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

Optimal. Leaf size=153 \[ -\frac{c^3 (A-6 B) \cos (e+f x)}{a^3 f}-\frac{a^3 c^3 (A-B) \cos ^7(e+f x)}{5 f (a \sin (e+f x)+a)^6}-\frac{2 a^3 c^3 (A-6 B) \cos ^3(e+f x)}{3 f \left (a^3 \sin (e+f x)+a^3\right )^2}-\frac{c^3 x (A-6 B)}{a^3}+\frac{2 a c^3 (A-6 B) \cos ^5(e+f x)}{15 f (a \sin (e+f x)+a)^4} \]

[Out]

-(((A - 6*B)*c^3*x)/a^3) - ((A - 6*B)*c^3*Cos[e + f*x])/(a^3*f) - (a^3*(A - B)*c^3*Cos[e + f*x]^7)/(5*f*(a + a
*Sin[e + f*x])^6) + (2*a*(A - 6*B)*c^3*Cos[e + f*x]^5)/(15*f*(a + a*Sin[e + f*x])^4) - (2*a^3*(A - 6*B)*c^3*Co
s[e + f*x]^3)/(3*f*(a^3 + a^3*Sin[e + f*x])^2)

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Rubi [A]  time = 0.331005, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2967, 2859, 2680, 2682, 8} \[ -\frac{c^3 (A-6 B) \cos (e+f x)}{a^3 f}-\frac{a^3 c^3 (A-B) \cos ^7(e+f x)}{5 f (a \sin (e+f x)+a)^6}-\frac{2 a^3 c^3 (A-6 B) \cos ^3(e+f x)}{3 f \left (a^3 \sin (e+f x)+a^3\right )^2}-\frac{c^3 x (A-6 B)}{a^3}+\frac{2 a c^3 (A-6 B) \cos ^5(e+f x)}{15 f (a \sin (e+f x)+a)^4} \]

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])^3,x]

[Out]

-(((A - 6*B)*c^3*x)/a^3) - ((A - 6*B)*c^3*Cos[e + f*x])/(a^3*f) - (a^3*(A - B)*c^3*Cos[e + f*x]^7)/(5*f*(a + a
*Sin[e + f*x])^6) + (2*a*(A - 6*B)*c^3*Cos[e + f*x]^5)/(15*f*(a + a*Sin[e + f*x])^4) - (2*a^3*(A - 6*B)*c^3*Co
s[e + f*x]^3)/(3*f*(a^3 + a^3*Sin[e + f*x])^2)

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

Mathematica [B]  time = 1.04783, size = 308, normalized size = 2.01 \[ \frac{(c-c \sin (e+f x))^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (48 (A-B) \sin \left (\frac{1}{2} (e+f x)\right )-15 (A-6 B) (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+4 (23 A-93 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+4 (11 A-21 B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-8 (11 A-21 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-24 (A-B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+15 B \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5\right )}{15 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

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])^3,x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(48*(A - B)*Sin[(e + f*x)/2] - 24*(A - B)*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]) - 8*(11*A - 21*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 4*(11*A - 21*B)*(Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2])^3 + 4*(23*A - 93*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 1
5*(A - 6*B)*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + 15*B*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2])^5)*(c - c*Sin[e + f*x])^3)/(15*a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*(1 + Sin[e + f*x])^3)

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Maple [B]  time = 0.141, size = 323, normalized size = 2.1 \begin{align*} 2\,{\frac{B{c}^{3}}{f{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{{c}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) A}{f{a}^{3}}}+12\,{\frac{{c}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) B}{f{a}^{3}}}+32\,{\frac{A{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-32\,{\frac{B{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}+8\,{\frac{A{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+8\,{\frac{B{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-4\,{\frac{A{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+12\,{\frac{B{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-{\frac{64\,A{c}^{3}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}+{\frac{64\,B{c}^{3}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}-{\frac{80\,A{c}^{3}}{3\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+16\,{\frac{B{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+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))^3,x)

[Out]

2/f*c^3/a^3*B/(1+tan(1/2*f*x+1/2*e)^2)-2/f*c^3/a^3*arctan(tan(1/2*f*x+1/2*e))*A+12/f*c^3/a^3*arctan(tan(1/2*f*
x+1/2*e))*B+32/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^4*A-32/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^4*B+8/f*c^3/a^3/(tan(1
/2*f*x+1/2*e)+1)^2*A+8/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^2*B-4/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)*A+12/f*c^3/a^3/
(tan(1/2*f*x+1/2*e)+1)*B-64/5/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^5*A+64/5/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^5*B-8
0/3/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^3*A+16/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^3*B

