Integrand size = 7, antiderivative size = 71 \[ \int \cos (x) \csc (5 x) \, dx=\frac {1}{5} \log (\sin (x))-\frac {\log \left (5-\sqrt {5}-8 \sin ^2(x)\right )}{\sqrt {5} \left (5-\sqrt {5}\right )}+\frac {\log \left (5+\sqrt {5}-8 \sin ^2(x)\right )}{\sqrt {5} \left (5+\sqrt {5}\right )} \] Output:
1/5*ln(sin(x))-1/5*ln(5-5^(1/2)-8*sin(x)^2)*5^(1/2)/(5-5^(1/2))+1/5*ln(5+5 ^(1/2)-8*sin(x)^2)*5^(1/2)/(5+5^(1/2))
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.80 \[ \int \cos (x) \csc (5 x) \, dx=\frac {1}{20} \left (-\left (\left (1+\sqrt {5}\right ) \log \left (1-\sqrt {5}+4 \cos (2 x)\right )\right )+\left (-1+\sqrt {5}\right ) \log \left (1+\sqrt {5}+4 \cos (2 x)\right )+4 \log (\sin (x))\right ) \] Input:
Integrate[Cos[x]*Csc[5*x],x]
Output:
(-((1 + Sqrt[5])*Log[1 - Sqrt[5] + 4*Cos[2*x]]) + (-1 + Sqrt[5])*Log[1 + S qrt[5] + 4*Cos[2*x]] + 4*Log[Sin[x]])/20
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4856, 1434, 1141, 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.
\(\displaystyle \int \cos (x) \csc (5 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (x)}{\sin (5 x)}dx\) |
\(\Big \downarrow \) 4856 |
\(\displaystyle \int \frac {\csc (x)}{16 \sin ^4(x)-20 \sin ^2(x)+5}d\sin (x)\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {1}{2} \int \frac {\csc (x)}{16 \sin ^4(x)-20 \sin ^2(x)+5}d\sin ^2(x)\) |
\(\Big \downarrow \) 1141 |
\(\displaystyle 8 \int \left (\frac {\csc (x)}{80}+\frac {1}{\sqrt {5} \left (5-\sqrt {5}\right ) \left (-8 \sin ^2(x)-\sqrt {5}+5\right )}-\frac {1}{\sqrt {5} \left (5+\sqrt {5}\right ) \left (-8 \sin ^2(x)+\sqrt {5}+5\right )}\right )d\sin ^2(x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 8 \left (\frac {1}{80} \log \left (\sin ^2(x)\right )-\frac {\log \left (-8 \sin ^2(x)-\sqrt {5}+5\right )}{8 \sqrt {5} \left (5-\sqrt {5}\right )}+\frac {\log \left (-8 \sin ^2(x)+\sqrt {5}+5\right )}{8 \sqrt {5} \left (5+\sqrt {5}\right )}\right )\) |
Input:
Int[Cos[x]*Csc[5*x],x]
Output:
8*(Log[Sin[x]^2]/80 - Log[5 - Sqrt[5] - 8*Sin[x]^2]/(8*Sqrt[5]*(5 - Sqrt[5 ])) + Log[5 + Sqrt[5] - 8*Sin[x]^2]/(8*Sqrt[5]*(5 + Sqrt[5])))
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[ (d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 1] || !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Sin[c*(a + b*x)], x]}, Simp[d/(b*c) Subst[Int[SubstFor[1, Sin[c*(a + b *x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a + b*x )]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cos] || EqQ[F, cos])
Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13
method | result | size |
