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

Optimal. Leaf size=64 \[ -\frac {x \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2}+\frac {a}{\left (a^2+b^2\right ) (a \cot (x)+b)}-\frac {2 a b \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3085, 3483, 3531, 3530} \[ -\frac {x \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2}+\frac {a}{\left (a^2+b^2\right ) (a \cot (x)+b)}-\frac {2 a b \log (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3085

Int[sin[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symb
ol] :> Int[(b + a*Cot[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b
^2, 0]

Rule 3483

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rubi steps

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

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Mathematica [C]  time = 0.26, size = 121, normalized size = 1.89 \[ \frac {\sin (x) \left (a^3-a^2 b x+a b^2 (1-2 i x)-a b^2 \log \left ((a \cos (x)+b \sin (x))^2\right )+b^3 x\right )-a \cos (x) \left (a b \log \left ((a \cos (x)+b \sin (x))^2\right )+x (a+i b)^2\right )+2 i a b \tan ^{-1}(\tan (x)) (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.71, size = 132, normalized size = 2.06 \[ -\frac {{\left (a^{2} b + {\left (a^{3} - a b^{2}\right )} x\right )} \cos \relax (x) + {\left (a^{2} b \cos \relax (x) + a b^{2} \sin \relax (x)\right )} \log \left (2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}\right ) - {\left (a^{3} - {\left (a^{2} b - b^{3}\right )} x\right )} \sin \relax (x)}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \relax (x) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2*b + (a^3 - a*b^2)*x)*cos(x) + (a^2*b*cos(x) + a*b^2*sin(x))*log(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x
)^2 + b^2) - (a^3 - (a^2*b - b^3)*x)*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))

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giac [B]  time = 1.98, size = 139, normalized size = 2.17 \[ -\frac {2 \, a b^{2} \log \left ({\left | b \tan \relax (x) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac {a b \log \left (\tan \relax (x)^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a^{2} - b^{2}\right )} x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, a b^{3} \tan \relax (x) - a^{4} + a^{2} b^{2}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} {\left (b \tan \relax (x) + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.56, size = 99, normalized size = 1.55 \[ -\frac {a^{2}}{\left (a^{2}+b^{2}\right ) b \left (a +b \tan \relax (x )\right )}-\frac {2 a b \ln \left (a +b \tan \relax (x )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {a b \ln \left (1+\tan ^{2}\relax (x )\right )}{\left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \relax (x )\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {\arctan \left (\tan \relax (x )\right ) b^{2}}{\left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 117, normalized size = 1.83 \[ -\frac {2 \, a b \log \left (b \tan \relax (x) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a b \log \left (\tan \relax (x)^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {a^{2}}{a^{3} b + a b^{3} + {\left (a^{2} b^{2} + b^{4}\right )} \tan \relax (x)} - \frac {{\left (a^{2} - b^{2}\right )} x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.06, size = 626, normalized size = 9.78 \[ \frac {a^3\,\sin \relax (x)+a\,b^2\,\sin \relax (x)-2\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\cos \relax (x)+2\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sin \relax (x)+2\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\cos \relax (x)-2\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sin \relax (x)+2\,a^2\,b\,\cos \relax (x)\,\ln \left (\frac {1024\,a^{14}+26624\,a^{12}\,b^2+146432\,a^{10}\,b^4-348160\,a^8\,b^6+146432\,a^6\,b^8+26624\,a^4\,b^{10}+1024\,a^2\,b^{12}}{\frac {a^{16}}{2}+\frac {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+\frac {a^{16}\,\cos \relax (x)}{2}+\frac {b^{16}\,\cos \relax (x)}{2}+4\,a^2\,b^{14}\,\cos \relax (x)+14\,a^4\,b^{12}\,\cos \relax (x)+28\,a^6\,b^{10}\,\cos \relax (x)+35\,a^8\,b^8\,\cos \relax (x)+28\,a^{10}\,b^6\,\cos \relax (x)+14\,a^{12}\,b^4\,\cos \relax (x)+4\,a^{14}\,b^2\,\cos \relax (x)}\right )+2\,a\,b^2\,\ln \left (\frac {1024\,a^{14}+26624\,a^{12}\,b^2+146432\,a^{10}\,b^4-348160\,a^8\,b^6+146432\,a^6\,b^8+26624\,a^4\,b^{10}+1024\,a^2\,b^{12}}{\frac {a^{16}}{2}+\frac {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+\frac {a^{16}\,\cos \relax (x)}{2}+\frac {b^{16}\,\cos \relax (x)}{2}+4\,a^2\,b^{14}\,\cos \relax (x)+14\,a^4\,b^{12}\,\cos \relax (x)+28\,a^6\,b^{10}\,\cos \relax (x)+35\,a^8\,b^8\,\cos \relax (x)+28\,a^{10}\,b^6\,\cos \relax (x)+14\,a^{12}\,b^4\,\cos \relax (x)+4\,a^{14}\,b^2\,\cos \relax (x)}\right )\,\sin \relax (x)-2\,a^2\,b\,\ln \left (\frac {a\,\cos \relax (x)+b\,\sin \relax (x)}{{\cos \left (\frac {x}{2}\right )}^2}\right )\,\cos \relax (x)-2\,a\,b^2\,\ln \left (\frac {a\,\cos \relax (x)+b\,\sin \relax (x)}{{\cos \left (\frac {x}{2}\right )}^2}\right )\,\sin \relax (x)}{\cos \relax (x)\,a^5+\sin \relax (x)\,a^4\,b+2\,\cos \relax (x)\,a^3\,b^2+2\,\sin \relax (x)\,a^2\,b^3+\cos \relax (x)\,a\,b^4+\sin \relax (x)\,b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*sin(x) + a*b^2*sin(x) - 2*a^3*atan(sin(x/2)/cos(x/2))*cos(x) + 2*b^3*atan(sin(x/2)/cos(x/2))*sin(x) + 2*a
*b^2*atan(sin(x/2)/cos(x/2))*cos(x) - 2*a^2*b*atan(sin(x/2)/cos(x/2))*sin(x) + 2*a^2*b*cos(x)*log((1024*a^14 +
1024*a^2*b^12 + 26624*a^4*b^10 + 146432*a^6*b^8 - 348160*a^8*b^6 + 146432*a^10*b^4 + 26624*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^1
6*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))) + 2*a*b^2*log((1024*a^14 + 1024*a^2*b^12 + 2
6624*a^4*b^10 + 146432*a^6*b^8 - 348160*a^8*b^6 + 146432*a^10*b^4 + 26624*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^1
6*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*co
s(x) + 14*a^12*b^4*cos(x) + 4*a^14*b^2*cos(x)))*sin(x) - 2*a^2*b*log((a*cos(x) + b*sin(x))/cos(x/2)^2)*cos(x)
- 2*a*b^2*log((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))

