\(\int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx\) [26]

Optimal result

Integrand size = 11, antiderivative size = 50 \[ \int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx=\frac {\cos (x)}{5 (a+a \sin (x))^3}-\frac {\cos (x)}{5 a (a+a \sin (x))^2}-\frac {\cos (x)}{5 \left (a^3+a^3 \sin (x)\right )} \]

[Out]

1/5*cos(x)/(a+a*sin(x))^3-1/5*cos(x)/a/(a+a*sin(x))^2-1/5*cos(x)/(a^3+a^3*sin(x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2829, 2729, 2727} \[ \int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx=-\frac {\cos (x)}{5 \left (a^3 \sin (x)+a^3\right )}-\frac {\cos (x)}{5 a (a \sin (x)+a)^2}+\frac {\cos (x)}{5 (a \sin (x)+a)^3} \]

[In]

Int[Sin[x]/(a + a*Sin[x])^3,x]

[Out]

Cos[x]/(5*(a + a*Sin[x])^3) - Cos[x]/(5*a*(a + a*Sin[x])^2) - Cos[x]/(5*(a^3 + a^3*Sin[x]))

Rule 2727

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]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2829

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = \frac {\cos (x)}{5 (a+a \sin (x))^3}+\frac {3 \int \frac {1}{(a+a \sin (x))^2} \, dx}{5 a} \\ & = \frac {\cos (x)}{5 (a+a \sin (x))^3}-\frac {\cos (x)}{5 a (a+a \sin (x))^2}+\frac {\int \frac {1}{a+a \sin (x)} \, dx}{5 a^2} \\ & = \frac {\cos (x)}{5 (a+a \sin (x))^3}-\frac {\cos (x)}{5 a (a+a \sin (x))^2}-\frac {\cos (x)}{5 \left (a^3+a^3 \sin (x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.50 \[ \int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx=-\frac {\cos (x) \left (1+3 \sin (x)+\sin ^2(x)\right )}{5 a^3 (1+\sin (x))^3} \]

[In]

Integrate[Sin[x]/(a + a*Sin[x])^3,x]

[Out]

-1/5*(Cos[x]*(1 + 3*Sin[x] + Sin[x]^2))/(a^3*(1 + Sin[x])^3)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(\frac {-\frac {2}{5}-2 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )-2 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right )}{a^{3} \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}\)	\(38\)
risch	\(-\frac {2 i \left (5 i {\mathrm e}^{2 i x}+5 \,{\mathrm e}^{3 i x}-i-5 \,{\mathrm e}^{i x}\right )}{5 \left ({\mathrm e}^{i x}+i\right )^{5} a^{3}}\)	\(42\)
default	\(\frac {-\frac {2}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {4}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {4}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {8}{5 \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}}{a^{3}}\)	\(45\)
norman	\(\frac {-\frac {2 \tan \left (\frac {x}{2}\right )}{a}-\frac {2 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{a}-\frac {4 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{a}-\frac {2 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{a}-\frac {2}{5 a}-\frac {12 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{5 a}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right ) a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}\)	\(82\)

[In]

int(sin(x)/(a+a*sin(x))^3,x,method=_RETURNVERBOSE)

[Out]

2/5*(-1-5*tan(1/2*x)^3-5*tan(1/2*x)^2-5*tan(1/2*x))/a^3/(tan(1/2*x)+1)^5

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.76 \[ \int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx=-\frac {\cos \left (x\right )^{3} - 2 \, \cos \left (x\right )^{2} - {\left (\cos \left (x\right )^{2} + 3 \, \cos \left (x\right ) + 1\right )} \sin \left (x\right ) - 2 \, \cos \left (x\right ) + 1}{5 \, {\left (a^{3} \cos \left (x\right )^{3} + 3 \, a^{3} \cos \left (x\right )^{2} - 2 \, a^{3} \cos \left (x\right ) - 4 \, a^{3} + {\left (a^{3} \cos \left (x\right )^{2} - 2 \, a^{3} \cos \left (x\right ) - 4 \, a^{3}\right )} \sin \left (x\right )\right )}} \]

[In]

integrate(sin(x)/(a+a*sin(x))^3,x, algorithm="fricas")

[Out]

-1/5*(cos(x)^3 - 2*cos(x)^2 - (cos(x)^2 + 3*cos(x) + 1)*sin(x) - 2*cos(x) + 1)/(a^3*cos(x)^3 + 3*a^3*cos(x)^2
- 2*a^3*cos(x) - 4*a^3 + (a^3*cos(x)^2 - 2*a^3*cos(x) - 4*a^3)*sin(x))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (44) = 88\).

