\(\int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx\) [284]

Optimal result

Integrand size = 16, antiderivative size = 70 \[ \int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 a b x}{\left (a^2+b^2\right )^2}-\frac {\left (a^2-b^2\right ) \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2}-\frac {b \sin (x)}{\left (a^2+b^2\right ) (a \cos (x)+b \sin (x))} \]

[Out]

2*a*b*x/(a^2+b^2)^2-(a^2-b^2)*ln(a*cos(x)+b*sin(x))/(a^2+b^2)^2-b*sin(x)/(a^2+b^2)/(a*cos(x)+b*sin(x))

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3190, 3177, 3212, 3176, 3154} \[ \int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 a b x}{\left (a^2+b^2\right )^2}-\frac {b \sin (x)}{\left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac {a^2 \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2}+\frac {b^2 \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

[In]

Int[(Cos[x]*Sin[x])/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

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

Rule 3154

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-2), x_Symbol] :> Simp[Sin[c + d*x]/(a*d*
(a*Cos[c + d*x] + b*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3176

Int[sin[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[b*(x/(a^2 + b^2)), x] - Dist[a/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3177

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[a*(x/(a^2 + b^2)), x] + Dist[b/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3190

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1)*(a*Cos[c + d*
x] + b*Sin[c + d*x])^(p + 1), x], x] + (Dist[a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n*(a*Cos[c +
d*x] + b*Sin[c + d*x])^(p + 1), x], x] - Dist[a*(b/(a^2 + b^2)), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^(n - 1
)*(a*Cos[c + d*x] + b*Sin[c + d*x])^p, x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 0] &&
IGtQ[n, 0] && ILtQ[p, 0]

Rule 3212

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[(b*B + c*C)*(x/(b^2 + c^2)), x] + Simp[(c*B - b*C)*(L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/(e*(b^2 + c^2))), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.06 \[ \int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {a \cos (x) \left (-2 i (a+i b)^2 x+\left (-a^2+b^2\right ) \log \left ((a \cos (x)+b \sin (x))^2\right )\right )+b \left (2 (a+i b) (a (-1-i x)+b (i+x))+\left (-a^2+b^2\right ) \log \left ((a \cos (x)+b \sin (x))^2\right )\right ) \sin (x)+2 i \left (a^2-b^2\right ) \arctan (\tan (x)) (a \cos (x)+b \sin (x))}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

[In]

Integrate[(Cos[x]*Sin[x])/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.20

