Integrand size = 7, antiderivative size = 71 \[ \int \sin (x) \tan (4 x) \, dx=\frac {1}{4} \sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{4} \sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )-\sin (x) \]
-sin(x)+1/4*arctanh(2*sin(x)/(2-2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)+1/4*arct anh(2*sin(x)/(2+2^(1/2))^(1/2))*(2+2^(1/2))^(1/2)
Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97 \[ \int \sin (x) \tan (4 x) \, dx=\frac {1}{4} \left (\sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )+\sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )-4 \sin (x)\right ) \]
(Sqrt[2 - Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[2]]] + Sqrt[2 + Sqrt[2 ]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[2]]] - 4*Sin[x])/4
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4878, 27, 1602, 27, 1480, 220}
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.
\(\displaystyle \int \sin (x) \tan (4 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (x) \tan (4 x)dx\) |
\(\Big \downarrow \) 4878 |
\(\displaystyle \int \frac {4 \sin ^2(x) \left (1-2 \sin ^2(x)\right )}{8 \sin ^4(x)-8 \sin ^2(x)+1}d\sin (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {\sin ^2(x) \left (1-2 \sin ^2(x)\right )}{8 \sin ^4(x)-8 \sin ^2(x)+1}d\sin (x)\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle 4 \left (-\frac {1}{8} \int -\frac {2 \left (1-4 \sin ^2(x)\right )}{8 \sin ^4(x)-8 \sin ^2(x)+1}d\sin (x)-\frac {\sin (x)}{4}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \left (\frac {1}{4} \int \frac {1-4 \sin ^2(x)}{8 \sin ^4(x)-8 \sin ^2(x)+1}d\sin (x)-\frac {\sin (x)}{4}\right )\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle 4 \left (\frac {1}{4} \left (-\left (\left (2-\sqrt {2}\right ) \int \frac {1}{8 \sin ^2(x)-2 \left (2-\sqrt {2}\right )}d\sin (x)\right )-\left (2+\sqrt {2}\right ) \int \frac {1}{8 \sin ^2(x)-2 \left (2+\sqrt {2}\right )}d\sin (x)\right )-\frac {\sin (x)}{4}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{4} \sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{4} \sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )\right )-\frac {\sin (x)}{4}\right )\) |
4*(((Sqrt[2 - Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[2]]])/4 + (Sqrt[2 + Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[2]]])/4)/4 - Sin[x]/4)
3.1.76.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
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 = 1.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {i {\mathrm e}^{i x}}{2}-\frac {i {\mathrm e}^{-i x}}{2}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (128 \textit {\_Z}^{4}-32 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}-4 i \textit {\_R} \,{\mathrm e}^{i x}-1\right )\right )}{2}\) | \(55\) |
default | \(\frac {\sqrt {2}\, \sqrt {2+\sqrt {2}}\, \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\sqrt {2+\sqrt {2}}}\right )}{4}+\frac {\left (\sqrt {2}-2\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2-\sqrt {2}}}-\sin \left (x \right )-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2+\sqrt {2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2-\sqrt {2}}}\) | \(115\) |
1/2*I*exp(I*x)-1/2*I*exp(-I*x)-1/2*sum(_R*ln(exp(2*I*x)-4*I*_R*exp(I*x)-1) ,_R=RootOf(128*_Z^4-32*_Z^2+1))
Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.42 \[ \int \sin (x) \tan (4 x) \, dx=\frac {1}{8} \, \sqrt {\sqrt {2} + 2} \log \left (\sqrt {\sqrt {2} + 2} + 2 \, \sin \left (x\right )\right ) - \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \log \left (\sqrt {\sqrt {2} + 2} - 2 \, \sin \left (x\right )\right ) + \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \log \left (\sqrt {-\sqrt {2} + 2} + 2 \, \sin \left (x\right )\right ) - \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \log \left (\sqrt {-\sqrt {2} + 2} - 2 \, \sin \left (x\right )\right ) - \sin \left (x\right ) \]
1/8*sqrt(sqrt(2) + 2)*log(sqrt(sqrt(2) + 2) + 2*sin(x)) - 1/8*sqrt(sqrt(2) + 2)*log(sqrt(sqrt(2) + 2) - 2*sin(x)) + 1/8*sqrt(-sqrt(2) + 2)*log(sqrt( -sqrt(2) + 2) + 2*sin(x)) - 1/8*sqrt(-sqrt(2) + 2)*log(sqrt(-sqrt(2) + 2) - 2*sin(x)) - sin(x)
\[ \int \sin (x) \tan (4 x) \, dx=\int \sin {\left (x \right )} \tan {\left (4 x \right )}\, dx \]
\[ \int \sin (x) \tan (4 x) \, dx=\int { \sin \left (x\right ) \tan \left (4 \, x\right ) \,d x } \]
integrate(((cos(7*x) + cos(x))*cos(8*x) + (sin(7*x) + sin(x))*sin(8*x) + c os(7*x) + cos(x))/(cos(8*x)^2 + sin(8*x)^2 + 2*cos(8*x) + 1), x) - sin(x)
\[ \int \sin (x) \tan (4 x) \, dx=\int { \sin \left (x\right ) \tan \left (4 \, x\right ) \,d x } \]
Time = 25.84 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.45 \[ \int \sin (x) \tan (4 x) \, dx=\frac {\mathrm {atanh}\left (\frac {34\,\sin \left (x\right )\,\sqrt {\sqrt {2}+2}+24\,\sqrt {2}\,\sin \left (x\right )\,\sqrt {\sqrt {2}+2}}{41\,\sqrt {2}+58}\right )\,\sqrt {\sqrt {2}+2}}{4}-\sin \left (x\right )-\frac {\mathrm {atanh}\left (\frac {34\,\sin \left (x\right )\,\sqrt {2-\sqrt {2}}-24\,\sqrt {2}\,\sin \left (x\right )\,\sqrt {2-\sqrt {2}}}{41\,\sqrt {2}-58}\right )\,\sqrt {2-\sqrt {2}}}{4} \]