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

Optimal. Leaf size=61 \[ \frac {2 \sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a b}-\frac {x}{a}-\frac {\tanh ^{-1}(\cos (x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3894, 4051, 3770, 3919, 3831, 2660, 618, 206} \[ \frac {2 \sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a b}-\frac {x}{a}-\frac {\tanh ^{-1}(\cos (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^2/(a + b*Csc[x]),x]

[Out]

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

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3770

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3894

Int[cot[(c_.) + (d_.)*(x_)]^2*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(-1 + Csc[c + d*x]
^2)*(a + b*Csc[c + d*x])^n, x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4051

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[C/b, I
nt[Csc[e + f*x], x], x] + Dist[1/b, Int[(A*b - a*C*Csc[e + f*x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, A, C}, x]

Rubi steps

\begin {align*} \int \frac {\cot ^2(x)}{a+b \csc (x)} \, dx &=\int \frac {-1+\csc ^2(x)}{a+b \csc (x)} \, dx\\ &=\frac {\int \csc (x) \, dx}{b}+\frac {\int \frac {-b-a \csc (x)}{a+b \csc (x)} \, dx}{b}\\ &=-\frac {x}{a}-\frac {\tanh ^{-1}(\cos (x))}{b}-\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {\csc (x)}{a+b \csc (x)} \, dx\\ &=-\frac {x}{a}-\frac {\tanh ^{-1}(\cos (x))}{b}-\frac {\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {1}{1+\frac {a \sin (x)}{b}} \, dx}{b}\\ &=-\frac {x}{a}-\frac {\tanh ^{-1}(\cos (x))}{b}-\frac {\left (2 \left (\frac {a}{b}-\frac {b}{a}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {2 a x}{b}+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b}\\ &=-\frac {x}{a}-\frac {\tanh ^{-1}(\cos (x))}{b}+\frac {\left (4 \left (\frac {a}{b}-\frac {b}{a}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (1-\frac {a^2}{b^2}\right )-x^2} \, dx,x,\frac {2 a}{b}+2 \tan \left (\frac {x}{2}\right )\right )}{b}\\ &=-\frac {x}{a}-\frac {\tanh ^{-1}(\cos (x))}{b}+\frac {2 \sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}+\tan \left (\frac {x}{2}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a b}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 71, normalized size = 1.16 \[ \frac {2 \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )+a \log \left (\sin \left (\frac {x}{2}\right )\right )-a \log \left (\cos \left (\frac {x}{2}\right )\right )-b x}{a b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^2/(a + b*Csc[x]),x]

[Out]

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

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fricas [A]  time = 0.59, size = 206, normalized size = 3.38 \[ \left [-\frac {2 \, b x + a \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - a \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - \sqrt {a^{2} - b^{2}} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} + 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right )}{2 \, a b}, -\frac {2 \, b x + a \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - a \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - 2 \, \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right )}{2 \, a b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*b*x + a*log(1/2*cos(x) + 1/2) - a*log(-1/2*cos(x) + 1/2) - sqrt(a^2 - b^2)*log(((a^2 - 2*b^2)*cos(x)^
2 + 2*a*b*sin(x) + a^2 + b^2 + 2*(b*cos(x)*sin(x) + a*cos(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) -
a^2 - b^2)))/(a*b), -1/2*(2*b*x + a*log(1/2*cos(x) + 1/2) - a*log(-1/2*cos(x) + 1/2) - 2*sqrt(-a^2 + b^2)*arct
an(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x))))/(a*b)]

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giac [A]  time = 0.31, size = 80, normalized size = 1.31 \[ -\frac {x}{a} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{b} - \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} {\left (a^{2} - b^{2}\right )}}{\sqrt {-a^{2} + b^{2}} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/a + log(abs(tan(1/2*x)))/b - 2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2
)))*(a^2 - b^2)/(sqrt(-a^2 + b^2)*a*b)

