\(\int \sec (5 x) \sin (x) \, dx\) [88]
Optimal result
Integrand size = 7, antiderivative size = 62 \[
\int \sec (5 x) \sin (x) \, dx=-\frac {1}{5} \log (\cos (x))+\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (5-\sqrt {5}-8 \cos ^2(x)\right )+\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (5+\sqrt {5}-8 \cos ^2(x)\right )
\]
[Out]
-1/5*ln(cos(x))+1/20*ln(5-8*cos(x)^2+5^(1/2))*(-5^(1/2)+1)+1/20*ln(5-8*cos(x)^2-5^(1/2))*(5^(1/2)+1)
Rubi [A] (verified)
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {4442, 1128, 719, 29, 646, 31}
\[
\int \sec (5 x) \sin (x) \, dx=\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (-8 \cos ^2(x)-\sqrt {5}+5\right )+\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (-8 \cos ^2(x)+\sqrt {5}+5\right )-\frac {1}{5} \log (\cos (x))
\]
[In]
Int[Sec[5*x]*Sin[x],x]
[Out]
-1/5*Log[Cos[x]] + ((1 + Sqrt[5])*Log[5 - Sqrt[5] - 8*Cos[x]^2])/20 + ((1 - Sqrt[5])*Log[5 + Sqrt[5] - 8*Cos[x
]^2])/20
Rule 29
Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 646
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]
Rule 719
Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
Rule 1128
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Rule 4442
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, Dist[-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])
Rubi steps \begin{align*}
\text {integral}& = -\text {Subst}\left (\int \frac {1}{x \left (5-20 x^2+16 x^4\right )} \, dx,x,\cos (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \left (5-20 x+16 x^2\right )} \, dx,x,\cos ^2(x)\right )\right ) \\ & = -\left (\frac {1}{10} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\cos ^2(x)\right )\right )-\frac {1}{10} \text {Subst}\left (\int \frac {20-16 x}{5-20 x+16 x^2} \, dx,x,\cos ^2(x)\right ) \\ & = -\frac {1}{5} \log (\cos (x))+\frac {1}{5} \left (4 \left (1-\sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{-10-2 \sqrt {5}+16 x} \, dx,x,\cos ^2(x)\right )+\frac {1}{5} \left (4 \left (1+\sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{-10+2 \sqrt {5}+16 x} \, dx,x,\cos ^2(x)\right ) \\ & = -\frac {1}{5} \log (\cos (x))+\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (5-\sqrt {5}-8 \cos ^2(x)\right )+\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (5+\sqrt {5}-8 \cos ^2(x)\right ) \\
\end{align*}
Mathematica [A] (verified)
Time = 0.58 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92
\[
\int \sec (5 x) \sin (x) \, dx=\frac {1}{20} \left (-4 \log (\cos (x))-\left (-1+\sqrt {5}\right ) \log \left (-1-\sqrt {5}+4 \cos (2 x)\right )+\left (1+\sqrt {5}\right ) \log \left (-1+\sqrt {5}+4 \cos (2 x)\right )\right )
\]
[In]
Integrate[Sec[5*x]*Sin[x],x]
[Out]
(-4*Log[Cos[x]] - (-1 + Sqrt[5])*Log[-1 - Sqrt[5] + 4*Cos[2*x]] + (1 + Sqrt[5])*Log[-1 + Sqrt[5] + 4*Cos[2*x]]
)/20
Maple [A] (verified)
Time = 2.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.69
| | |
| 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 |
\(-\frac {\ln \left ({\mathrm e}^{2 i x}+1\right )}{5}+\frac {\ln \left ({\mathrm e}^{4 i x}+\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right ) {\mathrm e}^{2 i x}+1\right )}{20}+\frac {\ln \left ({\mathrm e}^{4 i x}+\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right ) {\mathrm e}^{2 i x}+1\right ) \sqrt {5}}{20}+\frac {\ln \left ({\mathrm e}^{4 i x}+\left (-\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) {\mathrm e}^{2 i x}+1\right )}{20}-\frac {\ln \left ({\mathrm e}^{4 i x}+\left (-\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) {\mathrm e}^{2 i x}+1\right ) \sqrt {5}}{20}\) |
\(110\) |
| | |
|
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[In]
int(sec(5*x)*sin(x),x,method=_RETURNVERBOSE)
[Out]
-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))
Fricas [A] (verification not implemented)
none
Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.16
\[
