\(\int x^2 \coth ^2(a+2 \log (x)) \, dx\) [159]

Optimal result

Integrand size = 13, antiderivative size = 68 \[ \int x^2 \coth ^2(a+2 \log (x)) \, dx=\frac {x^3}{3}+\frac {x^3}{1-e^{2 a} x^4}+\frac {3}{2} e^{-3 a/2} \arctan \left (e^{a/2} x\right )-\frac {3}{2} e^{-3 a/2} \text {arctanh}\left (e^{a/2} x\right ) \]

[Out]

1/3*x^3+x^3/(1-exp(2*a)*x^4)+3/2*arctan(exp(1/2*a)*x)/exp(3/2*a)-3/2*arctanh(exp(1/2*a)*x)/exp(3/2*a)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {5657, 474, 470, 304, 209, 212} \[ \int x^2 \coth ^2(a+2 \log (x)) \, dx=\frac {3}{2} e^{-3 a/2} \arctan \left (e^{a/2} x\right )-\frac {3}{2} e^{-3 a/2} \text {arctanh}\left (e^{a/2} x\right )+\frac {x^3}{1-e^{2 a} x^4}+\frac {x^3}{3} \]

[In]

Int[x^2*Coth[a + 2*Log[x]]^2,x]

[Out]

x^3/3 + x^3/(1 - E^(2*a)*x^4) + (3*ArcTan[E^(a/2)*x])/(2*E^((3*a)/2)) - (3*ArcTanh[E^(a/2)*x])/(2*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 474

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(-(b*c - a*
d)^2)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b^2*e*n*(p + 1))), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a +
b*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a
, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

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*} \text {integral}& = \int \frac {x^2 \left (-1-e^{2 a} x^4\right )^2}{\left (1-e^{2 a} x^4\right )^2} \, dx \\ & = \frac {x^3}{1-e^{2 a} x^4}-\frac {1}{4} e^{-4 a} \int \frac {x^2 \left (8 e^{4 a}+4 e^{6 a} x^4\right )}{1-e^{2 a} x^4} \, dx \\ & = \frac {x^3}{3}+\frac {x^3}{1-e^{2 a} x^4}-3 \int \frac {x^2}{1-e^{2 a} x^4} \, dx \\ & = \frac {x^3}{3}+\frac {x^3}{1-e^{2 a} x^4}-\frac {1}{2} \left (3 e^{-a}\right ) \int \frac {1}{1-e^a x^2} \, dx+\frac {1}{2} \left (3 e^{-a}\right ) \int \frac {1}{1+e^a x^2} \, dx \\ & = \frac {x^3}{3}+\frac {x^3}{1-e^{2 a} x^4}+\frac {3}{2} e^{-3 a/2} \arctan \left (e^{a/2} x\right )-\frac {3}{2} e^{-3 a/2} \text {arctanh}\left (e^{a/2} x\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.17 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.26 \[ \int x^2 \coth ^2(a+2 \log (x)) \, dx=\frac {e^{-4 a} \left (-9317-17825 e^{2 a} x^4-4787 e^{4 a} x^8+1481 e^{6 a} x^{12}+7 \left (1331+1976 e^{2 a} x^4-398 e^{4 a} x^8-632 e^{6 a} x^{12}+27 e^{8 a} x^{16}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},e^{2 a} x^4\right )\right )}{2688 x^5}+\frac {16 e^{2 a} x^7 \left (1+e^{2 a} x^4\right )^2 \, _4F_3\left (\frac {7}{4},2,2,2;1,1,\frac {19}{4};e^{2 a} x^4\right )}{1155} \]

[In]

Integrate[x^2*Coth[a + 2*Log[x]]^2,x]

[Out]

(-9317 - 17825*E^(2*a)*x^4 - 4787*E^(4*a)*x^8 + 1481*E^(6*a)*x^12 + 7*(1331 + 1976*E^(2*a)*x^4 - 398*E^(4*a)*x
^8 - 632*E^(6*a)*x^12 + 27*E^(8*a)*x^16)*Hypergeometric2F1[3/4, 1, 7/4, E^(2*a)*x^4])/(2688*E^(4*a)*x^5) + (16
*E^(2*a)*x^7*(1 + E^(2*a)*x^4)^2*HypergeometricPFQ[{7/4, 2, 2, 2}, {1, 1, 19/4}, E^(2*a)*x^4])/1155