method	result	size
default	\(\frac {a}{\left (a^{2}+b^{2}\right ) \left (a +b \tan \left (x \right )\right )}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (x \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (x \right )^{2}\right )}{2}+2 a b \arctan \left (\tan \left (x \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}\)	\(84\)
parallelrisch	\(\frac {\left (\left (-a^{4}+b^{4}\right ) \cos \left (2 x \right )-\left (a -b \right )^{2} \left (a +b \right )^{2}\right ) \ln \left (\frac {-a \cos \left (x \right )-b \sin \left (x \right )}{\cos \left (x \right )+1}\right )+\left (\left (a^{4}-b^{4}\right ) \cos \left (2 x \right )+\left (a -b \right )^{2} \left (a +b \right )^{2}\right ) \ln \left (\frac {1}{\cos \left (x \right )+1}\right )+2 b \left (\left (a^{2}+b^{2}\right ) \left (a x -\frac {b}{2}\right ) \cos \left (2 x \right )+\frac {\left (-a^{3}-a \,b^{2}\right ) \sin \left (2 x \right )}{2}+a^{3} x -a \,b^{2} x +\frac {a^{2} b}{2}+\frac {b^{3}}{2}\right )}{\left (a^{2}-b^{2}+\cos \left (2 x \right ) \left (a^{2}+b^{2}\right )\right ) \left (a^{2}+b^{2}\right )^{2}}\)	\(184\)
risch	\(\frac {i x}{2 i b a -a^{2}+b^{2}}+\frac {2 i x \,a^{2}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {2 i x \,b^{2}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {2 i a b}{\left (i b +a \right ) \left (-i b +a \right )^{2} \left (-i b \,{\mathrm e}^{2 i x}+a \,{\mathrm e}^{2 i x}+i b +a \right )}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i b +a}{i b -a}\right ) a^{2}}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i b +a}{i b -a}\right ) b^{2}}{a^{4}+2 a^{2} b^{2}+b^{4}}\)	\(208\)
norman	\(\frac {-\frac {2 a^{2} b x}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {4 a \,b^{2} x \tan \left (\frac {x}{2}\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {8 a \,b^{2} x \tan \left (\frac {x}{2}\right )^{3}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {4 a \,b^{2} x \tan \left (\frac {x}{2}\right )^{5}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {2 a^{2} b x \tan \left (\frac {x}{2}\right )^{2}}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {2 a^{2} b x \tan \left (\frac {x}{2}\right )^{4}}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {2 a^{2} b x \tan \left (\frac {x}{2}\right )^{6}}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {2 b \tan \left (\frac {x}{2}\right )}{a^{2}+b^{2}}+\frac {4 b \tan \left (\frac {x}{2}\right )^{3}}{a^{2}+b^{2}}+\frac {2 b \tan \left (\frac {x}{2}\right )^{5}}{a^{2}+b^{2}}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {x}{2}\right )^{2} a -2 b \tan \left (\frac {x}{2}\right )-a \right )}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2} a -2 b \tan \left (\frac {x}{2}\right )-a \right )}{a^{4}+2 a^{2} b^{2}+b^{4}}\)	\(373\)

[In]

int(cos(x)*sin(x)/(a*cos(x)+b*sin(x))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.97 \[ \int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 \, {\left (2 \, a^{2} b x + a b^{2}\right )} \cos \left (x\right ) - {\left ({\left (a^{3} - a b^{2}\right )} \cos \left (x\right ) + {\left (a^{2} b - b^{3}\right )} \sin \left (x\right )\right )} \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}\right ) + 2 \, {\left (2 \, a b^{2} x - a^{2} b\right )} \sin \left (x\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right ) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )}} \]

[In]

integrate(cos(x)*sin(x)/(a*cos(x)+b*sin(x))^2,x, algorithm="fricas")

[Out]

1/2*(2*(2*a^2*b*x + a*b^2)*cos(x) - ((a^3 - a*b^2)*cos(x) + (a^2*b - b^3)*sin(x))*log(2*a*b*cos(x)*sin(x) + (a
^2 - b^2)*cos(x)^2 + b^2) + 2*(2*a*b^2*x - a^2*b)*sin(x))/((a^5 + 2*a^3*b^2 + a*b^4)*cos(x) + (a^4*b + 2*a^2*b
^3 + b^5)*sin(x))

Sympy [F(-2)]

Exception generated. \[ \int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\text {Exception raised: AttributeError} \]

[In]

integrate(cos(x)*sin(x)/(a*cos(x)+b*sin(x))**2,x)

[Out]

Exception raised: AttributeError >> 'NoneType' object has no attribute 'primitive'

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.69 \[ \int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 \, a b x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (b \tan \left (x\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {a}{a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \tan \left (x\right )} \]

[In]

integrate(cos(x)*sin(x)/(a*cos(x)+b*sin(x))^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (70) = 140\).

Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.06 \[ \int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\frac {2 \, a b x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac {{\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \tan \left (x\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac {a^{2} b \tan \left (x\right ) - b^{3} \tan \left (x\right ) + 2 \, a^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \tan \left (x\right ) + a\right )}} \]

[In]

integrate(cos(x)*sin(x)/(a*cos(x)+b*sin(x))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 27.76 (sec) , antiderivative size = 1017, normalized size of antiderivative = 14.53 \[ \int \frac {\cos (x) \sin (x)}{(a \cos (x)+b \sin (x))^2} \, dx=\text {Too large to display} \]

