3.107 \(\int \frac{(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=266 \[ \frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}-\frac{a^3 (3 A-17 B) \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}+\frac{a^3 (3 A-17 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 (3 A-17 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{512 \sqrt{2} c^{11/2} f}+\frac{a^3 c (3 A-17 B) \cos ^5(e+f x)}{80 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (3 A-17 B) \cos ^3(e+f x)}{96 c f (c-c \sin (e+f x))^{9/2}} \]

[Out]

-(a^3*(3*A - 17*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(512*Sqrt[2]*c^(11/2)*f
) + (a^3*(A + B)*c^3*Cos[e + f*x]^7)/(10*f*(c - c*Sin[e + f*x])^(17/2)) + (a^3*(3*A - 17*B)*c*Cos[e + f*x]^5)/
(80*f*(c - c*Sin[e + f*x])^(13/2)) - (a^3*(3*A - 17*B)*Cos[e + f*x]^3)/(96*c*f*(c - c*Sin[e + f*x])^(9/2)) + (
a^3*(3*A - 17*B)*Cos[e + f*x])/(128*c^3*f*(c - c*Sin[e + f*x])^(5/2)) - (a^3*(3*A - 17*B)*Cos[e + f*x])/(512*c
^4*f*(c - c*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.587054, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2967, 2859, 2680, 2650, 2649, 206} \[ \frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}-\frac{a^3 (3 A-17 B) \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}+\frac{a^3 (3 A-17 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 (3 A-17 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{512 \sqrt{2} c^{11/2} f}+\frac{a^3 c (3 A-17 B) \cos ^5(e+f x)}{80 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (3 A-17 B) \cos ^3(e+f x)}{96 c f (c-c \sin (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

-(a^3*(3*A - 17*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(512*Sqrt[2]*c^(11/2)*f
) + (a^3*(A + B)*c^3*Cos[e + f*x]^7)/(10*f*(c - c*Sin[e + f*x])^(17/2)) + (a^3*(3*A - 17*B)*c*Cos[e + f*x]^5)/
(80*f*(c - c*Sin[e + f*x])^(13/2)) - (a^3*(3*A - 17*B)*Cos[e + f*x]^3)/(96*c*f*(c - c*Sin[e + f*x])^(9/2)) + (
a^3*(3*A - 17*B)*Cos[e + f*x])/(128*c^3*f*(c - c*Sin[e + f*x])^(5/2)) - (a^3*(3*A - 17*B)*Cos[e + f*x])/(512*c
^4*f*(c - c*Sin[e + f*x])^(3/2))

