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

________________________________________________________________________________________

Rubi [A]  time = 0.118366, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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]

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]

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*}

Mathematica [C]  time = 0.24657, size = 121, normalized size = 1.89 \[ \frac{\sin (x) \left (-a^2 b x+a^3+a b^2 (1-2 i x)-a b^2 \log \left ((a \cos (x)+b \sin (x))^2\right )+b^3 x\right )-a \cos (x) \left (a b \log \left ((a \cos (x)+b \sin (x))^2\right )+x (a+i b)^2\right )+2 i a b \tan ^{-1}(\tan (x)) (a \cos (x)+b \sin (x))}{\left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))} \]

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]))

________________________________________________________________________________________

Maple [A]  time = 0.087, size = 99, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) b \left ( a+b\tan \left ( x \right ) \right ) }}-2\,{\frac{ab\ln \left ( a+b\tan \left ( x \right ) \right ) }{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\arctan \left ( \tan \left ( x \right ) \right ){a}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( x \right ) \right ){b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{ab\ln \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) }{ \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

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

________________________________________________________________________________________

Maxima [A]  time = 1.69264, size = 158, normalized size = 2.47 \begin{align*} -\frac{2 \, a b \log \left (b \tan \left (x\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{a b \log \left (\tan \left (x\right )^{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 \left (x\right )} - \frac{{\left (a^{2} - b^{2}\right )} x}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} \end{align*}

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)

________________________________________________________________________________________

Fricas [B]  time = 0.508699, size = 305, normalized size = 4.77 \begin{align*} -\frac{{\left (a^{2} b +{\left (a^{3} - a b^{2}\right )} x\right )} \cos \left (x\right ) +{\left (a^{2} b \cos \left (x\right ) + a b^{2} \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 ) -{\left (a^{3} -{\left (a^{2} b - b^{3}\right )} x\right )} \sin \left (x\right )}{{\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 )} \end{align*}

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))

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [B]  time = 1.11171, size = 188, normalized size = 2.94 \begin{align*} -\frac{2 \, a b^{2} \log \left ({\left | b \tan \left (x\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac{a b \log \left (\tan \left (x\right )^{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 \left (x\right ) - a^{4} + a^{2} b^{2}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )}{\left (b \tan \left (x\right ) + a\right )}} \end{align*}

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))