3.988 \(\int \frac{\text{sech}^2(x)}{3-4 \tanh ^3(x)} \, dx\)

Optimal. Leaf size=102 \[ \frac{\log \left (2 \sqrt [3]{2} \tanh ^2(x)+2^{2/3} \sqrt [3]{3} \tanh (x)+3^{2/3}\right )}{6\ 6^{2/3}}-\frac{\log \left (\sqrt [3]{3}-2^{2/3} \tanh (x)\right )}{3\ 6^{2/3}}+\frac{\tan ^{-1}\left (\frac{2\ 2^{2/3} \tanh (x)+\sqrt [3]{3}}{3^{5/6}}\right )}{3\ 2^{2/3} \sqrt [6]{3}} \]

[Out]

ArcTan[(3^(1/3) + 2*2^(2/3)*Tanh[x])/3^(5/6)]/(3*2^(2/3)*3^(1/6)) - Log[3^(1/3) - 2^(2/3)*Tanh[x]]/(3*6^(2/3))
+ Log[3^(2/3) + 2^(2/3)*3^(1/3)*Tanh[x] + 2*2^(1/3)*Tanh[x]^2]/(6*6^(2/3))

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Rubi [A]  time = 0.112152, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {3675, 200, 31, 634, 617, 204, 628} \[ \frac{\log \left (2 \sqrt [3]{2} \tanh ^2(x)+2^{2/3} \sqrt [3]{3} \tanh (x)+3^{2/3}\right )}{6\ 6^{2/3}}-\frac{\log \left (\sqrt [3]{3}-2^{2/3} \tanh (x)\right )}{3\ 6^{2/3}}+\frac{\tan ^{-1}\left (\frac{2\ 2^{2/3} \tanh (x)+\sqrt [3]{3}}{3^{5/6}}\right )}{3\ 2^{2/3} \sqrt [6]{3}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(3 - 4*Tanh[x]^3),x]

[Out]

ArcTan[(3^(1/3) + 2*2^(2/3)*Tanh[x])/3^(5/6)]/(3*2^(2/3)*3^(1/6)) - Log[3^(1/3) - 2^(2/3)*Tanh[x]]/(3*6^(2/3))
+ Log[3^(2/3) + 2^(2/3)*3^(1/3)*Tanh[x] + 2*2^(1/3)*Tanh[x]^2]/(6*6^(2/3))

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), 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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\text{sech}^2(x)}{3-4 \tanh ^3(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{3-4 x^3} \, dx,x,\tanh (x)\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{3}-2^{2/3} x} \, dx,x,\tanh (x)\right )}{3\ 3^{2/3}}+\frac{\operatorname{Subst}\left (\int \frac{2 \sqrt [3]{3}+2^{2/3} x}{3^{2/3}+2^{2/3} \sqrt [3]{3} x+2 \sqrt [3]{2} x^2} \, dx,x,\tanh (x)\right )}{3\ 3^{2/3}}\\ &=-\frac{\log \left (\sqrt [3]{3}-2^{2/3} \tanh (x)\right )}{3\ 6^{2/3}}+\frac{\operatorname{Subst}\left (\int \frac{1}{3^{2/3}+2^{2/3} \sqrt [3]{3} x+2 \sqrt [3]{2} x^2} \, dx,x,\tanh (x)\right )}{2 \sqrt [3]{3}}+\frac{\operatorname{Subst}\left (\int \frac{2^{2/3} \sqrt [3]{3}+4 \sqrt [3]{2} x}{3^{2/3}+2^{2/3} \sqrt [3]{3} x+2 \sqrt [3]{2} x^2} \, dx,x,\tanh (x)\right )}{6\ 6^{2/3}}\\ &=-\frac{\log \left (\sqrt [3]{3}-2^{2/3} \tanh (x)\right )}{3\ 6^{2/3}}+\frac{\log \left (3^{2/3}+2^{2/3} \sqrt [3]{3} \tanh (x)+2 \sqrt [3]{2} \tanh ^2(x)\right )}{6\ 6^{2/3}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2\ 2^{2/3} \tanh (x)}{\sqrt [3]{3}}\right )}{6^{2/3}}\\ &=\frac{\tan ^{-1}\left (\frac{3+2\ 6^{2/3} \tanh (x)}{3 \sqrt{3}}\right )}{3\ 2^{2/3} \sqrt [6]{3}}-\frac{\log \left (\sqrt [3]{3}-2^{2/3} \tanh (x)\right )}{3\ 6^{2/3}}+\frac{\log \left (3^{2/3}+2^{2/3} \sqrt [3]{3} \tanh (x)+2 \sqrt [3]{2} \tanh ^2(x)\right )}{6\ 6^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.105728, size = 74, normalized size = 0.73 \[ \frac{\log \left (2 \sqrt [3]{6} \tanh ^2(x)+6^{2/3} \tanh (x)+3\right )-2 \log \left (3-6^{2/3} \tanh (x)\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{2\ 6^{2/3} \tanh (x)+3}{3 \sqrt{3}}\right )}{6\ 6^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(3 - 4*Tanh[x]^3),x]

