3.198 \(\int \frac{1}{(a+b \sin (x))^3} \, dx\)

Optimal. Leaf size=102 \[ \frac{\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0946794, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {2664, 2754, 12, 2660, 618, 204} \[ \frac{\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{3 a b \cos (x)}{2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{b \cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x])^(-3),x]

[Out]

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

Rule 2664

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] + Dist[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] && Integer
Q[2*n]

Rule 2754

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]

Rule 12

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

Rule 2660

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 618

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(a+b \sin (x))^3} \, dx &=\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 )}\\ &=\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))}+\frac{\int \frac{2 a^2+b^2}{a+b \sin (x)} \, dx}{2 \left (a^2-b^2\right )^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))}+\frac{\left (2 a^2+b^2\right ) \int \frac{1}{a+b \sin (x)} \, dx}{2 \left (a^2-b^2\right )^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))}+\frac{\left (2 a^2+b^2\right ) \operatorname{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{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))}-\frac{\left (2 \left (2 a^2+b^2\right )\right ) \operatorname{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{\left (2 a^2+b^2\right ) \tan ^{-1}\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))}\\ \end{align*}

Mathematica [A]  time = 0.181512, size = 93, normalized size = 0.91 \[ \frac{\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{b \cos (x) \left (4 a^2+3 a b \sin (x)-b^2\right )}{2 (a-b)^2 (a+b)^2 (a+b \sin (x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x])^(-3),x]

[Out]

((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*Si
n[x]))/(2*(a - b)^2*(a + b)^2*(a + b*Sin[x])^2)

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Maple [B]  time = 0.052, size = 300, normalized size = 2.9 \begin{align*} 2\,{\frac{1}{ \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) ^{2}} \left ( 1/2\,{\frac{{b}^{2} \left ( 5\,{a}^{2}-2\,{b}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{3}}{a \left ({a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/2\,{\frac{b \left ( 4\,{a}^{4}+7\,{a}^{2}{b}^{2}-2\,{b}^{4} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{2}}{ \left ({a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4} \right ){a}^{2}}}+1/2\,{\frac{{b}^{2} \left ( 11\,{a}^{2}-2\,{b}^{2} \right ) \tan \left ( x/2 \right ) }{a \left ({a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/2\,{\frac{b \left ( 4\,{a}^{2}-{b}^{2} \right ) }{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}} \right ) }+2\,{\frac{{a}^{2}}{ \left ({a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+{\frac{{b}^{2}}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}\arctan \left ({\frac{1}{2} \left ( 2\,a\tan \left ( x/2 \right ) +2\,b \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(x))^3,x)

[Out]

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)
)/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2+2*a^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))+1/(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))*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.12588, size = 1135, normalized size = 11.13 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x))^3,x, algorithm="fricas")

[Out]

[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))]

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

[Out]

Timed out

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Giac [B]  time = 1.58914, size = 290, normalized size = 2.84 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x))^3,x, algorithm="giac")

[Out]

(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*t
an(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)