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Maxima [B]  time = 1.6299, size = 2267, normalized size = 14.82 \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))^3,x, algorithm="maxima")

[Out]

2/15*(3*B*c^3*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 189*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 200*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3 + 160*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 75*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 +
15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 24)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 11*a^3*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 15*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*a^3*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + 11*a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 5*a^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^3*sin(f*x +
e)^7/(cos(f*x + e) + 1)^7) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) - A*c^3*((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*si
n(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + 3*B*c^3*((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^3*(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) - 6*A*c^3*(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) + 6*B*c^3*(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) + 9*A*c^3*(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*s
in(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*B*c^3*(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 = 1.73465, size = 821, normalized size = 5.37 \begin{align*} \frac{15 \, B c^{3} \cos \left (f x + e\right )^{4} + 60 \,{\left (A - 6 \, B\right )} c^{3} f x + 24 \,{\left (A - B\right )} c^{3} -{\left (15 \,{\left (A - 6 \, B\right )} c^{3} f x +{\left (46 \, A - 231 \, B\right )} c^{3}\right )} \cos \left (f x + e\right )^{3} -{\left (45 \,{\left (A - 6 \, B\right )} c^{3} f x - 2 \,{\left (A - 66 \, B\right )} c^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (5 \,{\left (A - 6 \, B\right )} c^{3} f x + 2 \,{\left (6 \, A - 31 \, B\right )} c^{3}\right )} \cos \left (f x + e\right ) +{\left (15 \, B c^{3} \cos \left (f x + e\right )^{3} + 60 \,{\left (A - 6 \, B\right )} c^{3} f x - 24 \,{\left (A - B\right )} c^{3} -{\left (15 \,{\left (A - 6 \, B\right )} c^{3} f x - 2 \,{\left (23 \, A - 108 \, B\right )} c^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (5 \,{\left (A - 6 \, B\right )} c^{3} f x + 2 \,{\left (4 \, A - 29 \, B\right )} c^{3}\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-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(15*B*c^3*cos(f*x + e)^4 + 60*(A - 6*B)*c^3*f*x + 24*(A - B)*c^3 - (15*(A - 6*B)*c^3*f*x + (46*A - 231*B)
*c^3)*cos(f*x + e)^3 - (45*(A - 6*B)*c^3*f*x - 2*(A - 66*B)*c^3)*cos(f*x + e)^2 + 6*(5*(A - 6*B)*c^3*f*x + 2*(
6*A - 31*B)*c^3)*cos(f*x + e) + (15*B*c^3*cos(f*x + e)^3 + 60*(A - 6*B)*c^3*f*x - 24*(A - B)*c^3 - (15*(A - 6*
B)*c^3*f*x - 2*(23*A - 108*B)*c^3)*cos(f*x + e)^2 + 6*(5*(A - 6*B)*c^3*f*x + 2*(4*A - 29*B)*c^3)*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-c*sin(f*x+e))**3/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28588, size = 305, normalized size = 1.99 \begin{align*} \frac{\frac{30 \, B c^{3}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} a^{3}} - \frac{15 \,{\left (A c^{3} - 6 \, B c^{3}\right )}{\left (f x + e\right )}}{a^{3}} - \frac{4 \,{\left (15 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 45 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 30 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 210 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 100 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 420 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 50 \, A c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 270 \, B c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 13 \, A c^{3} - 63 \, B c^{3}\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-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(30*B*c^3/((tan(1/2*f*x + 1/2*e)^2 + 1)*a^3) - 15*(A*c^3 - 6*B*c^3)*(f*x + e)/a^3 - 4*(15*A*c^3*tan(1/2*f
*x + 1/2*e)^4 - 45*B*c^3*tan(1/2*f*x + 1/2*e)^4 + 30*A*c^3*tan(1/2*f*x + 1/2*e)^3 - 210*B*c^3*tan(1/2*f*x + 1/
2*e)^3 + 100*A*c^3*tan(1/2*f*x + 1/2*e)^2 - 420*B*c^3*tan(1/2*f*x + 1/2*e)^2 + 50*A*c^3*tan(1/2*f*x + 1/2*e) -
270*B*c^3*tan(1/2*f*x + 1/2*e) + 13*A*c^3 - 63*B*c^3)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f