default | \(-\frac {\ln \left (4 \cos \left (x \right )^{2}-2 \cos \left (x \right )-1\right )}{20}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (8 \cos \left (x \right )-2\right ) \sqrt {5}}{10}\right )}{10}+\frac {\ln \left (1+\cos \left (x \right )\right )}{10}-\frac {\ln \left (4 \cos \left (x \right )^{2}+2 \cos \left (x \right )-1\right )}{20}-\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (8 \cos \left (x \right )+2\right ) \sqrt {5}}{10}\right )}{10}+\frac {\ln \left (\cos \left (x \right )-1\right )}{10}\) | \(80\) |
risch | \(\frac {\ln \left ({\mathrm e}^{2 i x}-1\right )}{5}-\frac {\ln \left ({\mathrm e}^{4 i x}+\left (\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) {\mathrm e}^{2 i x}+1\right )}{20}-\frac {\ln \left ({\mathrm e}^{4 i x}+\left (\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) {\mathrm e}^{2 i x}+1\right ) \sqrt {5}}{20}-\frac {\ln \left ({\mathrm e}^{4 i x}+\left (\frac {\sqrt {5}}{2}+\frac {1}{2}\right ) {\mathrm e}^{2 i x}+1\right )}{20}+\frac {\ln \left ({\mathrm e}^{4 i x}+\left (\frac {\sqrt {5}}{2}+\frac {1}{2}\right ) {\mathrm e}^{2 i x}+1\right ) \sqrt {5}}{20}\) | \(110\) |
Input:
int(cos(x)*csc(5*x),x,method=_RETURNVERBOSE)
Output:
-1/20*ln(4*cos(x)^2-2*cos(x)-1)+1/10*5^(1/2)*arctanh(1/10*(8*cos(x)-2)*5^( 1/2))+1/10*ln(1+cos(x))-1/20*ln(4*cos(x)^2+2*cos(x)-1)-1/10*5^(1/2)*arctan h(1/10*(8*cos(x)+2)*5^(1/2))+1/10*ln(cos(x)-1)
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int \cos (x) \csc (5 x) \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {32 \, \cos \left (x\right )^{4} + 8 \, {\left (\sqrt {5} - 3\right )} \cos \left (x\right )^{2} - 3 \, \sqrt {5} + 7}{16 \, \cos \left (x\right )^{4} - 12 \, \cos \left (x\right )^{2} + 1}\right ) - \frac {1}{20} \, \log \left (16 \, \cos \left (x\right )^{4} - 12 \, \cos \left (x\right )^{2} + 1\right ) + \frac {1}{5} \, \log \left (\frac {1}{2} \, \sin \left (x\right )\right ) \] Input:
integrate(cos(x)*csc(5*x),x, algorithm="fricas")
Output:
1/20*sqrt(5)*log((32*cos(x)^4 + 8*(sqrt(5) - 3)*cos(x)^2 - 3*sqrt(5) + 7)/ (16*cos(x)^4 - 12*cos(x)^2 + 1)) - 1/20*log(16*cos(x)^4 - 12*cos(x)^2 + 1) + 1/5*log(1/2*sin(x))
\[ \int \cos (x) \csc (5 x) \, dx=\int \cos {\left (x \right )} \csc {\left (5 x \right )}\, dx \] Input:
integrate(cos(x)*csc(5*x),x)
Output:
Integral(cos(x)*csc(5*x), x)
\[ \int \cos (x) \csc (5 x) \, dx=\int { \cos \left (x\right ) \csc \left (5 \, x\right ) \,d x } \] Input:
integrate(cos(x)*csc(5*x),x, algorithm="maxima")
Output:
