3.263 \(\int \frac{(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=283 \[ -\frac{a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f (c+d)}+\frac{2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^4 f (c+d) \sqrt{c^2-d^2}}-\frac{a^3 x \left (2 A d (2 c-3 d)-B \left (6 c^2-12 c d+7 d^2\right )\right )}{2 d^4}+\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d^2 f (c+d)}+\frac{a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.941206, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {2975, 2976, 2968, 3023, 2735, 2660, 618, 204} \[ -\frac{a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f (c+d)}+\frac{2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^4 f (c+d) \sqrt{c^2-d^2}}-\frac{a^3 x \left (2 A d (2 c-3 d)-B \left (6 c^2-12 c d+7 d^2\right )\right )}{2 d^4}+\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d^2 f (c+d)}+\frac{a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))} \]

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])^2,x]

[Out]

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

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 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))^2} \, dx &=\frac{a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac{\int \frac{(a+a \sin (e+f x))^2 (-a (B (2 c-d)-3 A d)-a (2 A d-B (3 c+d)) \sin (e+f x))}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac{a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac{\int \frac{(a+a \sin (e+f x)) \left (-a^2 \left (2 A (c-3 d) d-B \left (3 c^2-3 c d+2 d^2\right )\right )+a^2 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{2 d^2 (c+d)}\\ &=\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac{a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac{\int \frac{-a^3 \left (2 A (c-3 d) d-B \left (3 c^2-3 c d+2 d^2\right )\right )+\left (a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right )-a^3 \left (2 A (c-3 d) d-B \left (3 c^2-3 c d+2 d^2\right )\right )\right ) \sin (e+f x)+a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \sin ^2(e+f x)}{c+d \sin (e+f x)} \, dx}{2 d^2 (c+d)}\\ &=-\frac{a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac{a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac{\int \frac{-a^3 d \left (2 A (c-3 d) d-B \left (3 c^2-3 c d+2 d^2\right )\right )-a^3 (c+d) \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) \sin (e+f x)}{c+d \sin (e+f x)} \, dx}{2 d^3 (c+d)}\\ &=-\frac{a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}-\frac{a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac{a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac{\left (a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right )\right ) \int \frac{1}{c+d \sin (e+f x)} \, dx}{d^4 (c+d)}\\ &=-\frac{a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}-\frac{a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac{a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac{\left (2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right )\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 (c+d) f}\\ &=-\frac{a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}-\frac{a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac{a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}-\frac{\left (4 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right )\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 (c+d) f}\\ &=-\frac{a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}+\frac{2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \tan ^{-1}\left (\frac{d+c \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{d^4 (c+d) \sqrt{c^2-d^2} f}-\frac{a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac{(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac{a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 1.48677, size = 244, normalized size = 0.86 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (2 (e+f x) \left (2 A d (3 d-2 c)+B \left (6 c^2-12 c d+7 d^2\right )\right )-\frac{8 (c-d)^2 \left (B \left (3 c^2+3 c d-d^2\right )-A d (2 c+3 d)\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d) \sqrt{c^2-d^2}}-4 d (A d-2 B c+3 B d) \cos (e+f x)+\frac{4 d (c-d)^2 (B c-A d) \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}-B d^2 \sin (2 (e+f x))\right )}{4 d^4 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

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])^2,x]

[Out]

(a^3*(1 + Sin[e + f*x])^3*(2*(2*A*d*(-2*c + 3*d) + B*(6*c^2 - 12*c*d + 7*d^2))*(e + f*x) - (8*(c - d)^2*(-(A*d
*(2*c + 3*d)) + B*(3*c^2 + 3*c*d - d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)*Sqrt[c^2 -
d^2]) - 4*d*(-2*B*c + A*d + 3*B*d)*Cos[e + f*x] + (4*(c - d)^2*d*(B*c - A*d)*Cos[e + f*x])/((c + d)*(c + d*Si
n[e + f*x])) - B*d^2*Sin[2*(e + f*x)]))/(4*d^4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6)

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Maple [B]  time = 0.189, size = 1534, normalized size = 5.4 \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))^2,x)

[Out]

