3.473 \(\int \frac{(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=142 \[ -\frac{(c-d) \left (2 c^2+11 c d+29 d^2\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{d^3 x}{a^3}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}-\frac{(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

[Out]

(d^3*x)/a^3 - ((c - d)^2*(2*c + 7*d)*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])^2) - ((c - d)*(2*c^2 + 11*c*d
+ 29*d^2)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*(
a + a*Sin[e + f*x])^3)

________________________________________________________________________________________

Rubi [A]  time = 0.332717, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2765, 2968, 3019, 2735, 2648} \[ -\frac{(c-d) \left (2 c^2+11 c d+29 d^2\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{d^3 x}{a^3}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}-\frac{(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*x)/a^3 - ((c - d)^2*(2*c + 7*d)*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])^2) - ((c - d)*(2*c^2 + 11*c*d
+ 29*d^2)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*(
a + a*Sin[e + f*x])^3)

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(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 - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && 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{(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx &=-\frac{(c-d) \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)) \left (-a \left (2 c^2+5 c d-2 d^2\right )-5 a d^2 \sin (e+f x)\right )}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac{(c-d) \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 \left (2 c^2+5 c d-2 d^2\right )+\left (-5 a c d^2-a d \left (2 c^2+5 c d-2 d^2\right )\right ) \sin (e+f x)-5 a d^3 \sin ^2(e+f x)}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac{(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{(c-d) \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 (2 c^3+9 c^2 d+18 c d^2-14 d^3\right )+15 a^2 d^3 \sin (e+f x)}{a+a \sin (e+f x)} \, dx}{15 a^4}\\ &=\frac{d^3 x}{a^3}-\frac{(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac{\left ((c-d) \left (2 c^2+11 c d+29 d^2\right )\right ) \int \frac{1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=\frac{d^3 x}{a^3}-\frac{(c-d)^2 (2 c+7 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{(c-d) \left (2 c^2+11 c d+29 d^2\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{(c-d) \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 = 5.59349, size = 408, normalized size = 2.87 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (30 d \cos \left (\frac{1}{2} (e+f x)\right ) \left (3 c^2+6 c d+d^2 (5 e+5 f x-9)\right )-5 \cos \left (\frac{3}{2} (e+f x)\right ) \left (18 c^2 d+4 c^3+24 c d^2+d^3 (15 e+15 f x-46)\right )+90 c^2 d \sin \left (\frac{1}{2} (e+f x)\right )-18 c^2 d \sin \left (\frac{5}{2} (e+f x)\right )+40 c^3 \sin \left (\frac{1}{2} (e+f x)\right )-4 c^3 \sin \left (\frac{5}{2} (e+f x)\right )+240 c d^2 \sin \left (\frac{1}{2} (e+f x)\right )+90 c d^2 \sin \left (\frac{3}{2} (e+f x)\right )-42 c d^2 \sin \left (\frac{5}{2} (e+f x)\right )-370 d^3 \sin \left (\frac{1}{2} (e+f x)\right )+150 d^3 e \sin \left (\frac{1}{2} (e+f x)\right )+150 d^3 f x \sin \left (\frac{1}{2} (e+f x)\right )-90 d^3 \sin \left (\frac{3}{2} (e+f x)\right )+75 d^3 e \sin \left (\frac{3}{2} (e+f x)\right )+75 d^3 f x \sin \left (\frac{3}{2} (e+f x)\right )+64 d^3 \sin \left (\frac{5}{2} (e+f x)\right )-15 d^3 e \sin \left (\frac{5}{2} (e+f x)\right )-15 d^3 f x \sin \left (\frac{5}{2} (e+f x)\right )-15 d^3 e \cos \left (\frac{5}{2} (e+f x)\right )-15 d^3 f x \cos \left (\frac{5}{2} (e+f x)\right )\right )}{60 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(30*d*(3*c^2 + 6*c*d + d^2*(-9 + 5*e + 5*f*x))*Cos[(e + f*x)/2] - 5*(4*
c^3 + 18*c^2*d + 24*c*d^2 + d^3*(-46 + 15*e + 15*f*x))*Cos[(3*(e + f*x))/2] - 15*d^3*e*Cos[(5*(e + f*x))/2] -
15*d^3*f*x*Cos[(5*(e + f*x))/2] + 40*c^3*Sin[(e + f*x)/2] + 90*c^2*d*Sin[(e + f*x)/2] + 240*c*d^2*Sin[(e + f*x
)/2] - 370*d^3*Sin[(e + f*x)/2] + 150*d^3*e*Sin[(e + f*x)/2] + 150*d^3*f*x*Sin[(e + f*x)/2] + 90*c*d^2*Sin[(3*
(e + f*x))/2] - 90*d^3*Sin[(3*(e + f*x))/2] + 75*d^3*e*Sin[(3*(e + f*x))/2] + 75*d^3*f*x*Sin[(3*(e + f*x))/2]
- 4*c^3*Sin[(5*(e + f*x))/2] - 18*c^2*d*Sin[(5*(e + f*x))/2] - 42*c*d^2*Sin[(5*(e + f*x))/2] + 64*d^3*Sin[(5*(
e + f*x))/2] - 15*d^3*e*Sin[(5*(e + f*x))/2] - 15*d^3*f*x*Sin[(5*(e + f*x))/2]))/(60*a^3*f*(1 + Sin[e + f*x])^
3)

