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))/(a^2+b^2)^(1/2))/(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 = 0.08, antiderivative size = 68, 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 = {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 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]

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 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]

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*}

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Mathematica [A]  time = 0.16, 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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fricas [B]  time = 0.48, size = 144, normalized size = 2.12 \[ \frac {\sqrt {a^{2} + b^{2}} a^{2} \log \left (-\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) - 2 \, {\left (a^{2} b + b^{3}\right )} \cos \relax (x) - 2 \, {\left (a^{3} + a b^{2}\right )} \sin \relax (x)}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \]

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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giac [A]  time = 0.26, size = 94, normalized size = 1.38 \[ -\frac {a^{2} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, x\right ) + b\right )}}{{\left (a^{2} + b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}} \]

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))

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maple [A]  time = 0.49, size = 84, normalized size = 1.24 \[ \frac {8 a^{2} \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (4 a^{2}+4 b^{2}\right ) \sqrt {a^{2}+b^{2}}}+\frac {-2 a \tan \left (\frac {x}{2}\right )-2 b}{\left (a^{2}+b^{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 106, normalized size = 1.56 \[ -\frac {a^{2} \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b + \frac {a \sin \relax (x)}{\cos \relax (x) + 1}\right )}}{a^{2} + b^{2} + \frac {{\left (a^{2} + b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}} \]

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]

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

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mupad [B]  time = 0.58, size = 94, normalized size = 1.38 \[ -\frac {\frac {2\,b}{a^2+b^2}+\frac {2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2+b^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {2\,a^2\,\mathrm {atanh}\left (\frac {2\,a^2\,b+2\,b^3-2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+b^2\right )}{2\,{\left (a^2+b^2\right )}^{3/2}}\right )}{{\left (a^2+b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a*cos(x) + b*sin(x)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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