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

Optimal. Leaf size=72 \[ \frac{a x \left (a^2+3 b^2\right )}{2 \left (a^2+b^2\right )^2}-\frac{\sin ^2(x) (a \cot (x)+b)}{2 \left (a^2+b^2\right )}-\frac{b^3 \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.129704, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3506, 741, 801, 635, 203, 260} \[ \frac{a x \left (a^2+3 b^2\right )}{2 \left (a^2+b^2\right )^2}-\frac{\sin ^2(x) (a \cot (x)+b)}{2 \left (a^2+b^2\right )}-\frac{b^3 \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Cot[x]),x]

[Out]

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

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [C]  time = 0.172764, size = 94, normalized size = 1.31 \[ \frac{b \left (a^2+b^2\right ) \cos (2 x)+2 a^3 x-a^3 \sin (2 x)+6 a b^2 x-a b^2 \sin (2 x)-2 b^3 \log \left ((a \sin (x)+b \cos (x))^2\right )-4 i b^3 x+4 i b^3 \tan ^{-1}(\tan (x))}{4 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + b*Cot[x]),x]

[Out]

(2*a^3*x + 6*a*b^2*x - (4*I)*b^3*x + (4*I)*b^3*ArcTan[Tan[x]] + b*(a^2 + b^2)*Cos[2*x] - 2*b^3*Log[(b*Cos[x] +
a*Sin[x])^2] - a^3*Sin[2*x] - a*b^2*Sin[2*x])/(4*(a^2 + b^2)^2)

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Maple [B]  time = 0.12, size = 173, normalized size = 2.4 \begin{align*} -{\frac{\tan \left ( x \right ){a}^{3}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }}-{\frac{a\tan \left ( x \right ){b}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{b{a}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{3}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{3}\ln \left ( 1+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{3\,\arctan \left ( \tan \left ( x \right ) \right ) a{b}^{2}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( x \right ) \right ){a}^{3}}{2\, \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{b}^{3}\ln \left ( a\tan \left ( x \right ) +b \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+b*cot(x)),x)

[Out]

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

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Maxima [A]  time = 1.83733, size = 162, normalized size = 2.25 \begin{align*} -\frac{b^{3} \log \left (a \tan \left (x\right ) + b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{b^{3} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac{{\left (a^{3} + 3 \, a b^{2}\right )} x}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac{a \tan \left (x\right ) - b}{2 \,{\left ({\left (a^{2} + b^{2}\right )} \tan \left (x\right )^{2} + a^{2} + b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.77594, size = 223, normalized size = 3.1 \begin{align*} -\frac{b^{3} \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}\right ) -{\left (a^{2} b + b^{3}\right )} \cos \left (x\right )^{2} +{\left (a^{3} + a b^{2}\right )} \cos \left (x\right ) \sin \left (x\right ) -{\left (a^{3} + 3 \, a b^{2}\right )} x}{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+b*cot(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{2}{\left (x \right )}}{a + b \cot{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(x)**2/(a + b*cot(x)), x)

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Giac [B]  time = 1.32006, size = 200, normalized size = 2.78 \begin{align*} -\frac{a b^{3} \log \left ({\left | a \tan \left (x\right ) + b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac{b^{3} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac{{\left (a^{3} + 3 \, a b^{2}\right )} x}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac{b^{3} \tan \left (x\right )^{2} + a^{3} \tan \left (x\right ) + a b^{2} \tan \left (x\right ) - a^{2} b}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (x\right )^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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