Integrand size = 16, antiderivative size = 55 \[ \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {b \text {arctanh}(\cos (x))}{a^2}-\frac {\sqrt {a^2+b^2} \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2}-\frac {\csc (x)}{a} \] Output:
b*arctanh(cos(x))/a^2-(a^2+b^2)^(1/2)*arctanh((b*cos(x)-a*sin(x))/(a^2+b^2 )^(1/2))/a^2-csc(x)/a
Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {2 \sqrt {a^2+b^2} \text {arctanh}\left (\frac {-b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )-a \csc (x)+b \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )}{a^2} \] Input:
Integrate[Csc[x]^2/(a*Cos[x] + b*Sin[x]),x]
Output:
(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
Time = 0.34 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3582, 3042, 3553, 219, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (x)^2 (a \cos (x)+b \sin (x))}dx\) |
\(\Big \downarrow \) 3582 |
\(\displaystyle \frac {\left (a^2+b^2\right ) \int \frac {1}{a \cos (x)+b \sin (x)}dx}{a^2}-\frac {b \int \csc (x)dx}{a^2}-\frac {\csc (x)}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\left (a^2+b^2\right ) \int \frac {1}{a \cos (x)+b \sin (x)}dx}{a^2}-\frac {b \int \csc (x)dx}{a^2}-\frac {\csc (x)}{a}\) |
\(\Big \downarrow \) 3553 |
\(\displaystyle -\frac {\left (a^2+b^2\right ) \int \frac {1}{a^2+b^2-(b \cos (x)-a \sin (x))^2}d(b \cos (x)-a \sin (x))}{a^2}-\frac {b \int \csc (x)dx}{a^2}-\frac {\csc (x)}{a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {b \int \csc (x)dx}{a^2}-\frac {\sqrt {a^2+b^2} \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2}-\frac {\csc (x)}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {\sqrt {a^2+b^2} \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^2}+\frac {b \text {arctanh}(\cos (x))}{a^2}-\frac {\csc (x)}{a}\) |
Input:
Int[Csc[x]^2/(a*Cos[x] + b*Sin[x]),x]
Output:
(b*ArcTanh[Cos[x]])/a^2 - (Sqrt[a^2 + b^2]*ArcTanh[(b*Cos[x] - a*Sin[x])/S qrt[a^2 + b^2]])/a^2 - Csc[x]/a
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x _Symbol] :> Simp[-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]
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] + (-Simp[b/a^2 Int[Sin[c + d*x]^(m + 1), x], x] + Simp[(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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.47
method | result | size |
default | \(-\frac {\tan \left (\frac {x}{2}\right )}{2 a}-\frac {\left (-4 a^{2}-4 b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{2 a^{2} \sqrt {a^{2}+b^{2}}}-\frac {1}{2 a \tan \left (\frac {x}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}\) | \(81\) |
risch | \(-\frac {2 i {\mathrm e}^{i x}}{a \left ({\mathrm e}^{2 i x}-1\right )}+\frac {b \ln \left ({\mathrm e}^{i x}+1\right )}{a^{2}}-\frac {\sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right )}{a^{2}}+\frac {\sqrt {a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i a -b}{\sqrt {a^{2}+b^{2}}}\right )}{a^{2}}-\frac {b \ln \left ({\mathrm e}^{i x}-1\right )}{a^{2}}\) | \(127\) |
Input:
int(csc(x)^2/(a*cos(x)+b*sin(x)),x,method=_RETURNVERBOSE)
Output:
-1/2/a*tan(1/2*x)-1/2/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))-1/2/a/tan(1/2*x)-b/a^2*ln(tan(1/2*x))
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (51) = 102\).
Time = 0.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.42 \[ \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {b \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) - b \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right ) \sin \left (x\right ) - 2 \, a}{2 \, a^{2} \sin \left (x\right )} \] Input:
integrate(csc(x)^2/(a*cos(x)+b*sin(x)),x, algorithm="fricas")
Output:
1/2*(b*log(1/2*cos(x) + 1/2)*sin(x) - b*log(-1/2*cos(x) + 1/2)*sin(x) + sq rt(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))
\[ \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\int \frac {\csc ^{2}{\left (x \right )}}{a \cos {\left (x \right )} + b \sin {\left (x \right )}}\, dx \] Input:
integrate(csc(x)**2/(a*cos(x)+b*sin(x)),x)
Output:
Integral(csc(x)**2/(a*cos(x) + b*sin(x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (51) = 102\).
Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.95 \[ \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {b \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{2}} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{a^{2}} - \frac {\cos \left (x\right ) + 1}{2 \, a \sin \left (x\right )} - \frac {\sin \left (x\right )}{2 \, a {\left (\cos \left (x\right ) + 1\right )}} \] Input:
integrate(csc(x)^2/(a*cos(x)+b*sin(x)),x, algorithm="maxima")
Output:
-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))
Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (51) = 102\).
Time = 0.16 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.96 \[ \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} - \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, a} - \frac {\sqrt {a^{2} + b^{2}} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{a^{2}} + \frac {2 \, b \tan \left (\frac {1}{2} \, x\right ) - a}{2 \, a^{2} \tan \left (\frac {1}{2} \, x\right )} \] Input:
integrate(csc(x)^2/(a*cos(x)+b*sin(x)),x, algorithm="giac")
Output:
-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))
Time = 17.37 (sec) , antiderivative size = 170, normalized size of antiderivative = 3.09 \[ \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {a^3\,\cos \left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}+4\,b^3\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}+3\,a^2\,b\,\sin \left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}+2\,a\,b^2\,\cos \left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{\sin \left (\frac {x}{2}\right )\,a^4+2\,\cos \left (\frac {x}{2}\right )\,a^3\,b+5\,\sin \left (\frac {x}{2}\right )\,a^2\,b^2+2\,\cos \left (\frac {x}{2}\right )\,a\,b^3+4\,\sin \left (\frac {x}{2}\right )\,b^4}\right )\,\sqrt {a^2+b^2}}{a^2}-\frac {1}{a\,\sin \left (x\right )}-\frac {b\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{a^2} \] Input:
int(1/(sin(x)^2*(a*cos(x) + b*sin(x))),x)
Output:
(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
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {-2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a i -b i}{\sqrt {a^{2}+b^{2}}}\right ) \sin \left (x \right ) i -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right ) b -a}{\sin \left (x \right ) a^{2}} \] Input:
int(csc(x)^2/(a*cos(x)+b*sin(x)),x)
Output:
( - 2*sqrt(a**2 + b**2)*atan((tan(x/2)*a*i - b*i)/sqrt(a**2 + b**2))*sin(x )*i - log(tan(x/2))*sin(x)*b - a)/(sin(x)*a**2)