Time = 1.21 (sec) , antiderivative size = 277, normalized size of antiderivative = 5.54 \[ \int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx=- \frac {10 \tan ^{3}{\left (\frac {x}{2} \right )}}{5 a^{3} \tan ^{5}{\left (\frac {x}{2} \right )} + 25 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 50 a^{3} \tan ^{3}{\left (\frac {x}{2} \right )} + 50 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} + 25 a^{3} \tan {\left (\frac {x}{2} \right )} + 5 a^{3}} - \frac {10 \tan ^{2}{\left (\frac {x}{2} \right )}}{5 a^{3} \tan ^{5}{\left (\frac {x}{2} \right )} + 25 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 50 a^{3} \tan ^{3}{\left (\frac {x}{2} \right )} + 50 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} + 25 a^{3} \tan {\left (\frac {x}{2} \right )} + 5 a^{3}} - \frac {10 \tan {\left (\frac {x}{2} \right )}}{5 a^{3} \tan ^{5}{\left (\frac {x}{2} \right )} + 25 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 50 a^{3} \tan ^{3}{\left (\frac {x}{2} \right )} + 50 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} + 25 a^{3} \tan {\left (\frac {x}{2} \right )} + 5 a^{3}} - \frac {2}{5 a^{3} \tan ^{5}{\left (\frac {x}{2} \right )} + 25 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 50 a^{3} \tan ^{3}{\left (\frac {x}{2} \right )} + 50 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} + 25 a^{3} \tan {\left (\frac {x}{2} \right )} + 5 a^{3}} \]

[In]

integrate(sin(x)/(a+a*sin(x))**3,x)

[Out]

-10*tan(x/2)**3/(5*a**3*tan(x/2)**5 + 25*a**3*tan(x/2)**4 + 50*a**3*tan(x/2)**3 + 50*a**3*tan(x/2)**2 + 25*a**
3*tan(x/2) + 5*a**3) - 10*tan(x/2)**2/(5*a**3*tan(x/2)**5 + 25*a**3*tan(x/2)**4 + 50*a**3*tan(x/2)**3 + 50*a**
3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 10*tan(x/2)/(5*a**3*tan(x/2)**5 + 25*a**3*tan(x/2)**4 + 50*a**3*t
an(x/2)**3 + 50*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3) - 2/(5*a**3*tan(x/2)**5 + 25*a**3*tan(x/2)**4 +
50*a**3*tan(x/2)**3 + 50*a**3*tan(x/2)**2 + 25*a**3*tan(x/2) + 5*a**3)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (44) = 88\).

Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.32 \[ \int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx=-\frac {2 \, {\left (\frac {5 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {5 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + 1\right )}}{5 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {10 \, a^{3} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}}\right )}} \]

[In]

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

[Out]

-2/5*(5*sin(x)/(cos(x) + 1) + 5*sin(x)^2/(cos(x) + 1)^2 + 5*sin(x)^3/(cos(x) + 1)^3 + 1)/(a^3 + 5*a^3*sin(x)/(
cos(x) + 1) + 10*a^3*sin(x)^2/(cos(x) + 1)^2 + 10*a^3*sin(x)^3/(cos(x) + 1)^3 + 5*a^3*sin(x)^4/(cos(x) + 1)^4
+ a^3*sin(x)^5/(cos(x) + 1)^5)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.74 \[ \int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx=-\frac {2 \, {\left (5 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 5 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 5 \, \tan \left (\frac {1}{2} \, x\right ) + 1\right )}}{5 \, a^{3} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{5}} \]

[In]

integrate(sin(x)/(a+a*sin(x))^3,x, algorithm="giac")

[Out]

-2/5*(5*tan(1/2*x)^3 + 5*tan(1/2*x)^2 + 5*tan(1/2*x) + 1)/(a^3*(tan(1/2*x) + 1)^5)

Mupad [B] (verification not implemented)

Time = 6.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.74 \[ \int \frac {\sin (x)}{(a+a \sin (x))^3} \, dx=-\frac {2\,\left (5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+5\,\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{5\,a^3\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^5} \]

[In]

int(sin(x)/(a + a*sin(x))^3,x)

[Out]

-(2*(5*tan(x/2) + 5*tan(x/2)^2 + 5*tan(x/2)^3 + 1))/(5*a^3*(tan(x/2) + 1)^5)