Integrand size = 8, antiderivative size = 102 \[ \int \frac {1}{(a+b \sin (x))^3} \, dx=\frac {\left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))} \] Output:
(2*a^2+b^2)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)+1/2*b *cos(x)/(a^2-b^2)/(a+b*sin(x))^2+3/2*a*b*cos(x)/(a^2-b^2)^2/(a+b*sin(x))
Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(a+b \sin (x))^3} \, dx=\frac {\left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b \cos (x) \left (4 a^2-b^2+3 a b \sin (x)\right )}{2 (a-b)^2 (a+b)^2 (a+b \sin (x))^2} \] Input:
Integrate[(a + b*Sin[x])^(-3),x]
Output:
((2*a^2 + b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (b*Cos[x]*(4*a^2 - b^2 + 3*a*b*Sin[x]))/(2*(a - b)^2*(a + b)^2*(a + b*S in[x])^2)
Time = 0.49 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.375, Rules used = {3042, 3143, 25, 3042, 3233, 25, 27, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {1}{(a+b \sin (x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+b \sin (x))^3}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle \frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\int -\frac {2 a-b \sin (x)}{(a+b \sin (x))^2}dx}{2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {2 a-b \sin (x)}{(a+b \sin (x))^2}dx}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a-b \sin (x)}{(a+b \sin (x))^2}dx}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {\frac {3 a b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int -\frac {2 a^2+b^2}{a+b \sin (x)}dx}{a^2-b^2}}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {2 a^2+b^2}{a+b \sin (x)}dx}{a^2-b^2}+\frac {3 a b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (2 a^2+b^2\right ) \int \frac {1}{a+b \sin (x)}dx}{a^2-b^2}+\frac {3 a b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (2 a^2+b^2\right ) \int \frac {1}{a+b \sin (x)}dx}{a^2-b^2}+\frac {3 a b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2+b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{a^2-b^2}+\frac {3 a b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {3 a b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac {4 \left (2 a^2+b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2}}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2+b^2\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {3 a b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 \left (a^2-b^2\right )}+\frac {b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}\) |
Input:
Int[(a + b*Sin[x])^(-3),x]
Output:
(b*Cos[x])/(2*(a^2 - b^2)*(a + b*Sin[x])^2) + ((2*(2*a^2 + b^2)*ArcTan[(2* b + 2*a*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/(a^2 - b^2)^(3/2) + (3*a*b*Cos[x]) /((a^2 - b^2)*(a + b*Sin[x])))/(2*(a^2 - b^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_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
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 + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Leaf count of result is larger than twice the leaf count of optimal. \(247\) vs. \(2(92)=184\).
Time = 0.42 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.43
| method | result | size |
| default | \(\frac {\frac {b^{2} \left (5 a^{2}-2 b^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}}{\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{4}+7 b^{2} a^{2}-2 b^{4}\right ) \tan \left (\frac {x}{2}\right )^{2}}{\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}-2 b^{2}\right ) \tan \left (\frac {x}{2}\right )}{a \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {2 b \left (4 a^{2}-b^{2}\right )}{2 a^{4}-4 b^{2} a^{2}+2 b^{4}}}{\left (a \tan \left (\frac {x}{2}\right )^{2}+2 b \tan \left (\frac {x}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}\) | \(248\) |
| risch | \(-\frac {i \left (-2 i b \,a^{2} {\mathrm e}^{3 i x}-i {\mathrm e}^{3 i x} b^{3}+10 i a^{2} b \,{\mathrm e}^{i x}-i b^{3} {\mathrm e}^{i x}+6 a^{3} {\mathrm e}^{2 i x}+3 b^{2} a \,{\mathrm e}^{2 i x}-3 a \,b^{2}\right )}{\left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\) | \(409\) |
Input:
int(1/(a+b*sin(x))^3,x,method=_RETURNVERBOSE)
Output:
2*(1/2*b^2*(5*a^2-2*b^2)/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*x)^3+1/2*b*(4*a^4+7 *a^2*b^2-2*b^4)/(a^4-2*a^2*b^2+b^4)/a^2*tan(1/2*x)^2+1/2*b^2*(11*a^2-2*b^2 )/a/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)+1/2*b*(4*a^2-b^2)/(a^4-2*a^2*b^2+b^4))/ (a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^2+(2*a^2+b^2)/(a^4-2*a^2*b^2+b^4)/(a^2-b ^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (92) = 184\).
