Integrand size = 9, antiderivative size = 32 \[ \int \frac {1}{\sin (x)+\sin (5 x)} \, dx=-\frac {1}{6} \text {arctanh}(\cos (x))-\frac {2}{3} \text {arctanh}(2 \cos (x))+\frac {\text {arctanh}\left (\sqrt {2} \cos (x)\right )}{\sqrt {2}} \] Output:
-1/6*arctanh(cos(x))-2/3*arctanh(2*cos(x))+1/2*arctanh(cos(x)*2^(1/2))*2^( 1/2)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.72 \[ \int \frac {1}{\sin (x)+\sin (5 x)} \, dx=\frac {1}{6} \left ((3+3 i) (-1)^{3/4} \text {arctanh}\left (\frac {-1+\tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )+(3-3 i) \sqrt [4]{-1} \text {arctanh}\left (\frac {1+\tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )-\log \left (\cos \left (\frac {x}{2}\right )\right )+2 \log (1-2 \cos (x))-2 \log (1+2 \cos (x))+\log \left (\sin \left (\frac {x}{2}\right )\right )\right ) \] Input:
Integrate[(Sin[x] + Sin[5*x])^(-1),x]
Output:
((3 + 3*I)*(-1)^(3/4)*ArcTanh[(-1 + Tan[x/2])/Sqrt[2]] + (3 - 3*I)*(-1)^(1 /4)*ArcTanh[(1 + Tan[x/2])/Sqrt[2]] - Log[Cos[x/2]] + 2*Log[1 - 2*Cos[x]] - 2*Log[1 + 2*Cos[x]] + Log[Sin[x/2]])/6
Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4824, 27, 1484, 2009}
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.
\(\displaystyle \int \frac {1}{\sin (x)+\sin (5 x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (x)+\sin (5 x)}dx\) |
\(\Big \downarrow \) 4824 |
\(\displaystyle -\int \frac {1}{2 \left (1-\cos ^2(x)\right ) \left (8 \cos ^4(x)-6 \cos ^2(x)+1\right )}d\cos (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{2} \int \frac {1}{\left (1-\cos ^2(x)\right ) \left (8 \cos ^4(x)-6 \cos ^2(x)+1\right )}d\cos (x)\) |
\(\Big \downarrow \) 1484 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {2}{2 \cos ^2(x)-1}-\frac {8}{3 \left (4 \cos ^2(x)-1\right )}-\frac {1}{3 \left (\cos ^2(x)-1\right )}\right )d\cos (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{3} \text {arctanh}(\cos (x))-\frac {4}{3} \text {arctanh}(2 \cos (x))+\sqrt {2} \text {arctanh}\left (\sqrt {2} \cos (x)\right )\right )\) |
Input:
Int[(Sin[x] + Sin[5*x])^(-1),x]
Output:
(-1/3*ArcTanh[Cos[x]] - (4*ArcTanh[2*Cos[x]])/3 + Sqrt[2]*ArcTanh[Sqrt[2]* Cos[x]])/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]
Int[((a_.)*sin[(m_.)*((c_.) + (d_.)*(x_))] + (b_.)*sin[(n_.)*((c_.) + (d_.) *(x_))])^(p_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Simplify[TrigExpand[a* Sin[m*ArcCos[x]] + b*Sin[n*ArcCos[x]]]]^p/Sqrt[1 - x^2], x], x, Cos[c + d*x ]], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[(p - 1)/2, 0] && IntegerQ[(m - 1)/ 2] && IntegerQ[(n - 1)/2]
Time = 0.69 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44
method | result | size |
default | \(\frac {\ln \left (2 \cos \left (x \right )-1\right )}{3}+\frac {\operatorname {arctanh}\left (\sqrt {2}\, \cos \left (x \right )\right ) \sqrt {2}}{2}-\frac {\ln \left (2 \cos \left (x \right )+1\right )}{3}+\frac {\ln \left (\cos \left (x \right )-1\right )}{12}-\frac {\ln \left (1+\cos \left (x \right )\right )}{12}\) | \(46\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{6}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{6}+\frac {\ln \left ({\mathrm e}^{2 i x}-{\mathrm e}^{i x}+1\right )}{3}-\frac {\ln \left ({\mathrm e}^{2 i x}+{\mathrm e}^{i x}+1\right )}{3}+\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+\sqrt {2}\, {\mathrm e}^{i x}+1\right )}{4}-\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-\sqrt {2}\, {\mathrm e}^{i x}+1\right )}{4}\) | \(99\) |
