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

Optimal. Leaf size=73 \[ \frac {a x \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2}+\frac {\sin ^2(x) (a \cot (x)+b)}{2 \left (a^2+b^2\right )}-\frac {a^2 b \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^2} \]

[Out]

1/2*a*(a^2-b^2)*x/(a^2+b^2)^2-a^2*b*ln(b*cos(x)+a*sin(x))/(a^2+b^2)^2+1/2*(b+a*cot(x))*sin(x)^2/(a^2+b^2)

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Rubi [A]  time = 0.16, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 = {3516, 1647, 801, 635, 203, 260} \[ \frac {a x \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2}+\frac {\sin ^2(x) (a \cot (x)+b)}{2 \left (a^2+b^2\right )}-\frac {a^2 b \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2/(a + b*Cot[x]),x]

[Out]

(a*(a^2 - b^2)*x)/(2*(a^2 + b^2)^2) - (a^2*b*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 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]

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 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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 3516

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

Rubi steps

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

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Mathematica [C]  time = 0.32, size = 82, normalized size = 1.12 \[ \frac {-b \left (a^2+b^2\right ) \cos (2 x)+a \left (\left (a^2+b^2\right ) \sin (2 x)+2 x (a-i b)^2-2 a b \log \left ((a \sin (x)+b \cos (x))^2\right )\right )+4 i a^2 b \tan ^{-1}(\tan (x))}{4 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^2/(a + b*Cot[x]),x]

[Out]

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

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fricas [A]  time = 0.64, size = 95, normalized size = 1.30 \[ -\frac {a^{2} b \log \left (2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2}\right ) + {\left (a^{2} b + b^{3}\right )} \cos \relax (x)^{2} - {\left (a^{3} + a b^{2}\right )} \cos \relax (x) \sin \relax (x) - {\left (a^{3} - a b^{2}\right )} x}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.38, size = 155, normalized size = 2.12 \[ -\frac {a^{3} b \log \left ({\left | a \tan \relax (x) + b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac {a^{2} b \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {{\left (a^{3} - a b^{2}\right )} x}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac {a^{2} b \tan \relax (x)^{2} - a^{3} \tan \relax (x) - a b^{2} \tan \relax (x) + 2 \, a^{2} b + b^{3}}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.27, size = 175, normalized size = 2.40 \[ -\frac {a^{2} b \ln \left (a \tan \relax (x )+b \right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {\tan \relax (x ) a^{3}}{2 \left (a^{2}+b^{2}\right )^{2} \left (1+\tan ^{2}\relax (x )\right )}+\frac {\tan \relax (x ) b^{2} a}{2 \left (a^{2}+b^{2}\right )^{2} \left (1+\tan ^{2}\relax (x )\right )}-\frac {a^{2} b}{2 \left (a^{2}+b^{2}\right )^{2} \left (1+\tan ^{2}\relax (x )\right )}-\frac {b^{3}}{2 \left (a^{2}+b^{2}\right )^{2} \left (1+\tan ^{2}\relax (x )\right )}+\frac {\arctan \left (\tan \relax (x )\right ) a^{3}}{2 \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \relax (x )\right ) b^{2} a}{2 \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\relax (x )\right ) a^{2} b}{2 \left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.72, size = 122, normalized size = 1.67 \[ -\frac {a^{2} b \log \left (a \tan \relax (x) + b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a^{2} b \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {{\left (a^{3} - a b^{2}\right )} x}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {a \tan \relax (x) - b}{2 \, {\left ({\left (a^{2} + b^{2}\right )} \tan \relax (x)^{2} + a^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*b*log(a*tan(x) + b)/(a^4 + 2*a^2*b^2 + b^4) + 1/2*a^2*b*log(tan(x)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) + 1/2*(
a^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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mupad [B]  time = 0.37, size = 118, normalized size = 1.62 \[ -{\cos \relax (x)}^2\,\left (\frac {b}{2\,\left (a^2+b^2\right )}-\frac {a\,\mathrm {tan}\relax (x)}{2\,\left (a^2+b^2\right )}\right )+\frac {a\,\ln \left (\mathrm {tan}\relax (x)+1{}\mathrm {i}\right )}{4\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {a^2\,b\,\ln \left (b+a\,\mathrm {tan}\relax (x)\right )}{{\left (a^2+b^2\right )}^2}+\frac {a\,\ln \left (\mathrm {tan}\relax (x)-\mathrm {i}\right )\,1{}\mathrm {i}}{4\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/(a + b*cot(x)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\relax (x )}}{a + b \cot {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(x)**2/(a + b*cot(x)), x)

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