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maple [A]  time = 0.35, size = 105, normalized size = 1.72 \[ -\frac {2 a \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b \sqrt {-a^{2}+b^{2}}}+\frac {2 b \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{b}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 0.53, size = 697, normalized size = 11.43 \[ \frac {2\,\mathrm {atan}\left (\frac {64\,b^3}{-64\,a^3-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+64\,a\,b^2+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}+\frac {64\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{-64\,a^3-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+64\,a\,b^2+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}-\frac {64\,a^2\,b}{-64\,a^3-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+64\,a\,b^2+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}-\frac {64\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{-64\,a^3-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+64\,a\,b^2+64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}\right )}{a}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{b}-\frac {2\,\mathrm {atanh}\left (\frac {512\,a^4\,\sqrt {a^2-b^2}}{256\,a\,b^4-64\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )+512\,a^5-768\,a^3\,b^2+832\,a^2\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}-1792\,a^4\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {512\,a^2\,\sqrt {a^2-b^2}}{256\,a\,b^2-64\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )-768\,a^3+\frac {512\,a^5}{b^2}-\frac {1792\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b^3}+832\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {64\,b^2\,\sqrt {a^2-b^2}}{256\,a\,b^2-64\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )-768\,a^3+\frac {512\,a^5}{b^2}-\frac {1792\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b^3}+832\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {1280\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{256\,a\,b^3-64\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )-1792\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )-768\,a^3\,b+\frac {512\,a^5}{b}+832\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b^2}}+\frac {1024\,a^5\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{1024\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^6+512\,a^5\,b-1792\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^4\,b^2-768\,a^3\,b^3+832\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^4+256\,a\,b^5-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^6}+\frac {320\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{256\,a\,b^2-64\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )-768\,a^3+\frac {512\,a^5}{b^2}-\frac {1792\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{b}+\frac {1024\,a^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{b^3}+832\,a^2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}\right )\,\sqrt {a^2-b^2}}{a\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^2/(a + b/sin(x)),x)

[Out]

(2*atan((64*b^3)/(64*b^3*tan(x/2) + 64*a*b^2 - 64*a^3 - 64*a^2*b*tan(x/2)) + (64*a^3*tan(x/2))/(64*b^3*tan(x/2
) + 64*a*b^2 - 64*a^3 - 64*a^2*b*tan(x/2)) - (64*a^2*b)/(64*b^3*tan(x/2) + 64*a*b^2 - 64*a^3 - 64*a^2*b*tan(x/
2)) - (64*a*b^2*tan(x/2))/(64*b^3*tan(x/2) + 64*a*b^2 - 64*a^3 - 64*a^2*b*tan(x/2))))/a + log(tan(x/2))/b - (2
*atanh((512*a^4*(a^2 - b^2)^(1/2))/(256*a*b^4 - 64*b^5*tan(x/2) + 512*a^5 - 768*a^3*b^2 + 832*a^2*b^3*tan(x/2)
+ (1024*a^6*tan(x/2))/b - 1792*a^4*b*tan(x/2)) - (512*a^2*(a^2 - b^2)^(1/2))/(256*a*b^2 - 64*b^3*tan(x/2) - 7
68*a^3 + (512*a^5)/b^2 - (1792*a^4*tan(x/2))/b + (1024*a^6*tan(x/2))/b^3 + 832*a^2*b*tan(x/2)) + (64*b^2*(a^2
- b^2)^(1/2))/(256*a*b^2 - 64*b^3*tan(x/2) - 768*a^3 + (512*a^5)/b^2 - (1792*a^4*tan(x/2))/b + (1024*a^6*tan(x
/2))/b^3 + 832*a^2*b*tan(x/2)) - (1280*a^3*tan(x/2)*(a^2 - b^2)^(1/2))/(256*a*b^3 - 64*b^4*tan(x/2) - 1792*a^4
*tan(x/2) - 768*a^3*b + (512*a^5)/b + 832*a^2*b^2*tan(x/2) + (1024*a^6*tan(x/2))/b^2) + (1024*a^5*tan(x/2)*(a^
2 - b^2)^(1/2))/(1024*a^6*tan(x/2) - 64*b^6*tan(x/2) + 256*a*b^5 + 512*a^5*b - 768*a^3*b^3 + 832*a^2*b^4*tan(x
/2) - 1792*a^4*b^2*tan(x/2)) + (320*a*b*tan(x/2)*(a^2 - b^2)^(1/2))/(256*a*b^2 - 64*b^3*tan(x/2) - 768*a^3 + (
512*a^5)/b^2 - (1792*a^4*tan(x/2))/b + (1024*a^6*tan(x/2))/b^3 + 832*a^2*b*tan(x/2)))*(a^2 - b^2)^(1/2))/(a*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(x)**2/(a + b*csc(x)), x)

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