4/f*a^3/d/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*A*c+2/f*a^3/d^3/(c*tan(1/2*f*x+1/2*e)^2+2*ta
n(1/2*f*x+1/2*e)*d+c)/(c+d)*B*c^3-4/f*a^3/d^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*B*c^2+2/
f*a^3/d/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*B*c-2/f*a^3/d^2/(c+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))*A*c^2-8/f*a^3/d/(c+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))*A*c-6/f*a^3/d^4/(c+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))*c^4*B+2/f*a^3/d^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*c^2*tan(1/2*f*x+1
/2*e)*B-4/f*a^3/d/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*c*tan(1/2*f*x+1/2*e)*B+4/f*a^3/d^3/(
c+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))*A*c^3-10/f*a^3/d/(c+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/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/
2*f*x+1/2*e)*d+c)/(c+d)*c*tan(1/2*f*x+1/2*e)*A-2/f*a^3*d/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+
d)/c*tan(1/2*f*x+1/2*e)*A+6/f*a^3/d^3/(c+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^3+6/f*a^3/d^4*arctan(tan(1/2*f*x+1/2*e))*B*c^2+8/f*a^3/d^2/(c+d)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*ta
n(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))*B*c^2-12/f*a^3/d^3*arctan(tan(1/2*f*x+1/2*e))*B*c+6/f*a^3/(c+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))*A+2/f*a^3/(c+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+4/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)
*tan(1/2*f*x+1/2*e)*A+2/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*tan(1/2*f*x+1/2*e)*B-6/f
*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)^2-1/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*
x+1/2*e)+4/f*a^3/d^3/(1+tan(1/2*f*x+1/2*e)^2)^2*B*c-4/f*a^3/d^3*arctan(tan(1/2*f*x+1/2*e))*A*c+1/f*a^3/d^2/(1+
tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)^3-2/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^2*A*tan(1/2*f*x+1/2*e)^2+4
/f*a^3/d^3/(1+tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)^2*c-2/f*a^3/d^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f
*x+1/2*e)*d+c)/(c+d)*A*c^2-2/f*a^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*A-2/f*a^3/d^2/(1+ta
n(1/2*f*x+1/2*e)^2)^2*A-6/f*a^3/d^2/(1+tan(1/2*f*x+1/2*e)^2)^2*B+6/f*a^3/d^2*arctan(tan(1/2*f*x+1/2*e))*A+7/f*
a^3/d^2*arctan(tan(1/2*f*x+1/2*e))*B

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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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.86849, size = 2242, normalized size = 7.92 \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))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.31483, size = 794, normalized size = 2.81 \begin{align*} -\frac{\frac{4 \,{\left (3 \, B a^{3} c^{4} - 2 \, A a^{3} c^{3} d - 3 \, B a^{3} c^{3} d + A a^{3} c^{2} d^{2} - 4 \, B a^{3} c^{2} d^{2} + 4 \, A a^{3} c d^{3} + 5 \, B a^{3} c d^{3} - 3 \, A a^{3} d^{4} - B a^{3} d^{4}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d}{\sqrt{c^{2} - d^{2}}}\right )\right )}}{{\left (c d^{4} + d^{5}\right )} \sqrt{c^{2} - d^{2}}} - \frac{4 \,{\left (B a^{3} c^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - A a^{3} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, B a^{3} c^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, A a^{3} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + B a^{3} c d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - A a^{3} d^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + B a^{3} c^{4} - A a^{3} c^{3} d - 2 \, B a^{3} c^{3} d + 2 \, A a^{3} c^{2} d^{2} + B a^{3} c^{2} d^{2} - A a^{3} c d^{3}\right )}}{{\left (c^{2} d^{3} + c d^{4}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c\right )}} - \frac{{\left (6 \, B a^{3} c^{2} - 4 \, A a^{3} c d - 12 \, B a^{3} c d + 6 \, A a^{3} d^{2} + 7 \, B a^{3} d^{2}\right )}{\left (f x + e\right )}}{d^{4}} - \frac{2 \,{\left (B a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 4 \, B a^{3} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, A a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 6 \, B a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B a^{3} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, B a^{3} c - 2 \, A a^{3} d - 6 \, B a^{3} d\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} d^{3}}}{2 \, f} \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))^2,x, algorithm="giac")

[Out]

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