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{17/2}} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}+\frac{1}{20} \left (a^3 (3 A-17 B) c^2\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{15/2}} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}+\frac{a^3 (3 A-17 B) c \cos ^5(e+f x)}{80 f (c-c \sin (e+f x))^{13/2}}-\frac{1}{32} \left (a^3 (3 A-17 B)\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}+\frac{a^3 (3 A-17 B) c \cos ^5(e+f x)}{80 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (3 A-17 B) \cos ^3(e+f x)}{96 c f (c-c \sin (e+f x))^{9/2}}+\frac{\left (a^3 (3 A-17 B)\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx}{64 c^2}\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}+\frac{a^3 (3 A-17 B) c \cos ^5(e+f x)}{80 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (3 A-17 B) \cos ^3(e+f x)}{96 c f (c-c \sin (e+f x))^{9/2}}+\frac{a^3 (3 A-17 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{\left (a^3 (3 A-17 B)\right ) \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{256 c^4}\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}+\frac{a^3 (3 A-17 B) c \cos ^5(e+f x)}{80 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (3 A-17 B) \cos ^3(e+f x)}{96 c f (c-c \sin (e+f x))^{9/2}}+\frac{a^3 (3 A-17 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 (3 A-17 B) \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (a^3 (3 A-17 B)\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{1024 c^5}\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}+\frac{a^3 (3 A-17 B) c \cos ^5(e+f x)}{80 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (3 A-17 B) \cos ^3(e+f x)}{96 c f (c-c \sin (e+f x))^{9/2}}+\frac{a^3 (3 A-17 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 (3 A-17 B) \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (a^3 (3 A-17 B)\right ) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,-\frac{c \cos (e+f x)}{\sqrt{c-c \sin (e+f x)}}\right )}{512 c^5 f}\\ &=-\frac{a^3 (3 A-17 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{512 \sqrt{2} c^{11/2} f}+\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{10 f (c-c \sin (e+f x))^{17/2}}+\frac{a^3 (3 A-17 B) c \cos ^5(e+f x)}{80 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (3 A-17 B) \cos ^3(e+f x)}{96 c f (c-c \sin (e+f x))^{9/2}}+\frac{a^3 (3 A-17 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac{a^3 (3 A-17 B) \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 6.86234, size = 485, normalized size = 1.82 \[ \frac{(a \sin (e+f x)+a)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (56370 A \sin \left (\frac{1}{2} (e+f x)\right )+31140 A \sin \left (\frac{3}{2} (e+f x)\right )-10404 A \sin \left (\frac{5}{2} (e+f x)\right )-435 A \sin \left (\frac{7}{2} (e+f x)\right )-45 A \sin \left (\frac{9}{2} (e+f x)\right )+56370 A \cos \left (\frac{1}{2} (e+f x)\right )-31140 A \cos \left (\frac{3}{2} (e+f x)\right )-10404 A \cos \left (\frac{5}{2} (e+f x)\right )+435 A \cos \left (\frac{7}{2} (e+f x)\right )-45 A \cos \left (\frac{9}{2} (e+f x)\right )+38970 B \sin \left (\frac{1}{2} (e+f x)\right )+38580 B \sin \left (\frac{3}{2} (e+f x)\right )-12724 B \sin \left (\frac{5}{2} (e+f x)\right )-7775 B \sin \left (\frac{7}{2} (e+f x)\right )+255 B \sin \left (\frac{9}{2} (e+f x)\right )+38970 B \cos \left (\frac{1}{2} (e+f x)\right )-38580 B \cos \left (\frac{3}{2} (e+f x)\right )-12724 B \cos \left (\frac{5}{2} (e+f x)\right )+7775 B \cos \left (\frac{7}{2} (e+f x)\right )+255 B \cos \left (\frac{9}{2} (e+f x)\right )\right )}{122880 f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6}+\frac{\left (\frac{1}{512}+\frac{i}{512}\right ) \sqrt [4]{-1} (3 A-17 B) (a \sin (e+f x)+a)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{11} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \sec \left (\frac{1}{4} (e+f x)\right ) \left (\sin \left (\frac{1}{4} (e+f x)\right )+\cos \left (\frac{1}{4} (e+f x)\right )\right )\right )}{f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

((1/512 + I/512)*(-1)^(1/4)*(3*A - 17*B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*Sec[(e + f*x)/4]*(Cos[(e + f*x)/4] + Si
n[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^11*(a + a*Sin[e + f*x])^3)/(f*(Cos[(e + f*x)/2] + Sin[(
e + f*x)/2])^6*(c - c*Sin[e + f*x])^(11/2)) + ((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a + a*Sin[e + f*x])^3*(5
6370*A*Cos[(e + f*x)/2] + 38970*B*Cos[(e + f*x)/2] - 31140*A*Cos[(3*(e + f*x))/2] - 38580*B*Cos[(3*(e + f*x))/
2] - 10404*A*Cos[(5*(e + f*x))/2] - 12724*B*Cos[(5*(e + f*x))/2] + 435*A*Cos[(7*(e + f*x))/2] + 7775*B*Cos[(7*
(e + f*x))/2] - 45*A*Cos[(9*(e + f*x))/2] + 255*B*Cos[(9*(e + f*x))/2] + 56370*A*Sin[(e + f*x)/2] + 38970*B*Si
n[(e + f*x)/2] + 31140*A*Sin[(3*(e + f*x))/2] + 38580*B*Sin[(3*(e + f*x))/2] - 10404*A*Sin[(5*(e + f*x))/2] -
12724*B*Sin[(5*(e + f*x))/2] - 435*A*Sin[(7*(e + f*x))/2] - 7775*B*Sin[(7*(e + f*x))/2] - 45*A*Sin[(9*(e + f*x
))/2] + 255*B*Sin[(9*(e + f*x))/2]))/(122880*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(1
1/2))

