3.262 \(\int \frac{(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx\)

Optimal. Leaf size=246 \[ \frac{a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f}+\frac{2 a^3 (c-d)^3 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^4 f \sqrt{c^2-d^2}}+\frac{a^3 x \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (-6 c^2 d+2 c^3+7 c d^2-5 d^3\right )\right )}{2 d^4}+\frac{(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{6 d^2 f}-\frac{a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f} \]

[Out]

(a^3*(A*d*(2*c^2 - 6*c*d + 7*d^2) - B*(2*c^3 - 6*c^2*d + 7*c*d^2 - 5*d^3))*x)/(2*d^4) + (2*a^3*(c - d)^3*(B*c
- A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d^4*Sqrt[c^2 - d^2]*f) + (a^3*(A*(2*c - 5*d)*d - B*(
2*c^2 - 5*c*d + 5*d^2))*Cos[e + f*x])/(2*d^3*f) - (a*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(3*d*f) + ((3*B*c
- 3*A*d - 5*B*d)*Cos[e + f*x]*(a^3 + a^3*Sin[e + f*x]))/(6*d^2*f)

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Rubi [A]  time = 0.895251, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2976, 2968, 3023, 2735, 2660, 618, 204} \[ \frac{a^3 \left (A d (2 c-5 d)-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f}+\frac{2 a^3 (c-d)^3 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^4 f \sqrt{c^2-d^2}}+\frac{a^3 x \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (-6 c^2 d+2 c^3+7 c d^2-5 d^3\right )\right )}{2 d^4}+\frac{(-3 A d+3 B c-5 B d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{6 d^2 f}-\frac{a B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(A*d*(2*c^2 - 6*c*d + 7*d^2) - B*(2*c^3 - 6*c^2*d + 7*c*d^2 - 5*d^3))*x)/(2*d^4) + (2*a^3*(c - d)^3*(B*c
- A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d^4*Sqrt[c^2 - d^2]*f) + (a^3*(A*(2*c - 5*d)*d - B*(
2*c^2 - 5*c*d + 5*d^2))*Cos[e + f*x])/(2*d^3*f) - (a*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(3*d*f) + ((3*B*c
- 3*A*d - 5*B*d)*Cos[e + f*x]*(a^3 + a^3*Sin[e + f*x]))/(6*d^2*f)

Rule 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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+d \sin (e+f x)} \, dx &=-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac{\int \frac{(a+a \sin (e+f x))^2 (a (2 B c+3 A d)-a (3 B c-3 A d-5 B d) \sin (e+f x))}{c+d \sin (e+f x)} \, dx}{3 d}\\ &=-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac{(3 B c-3 A d-5 B d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 d^2 f}+\frac{\int \frac{(a+a \sin (e+f x)) \left (-3 a^2 (B c (c-3 d)-A d (c+2 d))-3 a^2 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{6 d^2}\\ &=-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac{(3 B c-3 A d-5 B d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 d^2 f}+\frac{\int \frac{-3 a^3 (B c (c-3 d)-A d (c+2 d))+\left (-3 a^3 (B c (c-3 d)-A d (c+2 d))-3 a^3 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right )\right ) \sin (e+f x)-3 a^3 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right ) \sin ^2(e+f x)}{c+d \sin (e+f x)} \, dx}{6 d^2}\\ &=\frac{a^3 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac{(3 B c-3 A d-5 B d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 d^2 f}+\frac{\int \frac{-3 a^3 d (B c (c-3 d)-A d (c+2 d))+3 a^3 \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right ) \sin (e+f x)}{c+d \sin (e+f x)} \, dx}{6 d^3}\\ &=\frac{a^3 \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right ) x}{2 d^4}+\frac{a^3 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac{(3 B c-3 A d-5 B d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 d^2 f}+\frac{\left (a^3 (c-d)^3 (B c-A d)\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{d^4}\\ &=\frac{a^3 \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right ) x}{2 d^4}+\frac{a^3 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac{(3 B c-3 A d-5 B d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 d^2 f}+\frac{\left (2 a^3 (c-d)^3 (B c-A d)\right ) \operatorname{Subst}\left (\int \frac{1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{d^4 f}\\ &=\frac{a^3 \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right ) x}{2 d^4}+\frac{a^3 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac{(3 B c-3 A d-5 B d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 d^2 f}-\frac{\left (4 a^3 (c-d)^3 (B c-A d)\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac{1}{2} (e+f x)\right )\right )}{d^4 f}\\ &=\frac{a^3 \left (A d \left (2 c^2-6 c d+7 d^2\right )-B \left (2 c^3-6 c^2 d+7 c d^2-5 d^3\right )\right ) x}{2 d^4}+\frac{2 a^3 (c-d)^3 (B c-A d) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{d^4 \sqrt{c^2-d^2} f}+\frac{a^3 \left (A (2 c-5 d) d-B \left (2 c^2-5 c d+5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f}-\frac{a B \cos (e+f x) (a+a \sin (e+f x))^2}{3 d f}+\frac{(3 B c-3 A d-5 B d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 d^2 f}\\ \end{align*}