[Out]

(2*Sqrt[3]*ArcTan[(3 + 2*6^(2/3)*Tanh[x])/(3*Sqrt[3])] - 2*Log[3 - 6^(2/3)*Tanh[x]] + Log[3 + 6^(2/3)*Tanh[x]
+ 2*6^(1/3)*Tanh[x]^2])/(6*6^(2/3))

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Maple [C]  time = 0.052, size = 34, normalized size = 0.3 \begin{align*}{\frac{1}{3}\sum _{{\it \_R}={\it RootOf} \left ( 36\,{{\it \_Z}}^{3}+1 \right ) }{\it \_R}\,\ln \left ( -24\,\tanh \left ( x/2 \right ){{\it \_R}}^{2}+ \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(3-4*tanh(x)^3),x)

[Out]

1/3*sum(_R*ln(-24*tanh(1/2*x)*_R^2+tanh(1/2*x)^2+1),_R=RootOf(36*_Z^3+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\operatorname{sech}\left (x\right )^{2}}{4 \, \tanh \left (x\right )^{3} - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(3-4*tanh(x)^3),x, algorithm="maxima")

[Out]

-integrate(sech(x)^2/(4*tanh(x)^3 - 3), x)

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Fricas [B]  time = 2.1398, size = 1177, normalized size = 11.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(3-4*tanh(x)^3),x, algorithm="fricas")

[Out]

-1/18*36^(1/6)*sqrt(3)*(-1)^(1/3)*arctan(1/54*36^(1/6)*((36^(2/3)*sqrt(3)*(-1)^(2/3) + 3*36^(1/3)*sqrt(3) - 9*
sqrt(3)*(-1)^(1/3))*cosh(x)^2 + 2*(36^(2/3)*sqrt(3)*(-1)^(2/3) + 3*36^(1/3)*sqrt(3) - 9*sqrt(3)*(-1)^(1/3))*co
sh(x)*sinh(x) + (36^(2/3)*sqrt(3)*(-1)^(2/3) + 3*36^(1/3)*sqrt(3) - 9*sqrt(3)*(-1)^(1/3))*sinh(x)^2 - 36^(2/3)
*sqrt(3)*(-1)^(2/3) - 9*sqrt(3)*(-1)^(1/3))) - 1/216*36^(2/3)*(-1)^(1/3)*log(2*((36^(2/3)*(-1)^(1/3) + 3*36^(1
/3)*(-1)^(2/3) + 3)*cosh(x)^2 - 2*(36^(2/3)*(-1)^(1/3) + 3*36^(1/3)*(-1)^(2/3))*cosh(x)*sinh(x) + (36^(2/3)*(-
1)^(1/3) + 3*36^(1/3)*(-1)^(2/3) + 3)*sinh(x)^2 - 36^(2/3)*(-1)^(1/3) + 3*36^(1/3)*(-1)^(2/3) - 21)/(cosh(x)^2
- 2*cosh(x)*sinh(x) + sinh(x)^2)) + 1/108*36^(2/3)*(-1)^(1/3)*log(2*((36^(2/3)*(-1)^(1/3) - 3*36^(1/3)*(-1)^(
2/3) - 9)*cosh(x) - (36^(2/3)*(-1)^(1/3) - 3*36^(1/3)*(-1)^(2/3) - 12)*sinh(x))/(cosh(x) - sinh(x)))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\operatorname{sech}^{2}{\left (x \right )}}{4 \tanh ^{3}{\left (x \right )} - 3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(3-4*tanh(x)**3),x)

[Out]

-Integral(sech(x)**2/(4*tanh(x)**3 - 3), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(3-4*tanh(x)^3),x, algorithm="giac")

[Out]

Exception raised: TypeError