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Maple [B]  time = 1.792, size = 526, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x)

[Out]

1/15360*a^3*(15*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^6*(3*A-17*B)*sin(f*x+e)*cos(f*x+
e)^4-180*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^6*(3*A-17*B)*cos(f*x+e)^2*sin(f*x+e)+24
0*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^6*(3*A-17*B)*sin(f*x+e)-75*2^(1/2)*arctanh(1/2
*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^6*(3*A-17*B)*cos(f*x+e)^4+300*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^
(1/2)*2^(1/2)/c^(1/2))*c^6*(3*A-17*B)*cos(f*x+e)^2-90*A*(c+c*sin(f*x+e))^(9/2)*c^(3/2)+840*A*(c+c*sin(f*x+e))^
(7/2)*c^(5/2)+3072*A*(c+c*sin(f*x+e))^(5/2)*c^(7/2)-3360*A*(c+c*sin(f*x+e))^(3/2)*c^(9/2)+1440*A*(c+c*sin(f*x+
e))^(1/2)*c^(11/2)+510*B*(c+c*sin(f*x+e))^(9/2)*c^(3/2)+5480*B*(c+c*sin(f*x+e))^(7/2)*c^(5/2)-17408*B*(c+c*sin
(f*x+e))^(5/2)*c^(7/2)+19040*B*(c+c*sin(f*x+e))^(3/2)*c^(9/2)-8160*B*(c+c*sin(f*x+e))^(1/2)*c^(11/2)-720*A*2^(
1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^6+4080*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)
*2^(1/2)/c^(1/2))*c^6)*(c*(1+sin(f*x+e)))^(1/2)/c^(23/2)/(-1+sin(f*x+e))^4/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(11/2), x)

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Fricas [B]  time = 1.71799, size = 1971, normalized size = 7.41 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-1/30720*(15*sqrt(2)*((3*A - 17*B)*a^3*cos(f*x + e)^6 - 5*(3*A - 17*B)*a^3*cos(f*x + e)^5 - 18*(3*A - 17*B)*a^
3*cos(f*x + e)^4 + 20*(3*A - 17*B)*a^3*cos(f*x + e)^3 + 48*(3*A - 17*B)*a^3*cos(f*x + e)^2 - 16*(3*A - 17*B)*a
^3*cos(f*x + e) - 32*(3*A - 17*B)*a^3 + ((3*A - 17*B)*a^3*cos(f*x + e)^5 + 6*(3*A - 17*B)*a^3*cos(f*x + e)^4 -
12*(3*A - 17*B)*a^3*cos(f*x + e)^3 - 32*(3*A - 17*B)*a^3*cos(f*x + e)^2 + 16*(3*A - 17*B)*a^3*cos(f*x + e) +
32*(3*A - 17*B)*a^3)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c
)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x +
e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(15*(3*A - 17*B)*a^3*cos(f*x + e)^5 - 5*(39*A
+ 803*B)*a^3*cos(f*x + e)^4 + 4*(609*A + 389*B)*a^3*cos(f*x + e)^3 + 12*(449*A + 869*B)*a^3*cos(f*x + e)^2 -
24*(143*A + 43*B)*a^3*cos(f*x + e) - 6144*(A + B)*a^3 + (15*(3*A - 17*B)*a^3*cos(f*x + e)^4 + 80*(3*A + 47*B)*
a^3*cos(f*x + e)^3 + 12*(223*A + 443*B)*a^3*cos(f*x + e)^2 - 24*(113*A + 213*B)*a^3*cos(f*x + e) - 6144*(A + B
)*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^6*f*cos(f*x + e)^6 - 5*c^6*f*cos(f*x + e)^5 - 18*c^6*f*cos(
f*x + e)^4 + 20*c^6*f*cos(f*x + e)^3 + 48*c^6*f*cos(f*x + e)^2 - 16*c^6*f*cos(f*x + e) - 32*c^6*f + (c^6*f*cos
(f*x + e)^5 + 6*c^6*f*cos(f*x + e)^4 - 12*c^6*f*cos(f*x + e)^3 - 32*c^6*f*cos(f*x + e)^2 + 16*c^6*f*cos(f*x +
e) + 32*c^6*f)*sin(f*x + e))

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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+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

sage2