Mathematica [A]  time = 0.964217, size = 233, normalized size = 0.95 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (6 (e+f x) \left (A d \left (2 c^2-6 c d+7 d^2\right )+B \left (6 c^2 d-2 c^3-7 c d^2+5 d^3\right )\right )-3 d \left (4 A d (3 d-c)+B \left (4 c^2-12 c d+15 d^2\right )\right ) \cos (e+f x)+\frac{24 (c-d)^3 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}-3 d^2 (A d-B c+3 B d) \sin (2 (e+f x))+B d^3 \cos (3 (e+f x))\right )}{12 d^4 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Sin[e + f*x])^3*(6*(A*d*(2*c^2 - 6*c*d + 7*d^2) + B*(-2*c^3 + 6*c^2*d - 7*c*d^2 + 5*d^3))*(e + f*x)
+ (24*(c - d)^3*(B*c - A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] - 3*d*(4*A*d*(-c
+ 3*d) + B*(4*c^2 - 12*c*d + 15*d^2))*Cos[e + f*x] + B*d^3*Cos[3*(e + f*x)] - 3*d^2*(-(B*c) + A*d + 3*B*d)*Si
n[2*(e + f*x)]))/(12*d^4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6)

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Maple [B]  time = 0.165, size = 1357, normalized size = 5.5 \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+d*sin(f*x+e)),x)

[Out]

2/f*a^3/d^4/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*B*c^4-4/f*a^3/d^3/(1+tan(
1/2*f*x+1/2*e)^2)^3*B*tan(1/2*f*x+1/2*e)^2*c^2+2/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^3*A*tan(1/2*f*x+1/2*e)^4*c
-2/f*a^3/d^3/(1+tan(1/2*f*x+1/2*e)^2)^3*B*tan(1/2*f*x+1/2*e)^4*c^2+6/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^3*B*ta
n(1/2*f*x+1/2*e)^4*c+4/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^3*A*tan(1/2*f*x+1/2*e)^2*c+12/f*a^3/d^2/(1+tan(1/2*f
*x+1/2*e)^2)^3*B*tan(1/2*f*x+1/2*e)^2*c+1/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)*B*c-1/f*a^3/
d^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^5*B*c-6/f*a^3/d^3/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*
x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*B*c^3+6/f*a^3/d^2/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-
d^2)^(1/2))*B*c^2-2/f*a^3/d/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*B*c-2/f*a
^3/d^3/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*A*c^3+6/f*a^3/d^2/(c^2-d^2)^(1
/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*A*c^2-6/f*a^3/d/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*t
an(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*A*c+3/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^5*B-12/f*a
^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*A*tan(1/2*f*x+1/2*e)^2-16/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*B*tan(1/2*f*x+1/2
*e)^2-2/f*a^3/d^4*arctan(tan(1/2*f*x+1/2*e))*B*c^3-6/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*A*tan(1/2*f*x+1/2*e)^4
-6/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*B*tan(1/2*f*x+1/2*e)^4-1/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+
1/2*e)*A-3/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)*B+2/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^3*A*c-
2/f*a^3/d^3/(1+tan(1/2*f*x+1/2*e)^2)^3*B*c^2+6/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^3*B*c+2/f*a^3/d^3*arctan(tan
(1/2*f*x+1/2*e))*A*c^2-6/f*a^3/d^2*arctan(tan(1/2*f*x+1/2*e))*A*c+6/f*a^3/d^3*arctan(tan(1/2*f*x+1/2*e))*B*c^2
-7/f*a^3/d^2*arctan(tan(1/2*f*x+1/2*e))*B*c+1/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^5*A+5/f*a^
3/d*arctan(tan(1/2*f*x+1/2*e))*B+2/f*a^3/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/
2))*A-6/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*A-22/3/f*a^3/d/(1+tan(1/2*f*x+1/2*e)^2)^3*B+7/f*a^3/d*arctan(tan(1/
2*f*x+1/2*e))*A

