3.75 \(\int \sin (x) \tan (3 x) \, dx\)

Optimal. Leaf size=47 \[ -\sin (x)-\frac{1}{6} \log (1-2 \sin (x))-\frac{1}{6} \log (1-\sin (x))+\frac{1}{6} \log (\sin (x)+1)+\frac{1}{6} \log (2 \sin (x)+1) \]

[Out]

-Log[1 - 2*Sin[x]]/6 - Log[1 - Sin[x]]/6 + Log[1 + Sin[x]]/6 + Log[1 + 2*Sin[x]]/6 - Sin[x]

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Rubi [A]  time = 0.0519936, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {1279, 1161, 616, 31} \[ -\sin (x)-\frac{1}{6} \log (1-2 \sin (x))-\frac{1}{6} \log (1-\sin (x))+\frac{1}{6} \log (\sin (x)+1)+\frac{1}{6} \log (2 \sin (x)+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[1 - 2*Sin[x]]/6 - Log[1 - Sin[x]]/6 + Log[1 + Sin[x]]/6 + Log[1 + 2*Sin[x]]/6 - Sin[x]

Rule 1279

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f
*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m -
2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[
{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (Inte
gerQ[p] || IntegerQ[m])

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[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 \sin (x) \tan (3 x) \, dx &=\operatorname{Subst}\left (\int \frac{x^2 \left (3-4 x^2\right )}{1-5 x^2+4 x^4} \, dx,x,\sin (x)\right )\\ &=-\sin (x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{-4+8 x^2}{1-5 x^2+4 x^4} \, dx,x,\sin (x)\right )\\ &=-\sin (x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2}-\frac{x}{2}+x^2} \, dx,x,\sin (x)\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2}+\frac{x}{2}+x^2} \, dx,x,\sin (x)\right )\\ &=-\sin (x)-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{-1+x} \, dx,x,\sin (x)\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2}+x} \, dx,x,\sin (x)\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}+x} \, dx,x,\sin (x)\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\sin (x)\right )\\ &=-\frac{1}{6} \log (1-2 \sin (x))-\frac{1}{6} \log (1-\sin (x))+\frac{1}{6} \log (1+\sin (x))+\frac{1}{6} \log (1+2 \sin (x))-\sin (x)\\ \end{align*}

Mathematica [A]  time = 0.0284489, size = 21, normalized size = 0.45 \[ -\sin (x)+\frac{1}{3} \tanh ^{-1}(\sin (x))+\frac{1}{3} \tanh ^{-1}(2 \sin (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[x]]/3 + ArcTanh[2*Sin[x]]/3 - Sin[x]

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Maple [A]  time = 0.094, size = 38, normalized size = 0.8 \begin{align*}{\frac{\ln \left ( 1+\sin \left ( x \right ) \right ) }{6}}-{\frac{\ln \left ( \sin \left ( x \right ) -1 \right ) }{6}}+{\frac{\ln \left ( 1+2\,\sin \left ( x \right ) \right ) }{6}}-{\frac{\ln \left ( -1+2\,\sin \left ( x \right ) \right ) }{6}}-\sin \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*ln(1+sin(x))-1/6*ln(sin(x)-1)+1/6*ln(1+2*sin(x))-1/6*ln(-1+2*sin(x))-sin(x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (\cos \left (3 \, x\right ) + \cos \left (x\right )\right )} \cos \left (4 \, x\right ) -{\left (\cos \left (2 \, x\right ) - 1\right )} \cos \left (3 \, x\right ) - \cos \left (2 \, x\right ) \cos \left (x\right ) +{\left (\sin \left (3 \, x\right ) + \sin \left (x\right )\right )} \sin \left (4 \, x\right ) - \sin \left (3 \, x\right ) \sin \left (2 \, x\right ) - \sin \left (2 \, x\right ) \sin \left (x\right ) + \cos \left (x\right )}{3 \,{\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 )}}\,{d x} + \frac{1}{6} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) - \frac{1}{6} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right ) - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.54101, size = 138, normalized size = 2.94 \begin{align*} \frac{1}{6} \, \log \left (2 \, \sin \left (x\right ) + 1\right ) + \frac{1}{6} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{1}{6} \, \log \left (-\sin \left (x\right ) + 1\right ) - \frac{1}{6} \, \log \left (-2 \, \sin \left (x\right ) + 1\right ) - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (x \right )} \tan{\left (3 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(x)*tan(3*x), x)

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Giac [B]  time = 1.25093, size = 491, normalized size = 10.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(log((tan(1/2*x)^4 + 8*tan(1/2*x)^3 + 18*tan(1/2*x)^2 + 8*tan(1/2*x) + 1)/(tan(1/2*x)^4 + 2*tan(1/2*x)^2
+ 1))*tan(1/2*x)^2 - log((tan(1/2*x)^4 - 8*tan(1/2*x)^3 + 18*tan(1/2*x)^2 - 8*tan(1/2*x) + 1)/(tan(1/2*x)^4 +
2*tan(1/2*x)^2 + 1))*tan(1/2*x)^2 + 2*log(2*(tan(1/2*x)^2 + 2*tan(1/2*x) + 1)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^2
- 2*log(2*(tan(1/2*x)^2 - 2*tan(1/2*x) + 1)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^2 + log((tan(1/2*x)^4 + 8*tan(1/2*
x)^3 + 18*tan(1/2*x)^2 + 8*tan(1/2*x) + 1)/(tan(1/2*x)^4 + 2*tan(1/2*x)^2 + 1)) - log((tan(1/2*x)^4 - 8*tan(1/
2*x)^3 + 18*tan(1/2*x)^2 - 8*tan(1/2*x) + 1)/(tan(1/2*x)^4 + 2*tan(1/2*x)^2 + 1)) + 2*log(2*(tan(1/2*x)^2 + 2*
tan(1/2*x) + 1)/(tan(1/2*x)^2 + 1)) - 2*log(2*(tan(1/2*x)^2 - 2*tan(1/2*x) + 1)/(tan(1/2*x)^2 + 1)) - 24*tan(1
/2*x))/(tan(1/2*x)^2 + 1)