3.88 \(\int \sec (5 x) \sin (x) \, dx\)
Optimal. Leaf size=62 \[ \frac{1}{20} \left (1+\sqrt{5}\right ) \log \left (-8 \cos ^2(x)-\sqrt{5}+5\right )+\frac{1}{20} \left (1-\sqrt{5}\right ) \log \left (-8 \cos ^2(x)+\sqrt{5}+5\right )-\frac{1}{5} \log (\cos (x)) \]
[Out]
-Log[Cos[x]]/5 + ((1 + Sqrt[5])*Log[5 - Sqrt[5] - 8*Cos[x]^2])/20 + ((1 - Sqrt[5])*Log[5 + Sqrt[5] - 8*Cos[x]^
2])/20
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Rubi [A] time = 0.0725292, antiderivative size = 62, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used =
{4357, 1114, 705, 29, 632, 31} \[ \frac{1}{20} \left (1+\sqrt{5}\right ) \log \left (-8 \cos ^2(x)-\sqrt{5}+5\right )+\frac{1}{20} \left (1-\sqrt{5}\right ) \log \left (-8 \cos ^2(x)+\sqrt{5}+5\right )-\frac{1}{5} \log (\cos (x)) \]
Antiderivative was successfully verified.
[In]
Int[Sec[5*x]*Sin[x],x]
[Out]
-Log[Cos[x]]/5 + ((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 4357
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])
Rule 1114
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 705
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 29
Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]
Rule 632
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 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rubi steps
\begin{align*} \int \sec (5 x) \sin (x) \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x \left (5-20 x^2+16 x^4\right )} \, dx,x,\cos (x)\right )\\ &=-\left (\frac{1}{2} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\cos ^2(x)\right )\right )-\frac{1}{10} \operatorname{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 ) \operatorname{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 ) \operatorname{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] time = 0.0969927, size = 57, normalized size = 0.92 \[ \frac{1}{20} \left (-4 \log (\cos (x))-\left (\sqrt{5}-1\right ) \log \left (4 \cos (2 x)-\sqrt{5}-1\right )+\left (1+\sqrt{5}\right ) \log \left (4 \cos (2 x)+\sqrt{5}-1\right )\right ) \]
Antiderivative was successfully verified.
[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
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Maple [A] time = 0.079, size = 43, normalized size = 0.7 \begin{align*} -{\frac{\ln \left ( \cos \left ( x \right ) \right ) }{5}}+{\frac{\ln \left ( 16\, \left ( \cos \left ( x \right ) \right ) ^{4}-20\, \left ( \cos \left ( x \right ) \right ) ^{2}+5 \right ) }{20}}+{\frac{\sqrt{5}}{10}{\it Artanh} \left ({\frac{ \left ( 32\, \left ( \cos \left ( x \right ) \right ) ^{2}-20 \right ) \sqrt{5}}{20}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(5*x)*sin(x),x)
[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))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Fricas [A] time = 2.78257, size = 230, normalized size = 3.71 \begin{align*} \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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (x \right )} \sec{\left (5 x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(5*x)*sin(x),x)
[Out]
Integral(sin(x)*sec(5*x), x)
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Giac [B] time = 1.26142, size = 159, normalized size = 2.56 \begin{align*} \frac{1}{20} \, \sqrt{5} \log \left ({\left | 2560 \, \cos \left (x\right )^{2} + 320 \, \sqrt{5} - 1600 \right |}\right ) - \frac{1}{20} \, \sqrt{5} \log \left ({\left | 2560 \, \cos \left (x\right )^{2} - 320 \, \sqrt{5} - 1600 \right |}\right ) + \frac{1}{20} \, \log \left ({\left | \frac{44 \,{\left (\cos \left (x\right ) - 1\right )}}{\cos \left (x\right ) + 1} + \frac{166 \,{\left (\cos \left (x\right ) - 1\right )}^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac{44 \,{\left (\cos \left (x\right ) - 1\right )}^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac{{\left (\cos \left (x\right ) - 1\right )}^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + 1 \right |}\right ) - \frac{1}{5} \, \log \left ({\left | -\frac{\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1} - 1 \right |}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(5*x)*sin(x),x, algorithm="giac")
[Out]
1/20*sqrt(5)*log(abs(2560*cos(x)^2 + 320*sqrt(5) - 1600)) - 1/20*sqrt(5)*log(abs(2560*cos(x)^2 - 320*sqrt(5) -
1600)) + 1/20*log(abs(44*(cos(x) - 1)/(cos(x) + 1) + 166*(cos(x) - 1)^2/(cos(x) + 1)^2 + 44*(cos(x) - 1)^3/(c
os(x) + 1)^3 + (cos(x) - 1)^4/(cos(x) + 1)^4 + 1)) - 1/5*log(abs(-(cos(x) - 1)/(cos(x) + 1) - 1))