Time = 0.10 (sec) , antiderivative size = 516, normalized size of antiderivative = 5.06 \[ \int \frac {1}{(a+b \sin (x))^3} \, dx=\left [\frac {6 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (x\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )}{4 \, {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (x\right )\right )}}, \frac {3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (x\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) + {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )}{2 \, {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (x\right )\right )}}\right ] \] Input:
integrate(1/(a+b*sin(x))^3,x, algorithm="fricas")
Output:
[1/4*(6*(a^3*b^2 - a*b^4)*cos(x)*sin(x) - (2*a^4 + 3*a^2*b^2 + b^4 - (2*a^ 2*b^2 + b^4)*cos(x)^2 + 2*(2*a^3*b + a*b^3)*sin(x))*sqrt(-a^2 + b^2)*log(( (2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 + 2*(a*cos(x)*sin(x) + b *cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2* (4*a^4*b - 5*a^2*b^3 + b^5)*cos(x))/(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 - ( a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b - 3*a^5*b^3 + 3 *a^3*b^5 - a*b^7)*sin(x)), 1/2*(3*(a^3*b^2 - a*b^4)*cos(x)*sin(x) - (2*a^4 + 3*a^2*b^2 + b^4 - (2*a^2*b^2 + b^4)*cos(x)^2 + 2*(2*a^3*b + a*b^3)*sin( x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) + (4* a^4*b - 5*a^2*b^3 + b^5)*cos(x))/(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 - (a^6 *b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^ 3*b^5 - a*b^7)*sin(x))]
Timed out. \[ \int \frac {1}{(a+b \sin (x))^3} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sin(x))**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(a+b \sin (x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+b*sin(x))^3,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (92) = 184\).
Time = 0.12 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.11 \[ \int \frac {1}{(a+b \sin (x))^3} \, dx=\frac {{\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (2 \, a^{2} + b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, a b^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, x\right )^{2} + 7 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b^{5} \tan \left (\frac {1}{2} \, x\right )^{2} + 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, x\right ) - 2 \, a b^{4} \tan \left (\frac {1}{2} \, x\right ) + 4 \, a^{4} b - a^{2} b^{3}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}^{2}} \] Input:
integrate(1/(a+b*sin(x))^3,x, algorithm="giac")
Output:
(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^ 2)))*(2*a^2 + b^2)/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (5*a^3*b^2* tan(1/2*x)^3 - 2*a*b^4*tan(1/2*x)^3 + 4*a^4*b*tan(1/2*x)^2 + 7*a^2*b^3*tan (1/2*x)^2 - 2*b^5*tan(1/2*x)^2 + 11*a^3*b^2*tan(1/2*x) - 2*a*b^4*tan(1/2*x ) + 4*a^4*b - a^2*b^3)/((a^6 - 2*a^4*b^2 + a^2*b^4)*(a*tan(1/2*x)^2 + 2*b* tan(1/2*x) + a)^2)
