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

Optimal. Leaf size=164 \[ -\frac{\left (2 A \left (c^2+3 c d+2 d^2\right )+B \left (3 c^2+14 c d-29 d^2\right )\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{B d^2 x}{a^3}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}-\frac{(c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

[Out]

(B*d^2*x)/a^3 - ((c - d)*(B*(3*c - 7*d) + 2*A*(c + d))*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])^2) - ((B*(3*
c^2 + 14*c*d - 29*d^2) + 2*A*(c^2 + 3*c*d + 2*d^2))*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x])) - ((A - B)*C
os[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*(a + a*Sin[e + f*x])^3)

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Rubi [A]  time = 0.461075, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2977, 2968, 3019, 2735, 2648} \[ -\frac{\left (2 A \left (c^2+3 c d+2 d^2\right )+B \left (3 c^2+14 c d-29 d^2\right )\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{B d^2 x}{a^3}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}-\frac{(c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*d^2*x)/a^3 - ((c - d)*(B*(3*c - 7*d) + 2*A*(c + d))*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])^2) - ((B*(3*
c^2 + 14*c*d - 29*d^2) + 2*A*(c^2 + 3*c*d + 2*d^2))*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x])) - ((A - B)*C
os[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*(a + a*Sin[e + f*x])^3)

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && 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 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

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 2648

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

Rubi steps

\begin{align*} \int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx &=-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac{\int \frac{(c+d \sin (e+f x)) (a (B (3 c-2 d)+2 A (c+d))+5 a B d \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac{\int \frac{a c (B (3 c-2 d)+2 A (c+d))+(5 a B c d+a d (B (3 c-2 d)+2 A (c+d))) \sin (e+f x)+5 a B d^2 \sin ^2(e+f x)}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac{(c-d) (B (3 c-7 d)+2 A (c+d)) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}-\frac{\int \frac{-a^2 \left (B \left (3 c^2+14 c d-14 d^2\right )+2 A \left (c^2+3 c d+2 d^2\right )\right )-15 a^2 B d^2 \sin (e+f x)}{a+a \sin (e+f x)} \, dx}{15 a^4}\\ &=\frac{B d^2 x}{a^3}-\frac{(c-d) (B (3 c-7 d)+2 A (c+d)) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac{\left (B \left (3 c^2+14 c d-29 d^2\right )+2 A \left (c^2+3 c d+2 d^2\right )\right ) \int \frac{1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=\frac{B d^2 x}{a^3}-\frac{(c-d) (B (3 c-7 d)+2 A (c+d)) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{\left (B \left (3 c^2+14 c d-29 d^2\right )+2 A \left (c^2+3 c d+2 d^2\right )\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}\\ \end{align*}

