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

Optimal. Leaf size=162 \[ \frac{5 c^3 (2 A-5 B) \cos (e+f x)}{2 a^2 f}-\frac{a^3 c^3 (A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}+\frac{5 c^3 (2 A-5 B) \cos ^3(e+f x)}{6 f \left (a^2 \sin (e+f x)+a^2\right )}+\frac{5 c^3 x (2 A-5 B)}{2 a^2}+\frac{2 a c^3 (2 A-5 B) \cos ^5(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.332406, antiderivative size = 162, 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} \[ \frac{5 c^3 (2 A-5 B) \cos (e+f x)}{2 a^2 f}-\frac{a^3 c^3 (A-B) \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}+\frac{5 c^3 (2 A-5 B) \cos ^3(e+f x)}{6 f \left (a^2 \sin (e+f x)+a^2\right )}+\frac{5 c^3 x (2 A-5 B)}{2 a^2}+\frac{2 a c^3 (2 A-5 B) \cos ^5(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.845507, size = 274, normalized size = 1.69 \[ \frac{(c-c \sin (e+f x))^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (64 (A-B) \sin \left (\frac{1}{2} (e+f x)\right )+30 (2 A-5 B) (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+12 (A-5 B) \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-32 (7 A-13 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-32 (A-B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+3 B \sin (2 (e+f x)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{12 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^3*(64*(A - B)*Sin[(e + f*x)/2] - 32*(A - B)*(Cos[(
e + f*x)/2] + Sin[(e + f*x)/2]) - 32*(7*A - 13*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 3
0*(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 + 3*B*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sin[2*(e + f*x)]))/(12*a^2*f*(Cos[(e + f*x
)/2] - Sin[(e + f*x)/2])^6*(1 + Sin[e + f*x])^2)

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Maple [B]  time = 0.129, size = 399, normalized size = 2.5 \begin{align*} -{\frac{B{c}^{3}}{{a}^{2}f} \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{{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}A}{{a}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-10\,{\frac{{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}B}{{a}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{B{c}^{3}}{{a}^{2}f}\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{c}^{3}}{{a}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-10\,{\frac{B{c}^{3}}{{a}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-25\,{\frac{{c}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) B}{{a}^{2}f}}+10\,{\frac{{c}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) A}{{a}^{2}f}}+16\,{\frac{A{c}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-16\,{\frac{B{c}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+8\,{\frac{A{c}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-24\,{\frac{B{c}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-{\frac{32\,A{c}^{3}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+{\frac{32\,B{c}^{3}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.58791, size = 1860, 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+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-1/3*(B*c^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*s
in(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/(
cos(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/(co
s(f*x + e) + 1)^7) + 21*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - 4*A*c^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) + 12*B*c^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*A*c^3*((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*si
n(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*B*c^3*((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/(co
s(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) - 6*A*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) + 2*B*c^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x + e)/(c
os(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 [A]  time = 1.81425, size = 687, normalized size = 4.24 \begin{align*} \frac{3 \, B c^{3} \cos \left (f x + e\right )^{4} + 6 \,{\left (A - 4 \, B\right )} c^{3} \cos \left (f x + e\right )^{3} - 30 \,{\left (2 \, A - 5 \, B\right )} c^{3} f x + 16 \,{\left (A - B\right )} c^{3} +{\left (15 \,{\left (2 \, A - 5 \, B\right )} c^{3} f x -{\left (62 \, A - 131 \, B\right )} c^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (15 \,{\left (2 \, A - 5 \, B\right )} c^{3} f x + 2 \,{\left (26 \, A - 71 \, B\right )} c^{3}\right )} \cos \left (f x + e\right ) +{\left (3 \, B c^{3} \cos \left (f x + e\right )^{3} - 30 \,{\left (2 \, A - 5 \, B\right )} c^{3} f x - 3 \,{\left (2 \, A - 9 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 16 \,{\left (A - B\right )} c^{3} -{\left (15 \,{\left (2 \, A - 5 \, B\right )} c^{3} f x + 2 \,{\left (34 \, A - 79 \, B\right )} c^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{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+B*sin(f*x+e))*(c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.19919, size = 315, normalized size = 1.94 \begin{align*} \frac{\frac{15 \,{\left (2 \, A c^{3} - 5 \, B c^{3}\right )}{\left (f x + e\right )}}{a^{2}} - \frac{6 \,{\left (B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 10 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, A c^{3} + 10 \, B c^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} a^{2}} + \frac{16 \,{\left (3 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 9 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 12 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 24 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 5 \, A c^{3} - 11 \, B c^{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-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

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