Integrand size = 7, antiderivative size = 71 \[ \int \sec (5 x) \sin (x) \, dx=-\frac {1}{5} \log (\cos (x))+\frac {\log \left (5-\sqrt {5}-8 \cos ^2(x)\right )}{\sqrt {5} \left (5-\sqrt {5}\right )}-\frac {\log \left (5+\sqrt {5}-8 \cos ^2(x)\right )}{\sqrt {5} \left (5+\sqrt {5}\right )} \] Output:
-1/5*ln(cos(x))+1/5*ln(5-5^(1/2)-8*cos(x)^2)*5^(1/2)/(5-5^(1/2))-1/5*ln(5+ 5^(1/2)-8*cos(x)^2)*5^(1/2)/(5+5^(1/2))
Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.80 \[ \int \sec (5 x) \sin (x) \, dx=\frac {1}{20} \left (-4 \log (\cos (x))-\left (-1+\sqrt {5}\right ) \log \left (-1-\sqrt {5}+4 \cos (2 x)\right )+\left (1+\sqrt {5}\right ) \log \left (-1+\sqrt {5}+4 \cos (2 x)\right )\right ) \] Input:
Integrate[Sec[5*x]*Sin[x],x]
Output:
(-4*Log[Cos[x]] - (-1 + Sqrt[5])*Log[-1 - Sqrt[5] + 4*Cos[2*x]] + (1 + Sqr t[5])*Log[-1 + Sqrt[5] + 4*Cos[2*x]])/20
Time = 0.31 (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, 4857, 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.
\(\displaystyle \int \sin (x) \sec (5 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)}{\cos (5 x)}dx\) |
\(\Big \downarrow \) 4857 |
\(\displaystyle -\int \frac {\sec (x)}{16 \cos ^4(x)-20 \cos ^2(x)+5}d\cos (x)\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle -\frac {1}{2} \int \frac {\sec (x)}{16 \cos ^4(x)-20 \cos ^2(x)+5}d\cos ^2(x)\) |
\(\Big \downarrow \) 1141 |
\(\displaystyle -8 \int \left (\frac {\sec (x)}{80}+\frac {1}{\sqrt {5} \left (5-\sqrt {5}\right ) \left (-8 \cos ^2(x)-\sqrt {5}+5\right )}-\frac {1}{\sqrt {5} \left (5+\sqrt {5}\right ) \left (-8 \cos ^2(x)+\sqrt {5}+5\right )}\right )d\cos ^2(x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \left (\frac {1}{80} \log \left (\cos ^2(x)\right )-\frac {\log \left (-8 \cos ^2(x)-\sqrt {5}+5\right )}{8 \sqrt {5} \left (5-\sqrt {5}\right )}+\frac {\log \left (-8 \cos ^2(x)+\sqrt {5}+5\right )}{8 \sqrt {5} \left (5+\sqrt {5}\right )}\right )\) |
Input:
Int[Sec[5*x]*Sin[x],x]
Output:
-8*(Log[Cos[x]^2]/80 - Log[5 - Sqrt[5] - 8*Cos[x]^2]/(8*Sqrt[5]*(5 - Sqrt[ 5])) + Log[5 + Sqrt[5] - 8*Cos[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[Cos[c*(a + b*x)], x]}, Simp[-d/(b*c) Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b* x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])
Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.61
method | result | size |
default | \(-\frac {\ln \left (\cos \left (x \right )\right )}{5}+\frac {\ln \left (16 \cos \left (x \right )^{4}-20 \cos \left (x \right )^{2}+5\right )}{20}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (32 \cos \left (x \right )^{2}-20\right ) \sqrt {5}}{20}\right )}{10}\) | \(43\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{2 i x}+1\right )}{5}+\frac {\ln \left ({\mathrm e}^{4 i x}+\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right ) {\mathrm e}^{2 i x}+1\right )}{20}+\frac {\ln \left ({\mathrm e}^{4 i x}+\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right ) {\mathrm e}^{2 i x}+1\right ) \sqrt {5}}{20}+\frac {\ln \left ({\mathrm e}^{4 i x}+\left (-\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) {\mathrm e}^{2 i x}+1\right )}{20}-\frac {\ln \left ({\mathrm e}^{4 i x}+\left (-\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) {\mathrm e}^{2 i x}+1\right ) \sqrt {5}}{20}\) | \(110\) |
Input:
int(sec(5*x)*sin(x),x,method=_RETURNVERBOSE)
Output:
-1/5*ln(cos(x))+1/20*ln(16*cos(x)^4-20*cos(x)^2+5)+1/10*5^(1/2)*arctanh(1/ 20*(32*cos(x)^2-20)*5^(1/2))
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int \sec (5 x) \sin (x) \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {32 \, \cos \left (x\right )^{4} + 8 \, {\left (\sqrt {5} - 5\right )} \cos \left (x\right )^{2} - 5 \, \sqrt {5} + 15}{16 \, \cos \left (x\right )^{4} - 20 \, \cos \left (x\right )^{2} + 5}\right ) + \frac {1}{20} \, \log \left (16 \, \cos \left (x\right )^{4} - 20 \, \cos \left (x\right )^{2} + 5\right ) - \frac {1}{5} \, \log \left (-\cos \left (x\right )\right ) \] Input:
integrate(sec(5*x)*sin(x),x, algorithm="fricas")
Output:
1/20*sqrt(5)*log((32*cos(x)^4 + 8*(sqrt(5) - 5)*cos(x)^2 - 5*sqrt(5) + 15) /(16*cos(x)^4 - 20*cos(x)^2 + 5)) + 1/20*log(16*cos(x)^4 - 20*cos(x)^2 + 5 ) - 1/5*log(-cos(x))
\[ \int \sec (5 x) \sin (x) \, dx=\int \sin {\left (x \right )} \sec {\left (5 x \right )}\, dx \] Input:
integrate(sec(5*x)*sin(x),x)
Output:
Integral(sin(x)*sec(5*x), x)
\[ \int \sec (5 x) \sin (x) \, dx=\int { \sec \left (5 \, x\right ) \sin \left (x\right ) \,d x } \] Input:
integrate(sec(5*x)*sin(x),x, algorithm="maxima")
Output:
1/5*integrate(-(cos(4*x)*sin(8*x) - cos(8*x)*sin(4*x) + cos(3/2*arctan2(si n(4*x), cos(4*x)))*sin(4*x) + cos(1/2*arctan2(sin(4*x), cos(4*x)))*sin(4*x ) - cos(4*x)*sin(3/2*arctan2(sin(4*x), cos(4*x))) - cos(4*x)*sin(1/2*arcta n2(sin(4*x), cos(4*x))) - sin(4*x))/(2*(cos(4*x) + 1)*cos(8*x) + cos(8*x)^ 2 + cos(4*x)^2 - 2*(cos(8*x) + cos(4*x) - cos(1/2*arctan2(sin(4*x), cos(4* x))) + 1)*cos(3/2*arctan2(sin(4*x), cos(4*x))) + cos(3/2*arctan2(sin(4*x), cos(4*x)))^2 - 2*(cos(8*x) + cos(4*x) + 1)*cos(1/2*arctan2(sin(4*x), cos( 4*x))) + cos(1/2*arctan2(sin(4*x), cos(4*x)))^2 + sin(8*x)^2 + 2*sin(8*x)* sin(4*x) + sin(4*x)^2 - 2*(sin(8*x) + sin(4*x) - sin(1/2*arctan2(sin(4*x), cos(4*x))))*sin(3/2*arctan2(sin(4*x), cos(4*x))) + sin(3/2*arctan2(sin(4* x), cos(4*x)))^2 - 2*(sin(8*x) + sin(4*x))*sin(1/2*arctan2(sin(4*x), cos(4 *x))) + sin(1/2*arctan2(sin(4*x), cos(4*x)))^2 + 2*cos(4*x) + 1), x) + 4/5 *integrate(-(cos(2*x)*sin(8*x) - cos(2*x)*sin(6*x) + cos(2*x)*sin(4*x) - c os(8*x)*sin(2*x) + cos(6*x)*sin(2*x) - cos(4*x)*sin(2*x) - sin(2*x))/(2*(c os(6*x) - cos(4*x) + cos(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 2*(cos(4*x) - c os(2*x) + 1)*cos(6*x) - cos(6*x)^2 + 2*(cos(2*x) - 1)*cos(4*x) - cos(4*x)^ 2 - cos(2*x)^2 + 2*(sin(6*x) - sin(4*x) + sin(2*x))*sin(8*x) - sin(8*x)^2 + 2*(sin(4*x) - sin(2*x))*sin(6*x) - sin(6*x)^2 - sin(4*x)^2 + 2*sin(4*x)* sin(2*x) - sin(2*x)^2 + 2*cos(2*x) - 1), x) - 2/5*integrate(-(cos(4/3*arct an2(sin(6*x), cos(6*x)))*sin(6*x) + cos(2/3*arctan2(sin(6*x), cos(6*x))...
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.94 \[ \int \sec (5 x) \sin (x) \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {{\left | 32 \, \sin \left (x\right )^{2} - 4 \, \sqrt {5} - 12 \right |}}{{\left | 32 \, \sin \left (x\right )^{2} + 4 \, \sqrt {5} - 12 \right |}}\right ) - \frac {1}{10} \, \log \left (-\sin \left (x\right )^{2} + 1\right ) + \frac {1}{20} \, \log \left ({\left | 16 \, \sin \left (x\right )^{4} - 12 \, \sin \left (x\right )^{2} + 1 \right |}\right ) \] Input:
integrate(sec(5*x)*sin(x),x, algorithm="giac")
Output:
1/20*sqrt(5)*log(abs(32*sin(x)^2 - 4*sqrt(5) - 12)/abs(32*sin(x)^2 + 4*sqr t(5) - 12)) - 1/10*log(-sin(x)^2 + 1) + 1/20*log(abs(16*sin(x)^4 - 12*sin( x)^2 + 1))
Time = 17.84 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.66 \[ \int \sec (5 x) \sin (x) \, dx=\ln \left ({\cos \left (x\right )}^2+\frac {\sqrt {5}}{8}-\frac {5}{8}\right )\,\left (\frac {\sqrt {5}}{20}+\frac {1}{20}\right )-\ln \left ({\cos \left (x\right )}^2-\frac {\sqrt {5}}{8}-\frac {5}{8}\right )\,\left (\frac {\sqrt {5}}{20}-\frac {1}{20}\right )-\frac {\ln \left (\cos \left (x\right )\right )}{5} \] Input:
int(sin(x)/cos(5*x),x)
Output:
log(cos(x)^2 + 5^(1/2)/8 - 5/8)*(5^(1/2)/20 + 1/20) - log(cos(x)^2 - 5^(1/ 2)/8 - 5/8)*(5^(1/2)/20 - 1/20) - log(cos(x))/5
\[ \int \sec (5 x) \sin (x) \, dx=\int \frac {\sin \left (x \right )}{\cos \left (5 x \right )}d x -1 \] Input:
int(sec(5*x)*sin(x),x)
Output:
int(sin(x)/cos(5*x),x) - 1