Input:
int(1/(sin(x)+sin(5*x)),x,method=_RETURNVERBOSE)
Output:
1/3*ln(2*cos(x)-1)+1/2*arctanh(2^(1/2)*cos(x))*2^(1/2)-1/3*ln(2*cos(x)+1)+ 1/12*ln(cos(x)-1)-1/12*ln(1+cos(x))
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.19 \[ \int \frac {1}{\sin (x)+\sin (5 x)} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (-\frac {2 \, \cos \left (x\right )^{2} + 2 \, \sqrt {2} \cos \left (x\right ) + 1}{2 \, \cos \left (x\right )^{2} - 1}\right ) - \frac {1}{12} \, \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + \frac {1}{12} \, \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + \frac {1}{3} \, \log \left (-2 \, \cos \left (x\right ) + 1\right ) - \frac {1}{3} \, \log \left (-2 \, \cos \left (x\right ) - 1\right ) \] Input:
integrate(1/(sin(x)+sin(5*x)),x, algorithm="fricas")
Output:
1/4*sqrt(2)*log(-(2*cos(x)^2 + 2*sqrt(2)*cos(x) + 1)/(2*cos(x)^2 - 1)) - 1 /12*log(1/2*cos(x) + 1/2) + 1/12*log(-1/2*cos(x) + 1/2) + 1/3*log(-2*cos(x ) + 1) - 1/3*log(-2*cos(x) - 1)
\[ \int \frac {1}{\sin (x)+\sin (5 x)} \, dx=\int \frac {1}{\sin {\left (x \right )} + \sin {\left (5 x \right )}}\, dx \] Input:
integrate(1/(sin(x)+sin(5*x)),x)
Output:
Integral(1/(sin(x) + sin(5*x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (25) = 50\).
Time = 0.14 (sec) , antiderivative size = 257, normalized size of antiderivative = 8.03 \[ \int \frac {1}{\sin (x)+\sin (5 x)} \, dx=\frac {1}{8} \, \sqrt {2} \log \left (2 \, \sqrt {2} \sin \left (2 \, x\right ) \sin \left (x\right ) + 2 \, {\left (\sqrt {2} \cos \left (x\right ) + 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (x\right )^{2} + 2 \, \sqrt {2} \cos \left (x\right ) + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (-2 \, \sqrt {2} \sin \left (2 \, x\right ) \sin \left (x\right ) - 2 \, {\left (\sqrt {2} \cos \left (x\right ) - 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + 2 \, \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (x\right )^{2} - 2 \, \sqrt {2} \cos \left (x\right ) + 1\right ) - \frac {1}{6} \, \log \left (2 \, {\left (\cos \left (x\right ) + 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \sin \left (2 \, x\right ) \sin \left (x\right ) + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) + \frac {1}{6} \, \log \left (-2 \, {\left (\cos \left (x\right ) - 1\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + \cos \left (x\right )^{2} + \sin \left (2 \, x\right )^{2} - 2 \, \sin \left (2 \, x\right ) \sin \left (x\right ) + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) - \frac {1}{12} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) + \frac {1}{12} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) \] Input:
integrate(1/(sin(x)+sin(5*x)),x, algorithm="maxima")
Output:
1/8*sqrt(2)*log(2*sqrt(2)*sin(2*x)*sin(x) + 2*(sqrt(2)*cos(x) + 1)*cos(2*x ) + cos(2*x)^2 + 2*cos(x)^2 + sin(2*x)^2 + 2*sin(x)^2 + 2*sqrt(2)*cos(x) + 1) - 1/8*sqrt(2)*log(-2*sqrt(2)*sin(2*x)*sin(x) - 2*(sqrt(2)*cos(x) - 1)* cos(2*x) + cos(2*x)^2 + 2*cos(x)^2 + sin(2*x)^2 + 2*sin(x)^2 - 2*sqrt(2)*c os(x) + 1) - 1/6*log(2*(cos(x) + 1)*cos(2*x) + cos(2*x)^2 + cos(x)^2 + sin (2*x)^2 + 2*sin(2*x)*sin(x) + sin(x)^2 + 2*cos(x) + 1) + 1/6*log(-2*(cos(x ) - 1)*cos(2*x) + cos(2*x)^2 + cos(x)^2 + sin(2*x)^2 - 2*sin(2*x)*sin(x) + sin(x)^2 - 2*cos(x) + 1) - 1/12*log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1) + 1/12*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (25) = 50\).
