\(\int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx\) [197]

Optimal result

Integrand size = 11, antiderivative size = 103 \[ \int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx=-\frac {3 a b \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {a \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\left (a^2+2 b^2\right ) \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {2833, 12, 2739, 632, 210} \[ \int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx=-\frac {3 a b \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\left (a^2+2 b^2\right ) \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac {a \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2} \]

[In]

Int[Sin[x]/(a + b*Sin[x])^3,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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]

Rule 2833

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {a \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\int \frac {2 b-a \sin (x)}{(a+b \sin (x))^2} \, dx}{2 \left (a^2-b^2\right )} \\ & = -\frac {a \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\left (a^2+2 b^2\right ) \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {\int -\frac {3 a b}{a+b \sin (x)} \, dx}{2 \left (a^2-b^2\right )^2} \\ & = -\frac {a \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\left (a^2+2 b^2\right ) \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac {(3 a b) \int \frac {1}{a+b \sin (x)} \, dx}{2 \left (a^2-b^2\right )^2} \\ & = -\frac {a \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\left (a^2+2 b^2\right ) \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac {(3 a b) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2} \\ & = -\frac {a \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\left (a^2+2 b^2\right ) \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {(6 a b) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2} \\ & = -\frac {3 a b \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 {a \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {\left (a^2+2 b^2\right ) \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.91 \[ \int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx=-\frac {3 a b \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\cos (x) \left (a \left (2 a^2+b^2\right )+b \left (a^2+2 b^2\right ) \sin (x)\right )}{2 (a-b)^2 (a+b)^2 (a+b \sin (x))^2} \]

[In]

Integrate[Sin[x]/(a + b*Sin[x])^3,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(220\) vs. \(2(93)=186\).

Time = 0.64 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.15

method	result	size
default	\(\frac {-\frac {3 a^{2} b \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {\left (2 a^{4}+5 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}-\frac {\left (5 a^{2}+4 b^{2}\right ) b \tan \left (\frac {x}{2}\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {\left (2 a^{2}+b^{2}\right ) a}{a^{4}-2 a^{2} b^{2}+b^{4}}}{{\left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )}^{2}}-\frac {3 a b \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}\)	\(221\)
risch	\(\frac {i \left (-3 i a \,b^{3} {\mathrm e}^{3 i x}+4 i a^{3} b \,{\mathrm e}^{i x}+5 i a \,b^{3} {\mathrm e}^{i x}+2 a^{4} {\mathrm e}^{2 i x}+5 a^{2} b^{2} {\mathrm e}^{2 i x}+2 b^{4} {\mathrm e}^{2 i x}-a^{2} b^{2}-2 b^{4}\right )}{\left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right )^{2} b}-\frac {3 i b a \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {3 i a b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\)	\(274\)

[In]

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

[Out]

4*(-3/4*a^2*b/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3-1/4*(2*a^4+5*a^2*b^2+2*b^4)/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*x)^2-
1/4*(5*a^2+4*b^2)*b/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)-1/4*(2*a^2+b^2)*a/(a^4-2*a^2*b^2+b^4))/(a*tan(1/2*x)^2+2*b*
tan(1/2*x)+a)^2-3*a*b/(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))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (93) = 186\).

Time = 0.33 (sec) , antiderivative size = 490, normalized size of antiderivative = 4.76 \[ \int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx=\left [-\frac {2 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (x\right ) \sin \left (x\right ) - 3 \, {\left (a b^{3} \cos \left (x\right )^{2} - 2 \, a^{2} b^{2} \sin \left (x\right ) - a^{3} b - a b^{3}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right )}{4 \, {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (x\right )\right )}}, -\frac {{\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (x\right ) \sin \left (x\right ) + 3 \, {\left (a b^{3} \cos \left (x\right )^{2} - 2 \, a^{2} b^{2} \sin \left (x\right ) - a^{3} b - a b^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) + {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right )}{2 \, {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (x\right )\right )}}\right ] \]

[In]

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

[Out]

[-1/4*(2*(a^4*b + a^2*b^3 - 2*b^5)*cos(x)*sin(x) - 3*(a*b^3*cos(x)^2 - 2*a^2*b^2*sin(x) - a^3*b - a*b^3)*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*(2*a^5 - a^3*b^2 - a*b^4)*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*((a^4*b + a^2*b^3 - 2*b^5)*cos(x)*sin(x) + 3*(a*b^3*cos(x)^2 - 2*a^2*b^2*sin(x) - a^3*b - a*b^3)
*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) + (2*a^5 - a^3*b^2 - a*b^4)*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))]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (93) = 186\).

Time = 0.30 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.83 \[ \int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx=-\frac {3 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{2} + 5 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{2} + 5 \, a^{3} b \tan \left (\frac {1}{2} \, x\right ) + 4 \, a b^{3} \tan \left (\frac {1}{2} \, x\right ) + 2 \, a^{4} + a^{2} b^{2}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}^{2}} \]

[In]

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

[Out]

-3*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*a*b/((a^4 - 2*a^2*b^2 + b^4)
*sqrt(a^2 - b^2)) - (3*a^3*b*tan(1/2*x)^3 + 2*a^4*tan(1/2*x)^2 + 5*a^2*b^2*tan(1/2*x)^2 + 2*b^4*tan(1/2*x)^2 +
5*a^3*b*tan(1/2*x) + 4*a*b^3*tan(1/2*x) + 2*a^4 + a^2*b^2)/((a^5 - 2*a^3*b^2 + a*b^4)*(a*tan(1/2*x)^2 + 2*b*t
an(1/2*x) + a)^2)

Mupad [B] (verification not implemented)

Time = 6.99 (sec) , antiderivative size = 310, normalized size of antiderivative = 3.01 \[ \int \frac {\sin (x)}{(a+b \sin (x))^3} \, dx=-\frac {\frac {2\,a^3+a\,b^2}{a^4-2\,a^2\,b^2+b^4}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{a^4-2\,a^2\,b^2+b^4}+\frac {b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (5\,a^2+4\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2+b^2\right )\,\left (a^2+2\,b^2\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2+a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}-\frac {3\,a\,b\,\mathrm {atan}\left (\frac {\left (\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {3\,a\,b^2\,\left (2\,a^4-4\,a^2\,b^2+2\,b^4\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}\,\left (a^4-2\,a^2\,b^2+b^4\right )}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{3\,a\,b}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

[In]

int(sin(x)/(a + b*sin(x))^3,x)

[Out]

- ((a*b^2 + 2*a^3)/(a^4 + b^4 - 2*a^2*b^2) + (3*a^2*b*tan(x/2)^3)/(a^4 + b^4 - 2*a^2*b^2) + (b*tan(x/2)*(5*a^2
+ 4*b^2))/(a^4 + b^4 - 2*a^2*b^2) + (tan(x/2)^2*(2*a^2 + b^2)*(a^2 + 2*b^2))/(a*(a^4 + b^4 - 2*a^2*b^2)))/(ta
n(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) - (3*a*b*atan((((3*a^2*b*
tan(x/2))/((a + b)^(5/2)*(a - b)^(5/2)) + (3*a*b^2*(2*a^4 + 2*b^4 - 4*a^2*b^2))/(2*(a + b)^(5/2)*(a - b)^(5/2)
*(a^4 + b^4 - 2*a^2*b^2)))*(a^4 + b^4 - 2*a^2*b^2))/(3*a*b)))/((a + b)^(5/2)*(a - b)^(5/2))