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

Optimal. Leaf size=163 \[ -\frac{5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}+\frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac{5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}+\frac{5 a^3 x (2 A+5 B)}{2 c^2}-\frac{2 a^3 c (2 A+5 B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.34774, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2967, 2859, 2680, 2679, 2682, 8} \[ -\frac{5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}+\frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac{5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}+\frac{5 a^3 x (2 A+5 B)}{2 c^2}-\frac{2 a^3 c (2 A+5 B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

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])^2,x]

[Out]

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

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 2679

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

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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))^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))^5} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac{1}{3} \left (a^3 (2 A+5 B) c^2\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac{2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac{1}{3} \left (5 a^3 (2 A+5 B)\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac{2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac{5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}+\frac{\left (5 a^3 (2 A+5 B)\right ) \int \frac{\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{2 c}\\ &=-\frac{5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}+\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac{2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac{5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}+\frac{\left (5 a^3 (2 A+5 B)\right ) \int 1 \, dx}{2 c^2}\\ &=\frac{5 a^3 (2 A+5 B) x}{2 c^2}-\frac{5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}+\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac{2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac{5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.848578, size = 280, normalized size = 1.72 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (64 (A+B) \sin \left (\frac{1}{2} (e+f x)\right )+30 (2 A+5 B) (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-12 (A+5 B) \cos (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-32 (7 A+13 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+32 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-3 B \sin (2 (e+f x)) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{12 f (c-c \sin (e+f x))^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

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])^2,x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(32*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2]) + 30*(2*A + 5*B)*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - 12*(A + 5*B)*Cos[e + f*x]*(Cos[(e + f
*x)/2] - Sin[(e + f*x)/2])^3 + 64*(A + B)*Sin[(e + f*x)/2] - 32*(7*A + 13*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)
/2])^2*Sin[(e + f*x)/2] - 3*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sin[2*(e + f*x)]))/(12*f*(Cos[(e + f*x)/
2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^2)

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Maple [B]  time = 0.131, size = 399, normalized size = 2.5 \begin{align*} 8\,{\frac{A{a}^{3}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}+24\,{\frac{B{a}^{3}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-{\frac{32\,A{a}^{3}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-{\frac{32\,B{a}^{3}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-16\,{\frac{A{a}^{3}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-16\,{\frac{B{a}^{3}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}+{\frac{B{a}^{3}}{f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}A}{f{c}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-10\,{\frac{{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}B}{f{c}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{B{a}^{3}}{f{c}^{2}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{A{a}^{3}}{f{c}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-10\,{\frac{B{a}^{3}}{f{c}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+25\,{\frac{{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) B}{f{c}^{2}}}+10\,{\frac{{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) A}{f{c}^{2}}} \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))^2,x)

[Out]

8/f*a^3/c^2/(tan(1/2*f*x+1/2*e)-1)*A+24/f*a^3/c^2/(tan(1/2*f*x+1/2*e)-1)*B-32/3/f*a^3/c^2/(tan(1/2*f*x+1/2*e)-
1)^3*A-32/3/f*a^3/c^2/(tan(1/2*f*x+1/2*e)-1)^3*B-16/f*a^3/c^2/(tan(1/2*f*x+1/2*e)-1)^2*A-16/f*a^3/c^2/(tan(1/2
*f*x+1/2*e)-1)^2*B+1/f*a^3/c^2/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3*B-2/f*a^3/c^2/(1+tan(1/2*f*x+1/
2*e)^2)^2*tan(1/2*f*x+1/2*e)^2*A-10/f*a^3/c^2/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^2*B-1/f*a^3/c^2/(1
+tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)-2/f*a^3/c^2/(1+tan(1/2*f*x+1/2*e)^2)^2*A-10/f*a^3/c^2/(1+tan(1/2
*f*x+1/2*e)^2)^2*B+25/f*a^3/c^2*arctan(tan(1/2*f*x+1/2*e))*B+10/f*a^3/c^2*arctan(tan(1/2*f*x+1/2*e))*A

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Maxima [B]  time = 1.61027, size = 1871, normalized size = 11.48 \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))^2,x, algorithm="maxima")

[Out]

