Integrand size = 7, antiderivative size = 89 \[ \int \sin (x) \tan (6 x) \, dx=\frac {\text {arctanh}\left (\sqrt {2} \sin (x)\right )}{3 \sqrt {2}}+\frac {1}{6} \sqrt {2-\sqrt {3}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {3}}}\right )+\frac {1}{6} \sqrt {2+\sqrt {3}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {3}}}\right )-\sin (x) \]
-sin(x)+1/6*arctanh(sin(x)*2^(1/2))*2^(1/2)+1/6*arctanh(2*sin(x)/(1/2*6^(1 /2)-1/2*2^(1/2)))*(1/2*6^(1/2)-1/2*2^(1/2))+1/6*arctanh(2*sin(x)/(1/2*6^(1 /2)+1/2*2^(1/2)))*(1/2*6^(1/2)+1/2*2^(1/2))
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \sin (x) \tan (6 x) \, dx=\frac {1}{6} \left (\sqrt {2} \text {arctanh}\left (\sqrt {2} \sin (x)\right )+\sqrt {2-\sqrt {3}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {3}}}\right )+\sqrt {2+\sqrt {3}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {3}}}\right )-6 \sin (x)\right ) \]
(Sqrt[2]*ArcTanh[Sqrt[2]*Sin[x]] + Sqrt[2 - Sqrt[3]]*ArcTanh[(2*Sin[x])/Sq rt[2 - Sqrt[3]]] + Sqrt[2 + Sqrt[3]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[3]]] - 6*Sin[x])/6
Time = 0.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4878, 27, 2460, 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.
\(\displaystyle \int \sin (x) \tan (6 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (x) \tan (6 x)dx\) |
\(\Big \downarrow \) 4878 |
\(\displaystyle \int \frac {2 \sin ^2(x) \left (16 \sin ^4(x)-16 \sin ^2(x)+3\right )}{-32 \sin ^6(x)+48 \sin ^4(x)-18 \sin ^2(x)+1}d\sin (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {\sin ^2(x) \left (16 \sin ^4(x)-16 \sin ^2(x)+3\right )}{-32 \sin ^6(x)+48 \sin ^4(x)-18 \sin ^2(x)+1}d\sin (x)\) |
\(\Big \downarrow \) 2460 |
\(\displaystyle 2 \int \left (\frac {1-8 \sin ^2(x)}{3 \left (16 \sin ^4(x)-16 \sin ^2(x)+1\right )}-\frac {1}{6 \left (2 \sin ^2(x)-1\right )}-\frac {1}{2}\right )d\sin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {\text {arctanh}\left (\sqrt {2} \sin (x)\right )}{6 \sqrt {2}}+\frac {1}{12} \sqrt {2-\sqrt {3}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {3}}}\right )+\frac {1}{12} \sqrt {2+\sqrt {3}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {3}}}\right )-\frac {\sin (x)}{2}\right )\) |
2*(ArcTanh[Sqrt[2]*Sin[x]]/(6*Sqrt[2]) + (Sqrt[2 - Sqrt[3]]*ArcTanh[(2*Sin [x])/Sqrt[2 - Sqrt[3]]])/12 + (Sqrt[2 + Sqrt[3]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[3]]])/12 - Sin[x]/2)
3.1.78.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px /. x -> Sqrt[x]]}, Int[ExpandIntegrand[u*(Qx /. x -> x^2)^p, x], x] /; !SumQ[NonfreeFactors[Q x, x]]] /; PolyQ[Px, x^2] && GtQ[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0] && RationalFunctionQ[u, x]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Sin[v], x]}, d/Coefficient[v, x, 1] Subst[Int[SubstFor[1, Sin[v]/d, u/Cos[v], x], x], x, Sin[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[NonfreeF actors[Sin[v], x], u/Cos[v], x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.79 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.16
method | result | size |
risch | \(\frac {i {\mathrm e}^{i x}}{2}-\frac {i {\mathrm e}^{-i x}}{2}-\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {2}\, {\mathrm e}^{i x}-1\right )}{12}+\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {2}\, {\mathrm e}^{i x}-1\right )}{12}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1296 \textit {\_Z}^{4}-144 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}-6 i \textit {\_R} \,{\mathrm e}^{i x}-1\right )\right )}{2}\) | \(103\) |