Time = 17.08 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.42 \[ \int \frac {1}{(a+b \sin (x))^3} \, dx=\frac {\frac {4\,a^2\,b-b^3}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (4\,a^2\,b-b^3\right )\,\left (a^2+2\,b^2\right )}{a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (5\,a^2\,b-2\,b^3\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (11\,a^2\,b-2\,b^3\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2+a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}+\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,a^2+b^2\right )\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^2+b^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{2\,a^2+b^2}\right )\,\left (2\,a^2+b^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \] Input:
int(1/(a + b*sin(x))^3,x)
Output:
((4*a^2*b - b^3)/(a^4 + b^4 - 2*a^2*b^2) + (tan(x/2)^2*(4*a^2*b - b^3)*(a^ 2 + 2*b^2))/(a^2*(a^4 + b^4 - 2*a^2*b^2)) + (b*tan(x/2)^3*(5*a^2*b - 2*b^3 ))/(a*(a^4 + b^4 - 2*a^2*b^2)) + (b*tan(x/2)*(11*a^2*b - 2*b^3))/(a*(a^4 + b^4 - 2*a^2*b^2)))/(tan(x/2)^2*(2*a^2 + 4*b^2) + a^2 + a^2*tan(x/2)^4 + 4 *a*b*tan(x/2) + 4*a*b*tan(x/2)^3) + (atan(((((2*a^2 + b^2)*(2*a^4*b + 2*b^ 5 - 4*a^2*b^3))/(2*(a + b)^(5/2)*(a - b)^(5/2)*(a^4 + b^4 - 2*a^2*b^2)) + (a*tan(x/2)*(2*a^2 + b^2))/((a + b)^(5/2)*(a - b)^(5/2)))*(a^4 + b^4 - 2*a ^2*b^2))/(2*a^2 + b^2))*(2*a^2 + b^2))/((a + b)^(5/2)*(a - b)^(5/2))
Time = 0.22 (sec) , antiderivative size = 457, normalized size of antiderivative = 4.48 \[ \int \frac {1}{(a+b \sin (x))^3} \, dx=\frac {8 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right )^{2} a^{2} b^{2}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right )^{2} b^{4}+16 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) a^{3} b +8 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) a \,b^{3}+8 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} b^{2}+6 \cos \left (x \right ) \sin \left (x \right ) a^{3} b^{2}-6 \cos \left (x \right ) \sin \left (x \right ) a \,b^{4}+8 \cos \left (x \right ) a^{4} b -10 \cos \left (x \right ) a^{2} b^{3}+2 \cos \left (x \right ) b^{5}+3 \sin \left (x \right )^{2} a^{2} b^{3}-3 \sin \left (x \right )^{2} b^{5}+6 \sin \left (x \right ) a^{3} b^{2}-6 \sin \left (x \right ) a \,b^{4}+3 a^{4} b -3 a^{2} b^{3}}{4 \sin \left (x \right )^{2} a^{6} b^{2}-12 \sin \left (x \right )^{2} a^{4} b^{4}+12 \sin \left (x \right )^{2} a^{2} b^{6}-4 \sin \left (x \right )^{2} b^{8}+8 \sin \left (x \right ) a^{7} b -24 \sin \left (x \right ) a^{5} b^{3}+24 \sin \left (x \right ) a^{3} b^{5}-8 \sin \left (x \right ) a \,b^{7}+4 a^{8}-12 a^{6} b^{2}+12 a^{4} b^{4}-4 a^{2} b^{6}} \] Input:
int(1/(a+b*sin(x))^3,x)
Output:
(8*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)**2*a* *2*b**2 + 4*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin (x)**2*b**4 + 16*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2) )*sin(x)*a**3*b + 8*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b* *2))*sin(x)*a*b**3 + 8*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a**4 + 4*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2) )*a**2*b**2 + 6*cos(x)*sin(x)*a**3*b**2 - 6*cos(x)*sin(x)*a*b**4 + 8*cos(x )*a**4*b - 10*cos(x)*a**2*b**3 + 2*cos(x)*b**5 + 3*sin(x)**2*a**2*b**3 - 3 *sin(x)**2*b**5 + 6*sin(x)*a**3*b**2 - 6*sin(x)*a*b**4 + 3*a**4*b - 3*a**2 *b**3)/(4*(sin(x)**2*a**6*b**2 - 3*sin(x)**2*a**4*b**4 + 3*sin(x)**2*a**2* b**6 - sin(x)**2*b**8 + 2*sin(x)*a**7*b - 6*sin(x)*a**5*b**3 + 6*sin(x)*a* *3*b**5 - 2*sin(x)*a*b**7 + a**8 - 3*a**6*b**2 + 3*a**4*b**4 - a**2*b**6))