1/3*(B*a^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)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 5*c^2*sin(f*x + e)^2/(c
os(f*x + e) + 1)^2 - 7*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 7*c^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5
*c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3*c^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^2*sin(f*x + e)^7/(cos
(f*x + e) + 1)^7) + 21*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) + 4*A*a^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)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
- 4*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^2*sin(f*x + e)^5/(
cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) + 12*B*a^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/(c
os(f*x + e) + 1)^4 - 5)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 - 4*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^2*sin(f*x + e)^
5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) + 6*A*a^3*((9*sin(f*x + e)/(cos(f*x +
e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin
(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x +
e) + 1))/c^2) + 6*B*a^3*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2
- 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos
(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) - 2*A*a^3*(3*sin(f*x + e)/(cos(f*x + e) + 1
) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 6*A*a^3*(3*sin(f*x + e)/(cos(f*x + e) +
1) - 1)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*
x + e)^3/(cos(f*x + e) + 1)^3) + 2*B*a^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^2*sin(f*x + e)/(co
s(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3))/f

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Fricas [A]  time = 1.42961, size = 687, normalized size = 4.21 \begin{align*} \frac{3 \, B a^{3} \cos \left (f x + e\right )^{4} - 6 \,{\left (A + 4 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} - 30 \,{\left (2 \, A + 5 \, B\right )} a^{3} f x - 16 \,{\left (A + B\right )} a^{3} +{\left (15 \,{\left (2 \, A + 5 \, B\right )} a^{3} f x +{\left (62 \, A + 131 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (15 \,{\left (2 \, A + 5 \, B\right )} a^{3} f x - 2 \,{\left (26 \, A + 71 \, B\right )} a^{3}\right )} \cos \left (f x + e\right ) -{\left (3 \, B a^{3} \cos \left (f x + e\right )^{3} - 30 \,{\left (2 \, A + 5 \, B\right )} a^{3} f x + 3 \,{\left (2 \, A + 9 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 16 \,{\left (A + B\right )} a^{3} -{\left (15 \,{\left (2 \, A + 5 \, B\right )} a^{3} f x - 2 \,{\left (34 \, A + 79 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f +{\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\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-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

1/6*(3*B*a^3*cos(f*x + e)^4 - 6*(A + 4*B)*a^3*cos(f*x + e)^3 - 30*(2*A + 5*B)*a^3*f*x - 16*(A + B)*a^3 + (15*(
2*A + 5*B)*a^3*f*x + (62*A + 131*B)*a^3)*cos(f*x + e)^2 - (15*(2*A + 5*B)*a^3*f*x - 2*(26*A + 71*B)*a^3)*cos(f
*x + e) - (3*B*a^3*cos(f*x + e)^3 - 30*(2*A + 5*B)*a^3*f*x + 3*(2*A + 9*B)*a^3*cos(f*x + e)^2 + 16*(A + B)*a^3
- (15*(2*A + 5*B)*a^3*f*x - 2*(34*A + 79*B)*a^3)*cos(f*x + e))*sin(f*x + e))/(c^2*f*cos(f*x + e)^2 - c^2*f*co
s(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^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+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.23941, size = 315, normalized size = 1.93 \begin{align*} \frac{\frac{15 \,{\left (2 \, A a^{3} + 5 \, B a^{3}\right )}{\left (f x + e\right )}}{c^{2}} + \frac{6 \,{\left (B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 10 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, A a^{3} - 10 \, B a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} c^{2}} + \frac{16 \,{\left (3 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 9 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, A a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 24 \, B a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 5 \, A a^{3} + 11 \, B a^{3}\right )}}{c^{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+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

1/6*(15*(2*A*a^3 + 5*B*a^3)*(f*x + e)/c^2 + 6*(B*a^3*tan(1/2*f*x + 1/2*e)^3 - 2*A*a^3*tan(1/2*f*x + 1/2*e)^2 -
10*B*a^3*tan(1/2*f*x + 1/2*e)^2 - B*a^3*tan(1/2*f*x + 1/2*e) - 2*A*a^3 - 10*B*a^3)/((tan(1/2*f*x + 1/2*e)^2 +
1)^2*c^2) + 16*(3*A*a^3*tan(1/2*f*x + 1/2*e)^2 + 9*B*a^3*tan(1/2*f*x + 1/2*e)^2 - 12*A*a^3*tan(1/2*f*x + 1/2*
e) - 24*B*a^3*tan(1/2*f*x + 1/2*e) + 5*A*a^3 + 11*B*a^3)/(c^2*(tan(1/2*f*x + 1/2*e) - 1)^3))/f