3.86 \(\int \sec (3 x) \sin (x) \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{3} \log (\cos (x))-\frac {1}{6} \log \left (3-4 \cos ^2(x)\right ) \]

[Out]

1/3*ln(cos(x))-1/6*ln(3-4*cos(x)^2)

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {4357, 266, 36, 29, 31} \[ \frac {1}{3} \log (\cos (x))-\frac {1}{6} \log \left (3-4 \cos ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sec[3*x]*Sin[x],x]

[Out]

Log[Cos[x]]/3 - Log[3 - 4*Cos[x]^2]/6

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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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])

Rubi steps

\begin {align*} \int \sec (3 x) \sin (x) \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x \left (-3+4 x^2\right )} \, dx,x,\cos (x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x (-3+4 x)} \, dx,x,\cos ^2(x)\right )\right )\\ &=\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\cos ^2(x)\right )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-3+4 x} \, dx,x,\cos ^2(x)\right )\\ &=\frac {1}{3} \log (\cos (x))-\frac {1}{6} \log \left (3-4 \cos ^2(x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.81 \[ -\frac {1}{3} \tanh ^{-1}\left (\frac {1}{3} \left (8 \sin ^2(x)-5\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[3*x]*Sin[x],x]

[Out]

-1/3*ArcTanh[(-5 + 8*Sin[x]^2)/3]

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fricas [A]  time = 0.88, size = 19, normalized size = 0.90 \[ -\frac {1}{6} \, \log \left (4 \, \cos \relax (x)^{2} - 3\right ) + \frac {1}{3} \, \log \left (-\cos \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(3*x)*sin(x),x, algorithm="fricas")

[Out]

-1/6*log(4*cos(x)^2 - 3) + 1/3*log(-cos(x))

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giac [A]  time = 0.14, size = 24, normalized size = 1.14 \[ \frac {1}{6} \, \log \left (-\sin \relax (x)^{2} + 1\right ) - \frac {1}{6} \, \log \left ({\left | 4 \, \sin \relax (x)^{2} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(3*x)*sin(x),x, algorithm="giac")

[Out]

1/6*log(-sin(x)^2 + 1) - 1/6*log(abs(4*sin(x)^2 - 1))

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maple [A]  time = 0.19, size = 18, normalized size = 0.86 \[ -\frac {\ln \left (4 \left (\cos ^{2}\relax (x )\right )-3\right )}{6}+\frac {\ln \left (\cos \relax (x )\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(3*x)*sin(x),x)

[Out]

-1/6*ln(4*cos(x)^2-3)+1/3*ln(cos(x))

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maxima [B]  time = 0.42, size = 81, normalized size = 3.86 \[ -\frac {1}{12} \, \log \left (-2 \, {\left (\cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} - 2 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + \sin \left (2 \, x\right )^{2} - 2 \, \cos \left (2 \, x\right ) + 1\right ) + \frac {1}{6} \, \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(3*x)*sin(x),x, algorithm="maxima")

[Out]

-1/12*log(-2*(cos(2*x) - 1)*cos(4*x) + cos(4*x)^2 + cos(2*x)^2 + sin(4*x)^2 - 2*sin(4*x)*sin(2*x) + sin(2*x)^2
- 2*cos(2*x) + 1) + 1/6*log(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1)

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mupad [B]  time = 2.27, size = 15, normalized size = 0.71 \[ \frac {\ln \left (\cos \relax (x)\right )}{3}-\frac {\ln \left ({\cos \relax (x)}^2-\frac {3}{4}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/cos(3*x),x)

[Out]

log(cos(x))/3 - log(cos(x)^2 - 3/4)/6

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin {\relax (x )} \sec {\left (3 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(3*x)*sin(x),x)

[Out]

Integral(sin(x)*sec(3*x), x)

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