Integrand size = 13, antiderivative size = 47 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {2 x}{a^2}-\frac {4 \cos (x)}{3 a^2}-\frac {2 \cos (x)}{a^2 (1+\sin (x))}+\frac {\cos (x) \sin ^2(x)}{3 (a+a \sin (x))^2} \] Output:
-2*x/a^2-4/3*cos(x)/a^2-2*cos(x)/a^2/(1+sin(x))+1/3*cos(x)*sin(x)^2/(a+a*s in(x))^2
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.79 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right ) \left (6 (-5+6 x) \cos \left (\frac {x}{2}\right )+(41-12 x) \cos \left (\frac {3 x}{2}\right )-3 \cos \left (\frac {5 x}{2}\right )+6 (-9+8 x+4 (1+x) \cos (x)+\cos (2 x)) \sin \left (\frac {x}{2}\right )\right )}{12 a^2 (1+\sin (x))^2} \] Input:
Integrate[Sin[x]^3/(a + a*Sin[x])^2,x]
Output:
-1/12*((Cos[x/2] + Sin[x/2])*(6*(-5 + 6*x)*Cos[x/2] + (41 - 12*x)*Cos[(3*x )/2] - 3*Cos[(5*x)/2] + 6*(-9 + 8*x + 4*(1 + x)*Cos[x] + Cos[2*x])*Sin[x/2 ]))/(a^2*(1 + Sin[x])^2)
Time = 0.54 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {3042, 3244, 27, 3042, 3447, 3042, 3502, 27, 3042, 3214, 3042, 3127}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sin ^3(x)}{(a \sin (x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)^3}{(a \sin (x)+a)^2}dx\) |
\(\Big \downarrow \) 3244 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {\int \frac {2 \sin (x) (a-2 a \sin (x))}{\sin (x) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \int \frac {\sin (x) (a-2 a \sin (x))}{\sin (x) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \int \frac {\sin (x) (a-2 a \sin (x))}{\sin (x) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \int \frac {a \sin (x)-2 a \sin ^2(x)}{\sin (x) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \int \frac {a \sin (x)-2 a \sin (x)^2}{\sin (x) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \left (\frac {\int \frac {3 a^2 \sin (x)}{\sin (x) a+a}dx}{a}+2 \cos (x)\right )}{3 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \left (3 a \int \frac {\sin (x)}{\sin (x) a+a}dx+2 \cos (x)\right )}{3 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \left (3 a \int \frac {\sin (x)}{\sin (x) a+a}dx+2 \cos (x)\right )}{3 a^2}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \left (3 a \left (\frac {x}{a}-\int \frac {1}{\sin (x) a+a}dx\right )+2 \cos (x)\right )}{3 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \left (3 a \left (\frac {x}{a}-\int \frac {1}{\sin (x) a+a}dx\right )+2 \cos (x)\right )}{3 a^2}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {\sin ^2(x) \cos (x)}{3 (a \sin (x)+a)^2}-\frac {2 \left (3 a \left (\frac {x}{a}+\frac {\cos (x)}{a \sin (x)+a}\right )+2 \cos (x)\right )}{3 a^2}\) |
Input:
Int[Sin[x]^3/(a + a*Sin[x])^2,x]
Output:
(Cos[x]*Sin[x]^2)/(3*(a + a*Sin[x])^2) - (2*(2*Cos[x] + 3*a*(x/a + Cos[x]/ (a + a*Sin[x]))))/(3*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] + Simp[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]))
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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.35 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.17
