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

Optimal. Leaf size=68 \[ -\frac{a \sin (x)}{a^2+b^2}-\frac{b \cos (x)}{a^2+b^2}-\frac{a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0778495, antiderivative size = 68, 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 = {3099, 3074, 206, 2638} \[ -\frac{a \sin (x)}{a^2+b^2}-\frac{b \cos (x)}{a^2+b^2}-\frac{a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a*Cos[x] + b*Sin[x]),x]

[Out]

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

Rule 3099

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

Rule 3074

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx &=-\frac{a \sin (x)}{a^2+b^2}+\frac{a^2 \int \frac{1}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2}+\frac{b \int \sin (x) \, dx}{a^2+b^2}\\ &=-\frac{b \cos (x)}{a^2+b^2}-\frac{a \sin (x)}{a^2+b^2}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{a^2+b^2}\\ &=-\frac{a^2 \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac{b \cos (x)}{a^2+b^2}-\frac{a \sin (x)}{a^2+b^2}\\ \end{align*}

Mathematica [A]  time = 0.155249, size = 62, normalized size = 0.91 \[ \frac{2 a^2 \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac{a \sin (x)+b \cos (x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.081, size = 84, normalized size = 1.2 \begin{align*} 2\,{\frac{-a\tan \left ( x/2 \right ) -b}{ \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+8\,{\frac{{a}^{2}}{ \left ( 4\,{a}^{2}+4\,{b}^{2} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a^2+b^2)*(-a*tan(1/2*x)-b)/(tan(1/2*x)^2+1)+8*a^2/(4*a^2+4*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*x)
-2*b)/(a^2+b^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.515675, size = 350, normalized size = 5.15 \begin{align*} \frac{\sqrt{a^{2} + b^{2}} a^{2} \log \left (-\frac{2 \, a b \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{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 (a^{2} b + b^{3}\right )} \cos \left (x\right ) - 2 \,{\left (a^{3} + a b^{2}\right )} \sin \left (x\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)),x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.25161, size = 127, normalized size = 1.87 \begin{align*} -\frac{a^{2} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (a \tan \left (\frac{1}{2} \, x\right ) + b\right )}}{{\left (a^{2} + b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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