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

Optimal. Leaf size=55 \[ -\frac {\sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2}+\frac {b \tanh ^{-1}(\cos (x))}{a^2}-\frac {\csc (x)}{a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 55, 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 = {3103, 3770, 3074, 206} \[ -\frac {\sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2}+\frac {b \tanh ^{-1}(\cos (x))}{a^2}-\frac {\csc (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a*Cos[x] + b*Sin[x]),x]

[Out]

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

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 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 3103

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

Rule 3770

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 67, normalized size = 1.22 \[ \frac {2 \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )-a \csc (x)+b \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.49, size = 133, normalized size = 2.42 \[ \frac {b \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) - b \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) \sin \relax (x) - 2 \, a}{2 \, a^{2} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 2.97, size = 108, normalized size = 1.96 \[ -\frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} - \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, a} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{a^{2}} + \frac {2 \, b \tan \left (\frac {1}{2} \, x\right ) - a}{2 \, a^{2} \tan \left (\frac {1}{2} \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.79, size = 107, normalized size = 1.95 \[ -\frac {\tan \left (\frac {x}{2}\right )}{2 a}-\frac {1}{2 a \tan \left (\frac {x}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}+\frac {2 \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}+\frac {2 \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right ) b^{2}}{a^{2} \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.43, size = 107, normalized size = 1.95 \[ -\frac {b \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2}} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{a^{2}} - \frac {\cos \relax (x) + 1}{2 \, a \sin \relax (x)} - \frac {\sin \relax (x)}{2 \, a {\left (\cos \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.68, size = 170, normalized size = 3.09 \[ \frac {2\,\mathrm {atanh}\left (\frac {a^3\,\cos \left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}+4\,b^3\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}+3\,a^2\,b\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}+2\,a\,b^2\,\cos \left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{\sin \left (\frac {x}{2}\right )\,a^4+2\,\cos \left (\frac {x}{2}\right )\,a^3\,b+5\,\sin \left (\frac {x}{2}\right )\,a^2\,b^2+2\,\cos \left (\frac {x}{2}\right )\,a\,b^3+4\,\sin \left (\frac {x}{2}\right )\,b^4}\right )\,\sqrt {a^2+b^2}}{a^2}-\frac {1}{a\,\sin \relax (x)}-\frac {b\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^2*(a*cos(x) + b*sin(x))),x)

[Out]

(2*atanh((a^3*cos(x/2)*(a^2 + b^2)^(1/2) + 4*b^3*sin(x/2)*(a^2 + b^2)^(1/2) + 3*a^2*b*sin(x/2)*(a^2 + b^2)^(1/
2) + 2*a*b^2*cos(x/2)*(a^2 + b^2)^(1/2))/(a^4*sin(x/2) + 4*b^4*sin(x/2) + 5*a^2*b^2*sin(x/2) + 2*a*b^3*cos(x/2
) + 2*a^3*b*cos(x/2)))*(a^2 + b^2)^(1/2))/a^2 - 1/(a*sin(x)) - (b*log(sin(x/2)/cos(x/2)))/a^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________