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.56341, size = 1393, normalized size = 5.66 \begin{align*} \left [\frac{2 \, B a^{3} d^{3} \cos \left (f x + e\right )^{3} - 3 \,{\left (2 \, B a^{3} c^{3} - 2 \,{\left (A + 3 \, B\right )} a^{3} c^{2} d +{\left (6 \, A + 7 \, B\right )} a^{3} c d^{2} -{\left (7 \, A + 5 \, B\right )} a^{3} d^{3}\right )} f x + 3 \,{\left (B a^{3} c d^{2} -{\left (A + 3 \, B\right )} a^{3} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \,{\left (B a^{3} c^{3} -{\left (A + 2 \, B\right )} a^{3} c^{2} d +{\left (2 \, A + B\right )} a^{3} c d^{2} - A a^{3} d^{3}\right )} \sqrt{-\frac{c - d}{c + d}} \log \left (-\frac{{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} - 2 \,{\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) +{\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{-\frac{c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) - 6 \,{\left (B a^{3} c^{2} d -{\left (A + 3 \, B\right )} a^{3} c d^{2} +{\left (3 \, A + 4 \, B\right )} a^{3} d^{3}\right )} \cos \left (f x + e\right )}{6 \, d^{4} f}, \frac{2 \, B a^{3} d^{3} \cos \left (f x + e\right )^{3} - 3 \,{\left (2 \, B a^{3} c^{3} - 2 \,{\left (A + 3 \, B\right )} a^{3} c^{2} d +{\left (6 \, A + 7 \, B\right )} a^{3} c d^{2} -{\left (7 \, A + 5 \, B\right )} a^{3} d^{3}\right )} f x + 3 \,{\left (B a^{3} c d^{2} -{\left (A + 3 \, B\right )} a^{3} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 6 \,{\left (B a^{3} c^{3} -{\left (A + 2 \, B\right )} a^{3} c^{2} d +{\left (2 \, A + B\right )} a^{3} c d^{2} - A a^{3} d^{3}\right )} \sqrt{\frac{c - d}{c + d}} \arctan \left (-\frac{{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt{\frac{c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) - 6 \,{\left (B a^{3} c^{2} d -{\left (A + 3 \, B\right )} a^{3} c d^{2} +{\left (3 \, A + 4 \, B\right )} a^{3} d^{3}\right )} \cos \left (f x + e\right )}{6 \, d^{4} f}\right ] \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+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/6*(2*B*a^3*d^3*cos(f*x + e)^3 - 3*(2*B*a^3*c^3 - 2*(A + 3*B)*a^3*c^2*d + (6*A + 7*B)*a^3*c*d^2 - (7*A + 5*B
)*a^3*d^3)*f*x + 3*(B*a^3*c*d^2 - (A + 3*B)*a^3*d^3)*cos(f*x + e)*sin(f*x + e) + 3*(B*a^3*c^3 - (A + 2*B)*a^3*
c^2*d + (2*A + B)*a^3*c*d^2 - A*a^3*d^3)*sqrt(-(c - d)/(c + d))*log(-((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin
(f*x + e) - c^2 - d^2 - 2*((c^2 + c*d)*cos(f*x + e)*sin(f*x + e) + (c*d + d^2)*cos(f*x + e))*sqrt(-(c - d)/(c
+ d)))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) - 6*(B*a^3*c^2*d - (A + 3*B)*a^3*c*d^2 + (3*A +
4*B)*a^3*d^3)*cos(f*x + e))/(d^4*f), 1/6*(2*B*a^3*d^3*cos(f*x + e)^3 - 3*(2*B*a^3*c^3 - 2*(A + 3*B)*a^3*c^2*d
+ (6*A + 7*B)*a^3*c*d^2 - (7*A + 5*B)*a^3*d^3)*f*x + 3*(B*a^3*c*d^2 - (A + 3*B)*a^3*d^3)*cos(f*x + e)*sin(f*x
+ e) - 6*(B*a^3*c^3 - (A + 2*B)*a^3*c^2*d + (2*A + B)*a^3*c*d^2 - A*a^3*d^3)*sqrt((c - d)/(c + d))*arctan(-(c*
sin(f*x + e) + d)*sqrt((c - d)/(c + d))/((c - d)*cos(f*x + e))) - 6*(B*a^3*c^2*d - (A + 3*B)*a^3*c*d^2 + (3*A
+ 4*B)*a^3*d^3)*cos(f*x + e))/(d^4*f)]

