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

Optimal. Leaf size=85 \[ \frac {a \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2}}+\frac {\cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {x}{2 b (a+b \sin (x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4422, 2664, 12, 2660, 618, 204} \[ \frac {a \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2}}+\frac {\cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {x}{2 b (a+b \sin (x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*Cos[x])/(a + b*Sin[x])^3,x]

[Out]

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

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 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 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 4422

Int[Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol]
:> Simp[((e + f*x)^m*(a + b*Sin[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*Sin[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 \cos (x)}{(a+b \sin (x))^3} \, dx &=-\frac {x}{2 b (a+b \sin (x))^2}+\frac {\int \frac {1}{(a+b \sin (x))^2} \, dx}{2 b}\\ &=-\frac {x}{2 b (a+b \sin (x))^2}+\frac {\cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \frac {a}{a+b \sin (x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {x}{2 b (a+b \sin (x))^2}+\frac {\cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {a \int \frac {1}{a+b \sin (x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {x}{2 b (a+b \sin (x))^2}+\frac {\cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {a \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b \left (a^2-b^2\right )}\\ &=-\frac {x}{2 b (a+b \sin (x))^2}+\frac {\cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {(2 a) \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 )}{b \left (a^2-b^2\right )}\\ &=\frac {a \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2}}-\frac {x}{2 b (a+b \sin (x))^2}+\frac {\cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.26, size = 84, normalized size = 0.99 \[ \frac {a \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2}}+\frac {\frac {\cos (x) (a+b \sin (x))}{(a-b) (a+b)}-\frac {x}{b}}{2 (a+b \sin (x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.63, size = 459, normalized size = 5.40 \[ \left [\frac {2 \, {\left (a^{2} b^{2} - b^{4}\right )} \cos \relax (x) \sin \relax (x) - {\left (a b^{2} \cos \relax (x)^{2} - 2 \, a^{2} b \sin \relax (x) - a^{3} - a b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} - 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x)}{4 \, {\left (a^{6} b - a^{4} b^{3} - a^{2} b^{5} + b^{7} - {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \sin \relax (x)\right )}}, \frac {{\left (a^{2} b^{2} - b^{4}\right )} \cos \relax (x) \sin \relax (x) + {\left (a b^{2} \cos \relax (x)^{2} - 2 \, a^{2} b \sin \relax (x) - a^{3} - a b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x + {\left (a^{3} b - a b^{3}\right )} \cos \relax (x)}{2 \, {\left (a^{6} b - a^{4} b^{3} - a^{2} b^{5} + b^{7} - {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \sin \relax (x)\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.83, size = 257, normalized size = 3.02 \[ \frac {2 i a^{2} {\mathrm e}^{2 i x}+i b^{2} {\mathrm e}^{2 i x}+2 x \,a^{2} {\mathrm e}^{2 i x}+b a \,{\mathrm e}^{3 i x}-2 b^{2} x \,{\mathrm e}^{2 i x}-i b^{2}-3 a b \,{\mathrm e}^{i x}}{\left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right ) b}-\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b}+\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*cos(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 details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*cos(x))/(a + b*sin(x))^3,x)

[Out]

int((x*cos(x))/(a + b*sin(x))^3, x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________