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

Optimal. Leaf size=228 \[ \frac{2 d \left (A \left (c^2+6 c d-5 d^2\right )+B \left (2 c^2-15 c d+8 d^2\right )\right ) \cos (e+f x)}{3 a^2 f}+\frac{d x \left (2 A d (3 c-2 d)+B \left (6 c^2-12 c d+7 d^2\right )\right )}{2 a^2}+\frac{d^2 (2 A (c+6 d)+B (4 c-21 d)) \sin (e+f x) \cos (e+f x)}{6 a^2 f}-\frac{(A (c+5 d)+2 B (c-4 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 a^2 f (\sin (e+f x)+1)}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{3 f (a \sin (e+f x)+a)^2} \]

[Out]

(d*(2*A*(3*c - 2*d)*d + B*(6*c^2 - 12*c*d + 7*d^2))*x)/(2*a^2) + (2*d*(A*(c^2 + 6*c*d - 5*d^2) + B*(2*c^2 - 15
*c*d + 8*d^2))*Cos[e + f*x])/(3*a^2*f) + (d^2*(B*(4*c - 21*d) + 2*A*(c + 6*d))*Cos[e + f*x]*Sin[e + f*x])/(6*a
^2*f) - ((2*B*(c - 4*d) + A*(c + 5*d))*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(3*a^2*f*(1 + Sin[e + f*x])) - ((A
- B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(3*f*(a + a*Sin[e + f*x])^2)

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Rubi [A]  time = 0.523275, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {2977, 2734} \[ \frac{2 d \left (A \left (c^2+6 c d-5 d^2\right )+B \left (2 c^2-15 c d+8 d^2\right )\right ) \cos (e+f x)}{3 a^2 f}+\frac{d x \left (2 A d (3 c-2 d)+B \left (6 c^2-12 c d+7 d^2\right )\right )}{2 a^2}+\frac{d^2 (2 A (c+6 d)+B (4 c-21 d)) \sin (e+f x) \cos (e+f x)}{6 a^2 f}-\frac{(A (c+5 d)+2 B (c-4 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 a^2 f (\sin (e+f x)+1)}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{3 f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(2*A*(3*c - 2*d)*d + B*(6*c^2 - 12*c*d + 7*d^2))*x)/(2*a^2) + (2*d*(A*(c^2 + 6*c*d - 5*d^2) + B*(2*c^2 - 15
*c*d + 8*d^2))*Cos[e + f*x])/(3*a^2*f) + (d^2*(B*(4*c - 21*d) + 2*A*(c + 6*d))*Cos[e + f*x]*Sin[e + f*x])/(6*a
^2*f) - ((2*B*(c - 4*d) + A*(c + 5*d))*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(3*a^2*f*(1 + Sin[e + f*x])) - ((A
- B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(3*f*(a + a*Sin[e + f*x])^2)

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 2734

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

Rubi steps

\begin{align*} \int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx &=-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{3 f (a+a \sin (e+f x))^2}+\frac{\int \frac{(c+d \sin (e+f x))^2 (a (A c+2 B c+3 A d-3 B d)-a (2 A-5 B) d \sin (e+f x))}{a+a \sin (e+f x)} \, dx}{3 a^2}\\ &=-\frac{(2 B (c-4 d)+A (c+5 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 a^2 f (1+\sin (e+f x))}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{3 f (a+a \sin (e+f x))^2}+\frac{\int (c+d \sin (e+f x)) \left (a^2 d (9 B c+10 A d-16 B d)-a^2 d (B (4 c-21 d)+2 A (c+6 d)) \sin (e+f x)\right ) \, dx}{3 a^4}\\ &=\frac{d \left (2 A (3 c-2 d) d+B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 a^2}+\frac{2 d \left (A \left (c^2+6 c d-5 d^2\right )+B \left (2 c^2-15 c d+8 d^2\right )\right ) \cos (e+f x)}{3 a^2 f}+\frac{d^2 (B (4 c-21 d)+2 A (c+6 d)) \cos (e+f x) \sin (e+f x)}{6 a^2 f}-\frac{(2 B (c-4 d)+A (c+5 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 a^2 f (1+\sin (e+f x))}-\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{3 f (a+a \sin (e+f x))^2}\\ \end{align*}