Mathematica [B]  time = 0.896707, size = 514, normalized size = 3.13 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (30 \cos \left (\frac{1}{2} (e+f x)\right ) \left (2 A d (c+d)+B \left (c^2+4 c d+d^2 (5 e+5 f x-9)\right )\right )-5 \cos \left (\frac{3}{2} (e+f x)\right ) \left (4 A \left (c^2+3 c d+2 d^2\right )+B \left (6 c^2+16 c d+d^2 (15 e+15 f x-46)\right )\right )+40 A c^2 \sin \left (\frac{1}{2} (e+f x)\right )-4 A c^2 \sin \left (\frac{5}{2} (e+f x)\right )+60 A c d \sin \left (\frac{1}{2} (e+f x)\right )-12 A c d \sin \left (\frac{5}{2} (e+f x)\right )+80 A d^2 \sin \left (\frac{1}{2} (e+f x)\right )+30 A d^2 \sin \left (\frac{3}{2} (e+f x)\right )-14 A d^2 \sin \left (\frac{5}{2} (e+f x)\right )+30 B c^2 \sin \left (\frac{1}{2} (e+f x)\right )-6 B c^2 \sin \left (\frac{5}{2} (e+f x)\right )+160 B c d \sin \left (\frac{1}{2} (e+f x)\right )+60 B c d \sin \left (\frac{3}{2} (e+f x)\right )-28 B c d \sin \left (\frac{5}{2} (e+f x)\right )-370 B d^2 \sin \left (\frac{1}{2} (e+f x)\right )+150 B d^2 e \sin \left (\frac{1}{2} (e+f x)\right )+150 B d^2 f x \sin \left (\frac{1}{2} (e+f x)\right )-90 B d^2 \sin \left (\frac{3}{2} (e+f x)\right )+75 B d^2 e \sin \left (\frac{3}{2} (e+f x)\right )+75 B d^2 f x \sin \left (\frac{3}{2} (e+f x)\right )+64 B d^2 \sin \left (\frac{5}{2} (e+f x)\right )-15 B d^2 e \sin \left (\frac{5}{2} (e+f x)\right )-15 B d^2 f x \sin \left (\frac{5}{2} (e+f x)\right )-15 B d^2 e \cos \left (\frac{5}{2} (e+f x)\right )-15 B d^2 f x \cos \left (\frac{5}{2} (e+f x)\right )\right )}{60 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(30*(2*A*d*(c + d) + B*(c^2 + 4*c*d + d^2*(-9 + 5*e + 5*f*x)))*Cos[(e +
f*x)/2] - 5*(4*A*(c^2 + 3*c*d + 2*d^2) + B*(6*c^2 + 16*c*d + d^2*(-46 + 15*e + 15*f*x)))*Cos[(3*(e + f*x))/2]
- 15*B*d^2*e*Cos[(5*(e + f*x))/2] - 15*B*d^2*f*x*Cos[(5*(e + f*x))/2] + 40*A*c^2*Sin[(e + f*x)/2] + 30*B*c^2*
Sin[(e + f*x)/2] + 60*A*c*d*Sin[(e + f*x)/2] + 160*B*c*d*Sin[(e + f*x)/2] + 80*A*d^2*Sin[(e + f*x)/2] - 370*B*
d^2*Sin[(e + f*x)/2] + 150*B*d^2*e*Sin[(e + f*x)/2] + 150*B*d^2*f*x*Sin[(e + f*x)/2] + 60*B*c*d*Sin[(3*(e + f*
x))/2] + 30*A*d^2*Sin[(3*(e + f*x))/2] - 90*B*d^2*Sin[(3*(e + f*x))/2] + 75*B*d^2*e*Sin[(3*(e + f*x))/2] + 75*
B*d^2*f*x*Sin[(3*(e + f*x))/2] - 4*A*c^2*Sin[(5*(e + f*x))/2] - 6*B*c^2*Sin[(5*(e + f*x))/2] - 12*A*c*d*Sin[(5
*(e + f*x))/2] - 28*B*c*d*Sin[(5*(e + f*x))/2] - 14*A*d^2*Sin[(5*(e + f*x))/2] + 64*B*d^2*Sin[(5*(e + f*x))/2]
- 15*B*d^2*e*Sin[(5*(e + f*x))/2] - 15*B*d^2*f*x*Sin[(5*(e + f*x))/2]))/(60*a^3*f*(1 + Sin[e + f*x])^3)

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Maple [B]  time = 0.098, size = 617, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/f/a^3/(tan(1/2*f*x+1/2*e)+1)^3*A*c*d-16/3/f/a^3/(tan(1/2*f*x+1/2*e)+1)^3*B*c*d-4/f/a^3/(tan(1/2*f*x+1/2*e)+1
)^2*A*c*d-8/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*A*c*d+8/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*B*c*d-2/f/a^3/(tan(1/2*f*x+1
/2*e)+1)*A*c^2+2/f/a^3/(tan(1/2*f*x+1/2*e)+1)*B*d^2+4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^2*A*c^2-2/f/a^3/(tan(1/2*f*
x+1/2*e)+1)^2*B*c^2+2/f/a^3/(tan(1/2*f*x+1/2*e)+1)^2*B*d^2+4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*A*c^2+4/f/a^3/(tan
(1/2*f*x+1/2*e)+1)^4*A*d^2-4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*B*c^2+16/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*A*c*d-16
/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*B*c*d-8/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*A*c^2-8/5/f/a^3/(tan(1/2*f*x+1/2*e)
+1)^5*A*d^2+8/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*B*c^2+8/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*B*d^2-16/3/f/a^3/(tan(
1/2*f*x+1/2*e)+1)^3*A*c^2-8/3/f/a^3/(tan(1/2*f*x+1/2*e)+1)^3*A*d^2+4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^3*B*c^2+4/3/
f/a^3/(tan(1/2*f*x+1/2*e)+1)^3*B*d^2+2/f/a^3*B*d^2*arctan(tan(1/2*f*x+1/2*e))-4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4
*B*d^2

