3.2.52 \(\int x^2 \coth (a+2 \log (x)) \, dx\) [152]

Optimal. Leaf size=45 \[ \frac {x^3}{3}+e^{-3 a/2} \text {ArcTan}\left (e^{a/2} x\right )-e^{-3 a/2} \tanh ^{-1}\left (e^{a/2} x\right ) \]

[Out]

1/3*x^3+arctan(exp(1/2*a)*x)/exp(3/2*a)-arctanh(exp(1/2*a)*x)/exp(3/2*a)

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Rubi [A]
time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {5657, 470, 304, 209, 212} \begin {gather*} e^{-3 a/2} \text {ArcTan}\left (e^{a/2} x\right )-e^{-3 a/2} \tanh ^{-1}\left (e^{a/2} x\right )+\frac {x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Coth[a + 2*Log[x]],x]

[Out]

x^3/3 + ArcTan[E^(a/2)*x]/E^((3*a)/2) - ArcTanh[E^(a/2)*x]/E^((3*a)/2)

Rule 209

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

Rule 212

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

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 5657

Int[Coth[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*((-1 - E^(2*a*d)*x^
(2*b*d))^p/(1 - E^(2*a*d)*x^(2*b*d))^p), x] /; FreeQ[{a, b, d, e, m, p}, x]

Rubi steps

\begin {align*} \int x^2 \coth (a+2 \log (x)) \, dx &=\int x^2 \coth (a+2 \log (x)) \, dx\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.17, size = 64, normalized size = 1.42 \begin {gather*} \frac {1}{6} \left (2 x^3+3 \text {RootSum}\left [-\cosh (a)+\sinh (a)+\cosh (a) \text {$\#$1}^4+\sinh (a) \text {$\#$1}^4\&,\frac {\log (x)-\log (x-\text {$\#$1})}{\text {$\#$1}}\&\right ] (-\cosh (2 a)+\sinh (2 a))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Coth[a + 2*Log[x]],x]

[Out]

(2*x^3 + 3*RootSum[-Cosh[a] + Sinh[a] + Cosh[a]*#1^4 + Sinh[a]*#1^4 & , (Log[x] - Log[x - #1])/#1 & ]*(-Cosh[2
*a] + Sinh[2*a]))/6

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(82\) vs. \(2(35)=70\).
time = 0.71, size = 83, normalized size = 1.84

method	result	size
risch	\(\frac {x^{3}}{3}+\frac {\ln \left (-\sqrt {{\mathrm e}^{a}}\, x +1\right )}{2 \left ({\mathrm e}^{a}\right )^{\frac {3}{2}}}-\frac {\ln \left (\sqrt {{\mathrm e}^{a}}\, x +1\right )}{2 \left ({\mathrm e}^{a}\right )^{\frac {3}{2}}}+\frac {\ln \left (-{\mathrm e}^{2 a} x +\left (-{\mathrm e}^{a}\right )^{\frac {3}{2}}\right )}{2 \left (-{\mathrm e}^{a}\right )^{\frac {3}{2}}}-\frac {\ln \left ({\mathrm e}^{2 a} x +\left (-{\mathrm e}^{a}\right )^{\frac {3}{2}}\right )}{2 \left (-{\mathrm e}^{a}\right )^{\frac {3}{2}}}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*coth(a+2*ln(x)),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3+1/2/exp(a)^(3/2)*ln(-exp(a)^(1/2)*x+1)-1/2/exp(a)^(3/2)*ln(exp(a)^(1/2)*x+1)+1/2/(-exp(a))^(3/2)*ln(-e
xp(2*a)*x+(-exp(a))^(3/2))-1/2/(-exp(a))^(3/2)*ln(exp(2*a)*x+(-exp(a))^(3/2))

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Maxima [A]
time = 0.51, size = 48, normalized size = 1.07 \begin {gather*} \frac {1}{3} \, x^{3} + \arctan \left (x e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {1}{2} \, e^{\left (-\frac {3}{2} \, a\right )} \log \left (\frac {x e^{a} - e^{\left (\frac {1}{2} \, a\right )}}{x e^{a} + e^{\left (\frac {1}{2} \, a\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*coth(a+2*log(x)),x, algorithm="maxima")

[Out]

1/3*x^3 + arctan(x*e^(1/2*a))*e^(-3/2*a) + 1/2*e^(-3/2*a)*log((x*e^a - e^(1/2*a))/(x*e^a + e^(1/2*a)))

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Fricas [A]
time = 0.36, size = 62, normalized size = 1.38 \begin {gather*} \frac {1}{6} \, {\left (2 \, x^{3} e^{\left (2 \, a\right )} + 6 \, \arctan \left (x e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + 3 \, e^{\left (\frac {1}{2} \, a\right )} \log \left (\frac {x^{2} e^{a} - 2 \, x e^{\left (\frac {1}{2} \, a\right )} + 1}{x^{2} e^{a} - 1}\right )\right )} e^{\left (-2 \, a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*coth(a+2*log(x)),x, algorithm="fricas")

[Out]

1/6*(2*x^3*e^(2*a) + 6*arctan(x*e^(1/2*a))*e^(1/2*a) + 3*e^(1/2*a)*log((x^2*e^a - 2*x*e^(1/2*a) + 1)/(x^2*e^a
- 1)))*e^(-2*a)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \coth {\left (a + 2 \log {\left (x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*coth(a+2*ln(x)),x)

[Out]

Integral(x**2*coth(a + 2*log(x)), x)

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Giac [A]
time = 0.40, size = 54, normalized size = 1.20 \begin {gather*} \frac {1}{3} \, x^{3} + \arctan \left (x e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {1}{2} \, e^{\left (-\frac {3}{2} \, a\right )} \log \left (\frac {{\left | 2 \, x e^{a} - 2 \, e^{\left (\frac {1}{2} \, a\right )} \right |}}{{\left | 2 \, x e^{a} + 2 \, e^{\left (\frac {1}{2} \, a\right )} \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*coth(a+2*log(x)),x, algorithm="giac")

[Out]

1/3*x^3 + arctan(x*e^(1/2*a))*e^(-3/2*a) + 1/2*e^(-3/2*a)*log(abs(2*x*e^a - 2*e^(1/2*a))/abs(2*x*e^a + 2*e^(1/
2*a)))

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Mupad [B]
time = 1.21, size = 39, normalized size = 0.87 \begin {gather*} \frac {\mathrm {atan}\left (x\,{\left ({\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left ({\mathrm {e}}^{2\,a}\right )}^{3/4}}-\frac {\mathrm {atanh}\left (x\,{\left ({\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left ({\mathrm {e}}^{2\,a}\right )}^{3/4}}+\frac {x^3}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*coth(a + 2*log(x)),x)

[Out]

atan(x*exp(2*a)^(1/4))/exp(2*a)^(3/4) - atanh(x*exp(2*a)^(1/4))/exp(2*a)^(3/4) + x^3/3

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