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.47

method	result	size
risch	\(\frac {x^{3}}{3}-\frac {x^{3}}{-1+{\mathrm e}^{2 a} x^{4}}+\frac {3 \ln \left (-\sqrt {{\mathrm e}^{a}}\, x +1\right )}{4 \left ({\mathrm e}^{a}\right )^{\frac {3}{2}}}-\frac {3 \ln \left (\sqrt {{\mathrm e}^{a}}\, x +1\right )}{4 \left ({\mathrm e}^{a}\right )^{\frac {3}{2}}}+\frac {3 \ln \left (\left (-{\mathrm e}^{a}\right )^{\frac {3}{2}}-{\mathrm e}^{2 a} x \right )}{4 \left (-{\mathrm e}^{a}\right )^{\frac {3}{2}}}-\frac {3 \ln \left (\left (-{\mathrm e}^{a}\right )^{\frac {3}{2}}+{\mathrm e}^{2 a} x \right )}{4 \left (-{\mathrm e}^{a}\right )^{\frac {3}{2}}}\)	\(100\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (49) = 98\).

Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.53 \[ \int x^2 \coth ^2(a+2 \log (x)) \, dx=\frac {4 \, x^{7} e^{\left (4 \, a\right )} - 16 \, x^{3} e^{\left (2 \, a\right )} + 18 \, {\left (x^{4} e^{\left (2 \, a\right )} - 1\right )} \arctan \left (x e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + 9 \, {\left (x^{4} e^{\left (2 \, a\right )} - 1\right )} e^{\left (\frac {1}{2} \, a\right )} \log \left (\frac {x^{2} e^{a} - 2 \, x e^{\left (\frac {1}{2} \, a\right )} + 1}{x^{2} e^{a} - 1}\right )}{12 \, {\left (x^{4} e^{\left (4 \, a\right )} - e^{\left (2 \, a\right )}\right )}} \]

[In]

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

[Out]

1/12*(4*x^7*e^(4*a) - 16*x^3*e^(2*a) + 18*(x^4*e^(2*a) - 1)*arctan(x*e^(1/2*a))*e^(1/2*a) + 9*(x^4*e^(2*a) - 1
)*e^(1/2*a)*log((x^2*e^a - 2*x*e^(1/2*a) + 1)/(x^2*e^a - 1)))/(x^4*e^(4*a) - e^(2*a))

Sympy [F]

\[ \int x^2 \coth ^2(a+2 \log (x)) \, dx=\int x^{2} \coth ^{2}{\left (a + 2 \log {\left (x \right )} \right )}\, dx \]

[In]

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

[Out]

Integral(x**2*coth(a + 2*log(x))**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int x^2 \coth ^2(a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} - 1} + \frac {3}{2} \, \arctan \left (x e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{4} \, e^{\left (-\frac {3}{2} \, a\right )} \log \left (\frac {x e^{a} - e^{\left (\frac {1}{2} \, a\right )}}{x e^{a} + e^{\left (\frac {1}{2} \, a\right )}}\right ) \]

[In]

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

[Out]

1/3*x^3 - x^3/(x^4*e^(2*a) - 1) + 3/2*arctan(x*e^(1/2*a))*e^(-3/2*a) + 3/4*e^(-3/2*a)*log((x*e^a - e^(1/2*a))/
(x*e^a + e^(1/2*a)))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.06 \[ \int x^2 \coth ^2(a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} - 1} + \frac {3}{2} \, \arctan \left (x e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{4} \, e^{\left (-\frac {3}{2} \, a\right )} \log \left (\frac {{\left | 2 \, x e^{a} - 2 \, e^{\left (\frac {1}{2} \, a\right )} \right |}}{{\left | 2 \, x e^{a} + 2 \, e^{\left (\frac {1}{2} \, a\right )} \right |}}\right ) \]

[In]

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

[Out]

1/3*x^3 - x^3/(x^4*e^(2*a) - 1) + 3/2*arctan(x*e^(1/2*a))*e^(-3/2*a) + 3/4*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)))

Mupad [B] (verification not implemented)

Time = 1.98 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int x^2 \coth ^2(a+2 \log (x)) \, dx=\frac {3\,\mathrm {atan}\left (x\,{\left ({\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2\,{\left ({\mathrm {e}}^{2\,a}\right )}^{3/4}}-\frac {x^3}{x^4\,{\mathrm {e}}^{2\,a}-1}+\frac {x^3}{3}+\frac {\mathrm {atan}\left (x\,{\left ({\mathrm {e}}^{2\,a}\right )}^{1/4}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,{\left ({\mathrm {e}}^{2\,a}\right )}^{3/4}} \]

[In]

int(x^2*coth(a + 2*log(x))^2,x)

[Out]

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