1/2*I*exp(I*x)-1/2*I*exp(-I*x)-1/12*2^(1/2)*ln(exp(2*I*x)-I*2^(1/2)*exp(I* x)-1)+1/12*2^(1/2)*ln(exp(2*I*x)+I*2^(1/2)*exp(I*x)-1)-1/2*sum(_R*ln(exp(2 *I*x)-6*I*_R*exp(I*x)-1),_R=RootOf(1296*_Z^4-144*_Z^2+1))
Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.51 \[ \int \sin (x) \tan (6 x) \, dx=\frac {1}{12} \, \sqrt {\sqrt {3} + 2} \log \left (\sqrt {\sqrt {3} + 2} + 2 \, \sin \left (x\right )\right ) - \frac {1}{12} \, \sqrt {\sqrt {3} + 2} \log \left (\sqrt {\sqrt {3} + 2} - 2 \, \sin \left (x\right )\right ) + \frac {1}{12} \, \sqrt {-\sqrt {3} + 2} \log \left (\sqrt {-\sqrt {3} + 2} + 2 \, \sin \left (x\right )\right ) - \frac {1}{12} \, \sqrt {-\sqrt {3} + 2} \log \left (\sqrt {-\sqrt {3} + 2} - 2 \, \sin \left (x\right )\right ) + \frac {1}{12} \, \sqrt {2} \log \left (-\frac {2 \, \cos \left (x\right )^{2} - 2 \, \sqrt {2} \sin \left (x\right ) - 3}{2 \, \cos \left (x\right )^{2} - 1}\right ) - \sin \left (x\right ) \]
1/12*sqrt(sqrt(3) + 2)*log(sqrt(sqrt(3) + 2) + 2*sin(x)) - 1/12*sqrt(sqrt( 3) + 2)*log(sqrt(sqrt(3) + 2) - 2*sin(x)) + 1/12*sqrt(-sqrt(3) + 2)*log(sq rt(-sqrt(3) + 2) + 2*sin(x)) - 1/12*sqrt(-sqrt(3) + 2)*log(sqrt(-sqrt(3) + 2) - 2*sin(x)) + 1/12*sqrt(2)*log(-(2*cos(x)^2 - 2*sqrt(2)*sin(x) - 3)/(2 *cos(x)^2 - 1)) - sin(x)
\[ \int \sin (x) \tan (6 x) \, dx=\int \sin {\left (x \right )} \tan {\left (6 x \right )}\, dx \]
\[ \int \sin (x) \tan (6 x) \, dx=\int { \sin \left (x\right ) \tan \left (6 \, x\right ) \,d x } \]
1/24*sqrt(2)*log(2*cos(x)^2 + 2*sin(x)^2 + 2*sqrt(2)*cos(x) + 2*sqrt(2)*si n(x) + 2) - 1/24*sqrt(2)*log(2*cos(x)^2 + 2*sin(x)^2 + 2*sqrt(2)*cos(x) - 2*sqrt(2)*sin(x) + 2) + 1/24*sqrt(2)*log(2*cos(x)^2 + 2*sin(x)^2 - 2*sqrt( 2)*cos(x) + 2*sqrt(2)*sin(x) + 2) - 1/24*sqrt(2)*log(2*cos(x)^2 + 2*sin(x) ^2 - 2*sqrt(2)*cos(x) - 2*sqrt(2)*sin(x) + 2) + integrate(-1/3*((2*cos(7*x ) - cos(5*x) - cos(3*x) + 2*cos(x))*cos(8*x) - 2*(cos(4*x) - 1)*cos(7*x) + (cos(4*x) - 1)*cos(5*x) + (cos(3*x) - 2*cos(x))*cos(4*x) + (2*sin(7*x) - sin(5*x) - sin(3*x) + 2*sin(x))*sin(8*x) + (sin(3*x) - 2*sin(x))*sin(4*x) - 2*sin(7*x)*sin(4*x) + sin(5*x)*sin(4*x) - cos(3*x) + 2*cos(x))/(2*(cos(4 *x) - 1)*cos(8*x) - cos(8*x)^2 - cos(4*x)^2 - sin(8*x)^2 + 2*sin(8*x)*sin( 4*x) - sin(4*x)^2 + 2*cos(4*x) - 1), x) - sin(x)
\[ \int \sin (x) \tan (6 x) \, dx=\int { \sin \left (x\right ) \tan \left (6 \, x\right ) \,d x } \]
Time = 27.67 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.47 \[ \int \sin (x) \tan (6 x) \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\sqrt {2}\,\sin \left (x\right )\right )}{6}-\sin \left (x\right )-\frac {\sqrt {2}\,\mathrm {atanh}\left (\sqrt {2}\,\sin \left (x\right )-\sqrt {6}\,\sin \left (x\right )\right )}{12}-\frac {\sqrt {6}\,\mathrm {atanh}\left (\sqrt {2}\,\sin \left (x\right )-\sqrt {6}\,\sin \left (x\right )\right )}{12}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sin \left (x\right )\,102818{}\mathrm {i}-\sqrt {6}\,\sin \left (x\right )\,59362{}\mathrm {i}}{40545\,\sqrt {2}\,\sqrt {6}-140452}\right )\,1{}\mathrm {i}}{12}-\frac {\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sin \left (x\right )\,102818{}\mathrm {i}-\sqrt {6}\,\sin \left (x\right )\,59362{}\mathrm {i}}{40545\,\sqrt {2}\,\sqrt {6}-140452}\right )\,1{}\mathrm {i}}{12} \]
(2^(1/2)*atan((2^(1/2)*sin(x)*102818i - 6^(1/2)*sin(x)*59362i)/(40545*2^(1 /2)*6^(1/2) - 140452))*1i)/12 - sin(x) - (6^(1/2)*atan((2^(1/2)*sin(x)*102 818i - 6^(1/2)*sin(x)*59362i)/(40545*2^(1/2)*6^(1/2) - 140452))*1i)/12 + ( 2^(1/2)*atanh(2^(1/2)*sin(x)))/6 - (2^(1/2)*atanh(2^(1/2)*sin(x) - 6^(1/2) *sin(x)))/12 - (6^(1/2)*atanh(2^(1/2)*sin(x) - 6^(1/2)*sin(x)))/12