
+ 2*e) + 6300*a^5)*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) + (6300*a^5*cos(4*f*x + 4*e) + 12600*a^5*cos(2*f*x + 2*e) + 6300*I*a^5*sin(4*f*x
+ 4*e) + 12600*I*a^5*sin(2*f*x + 2*e) + 6300*a^5)*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) - (320*a^5*cos(4*f*x + 4*e) + 640*a^5*cos(2*f*x
+ 2*e) + 320*I*a^5*sin(4*f*x + 4*e) + 640*I*a^5*sin(2*f*x + 2*e) + 320*a^5)*cos(5/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))) + (1600*a^5*cos(4*f*x + 4*e) + 3200*a^5*cos(2*f*x + 2*e) + 1600*I*a^5*sin(4*f*x + 4*e) + 32
00*I*a^5*sin(2*f*x + 2*e) - 1800*a^5)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (9600*a^5*cos(4*f
*x + 4*e) + 19200*a^5*cos(2*f*x + 2*e) + 9600*I*a^5*sin(4*f*x + 4*e) + 19200*I*a^5*sin(2*f*x + 2*e) + 12600*a^
5)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-3150*I*a^5*cos(4*f*x + 4*e) - 6300*I*a^5*cos(2*f*x
+ 2*e) + 3150*a^5*sin(4*f*x + 4*e) + 6300*a^5*sin(2*f*x + 2*e) - 3150*I*a^5)*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) - (3150*I*a^5*cos(4*f*x + 4*e) + 6300*I*a^5*cos(2*f*x + 2*e) - 3150*a^5*si
n(4*f*x + 4*e) - 6300*a^5*sin(2*f*x + 2*e) + 3150*I*a^5)*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) - (320*I*a^5*cos(4*f*x + 4*e) + 640*I*a^5*cos(2*f*x + 2*e) - 320*a^5*sin(4*f*x + 4*e) - 640*a^5
*sin(2*f*x + 2*e) + 320*I*a^5)*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-1600*I*a^5*cos(4*f*x +
4*e) - 3200*I*a^5*cos(2*f*x + 2*e) + 1600*a^5*sin(4*f*x + 4*e) + 3200*a^5*sin(2*f*x + 2*e) + 1800*I*a^5)*sin(
3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (9600*I*a^5*cos(4*f*x + 4*e) + 19200*I*a^5*cos(2*f*x + 2*e)
- 9600*a^5*sin(4*f*x + 4*e) - 19200*a^5*sin(2*f*x + 2*e) + 12600*I*a^5)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-200*I*c^3*cos(4*f*x + 4*e) - 400*I*c^3*cos(2*f*x + 2*e) + 200*c^3*sin(4*f*
x + 4*e) + 400*c^3*sin(2*f*x + 2*e) - 200*I*c^3)*f)

________________________________________________________________________________________

Fricas [A]  time = 1.62969, size = 1141, normalized size = 3.75 \begin{align*} \frac{2 \,{\left (-16 i \, a^{5} e^{\left (8 i \, f x + 8 i \, e\right )} + 48 i \, a^{5} e^{\left (6 i \, f x + 6 i \, e\right )} - 336 i \, a^{5} e^{\left (4 i \, f x + 4 i \, e\right )} - 1050 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 630 i \, a^{5}\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 )} + 315 \,{\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )} \sqrt{\frac{a^{11}}{c^{5} f^{2}}} \log \left (\frac{8 \,{\left (a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{5}\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 )} +{\left (4 i \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, c^{3} f\right )} \sqrt{\frac{a^{11}}{c^{5} f^{2}}}}{a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{5}}\right ) - 315 \,{\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )} \sqrt{\frac{a^{11}}{c^{5} f^{2}}} \log \left (\frac{8 \,{\left (a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{5}\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 )} +{\left (-4 i \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, c^{3} f\right )} \sqrt{\frac{a^{11}}{c^{5} f^{2}}}}{a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{5}}\right )}{20 \,{\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(11/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/20*(2*(-16*I*a^5*e^(8*I*f*x + 8*I*e) + 48*I*a^5*e^(6*I*f*x + 6*I*e) - 336*I*a^5*e^(4*I*f*x + 4*I*e) - 1050*I
*a^5*e^(2*I*f*x + 2*I*e) - 630*I*a^5)*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) + 315*(c^3*f*e^(2*I*f*x + 2*I*e) + c^3*f)*sqrt(a^11/(c^5*f^2))*log((8*(a^5*e^(2*I*f*x + 2*I*e) + a
^5)*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) + (4*I*c^3*f*e^(2*I*f*
x + 2*I*e) - 4*I*c^3*f)*sqrt(a^11/(c^5*f^2)))/(a^5*e^(2*I*f*x + 2*I*e) + a^5)) - 315*(c^3*f*e^(2*I*f*x + 2*I*e
) + c^3*f)*sqrt(a^11/(c^5*f^2))*log((8*(a^5*e^(2*I*f*x + 2*I*e) + a^5)*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) + (-4*I*c^3*f*e^(2*I*f*x + 2*I*e) + 4*I*c^3*f)*sqrt(a^11/(c^5*f^2
)))/(a^5*e^(2*I*f*x + 2*I*e) + a^5)))/(c^3*f*e^(2*I*f*x + 2*I*e) + c^3*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))**(11/2)/(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 \frac{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{11}{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))^(11/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(11/2)/(-I*c*tan(f*x + e) + c)^(5/2), x)