[In]

int((cos(x)*sin(x))/(a*cos(x) + b*sin(x))^2,x)

[Out]

-(b^3*sin(x) + a^3*log((a*cos(x) + b*sin(x))/cos(x/2)^2)*cos(x) - b^3*log((a*cos(x) + b*sin(x))/cos(x/2)^2)*si
n(x) + a^2*b*sin(x) - a^3*log(-(65536*a^4*b^10 - 131072*a^6*b^8 + 196608*a^8*b^6 - 131072*a^10*b^4 + 65536*a^1
2*b^2)/(a^16/2 + b^16/2 + 4*a^2*b^14 + 14*a^4*b^12 + 28*a^6*b^10 + 35*a^8*b^8 + 28*a^10*b^6 + 14*a^12*b^4 + 4*
a^14*b^2 + (a^16*cos(x))/2 + (b^16*cos(x))/2 + 4*a^2*b^14*cos(x) + 14*a^4*b^12*cos(x) + 28*a^6*b^10*cos(x) + 3
5*a^8*b^8*cos(x) + 28*a^10*b^6*cos(x) + 14*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*cos(x) + b^3*log(-(65536*a^4*
b^10 - 131072*a^6*b^8 + 196608*a^8*b^6 - 131072*a^10*b^4 + 65536*a^12*b^2)/(a^16/2 + b^16/2 + 4*a^2*b^14 + 14*
a^4*b^12 + 28*a^6*b^10 + 35*a^8*b^8 + 28*a^10*b^6 + 14*a^12*b^4 + 4*a^14*b^2 + (a^16*cos(x))/2 + (b^16*cos(x))
/2 + 4*a^2*b^14*cos(x) + 14*a^4*b^12*cos(x) + 28*a^6*b^10*cos(x) + 35*a^8*b^8*cos(x) + 28*a^10*b^6*cos(x) + 14
*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*sin(x) - 4*a^2*b*atan(sin(x/2)/cos(x/2))*cos(x) - 4*a*b^2*atan(sin(x/2)
/cos(x/2))*sin(x) + a*b^2*log(-(65536*a^4*b^10 - 131072*a^6*b^8 + 196608*a^8*b^6 - 131072*a^10*b^4 + 65536*a^1
2*b^2)/(a^16/2 + b^16/2 + 4*a^2*b^14 + 14*a^4*b^12 + 28*a^6*b^10 + 35*a^8*b^8 + 28*a^10*b^6 + 14*a^12*b^4 + 4*
a^14*b^2 + (a^16*cos(x))/2 + (b^16*cos(x))/2 + 4*a^2*b^14*cos(x) + 14*a^4*b^12*cos(x) + 28*a^6*b^10*cos(x) + 3
5*a^8*b^8*cos(x) + 28*a^10*b^6*cos(x) + 14*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*cos(x) - a^2*b*log(-(65536*a^
4*b^10 - 131072*a^6*b^8 + 196608*a^8*b^6 - 131072*a^10*b^4 + 65536*a^12*b^2)/(a^16/2 + b^16/2 + 4*a^2*b^14 + 1
4*a^4*b^12 + 28*a^6*b^10 + 35*a^8*b^8 + 28*a^10*b^6 + 14*a^12*b^4 + 4*a^14*b^2 + (a^16*cos(x))/2 + (b^16*cos(x
))/2 + 4*a^2*b^14*cos(x) + 14*a^4*b^12*cos(x) + 28*a^6*b^10*cos(x) + 35*a^8*b^8*cos(x) + 28*a^10*b^6*cos(x) +
14*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*sin(x) - a*b^2*log((a*cos(x) + b*sin(x))/cos(x/2)^2)*cos(x) + a^2*b*l
og((a*cos(x) + b*sin(x))/cos(x/2)^2)*sin(x))/(b^5*sin(x) + a^5*cos(x) + a*b^4*cos(x) + a^4*b*sin(x) + 2*a^3*b^
2*cos(x) + 2*a^2*b^3*sin(x))