\int \sec (5 x) \sin (x) \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {32 \, \cos \left (x\right )^{4} + 8 \, {\left (\sqrt {5} - 5\right )} \cos \left (x\right )^{2} - 5 \, \sqrt {5} + 15}{16 \, \cos \left (x\right )^{4} - 20 \, \cos \left (x\right )^{2} + 5}\right ) + \frac {1}{20} \, \log \left (16 \, \cos \left (x\right )^{4} - 20 \, \cos \left (x\right )^{2} + 5\right ) - \frac {1}{5} \, \log \left (-\cos \left (x\right )\right )
\]
[In]
integrate(sec(5*x)*sin(x),x, algorithm="fricas")
[Out]
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))
Sympy [F]
\[
\int \sec (5 x) \sin (x) \, dx=\int \sin {\left (x \right )} \sec {\left (5 x \right )}\, dx
\]
[In]
integrate(sec(5*x)*sin(x),x)
[Out]
Integral(sin(x)*sec(5*x), x)
Maxima [F]
\[
\int \sec (5 x) \sin (x) \, dx=\int { \sec \left (5 \, x\right ) \sin \left (x\right ) \,d x }
\]
[In]
integrate(sec(5*x)*sin(x),x, algorithm="maxima")
[Out]
1/5*integrate(-(cos(4*x)*sin(8*x) - cos(8*x)*sin(4*x) + cos(3/2*arctan2(sin(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*arc
tan2(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) - cos(8*x)*sin(2*x) + cos(6*x)*
sin(2*x) - cos(4*x)*sin(2*x) - sin(2*x))/(2*(cos(6*x) - cos(4*x) + cos(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 2*(co
s(4*x) - cos(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*arctan2(sin(6*x), cos(6*x)))
*sin(6*x) + cos(2/3*arctan2(sin(6*x), cos(6*x)))*sin(6*x) - cos(1/3*arctan2(sin(6*x), cos(6*x)))*sin(6*x) - co
s(6*x)*sin(4/3*arctan2(sin(6*x), cos(6*x))) - cos(6*x)*sin(2/3*arctan2(sin(6*x), cos(6*x))) + cos(6*x)*sin(1/3
*arctan2(sin(6*x), cos(6*x))) + sin(6*x))/(cos(6*x)^2 - 2*(cos(6*x) - cos(2/3*arctan2(sin(6*x), cos(6*x))) + c
os(1/3*arctan2(sin(6*x), cos(6*x))) - 1)*cos(4/3*arctan2(sin(6*x), cos(6*x))) + cos(4/3*arctan2(sin(6*x), cos(
6*x)))^2 - 2*(cos(6*x) + cos(1/3*arctan2(sin(6*x), cos(6*x))) - 1)*cos(2/3*arctan2(sin(6*x), cos(6*x))) + cos(
2/3*arctan2(sin(6*x), cos(6*x)))^2 + 2*(cos(6*x) - 1)*cos(1/3*arctan2(sin(6*x), cos(6*x))) + cos(1/3*arctan2(s
in(6*x), cos(6*x)))^2 + sin(6*x)^2 - 2*(sin(6*x) - sin(2/3*arctan2(sin(6*x), cos(6*x))) + sin(1/3*arctan2(sin(
6*x), cos(6*x))))*sin(4/3*arctan2(sin(6*x), cos(6*x))) + sin(4/3*arctan2(sin(6*x), cos(6*x)))^2 - 2*(sin(6*x)
+ sin(1/3*arctan2(sin(6*x), cos(6*x))))*sin(2/3*arctan2(sin(6*x), cos(6*x))) + sin(2/3*arctan2(sin(6*x), cos(6
*x)))^2 + 2*sin(6*x)*sin(1/3*arctan2(sin(6*x), cos(6*x))) + sin(1/3*arctan2(sin(6*x), cos(6*x)))^2 - 2*cos(6*x
) + 1), x) - 2/5*integrate(-(sin(8*x) - sin(6*x) + sin(4*x) - sin(2*x))/(2*(cos(6*x) - cos(4*x) + cos(2*x) - 1
)*cos(8*x) - cos(8*x)^2 + 2*(cos(4*x) - cos(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) - 1/10*log(cos(2*x)^2 +
sin(2*x)^2 + 2*cos(2*x) + 1)
Giac [A] (verification not implemented)
none
Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08
\[
\int \sec (5 x) \sin (x) \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {{\left | 32 \, \sin \left (x\right )^{2} - 4 \, \sqrt {5} - 12 \right |}}{{\left | 32 \, \sin \left (x\right )^{2} + 4 \, \sqrt {5} - 12 \right |}}\right ) - \frac {1}{10} \, \log \left (-\sin \left (x\right )^{2} + 1\right ) + \frac {1}{20} \, \log \left ({\left | 16 \, \sin \left (x\right )^{4} - 12 \, \sin \left (x\right )^{2} + 1 \right |}\right )
\]
[In]
integrate(sec(5*x)*sin(x),x, algorithm="giac")
[Out]
1/20*sqrt(5)*log(abs(32*sin(x)^2 - 4*sqrt(5) - 12)/abs(32*sin(x)^2 + 4*sqrt(5) - 12)) - 1/10*log(-sin(x)^2 + 1
) + 1/20*log(abs(16*sin(x)^4 - 12*sin(x)^2 + 1))
Mupad [B] (verification not implemented)
Time = 0.67 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.76
\[
\int \sec (5 x) \sin (x) \, dx=\ln \left ({\cos \left (x\right )}^2+\frac {\sqrt {5}}{8}-\frac {5}{8}\right )\,\left (\frac {\sqrt {5}}{20}+\frac {1}{20}\right )-\ln \left ({\cos \left (x\right )}^2-\frac {\sqrt {5}}{8}-\frac {5}{8}\right )\,\left (\frac {\sqrt {5}}{20}-\frac {1}{20}\right )-\frac {\ln \left (\cos \left (x\right )\right )}{5}
\]
[In]
int(sin(x)/cos(5*x),x)
[Out]
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) - lo
g(cos(x))/5