-1/10*integrate(-(cos(2*x)*sin(4*x) - cos(4*x)*sin(2*x) + cos(3/2*arctan2( sin(2*x), cos(2*x)))*sin(2*x) + cos(1/2*arctan2(sin(2*x), cos(2*x)))*sin(2 *x) - cos(2*x)*sin(3/2*arctan2(sin(2*x), cos(2*x))) - cos(2*x)*sin(1/2*arc tan2(sin(2*x), cos(2*x))) - sin(2*x))/(2*(cos(2*x) + 1)*cos(4*x) + cos(4*x )^2 + cos(2*x)^2 - 2*(cos(4*x) + cos(2*x) - cos(1/2*arctan2(sin(2*x), cos( 2*x))) + 1)*cos(3/2*arctan2(sin(2*x), cos(2*x))) + cos(3/2*arctan2(sin(2*x ), cos(2*x)))^2 - 2*(cos(4*x) + cos(2*x) + 1)*cos(1/2*arctan2(sin(2*x), co s(2*x))) + cos(1/2*arctan2(sin(2*x), cos(2*x)))^2 + sin(4*x)^2 + 2*sin(4*x )*sin(2*x) + sin(2*x)^2 - 2*(sin(4*x) + sin(2*x) - sin(1/2*arctan2(sin(2*x ), cos(2*x))))*sin(3/2*arctan2(sin(2*x), cos(2*x))) + sin(3/2*arctan2(sin( 2*x), cos(2*x)))^2 - 2*(sin(4*x) + sin(2*x))*sin(1/2*arctan2(sin(2*x), cos (2*x))) + sin(1/2*arctan2(sin(2*x), cos(2*x)))^2 + 2*cos(2*x) + 1), x) + 1 /10*integrate((cos(2*x)*sin(4*x) - cos(4*x)*sin(2*x) - cos(3/2*arctan2(sin (2*x), cos(2*x)))*sin(2*x) - cos(1/2*arctan2(sin(2*x), cos(2*x)))*sin(2*x) + cos(2*x)*sin(3/2*arctan2(sin(2*x), cos(2*x))) + cos(2*x)*sin(1/2*arctan 2(sin(2*x), cos(2*x))) - sin(2*x))/(2*(cos(2*x) + 1)*cos(4*x) + cos(4*x)^2 + cos(2*x)^2 + 2*(cos(4*x) + cos(2*x) + cos(1/2*arctan2(sin(2*x), cos(2*x ))) + 1)*cos(3/2*arctan2(sin(2*x), cos(2*x))) + cos(3/2*arctan2(sin(2*x), cos(2*x)))^2 + 2*(cos(4*x) + cos(2*x) + 1)*cos(1/2*arctan2(sin(2*x), cos(2 *x))) + cos(1/2*arctan2(sin(2*x), cos(2*x)))^2 + sin(4*x)^2 + 2*sin(4*x...
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.94 \[ \int \cos (x) \csc (5 x) \, dx=-\frac {1}{20} \, \sqrt {5} \log \left (\frac {{\left | 32 \, \cos \left (x\right )^{2} - 4 \, \sqrt {5} - 12 \right |}}{{\left | 32 \, \cos \left (x\right )^{2} + 4 \, \sqrt {5} - 12 \right |}}\right ) + \frac {1}{10} \, \log \left (-\cos \left (x\right )^{2} + 1\right ) - \frac {1}{20} \, \log \left ({\left | 16 \, \cos \left (x\right )^{4} - 12 \, \cos \left (x\right )^{2} + 1 \right |}\right ) \] Input:
integrate(cos(x)*csc(5*x),x, algorithm="giac")
Output:
-1/20*sqrt(5)*log(abs(32*cos(x)^2 - 4*sqrt(5) - 12)/abs(32*cos(x)^2 + 4*sq rt(5) - 12)) + 1/10*log(-cos(x)^2 + 1) - 1/20*log(abs(16*cos(x)^4 - 12*cos (x)^2 + 1))
Time = 0.52 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72 \[ \int \cos (x) \csc (5 x) \, dx=\frac {\ln \left (\sin \left (x\right )\right )}{5}+\ln \left (-{\cos \left (x\right )}^2-\frac {\sqrt {5}}{8}+\frac {3}{8}\right )\,\left (\frac {\sqrt {5}}{20}-\frac {1}{20}\right )-\ln \left (-{\cos \left (x\right )}^2+\frac {\sqrt {5}}{8}+\frac {3}{8}\right )\,\left (\frac {\sqrt {5}}{20}+\frac {1}{20}\right ) \] Input:
int(cos(x)/sin(5*x),x)
Output:
log(sin(x))/5 + log(3/8 - 5^(1/2)/8 - cos(x)^2)*(5^(1/2)/20 - 1/20) - log( 5^(1/2)/8 - cos(x)^2 + 3/8)*(5^(1/2)/20 + 1/20)
\[ \int \cos (x) \csc (5 x) \, dx=\int \cos \left (x \right ) \csc \left (5 x \right )d x \] Input:
int(cos(x)*csc(5*x),x)
Output:
int(cos(x)*csc(5*x),x)