________________________________________________________________________________________

Maple [B]  time = 0.075, size = 438, normalized size = 3.1 \begin{align*} 2\,{\frac{{d}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}}}-2\,{\frac{{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{{d}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+4\,{\frac{{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-6\,{\frac{{c}^{2}d}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+2\,{\frac{{d}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+4\,{\frac{{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-12\,{\frac{{c}^{2}d}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}+12\,{\frac{c{d}^{2}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-4\,{\frac{{d}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-{\frac{8\,{c}^{3}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}+{\frac{24\,{c}^{2}d}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}-{\frac{24\,c{d}^{2}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}+{\frac{8\,{d}^{3}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}-{\frac{16\,{c}^{3}}{3\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+12\,{\frac{{c}^{2}d}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}-8\,{\frac{c{d}^{2}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}+{\frac{4\,{d}^{3}}{3\,f{a}^{3}} \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((c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x)

[Out]

2/f/a^3*d^3*arctan(tan(1/2*f*x+1/2*e))-2/f/a^3/(tan(1/2*f*x+1/2*e)+1)*c^3+2/f/a^3/(tan(1/2*f*x+1/2*e)+1)*d^3+4
/f/a^3/(tan(1/2*f*x+1/2*e)+1)^2*c^3-6/f/a^3/(tan(1/2*f*x+1/2*e)+1)^2*c^2*d+2/f/a^3/(tan(1/2*f*x+1/2*e)+1)^2*d^
3+4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*c^3-12/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*c^2*d+12/f/a^3/(tan(1/2*f*x+1/2*e)+1)
^4*c*d^2-4/f/a^3/(tan(1/2*f*x+1/2*e)+1)^4*d^3-8/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*c^3+24/5/f/a^3/(tan(1/2*f*x+1
/2*e)+1)^5*c^2*d-24/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*c*d^2+8/5/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5*d^3-16/3/f/a^3/(
tan(1/2*f*x+1/2*e)+1)^3*c^3+12/f/a^3/(tan(1/2*f*x+1/2*e)+1)^3*c^2*d-8/f/a^3/(tan(1/2*f*x+1/2*e)+1)^3*c*d^2+4/3
/f/a^3/(tan(1/2*f*x+1/2*e)+1)^3*d^3

________________________________________________________________________________________

Maxima [B]  time = 2.6582, size = 1058, normalized size = 7.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(d^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) - 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*c*d^2*(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*c^2*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

________________________________________________________________________________________

Fricas [B]  time = 1.64782, size = 824, normalized size = 5.8 \begin{align*} -\frac{60 \, d^{3} f x -{\left (15 \, d^{3} f x - 2 \, c^{3} - 9 \, c^{2} d - 21 \, c d^{2} + 32 \, d^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, c^{3} + 9 \, c^{2} d - 9 \, c d^{2} + 3 \, d^{3} -{\left (45 \, d^{3} f x + 4 \, c^{3} + 18 \, c^{2} d - 3 \, c d^{2} - 19 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (10 \, d^{3} f x - 3 \, c^{3} - 6 \, c^{2} d - 9 \, c d^{2} + 18 \, d^{3}\right )} \cos \left (f x + e\right ) +{\left (60 \, d^{3} f x + 3 \, c^{3} - 9 \, c^{2} d + 9 \, c d^{2} - 3 \, d^{3} -{\left (15 \, d^{3} f x + 2 \, c^{3} + 9 \, c^{2} d + 21 \, c d^{2} - 32 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (10 \, d^{3} f x - 2 \, c^{3} - 9 \, c^{2} d - 6 \, c d^{2} + 17 \, d^{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((c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(60*d^3*f*x - (15*d^3*f*x - 2*c^3 - 9*c^2*d - 21*c*d^2 + 32*d^3)*cos(f*x + e)^3 - 3*c^3 + 9*c^2*d - 9*c*
d^2 + 3*d^3 - (45*d^3*f*x + 4*c^3 + 18*c^2*d - 3*c*d^2 - 19*d^3)*cos(f*x + e)^2 + 3*(10*d^3*f*x - 3*c^3 - 6*c^
2*d - 9*c*d^2 + 18*d^3)*cos(f*x + e) + (60*d^3*f*x + 3*c^3 - 9*c^2*d + 9*c*d^2 - 3*d^3 - (15*d^3*f*x + 2*c^3 +
9*c^2*d + 21*c*d^2 - 32*d^3)*cos(f*x + e)^2 + 3*(10*d^3*f*x - 2*c^3 - 9*c^2*d - 6*c*d^2 + 17*d^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))

________________________________________________________________________________________

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((c+d*sin(f*x+e))**3/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.32374, size = 378, normalized size = 2.66 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )} d^{3}}{a^{3}} - \frac{2 \,{\left (15 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 15 \, d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 30 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 45 \, c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 75 \, d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 40 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 45 \, c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 60 \, c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 145 \, d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 20 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 45 \, c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 30 \, c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 95 \, d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, c^{3} + 9 \, c^{2} d + 6 \, c d^{2} - 22 \, d^{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((c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(15*(f*x + e)*d^3/a^3 - 2*(15*c^3*tan(1/2*f*x + 1/2*e)^4 - 15*d^3*tan(1/2*f*x + 1/2*e)^4 + 30*c^3*tan(1/2
*f*x + 1/2*e)^3 + 45*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 75*d^3*tan(1/2*f*x + 1/2*e)^3 + 40*c^3*tan(1/2*f*x + 1/2*e
)^2 + 45*c^2*d*tan(1/2*f*x + 1/2*e)^2 + 60*c*d^2*tan(1/2*f*x + 1/2*e)^2 - 145*d^3*tan(1/2*f*x + 1/2*e)^2 + 20*
c^3*tan(1/2*f*x + 1/2*e) + 45*c^2*d*tan(1/2*f*x + 1/2*e) + 30*c*d^2*tan(1/2*f*x + 1/2*e) - 95*d^3*tan(1/2*f*x
+ 1/2*e) + 7*c^3 + 9*c^2*d + 6*c*d^2 - 22*d^3)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f