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Maxima [B]  time = 1.59682, size = 1528, normalized size = 9.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(B*d^2*((95*sin(f*x + e)/(cos(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3
/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) +
1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e)
+ 1))/a^3) - A*c^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x +
e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(
f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 4*B*c*d*(5*sin(f*x + e)/(cos(f*x
+ e) + 1) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f
*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 2*A*d^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 10*si
n(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(c
os(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 +
a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*B*c^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(
f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10
*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(
cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 6*A*c*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) +
5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(
cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 +
5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [B]  time = 2.18067, size = 1002, normalized size = 6.11 \begin{align*} -\frac{60 \, B d^{2} f x -{\left (15 \, B d^{2} f x -{\left (2 \, A + 3 \, B\right )} c^{2} - 2 \,{\left (3 \, A + 7 \, B\right )} c d -{\left (7 \, A - 32 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left (A - B\right )} c^{2} + 6 \,{\left (A - B\right )} c d - 3 \,{\left (A - B\right )} d^{2} -{\left (45 \, B d^{2} f x + 2 \,{\left (2 \, A + 3 \, B\right )} c^{2} + 2 \,{\left (6 \, A - B\right )} c d -{\left (A + 19 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (10 \, B d^{2} f x -{\left (3 \, A + 2 \, B\right )} c^{2} - 2 \,{\left (2 \, A + 3 \, B\right )} c d - 3 \,{\left (A - 6 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) +{\left (60 \, B d^{2} f x + 3 \,{\left (A - B\right )} c^{2} - 6 \,{\left (A - B\right )} c d + 3 \,{\left (A - B\right )} d^{2} -{\left (15 \, B d^{2} f x +{\left (2 \, A + 3 \, B\right )} c^{2} + 2 \,{\left (3 \, A + 7 \, B\right )} c d +{\left (7 \, A - 32 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (10 \, B d^{2} f x -{\left (2 \, A + 3 \, B\right )} c^{2} - 2 \,{\left (3 \, A + 2 \, B\right )} c d -{\left (2 \, A - 17 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(60*B*d^2*f*x - (15*B*d^2*f*x - (2*A + 3*B)*c^2 - 2*(3*A + 7*B)*c*d - (7*A - 32*B)*d^2)*cos(f*x + e)^3 -
3*(A - B)*c^2 + 6*(A - B)*c*d - 3*(A - B)*d^2 - (45*B*d^2*f*x + 2*(2*A + 3*B)*c^2 + 2*(6*A - B)*c*d - (A + 19
*B)*d^2)*cos(f*x + e)^2 + 3*(10*B*d^2*f*x - (3*A + 2*B)*c^2 - 2*(2*A + 3*B)*c*d - 3*(A - 6*B)*d^2)*cos(f*x + e
) + (60*B*d^2*f*x + 3*(A - B)*c^2 - 6*(A - B)*c*d + 3*(A - B)*d^2 - (15*B*d^2*f*x + (2*A + 3*B)*c^2 + 2*(3*A +
7*B)*c*d + (7*A - 32*B)*d^2)*cos(f*x + e)^2 + 3*(10*B*d^2*f*x - (2*A + 3*B)*c^2 - 2*(3*A + 2*B)*c*d - (2*A -
17*B)*d^2)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) -
4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*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+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.21794, size = 516, normalized size = 3.15 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )} B d^{2}}{a^{3}} - \frac{2 \,{\left (15 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 15 \, B d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 30 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 30 \, A c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 75 \, B d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 40 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 15 \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 30 \, A c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 40 \, B c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 20 \, A d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 145 \, B d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 20 \, A c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 15 \, B c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 30 \, A c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 20 \, B c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 10 \, A d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 95 \, B d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, A c^{2} + 3 \, B c^{2} + 6 \, A c d + 4 \, B c d + 2 \, A d^{2} - 22 \, B d^{2}\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(f*x + e)*B*d^2/a^3 - 2*(15*A*c^2*tan(1/2*f*x + 1/2*e)^4 - 15*B*d^2*tan(1/2*f*x + 1/2*e)^4 + 30*A*c^2
*tan(1/2*f*x + 1/2*e)^3 + 15*B*c^2*tan(1/2*f*x + 1/2*e)^3 + 30*A*c*d*tan(1/2*f*x + 1/2*e)^3 - 75*B*d^2*tan(1/2
*f*x + 1/2*e)^3 + 40*A*c^2*tan(1/2*f*x + 1/2*e)^2 + 15*B*c^2*tan(1/2*f*x + 1/2*e)^2 + 30*A*c*d*tan(1/2*f*x + 1
/2*e)^2 + 40*B*c*d*tan(1/2*f*x + 1/2*e)^2 + 20*A*d^2*tan(1/2*f*x + 1/2*e)^2 - 145*B*d^2*tan(1/2*f*x + 1/2*e)^2
+ 20*A*c^2*tan(1/2*f*x + 1/2*e) + 15*B*c^2*tan(1/2*f*x + 1/2*e) + 30*A*c*d*tan(1/2*f*x + 1/2*e) + 20*B*c*d*ta
n(1/2*f*x + 1/2*e) + 10*A*d^2*tan(1/2*f*x + 1/2*e) - 95*B*d^2*tan(1/2*f*x + 1/2*e) + 7*A*c^2 + 3*B*c^2 + 6*A*c
*d + 4*B*c*d + 2*A*d^2 - 22*B*d^2)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f