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

Optimal. Leaf size=88 \[ \frac {\sin (x)}{2 \left (a^2-b^2\right ) (a+b \cos (x))}+\frac {x}{2 b (a+b \cos (x))^2}-\frac {a \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b (a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4423, 2664, 12, 2659, 205} \[ \frac {\sin (x)}{2 \left (a^2-b^2\right ) (a+b \cos (x))}+\frac {x}{2 b (a+b \cos (x))^2}-\frac {a \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b (a-b)^{3/2} (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sin[x])/(a + b*Cos[x])^3,x]

[Out]

-((a*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*b*(a + b)^(3/2))) + x/(2*b*(a + b*Cos[x])^2) +
Sin[x]/(2*(a^2 - b^2)*(a + b*Cos[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 205

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 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 4423

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

Rubi steps

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

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Mathematica [A]  time = 0.32, size = 85, normalized size = 0.97 \[ \frac {\frac {\sin (x) (a+b \cos (x))}{(a-b) (a+b)}+\frac {x}{b}}{2 (a+b \cos (x))^2}-\frac {a \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{b \left (b^2-a^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Sin[x])/(a + b*Cos[x])^3,x]

[Out]

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

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fricas [B]  time = 0.66, size = 417, normalized size = 4.74 \[ \left [\frac {{\left (a b^{2} \cos \relax (x)^{2} + 2 \, a^{2} b \cos \relax (x) + a^{3}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \relax (x) + {\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \relax (x) + b\right )} \sin \relax (x) - a^{2} + 2 \, b^{2}}{b^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + a^{2}}\right ) + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x + 2 \, {\left (a^{3} b - a b^{3} + {\left (a^{2} b^{2} - b^{4}\right )} \cos \relax (x)\right )} \sin \relax (x)}{4 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \relax (x)\right )}}, -\frac {{\left (a b^{2} \cos \relax (x)^{2} + 2 \, a^{2} b \cos \relax (x) + a^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \relax (x) + b}{\sqrt {a^{2} - b^{2}} \sin \relax (x)}\right ) - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x - {\left (a^{3} b - a b^{3} + {\left (a^{2} b^{2} - b^{4}\right )} \cos \relax (x)\right )} \sin \relax (x)}{2 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \relax (x)\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((a*b^2*cos(x)^2 + 2*a^2*b*cos(x) + a^3)*sqrt(-a^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2*
sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2)) + 2*(a^4 - 2*a^2*b^
2 + b^4)*x + 2*(a^3*b - a*b^3 + (a^2*b^2 - b^4)*cos(x))*sin(x))/(a^6*b - 2*a^4*b^3 + a^2*b^5 + (a^4*b^3 - 2*a^
2*b^5 + b^7)*cos(x)^2 + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(x)), -1/2*((a*b^2*cos(x)^2 + 2*a^2*b*cos(x) + a^3)
*sqrt(a^2 - b^2)*arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x))) - (a^4 - 2*a^2*b^2 + b^4)*x - (a^3*b - a*b^3
+ (a^2*b^2 - b^4)*cos(x))*sin(x))/(a^6*b - 2*a^4*b^3 + a^2*b^5 + (a^4*b^3 - 2*a^2*b^5 + b^7)*cos(x)^2 + 2*(a^
5*b^2 - 2*a^3*b^4 + a*b^6)*cos(x))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sin \relax (x)}{{\left (b \cos \relax (x) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.44, size = 250, normalized size = 2.84 \[ \frac {i \left (-2 i a^{2} x \,{\mathrm e}^{2 i x}+2 i b^{2} x \,{\mathrm e}^{2 i x}+b a \,{\mathrm e}^{3 i x}+2 a^{2} {\mathrm e}^{2 i x}+b^{2} {\mathrm e}^{2 i x}+3 a b \,{\mathrm e}^{i x}+b^{2}\right )}{b \left (b \,{\mathrm e}^{2 i x}+2 a \,{\mathrm e}^{i x}+b \right )^{2} \left (a^{2}-b^{2}\right )}-\frac {i a \ln \left ({\mathrm e}^{i x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, b}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) b}+\frac {i a \ln \left ({\mathrm e}^{i x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, b}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(-2*I*a^2*x*exp(2*I*x)+2*I*b^2*x*exp(2*I*x)+b*a*exp(3*I*x)+2*a^2*exp(2*I*x)+b^2*exp(2*I*x)+3*a*b*exp(I*x)+b^
2)/b/(b*exp(2*I*x)+2*a*exp(I*x)+b)^2/(a^2-b^2)-1/2*I/(a^2-b^2)^(1/2)*a/(a+b)/(a-b)/b*ln(exp(I*x)+(a*(a^2-b^2)^
(1/2)+a^2-b^2)/(a^2-b^2)^(1/2)/b)+1/2*I/(a^2-b^2)^(1/2)*a/(a+b)/(a-b)/b*ln(exp(I*x)+(a*(a^2-b^2)^(1/2)-a^2+b^2
)/(a^2-b^2)^(1/2)/b)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(x)/(a+b*cos(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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\sin \relax (x)}{{\left (a+b\,\cos \relax (x)\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*sin(x))/(a + b*cos(x))^3,x)

[Out]

int((x*sin(x))/(a + b*cos(x))^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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