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sympy [A]  time = 2.20, size = 1017, normalized size = 15.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (2*x*sin(x)**2/(8*b**2*sin(x)**2 - 16*I*b**2*sin(x)*cos(x) - 8*b**2*co
s(x)**2) - 4*I*x*sin(x)*cos(x)/(8*b**2*sin(x)**2 - 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2) - 2*x*cos(x)**2
/(8*b**2*sin(x)**2 - 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**2) + 3*I*sin(x)**2/(8*b**2*sin(x)**2 - 16*I*b**2
*sin(x)*cos(x) - 8*b**2*cos(x)**2) + I*cos(x)**2/(8*b**2*sin(x)**2 - 16*I*b**2*sin(x)*cos(x) - 8*b**2*cos(x)**
2), Eq(a, -I*b)), (-2*I*x*sin(x)**2/(-8*I*b**2*sin(x)**2 + 16*b**2*sin(x)*cos(x) + 8*I*b**2*cos(x)**2) + 4*x*s
in(x)*cos(x)/(-8*I*b**2*sin(x)**2 + 16*b**2*sin(x)*cos(x) + 8*I*b**2*cos(x)**2) + 2*I*x*cos(x)**2/(-8*I*b**2*s
in(x)**2 + 16*b**2*sin(x)*cos(x) + 8*I*b**2*cos(x)**2) - 3*sin(x)**2/(-8*I*b**2*sin(x)**2 + 16*b**2*sin(x)*cos
(x) + 8*I*b**2*cos(x)**2) - cos(x)**2/(-8*I*b**2*sin(x)**2 + 16*b**2*sin(x)*cos(x) + 8*I*b**2*cos(x)**2), Eq(a
, I*b)), ((-x + sin(x)/cos(x))/a**2, Eq(b, 0)), (-a**4*cos(x)/(a**5*b*cos(x) + a**4*b**2*sin(x) + 2*a**3*b**3*
cos(x) + 2*a**2*b**4*sin(x) + a*b**5*cos(x) + b**6*sin(x)) - a**3*b*x*cos(x)/(a**5*b*cos(x) + a**4*b**2*sin(x)
+ 2*a**3*b**3*cos(x) + 2*a**2*b**4*sin(x) + a*b**5*cos(x) + b**6*sin(x)) - a**2*b**2*x*sin(x)/(a**5*b*cos(x)
+ a**4*b**2*sin(x) + 2*a**3*b**3*cos(x) + 2*a**2*b**4*sin(x) + a*b**5*cos(x) + b**6*sin(x)) - 2*a**2*b**2*log(
a*cos(x)/b + sin(x))*cos(x)/(a**5*b*cos(x) + a**4*b**2*sin(x) + 2*a**3*b**3*cos(x) + 2*a**2*b**4*sin(x) + a*b*
*5*cos(x) + b**6*sin(x)) - a**2*b**2*cos(x)/(a**5*b*cos(x) + a**4*b**2*sin(x) + 2*a**3*b**3*cos(x) + 2*a**2*b*
*4*sin(x) + a*b**5*cos(x) + b**6*sin(x)) + a*b**3*x*cos(x)/(a**5*b*cos(x) + a**4*b**2*sin(x) + 2*a**3*b**3*cos
(x) + 2*a**2*b**4*sin(x) + a*b**5*cos(x) + b**6*sin(x)) - 2*a*b**3*log(a*cos(x)/b + sin(x))*sin(x)/(a**5*b*cos
(x) + a**4*b**2*sin(x) + 2*a**3*b**3*cos(x) + 2*a**2*b**4*sin(x) + a*b**5*cos(x) + b**6*sin(x)) + b**4*x*sin(x
)/(a**5*b*cos(x) + a**4*b**2*sin(x) + 2*a**3*b**3*cos(x) + 2*a**2*b**4*sin(x) + a*b**5*cos(x) + b**6*sin(x)),
True))

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