method | result | size |
parallelrisch | \(\frac {\left (-12 x -20\right ) \cos \left (3 x \right )-48 \cos \left (2 x \right )-3 \cos \left (4 x \right )+16 \sin \left (3 x \right )-29+\left (-36 x -60\right ) \cos \left (x \right )}{6 a^{2} \left (\cos \left (3 x \right )+3 \cos \left (x \right )\right )}\) | \(55\) |
default | \(\frac {\frac {4}{3 \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {4}{\tan \left (\frac {x}{2}\right )+1}-\frac {2}{\tan \left (\frac {x}{2}\right )^{2}+1}-4 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\) | \(56\) |
risch | \(-\frac {2 x}{a^{2}}-\frac {{\mathrm e}^{i x}}{2 a^{2}}-\frac {{\mathrm e}^{-i x}}{2 a^{2}}-\frac {2 \left (15 i {\mathrm e}^{i x}+9 \,{\mathrm e}^{2 i x}-8\right )}{3 a^{2} \left ({\mathrm e}^{i x}+i\right )^{3}}\) | \(60\) |
norman | \(\frac {-\frac {16 \tan \left (\frac {x}{2}\right )}{a}-\frac {4 \tan \left (\frac {x}{2}\right )^{8}}{a}-\frac {2 x}{a}-\frac {20}{3 a}-\frac {6 x \tan \left (\frac {x}{2}\right )}{a}-\frac {12 x \tan \left (\frac {x}{2}\right )^{2}}{a}-\frac {20 x \tan \left (\frac {x}{2}\right )^{3}}{a}-\frac {24 x \tan \left (\frac {x}{2}\right )^{4}}{a}-\frac {24 x \tan \left (\frac {x}{2}\right )^{5}}{a}-\frac {20 x \tan \left (\frac {x}{2}\right )^{6}}{a}-\frac {12 x \tan \left (\frac {x}{2}\right )^{7}}{a}-\frac {6 x \tan \left (\frac {x}{2}\right )^{8}}{a}-\frac {2 x \tan \left (\frac {x}{2}\right )^{9}}{a}-\frac {12 \tan \left (\frac {x}{2}\right )^{7}}{a}-\frac {28 \tan \left (\frac {x}{2}\right )^{2}}{a}-\frac {40 \tan \left (\frac {x}{2}\right )^{4}}{a}-\frac {40 \tan \left (\frac {x}{2}\right )^{5}}{a}-\frac {44 \tan \left (\frac {x}{2}\right )^{3}}{a}-\frac {68 \tan \left (\frac {x}{2}\right )^{6}}{3 a}}{\left (\tan \left (\frac {x}{2}\right )^{2}+1\right )^{3} a \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}\) | \(227\) |
Input:
int(sin(x)^3/(a+a*sin(x))^2,x,method=_RETURNVERBOSE)
Output:
1/6*((-12*x-20)*cos(3*x)-48*cos(2*x)-3*cos(4*x)+16*sin(3*x)-29+(-36*x-60)* cos(x))/a^2/(cos(3*x)+3*cos(x))
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (43) = 86\).
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.02 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {{\left (6 \, x - 11\right )} \cos \left (x\right )^{2} + 3 \, \cos \left (x\right )^{3} - {\left (6 \, x + 13\right )} \cos \left (x\right ) - {\left (2 \, {\left (3 \, x + 7\right )} \cos \left (x\right ) + 3 \, \cos \left (x\right )^{2} + 12 \, x + 1\right )} \sin \left (x\right ) - 12 \, x + 1}{3 \, {\left (a^{2} \cos \left (x\right )^{2} - a^{2} \cos \left (x\right ) - 2 \, a^{2} - {\left (a^{2} \cos \left (x\right ) + 2 \, a^{2}\right )} \sin \left (x\right )\right )}} \] Input:
integrate(sin(x)^3/(a+a*sin(x))^2,x, algorithm="fricas")
Output:
-1/3*((6*x - 11)*cos(x)^2 + 3*cos(x)^3 - (6*x + 13)*cos(x) - (2*(3*x + 7)* cos(x) + 3*cos(x)^2 + 12*x + 1)*sin(x) - 12*x + 1)/(a^2*cos(x)^2 - a^2*cos (x) - 2*a^2 - (a^2*cos(x) + 2*a^2)*sin(x))
Leaf count of result is larger than twice the leaf count of optimal. 779 vs. \(2 (48) = 96\).
