3.76 \(\int \sin (x) \tan (4 x) \, dx\)

Optimal. Leaf size=71 \[ -\sin (x)+\frac{1}{4} \sqrt{2-\sqrt{2}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2-\sqrt{2}}}\right )+\frac{1}{4} \sqrt{2+\sqrt{2}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2+\sqrt{2}}}\right ) \]

[Out]

(Sqrt[2 - Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[2]]])/4 + (Sqrt[2 + Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 + S
qrt[2]]])/4 - Sin[x]

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Rubi [A]  time = 0.109319, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {1279, 1166, 207} \[ -\sin (x)+\frac{1}{4} \sqrt{2-\sqrt{2}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2-\sqrt{2}}}\right )+\frac{1}{4} \sqrt{2+\sqrt{2}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2+\sqrt{2}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]*Tan[4*x],x]

[Out]

(Sqrt[2 - Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[2]]])/4 + (Sqrt[2 + Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 + S
qrt[2]]])/4 - 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 1166

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \sin (x) \tan (4 x) \, dx &=\operatorname{Subst}\left (\int \frac{x^2 \left (4-8 x^2\right )}{1-8 x^2+8 x^4} \, dx,x,\sin (x)\right )\\ &=-\sin (x)-\frac{1}{8} \operatorname{Subst}\left (\int \frac{-8+32 x^2}{1-8 x^2+8 x^4} \, dx,x,\sin (x)\right )\\ &=-\sin (x)-\left (2-\sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{-4+2 \sqrt{2}+8 x^2} \, dx,x,\sin (x)\right )-\left (2+\sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{-4-2 \sqrt{2}+8 x^2} \, dx,x,\sin (x)\right )\\ &=\frac{1}{4} \sqrt{2-\sqrt{2}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2-\sqrt{2}}}\right )+\frac{1}{4} \sqrt{2+\sqrt{2}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2+\sqrt{2}}}\right )-\sin (x)\\ \end{align*}

Mathematica [A]  time = 0.0688958, size = 69, normalized size = 0.97 \[ \frac{1}{4} \left (-4 \sin (x)+\sqrt{2-\sqrt{2}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2-\sqrt{2}}}\right )+\sqrt{2+\sqrt{2}} \tanh ^{-1}\left (\frac{2 \sin (x)}{\sqrt{2+\sqrt{2}}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]*Tan[4*x],x]

[Out]

(Sqrt[2 - Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[2]]] + Sqrt[2 + Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[
2]]] - 4*Sin[x])/4

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Maple [B]  time = 0.166, size = 115, normalized size = 1.6 \begin{align*}{\frac{ \left ( \sqrt{2}-2 \right ) \sqrt{2}}{4\,\sqrt{2-\sqrt{2}}}{\it Artanh} \left ( 2\,{\frac{\sin \left ( x \right ) }{\sqrt{2-\sqrt{2}}}} \right ) }+{\frac{\sqrt{2}\sqrt{2+\sqrt{2}}}{4}{\it Artanh} \left ( 2\,{\frac{\sin \left ( x \right ) }{\sqrt{2+\sqrt{2}}}} \right ) }-\sin \left ( x \right ) +{\frac{\sqrt{2}}{4\,\sqrt{2-\sqrt{2}}}{\it Artanh} \left ( 2\,{\frac{\sin \left ( x \right ) }{\sqrt{2-\sqrt{2}}}} \right ) }-{\frac{\sqrt{2}}{4\,\sqrt{2+\sqrt{2}}}{\it Artanh} \left ( 2\,{\frac{\sin \left ( x \right ) }{\sqrt{2+\sqrt{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)*tan(4*x),x)

[Out]

1/4*(2^(1/2)-2)*2^(1/2)/(2-2^(1/2))^(1/2)*arctanh(2*sin(x)/(2-2^(1/2))^(1/2))+1/4*(2+2^(1/2))^(1/2)*2^(1/2)*ar
ctanh(2*sin(x)/(2+2^(1/2))^(1/2))-sin(x)+1/4*2^(1/2)/(2-2^(1/2))^(1/2)*arctanh(2*sin(x)/(2-2^(1/2))^(1/2))-1/4
*2^(1/2)/(2+2^(1/2))^(1/2)*arctanh(2*sin(x)/(2+2^(1/2))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (\cos \left (7 \, x\right ) + \cos \left (x\right )\right )} \cos \left (8 \, x\right ) +{\left (\sin \left (7 \, x\right ) + \sin \left (x\right )\right )} \sin \left (8 \, x\right ) + \cos \left (7 \, x\right ) + \cos \left (x\right )}{\cos \left (8 \, x\right )^{2} + \sin \left (8 \, x\right )^{2} + 2 \, \cos \left (8 \, x\right ) + 1}\,{d x} - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(4*x),x, algorithm="maxima")

[Out]

integrate(((cos(7*x) + cos(x))*cos(8*x) + (sin(7*x) + sin(x))*sin(8*x) + cos(7*x) + cos(x))/(cos(8*x)^2 + sin(
8*x)^2 + 2*cos(8*x) + 1), x) - sin(x)

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Fricas [A]  time = 2.45544, size = 329, normalized size = 4.63 \begin{align*} \frac{1}{8} \, \sqrt{\sqrt{2} + 2} \log \left (\sqrt{\sqrt{2} + 2} + 2 \, \sin \left (x\right )\right ) - \frac{1}{8} \, \sqrt{\sqrt{2} + 2} \log \left (\sqrt{\sqrt{2} + 2} - 2 \, \sin \left (x\right )\right ) + \frac{1}{8} \, \sqrt{-\sqrt{2} + 2} \log \left (\sqrt{-\sqrt{2} + 2} + 2 \, \sin \left (x\right )\right ) - \frac{1}{8} \, \sqrt{-\sqrt{2} + 2} \log \left (\sqrt{-\sqrt{2} + 2} - 2 \, \sin \left (x\right )\right ) - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(4*x),x, algorithm="fricas")

[Out]

1/8*sqrt(sqrt(2) + 2)*log(sqrt(sqrt(2) + 2) + 2*sin(x)) - 1/8*sqrt(sqrt(2) + 2)*log(sqrt(sqrt(2) + 2) - 2*sin(
x)) + 1/8*sqrt(-sqrt(2) + 2)*log(sqrt(-sqrt(2) + 2) + 2*sin(x)) - 1/8*sqrt(-sqrt(2) + 2)*log(sqrt(-sqrt(2) + 2
) - 2*sin(x)) - sin(x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(4*x),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (x\right ) \tan \left (4 \, x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(4*x),x, algorithm="giac")

[Out]

integrate(sin(x)*tan(4*x), x)