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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+d*sin(f*x+e)),x)

[Out]

Timed out

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Giac [B]  time = 1.28872, size = 833, normalized size = 3.39 \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+d*sin(f*x+e)),x, algorithm="giac")

[Out]

-1/6*(3*(2*B*a^3*c^3 - 2*A*a^3*c^2*d - 6*B*a^3*c^2*d + 6*A*a^3*c*d^2 + 7*B*a^3*c*d^2 - 7*A*a^3*d^3 - 5*B*a^3*d
^3)*(f*x + e)/d^4 - 12*(B*a^3*c^4 - A*a^3*c^3*d - 3*B*a^3*c^3*d + 3*A*a^3*c^2*d^2 + 3*B*a^3*c^2*d^2 - 3*A*a^3*
c*d^3 - B*a^3*c*d^3 + A*a^3*d^4)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d
)/sqrt(c^2 - d^2)))/(sqrt(c^2 - d^2)*d^4) + 2*(3*B*a^3*c*d*tan(1/2*f*x + 1/2*e)^5 - 3*A*a^3*d^2*tan(1/2*f*x +
1/2*e)^5 - 9*B*a^3*d^2*tan(1/2*f*x + 1/2*e)^5 + 6*B*a^3*c^2*tan(1/2*f*x + 1/2*e)^4 - 6*A*a^3*c*d*tan(1/2*f*x +
1/2*e)^4 - 18*B*a^3*c*d*tan(1/2*f*x + 1/2*e)^4 + 18*A*a^3*d^2*tan(1/2*f*x + 1/2*e)^4 + 18*B*a^3*d^2*tan(1/2*f
*x + 1/2*e)^4 + 12*B*a^3*c^2*tan(1/2*f*x + 1/2*e)^2 - 12*A*a^3*c*d*tan(1/2*f*x + 1/2*e)^2 - 36*B*a^3*c*d*tan(1
/2*f*x + 1/2*e)^2 + 36*A*a^3*d^2*tan(1/2*f*x + 1/2*e)^2 + 48*B*a^3*d^2*tan(1/2*f*x + 1/2*e)^2 - 3*B*a^3*c*d*ta
n(1/2*f*x + 1/2*e) + 3*A*a^3*d^2*tan(1/2*f*x + 1/2*e) + 9*B*a^3*d^2*tan(1/2*f*x + 1/2*e) + 6*B*a^3*c^2 - 6*A*a
^3*c*d - 18*B*a^3*c*d + 18*A*a^3*d^2 + 22*B*a^3*d^2)/((tan(1/2*f*x + 1/2*e)^2 + 1)^3*d^3))/f