Time = 2.03 (sec) , antiderivative size = 779, normalized size of antiderivative = 16.57 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=\text {Too large to display} \] Input:
integrate(sin(x)**3/(a+a*sin(x))**2,x)
Output:
-6*x*tan(x/2)**5/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/ 2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 18*x*tan(x/2)**4 /(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2* tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 24*x*tan(x/2)**3/(3*a**2*tan(x/2 )**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9* a**2*tan(x/2) + 3*a**2) - 24*x*tan(x/2)**2/(3*a**2*tan(x/2)**5 + 9*a**2*ta n(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 18*x*tan(x/2)/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2 *tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 6*x/(3*a* *2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/ 2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 12*tan(x/2)**4/(3*a**2*tan(x/2)**5 + 9 *a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan (x/2) + 3*a**2) - 36*tan(x/2)**3/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 44*tan(x/2)**2/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2) **3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a**2) - 48*tan(x/2)/(3*a** 2*tan(x/2)**5 + 9*a**2*tan(x/2)**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2 )**2 + 9*a**2*tan(x/2) + 3*a**2) - 20/(3*a**2*tan(x/2)**5 + 9*a**2*tan(x/2 )**4 + 12*a**2*tan(x/2)**3 + 12*a**2*tan(x/2)**2 + 9*a**2*tan(x/2) + 3*a** 2)
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (43) = 86\).
Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 3.06 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {4 \, {\left (\frac {12 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {11 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {9 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + 5\right )}}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {4 \, a^{2} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {4 \, a^{2} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, a^{2} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}}\right )}} - \frac {4 \, \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{2}} \] Input:
integrate(sin(x)^3/(a+a*sin(x))^2,x, algorithm="maxima")
Output:
-4/3*(12*sin(x)/(cos(x) + 1) + 11*sin(x)^2/(cos(x) + 1)^2 + 9*sin(x)^3/(co s(x) + 1)^3 + 3*sin(x)^4/(cos(x) + 1)^4 + 5)/(a^2 + 3*a^2*sin(x)/(cos(x) + 1) + 4*a^2*sin(x)^2/(cos(x) + 1)^2 + 4*a^2*sin(x)^3/(cos(x) + 1)^3 + 3*a^ 2*sin(x)^4/(cos(x) + 1)^4 + a^2*sin(x)^5/(cos(x) + 1)^5) - 4*arctan(sin(x) /(cos(x) + 1))/a^2
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {2 \, x}{a^{2}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )} a^{2}} - \frac {2 \, {\left (6 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, x\right ) + 7\right )}}{3 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{3}} \] Input:
integrate(sin(x)^3/(a+a*sin(x))^2,x, algorithm="giac")
Output:
-2*x/a^2 - 2/((tan(1/2*x)^2 + 1)*a^2) - 2/3*(6*tan(1/2*x)^2 + 15*tan(1/2*x ) + 7)/(a^2*(tan(1/2*x) + 1)^3)
Time = 17.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.32 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=-\frac {2\,x}{a^2}-\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+12\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {44\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{3}+16\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {20}{3}}{a^2\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^3} \] Input:
int(sin(x)^3/(a + a*sin(x))^2,x)
Output:
- (2*x)/a^2 - (16*tan(x/2) + (44*tan(x/2)^2)/3 + 12*tan(x/2)^3 + 4*tan(x/2 )^4 + 20/3)/(a^2*(tan(x/2)^2 + 1)*(tan(x/2) + 1)^3)
Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.89 \[ \int \frac {\sin ^3(x)}{(a+a \sin (x))^2} \, dx=\frac {3 \cos \left (x \right ) \sin \left (x \right )^{2}-6 \cos \left (x \right ) \sin \left (x \right ) x +8 \cos \left (x \right ) \sin \left (x \right )-6 \cos \left (x \right ) x +4 \cos \left (x \right )+3 \sin \left (x \right )^{3}+6 \sin \left (x \right )^{2} x +17 \sin \left (x \right )^{2}+12 \sin \left (x \right ) x +8 \sin \left (x \right )+6 x -4}{3 a^{2} \left (\cos \left (x \right ) \sin \left (x \right )+\cos \left (x \right )-\sin \left (x \right )^{2}-2 \sin \left (x \right )-1\right )} \] Input:
int(sin(x)^3/(a+a*sin(x))^2,x)
Output:
(3*cos(x)*sin(x)**2 - 6*cos(x)*sin(x)*x + 8*cos(x)*sin(x) - 6*cos(x)*x + 4 *cos(x) + 3*sin(x)**3 + 6*sin(x)**2*x + 17*sin(x)**2 + 12*sin(x)*x + 8*sin (x) + 6*x - 4)/(3*a**2*(cos(x)*sin(x) + cos(x) - sin(x)**2 - 2*sin(x) - 1) )