Mathematica [B]  time = 3.51596, size = 547, normalized size = 2.4 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 \cos \left (\frac{1}{2} (e+f x)\right ) \left (8 A d \left (6 c^2+3 c d (3 e+3 f x-4)+d^2 (-6 e-6 f x+5)\right )+B \left (24 c^2 d (3 e+3 f x-4)+16 c^3-24 c d^2 (6 e+6 f x-5)+7 d^3 (12 e+12 f x-7)\right )\right )-\cos \left (\frac{3}{2} (e+f x)\right ) \left (4 A \left (24 c^2 d+4 c^3+6 c d^2 (3 e+3 f x-10)+d^3 (-12 e-12 f x+41)\right )+B \left (24 c^2 d (3 e+3 f x-10)+32 c^3-12 c d^2 (12 e+12 f x-41)+d^3 (84 e+84 f x-239)\right )\right )+3 \left (2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (d \cos (e+f x) \left (8 A d (3 c (e+f x)-2 d (e+f x+1))+B \left (24 c^2 (e+f x)-48 c d (e+f x+1)+d^2 (28 e+28 f x+27)\right )\right )+2 d^2 (-2 A d-6 B c+3 B d) \cos (2 (e+f x))+24 A c^2 d+8 A c^3+48 A c d^2 e+48 A c d^2 f x-72 A c d^2-32 A d^3 e-32 A d^3 f x+36 A d^3+48 B c^2 d e+48 B c^2 d f x-72 B c^2 d+8 B c^3-96 B c d^2 e-96 B c d^2 f x+108 B c d^2+B d^3 \cos (3 (e+f x))+56 B d^3 e+56 B d^3 f x-50 B d^3\right )+d^2 (4 A d+12 B c-5 B d) \cos \left (\frac{5}{2} (e+f x)\right )+B d^3 \cos \left (\frac{7}{2} (e+f x)\right )\right )\right )}{48 a^2 f (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(3*(8*A*d*(6*c^2 + d^2*(5 - 6*e - 6*f*x) + 3*c*d*(-4 + 3*e + 3*f*x)) +
B*(16*c^3 + 24*c^2*d*(-4 + 3*e + 3*f*x) - 24*c*d^2*(-5 + 6*e + 6*f*x) + 7*d^3*(-7 + 12*e + 12*f*x)))*Cos[(e +
f*x)/2] - (4*A*(4*c^3 + 24*c^2*d + d^3*(41 - 12*e - 12*f*x) + 6*c*d^2*(-10 + 3*e + 3*f*x)) + B*(32*c^3 + 24*c^
2*d*(-10 + 3*e + 3*f*x) - 12*c*d^2*(-41 + 12*e + 12*f*x) + d^3*(-239 + 84*e + 84*f*x)))*Cos[(3*(e + f*x))/2] +
3*(d^2*(12*B*c + 4*A*d - 5*B*d)*Cos[(5*(e + f*x))/2] + B*d^3*Cos[(7*(e + f*x))/2] + 2*(8*A*c^3 + 8*B*c^3 + 24
*A*c^2*d - 72*B*c^2*d - 72*A*c*d^2 + 108*B*c*d^2 + 36*A*d^3 - 50*B*d^3 + 48*B*c^2*d*e + 48*A*c*d^2*e - 96*B*c*
d^2*e - 32*A*d^3*e + 56*B*d^3*e + 48*B*c^2*d*f*x + 48*A*c*d^2*f*x - 96*B*c*d^2*f*x - 32*A*d^3*f*x + 56*B*d^3*f
*x + d*(8*A*d*(3*c*(e + f*x) - 2*d*(1 + e + f*x)) + B*(24*c^2*(e + f*x) - 48*c*d*(1 + e + f*x) + d^2*(27 + 28*
e + 28*f*x)))*Cos[e + f*x] + 2*d^2*(-6*B*c - 2*A*d + 3*B*d)*Cos[2*(e + f*x)] + B*d^3*Cos[3*(e + f*x)])*Sin[(e
+ f*x)/2])))/(48*a^2*f*(1 + Sin[e + f*x])^2)

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Maple [B]  time = 0.105, size = 946, normalized size = 4.2 \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))^3/(a+a*sin(f*x+e))^2,x)

[Out]

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

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Maxima [B]  time = 1.5814, size = 1866, normalized size = 8.18 \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))^3/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

1/3*(B*d^3*((75*sin(f*x + e)/(cos(f*x + e) + 1) + 97*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 126*sin(f*x + e)^3/
(cos(f*x + e) + 1)^3 + 98*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 63*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 21*si
n(f*x + e)^6/(cos(f*x + e) + 1)^6 + 32)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a^2*sin(f*x + e)^2/(c
os(f*x + e) + 1)^2 + 7*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 7*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5
*a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3*a^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^2*sin(f*x + e)^7/(cos
(f*x + e) + 1)^7) + 21*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - 12*B*c*d^2*((12*sin(f*x + e)/(cos(f*x +
e) + 1) + 11*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(c
os(f*x + e) + 1)^4 + 5)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 + 4*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^
5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - 4*A*d^3*((12*sin(f*x + e)/(cos(f*x
+ e) + 1) + 11*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/
(cos(f*x + e) + 1)^4 + 5)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*a^2*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 4*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e
)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 6*B*c^2*d*((9*sin(f*x + e)/(cos(f
*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2
*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f
*x + e) + 1))/a^2) + 6*A*c*d^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4
)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^
3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - 2*A*c^3*(3*sin(f*x + e)/(cos(f*x +
e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(
f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) - 2*B*c^3*(3*sin(f*x + e)/(cos(f*x
+ e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*
sin(f*x + e)^3/(cos(f*x + e) + 1)^3) - 6*A*c^2*d*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x
+ e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)
)/f