Time = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.16 \[ \int \frac {1}{\sin (x)+\sin (5 x)} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (\frac {{\left | -4 \, \sqrt {2} - \frac {2 \, {\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - 6 \right |}}{{\left | 4 \, \sqrt {2} - \frac {2 \, {\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - 6 \right |}}\right ) + \frac {1}{12} \, \log \left (-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right ) - \frac {1}{3} \, \log \left ({\left | -\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1} - 3 \right |}\right ) + \frac {1}{3} \, \log \left ({\left | -\frac {3 \, {\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} - 1 \right |}\right ) \] Input:
integrate(1/(sin(x)+sin(5*x)),x, algorithm="giac")
Output:
1/4*sqrt(2)*log(abs(-4*sqrt(2) - 2*(cos(x) - 1)/(cos(x) + 1) - 6)/abs(4*sq rt(2) - 2*(cos(x) - 1)/(cos(x) + 1) - 6)) + 1/12*log(-(cos(x) - 1)/(cos(x) + 1)) - 1/3*log(abs(-(cos(x) - 1)/(cos(x) + 1) - 3)) + 1/3*log(abs(-3*(co s(x) - 1)/(cos(x) + 1) - 1))
Time = 21.70 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.31 \[ \int \frac {1}{\sin (x)+\sin (5 x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{6}-\frac {2\,\mathrm {atanh}\left (\frac {67108864}{193710123\,\left (\frac {13374142283776\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{129140163}-\frac {1486008352768}{43046721}\right )}-\frac {797162}{797161}\right )}{3}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {120967921664\,\sqrt {2}}{531441\,\left (\frac {110788345856\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{59049}-\frac {513223426048}{1594323}\right )}-\frac {2115158147072\,\sqrt {2}\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{1594323\,\left (\frac {110788345856\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{59049}-\frac {513223426048}{1594323}\right )}\right )}{2} \] Input:
int(1/(sin(5*x) + sin(x)),x)
Output:
log(tan(x/2))/6 - (2*atanh(67108864/(193710123*((13374142283776*tan(x/2)^2 )/129140163 - 1486008352768/43046721)) - 797162/797161))/3 + (2^(1/2)*atan h((120967921664*2^(1/2))/(531441*((110788345856*tan(x/2)^2)/59049 - 513223 426048/1594323)) - (2115158147072*2^(1/2)*tan(x/2)^2)/(1594323*((110788345 856*tan(x/2)^2)/59049 - 513223426048/1594323))))/2
\[ \int \frac {1}{\sin (x)+\sin (5 x)} \, dx=\int \frac {1}{\sin \left (5 x \right )+\sin \left (x \right )}d x \] Input:
int(1/(sin(x)+sin(5*x)),x)
Output:
int(1/(sin(5*x) + sin(x)),x)