\(\int \frac {x \cos (x)}{(a+b \sin (x))^3} \, dx\) [75]

Optimal result

Integrand size = 12, antiderivative size = 85 \[ \int \frac {x \cos (x)}{(a+b \sin (x))^3} \, dx=\frac {a \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2}}-\frac {x}{2 b (a+b \sin (x))^2}+\frac {\cos (x)}{2 \left (a^2-b^2\right ) (a+b \sin (x))} \] Output:

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))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99 \[ \int \frac {x \cos (x)}{(a+b \sin (x))^3} \, dx=\frac {a \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2}}+\frac {-\frac {x}{b}+\frac {\cos (x) (a+b \sin (x))}{(a-b) (a+b)}}{2 (a+b \sin (x))^2} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {4922, 3042, 3143, 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.

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

\(\Big \downarrow \) 4922

\(\displaystyle \frac {\int \frac {1}{(a+b \sin (x))^2}dx}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{(a+b \sin (x))^2}dx}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

\(\Big \downarrow \) 3143

\(\displaystyle \frac {\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\int -\frac {a}{a+b \sin (x)}dx}{a^2-b^2}}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {a}{a+b \sin (x)}dx}{a^2-b^2}+\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a \int \frac {1}{a+b \sin (x)}dx}{a^2-b^2}+\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a \int \frac {1}{a+b \sin (x)}dx}{a^2-b^2}+\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {2 a \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{a^2-b^2}+\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}-\frac {4 a \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2}}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}}{2 b}-\frac {x}{2 b (a+b \sin (x))^2}\)

Input:

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

Output:

-1/2*x/(b*(a + b*Sin[x])^2) + ((2*a*ArcTan[(2*b + 2*a*Tan[x/2])/(2*Sqrt[a^ 
2 - b^2])])/(a^2 - b^2)^(3/2) + (b*Cos[x])/((a^2 - b^2)*(a + b*Sin[x])))/( 
2*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 1083

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

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]
 

rule 3143

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]
 

rule 4922

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] - Simp[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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.02

method	result	size
risch	\(\frac {2 i a^{2} {\mathrm e}^{2 i x}+i b^{2} {\mathrm e}^{2 i x}+2 x \,a^{2} {\mathrm e}^{2 i x}+b a \,{\mathrm e}^{3 i x}-2 b^{2} x \,{\mathrm e}^{2 i x}-i b^{2}-3 a b \,{\mathrm e}^{i x}}{\left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right ) b}-\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b}+\frac {a \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) b}\)	\(257\)

Input:

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

Output:

(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*l 
n(exp(I*x)+(I*a*(-a^2+b^2)^(1/2)+a^2-b^2)/(-a^2+b^2)^(1/2)/b)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (75) = 150\).

Time = 0.09 (sec) , antiderivative size = 459, normalized size of antiderivative = 5.40 \[ \int \frac {x \cos (x)}{(a+b \sin (x))^3} \, dx=\left [\frac {2 \, {\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) - {\left (a b^{2} \cos \left (x\right )^{2} - 2 \, a^{2} b \sin \left (x\right ) - a^{3} - a b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right )}{4 \, {\left (a^{6} b - a^{4} b^{3} - a^{2} b^{5} + b^{7} - {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \sin \left (x\right )\right )}}, \frac {{\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right ) \sin \left (x\right ) + {\left (a b^{2} \cos \left (x\right )^{2} - 2 \, a^{2} b \sin \left (x\right ) - a^{3} - a b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x + {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right )}{2 \, {\left (a^{6} b - a^{4} b^{3} - a^{2} b^{5} + b^{7} - {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (x\right )^{2} + 2 \, {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} \sin \left (x\right )\right )}}\right ] \] Input:

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

Output:

[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)*si 
n(x))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(x*cos(x)/(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
 

Giac [F]

\[ \int \frac {x \cos (x)}{(a+b \sin (x))^3} \, dx=\int { \frac {x \cos \left (x\right )}{{\left (b \sin \left (x\right ) + a\right )}^{3}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x \cos (x)}{(a+b \sin (x))^3} \, dx=\int \frac {x\,\cos \left (x\right )}{{\left (a+b\,\sin \left (x\right )\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 321, normalized size of antiderivative = 3.78 \[ \int \frac {x \cos (x)}{(a+b \sin (x))^3} \, dx=\frac {4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right )^{2} a^{2} b^{2}+8 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) a^{3} b +4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4}+2 \cos \left (x \right ) \sin \left (x \right ) a^{3} b^{2}-2 \cos \left (x \right ) \sin \left (x \right ) a \,b^{4}+2 \cos \left (x \right ) a^{4} b -2 \cos \left (x \right ) a^{2} b^{3}+\sin \left (x \right )^{2} a^{2} b^{3}-\sin \left (x \right )^{2} b^{5}+2 \sin \left (x \right ) a^{3} b^{2}-2 \sin \left (x \right ) a \,b^{4}-2 a^{5} x +a^{4} b +4 a^{3} b^{2} x -a^{2} b^{3}-2 a \,b^{4} x}{4 a b \left (\sin \left (x \right )^{2} a^{4} b^{2}-2 \sin \left (x \right )^{2} a^{2} b^{4}+\sin \left (x \right )^{2} b^{6}+2 \sin \left (x \right ) a^{5} b -4 \sin \left (x \right ) a^{3} b^{3}+2 \sin \left (x \right ) a \,b^{5}+a^{6}-2 a^{4} b^{2}+a^{2} b^{4}\right )} \] Input:

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

Output:

(4*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)**2*a* 
*2*b**2 + 8*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin 
(x)*a**3*b + 4*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))* 
a**4 + 2*cos(x)*sin(x)*a**3*b**2 - 2*cos(x)*sin(x)*a*b**4 + 2*cos(x)*a**4* 
b - 2*cos(x)*a**2*b**3 + sin(x)**2*a**2*b**3 - sin(x)**2*b**5 + 2*sin(x)*a 
**3*b**2 - 2*sin(x)*a*b**4 - 2*a**5*x + a**4*b + 4*a**3*b**2*x - a**2*b**3 
 - 2*a*b**4*x)/(4*a*b*(sin(x)**2*a**4*b**2 - 2*sin(x)**2*a**2*b**4 + sin(x 
)**2*b**6 + 2*sin(x)*a**5*b - 4*sin(x)*a**3*b**3 + 2*sin(x)*a*b**5 + a**6 
- 2*a**4*b**2 + a**2*b**4))