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Fricas [B]  time = 2.20721, size = 1334, normalized size = 5.85 \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))^3/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/6*(3*B*d^3*cos(f*x + e)^4 - 2*(A - B)*c^3 + 6*(A - B)*c^2*d - 6*(A - B)*c*d^2 + 2*(A - B)*d^3 + 6*(3*B*c*d^
2 + (A - B)*d^3)*cos(f*x + e)^3 + 6*(6*B*c^2*d + 6*(A - 2*B)*c*d^2 - (4*A - 7*B)*d^3)*f*x - (2*(A + 2*B)*c^3 +
6*(2*A - 5*B)*c^2*d - 6*(5*A - 11*B)*c*d^2 + (22*A - 31*B)*d^3 + 3*(6*B*c^2*d + 6*(A - 2*B)*c*d^2 - (4*A - 7*
B)*d^3)*f*x)*cos(f*x + e)^2 - (2*(2*A + B)*c^3 + 6*(A - 4*B)*c^2*d - 6*(4*A - 13*B)*c*d^2 + 2*(13*A - 19*B)*d^
3 - 3*(6*B*c^2*d + 6*(A - 2*B)*c*d^2 - (4*A - 7*B)*d^3)*f*x)*cos(f*x + e) + (3*B*d^3*cos(f*x + e)^3 + 2*(A - B
)*c^3 - 6*(A - B)*c^2*d + 6*(A - B)*c*d^2 - 2*(A - B)*d^3 + 6*(6*B*c^2*d + 6*(A - 2*B)*c*d^2 - (4*A - 7*B)*d^3
)*f*x - 3*(6*B*c*d^2 + (2*A - 3*B)*d^3)*cos(f*x + e)^2 - (2*(A + 2*B)*c^3 + 6*(2*A - 5*B)*c^2*d - 6*(5*A - 14*
B)*c*d^2 + 4*(7*A - 10*B)*d^3 - 3*(6*B*c^2*d + 6*(A - 2*B)*c*d^2 - (4*A - 7*B)*d^3)*f*x)*cos(f*x + e))*sin(f*x
+ e))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*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))**3/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.322, size = 667, normalized size = 2.93 \begin{align*} \frac{\frac{3 \,{\left (6 \, B c^{2} d + 6 \, A c d^{2} - 12 \, B c d^{2} - 4 \, A d^{3} + 7 \, B d^{3}\right )}{\left (f x + e\right )}}{a^{2}} + \frac{6 \,{\left (B d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 \, B c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, A d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 4 \, B d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 \, B c d^{2} - 2 \, A d^{3} + 4 \, B d^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} a^{2}} - \frac{4 \,{\left (3 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 9 \, B c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 9 \, A c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 18 \, B c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 6 \, A d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 9 \, B d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 9 \, A c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 27 \, B c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 27 \, A c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 45 \, B c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 15 \, A d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 21 \, B d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, A c^{3} + B c^{3} + 3 \, A c^{2} d - 12 \, B c^{2} d - 12 \, A c d^{2} + 21 \, B c d^{2} + 7 \, A d^{3} - 10 \, B d^{3}\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}}{6 \, 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))^3/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

1/6*(3*(6*B*c^2*d + 6*A*c*d^2 - 12*B*c*d^2 - 4*A*d^3 + 7*B*d^3)*(f*x + e)/a^2 + 6*(B*d^3*tan(1/2*f*x + 1/2*e)^
3 - 6*B*c*d^2*tan(1/2*f*x + 1/2*e)^2 - 2*A*d^3*tan(1/2*f*x + 1/2*e)^2 + 4*B*d^3*tan(1/2*f*x + 1/2*e)^2 - B*d^3
*tan(1/2*f*x + 1/2*e) - 6*B*c*d^2 - 2*A*d^3 + 4*B*d^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*a^2) - 4*(3*A*c^3*tan(1
/2*f*x + 1/2*e)^2 - 9*B*c^2*d*tan(1/2*f*x + 1/2*e)^2 - 9*A*c*d^2*tan(1/2*f*x + 1/2*e)^2 + 18*B*c*d^2*tan(1/2*f
*x + 1/2*e)^2 + 6*A*d^3*tan(1/2*f*x + 1/2*e)^2 - 9*B*d^3*tan(1/2*f*x + 1/2*e)^2 + 3*A*c^3*tan(1/2*f*x + 1/2*e)
+ 3*B*c^3*tan(1/2*f*x + 1/2*e) + 9*A*c^2*d*tan(1/2*f*x + 1/2*e) - 27*B*c^2*d*tan(1/2*f*x + 1/2*e) - 27*A*c*d^
2*tan(1/2*f*x + 1/2*e) + 45*B*c*d^2*tan(1/2*f*x + 1/2*e) + 15*A*d^3*tan(1/2*f*x + 1/2*e) - 21*B*d^3*tan(1/2*f*
x + 1/2*e) + 2*A*c^3 + B*c^3 + 3*A*c^2*d - 12*B*c^2*d - 12*A*c*d^2 + 21*B*c*d^2 + 7*A*d^3 - 10*B*d^3)/(a^2*(ta
n(1/2*f*x + 1/2*e) + 1)^3))/f