3.2.63 \(\int \frac {\coth ^2(a+2 \log (x))}{x^2} \, dx\) [163]

Optimal. Leaf size=86 \[ -\frac {1}{x \left (1-e^{2 a} x^4\right )}+\frac {2 e^{2 a} x^3}{1-e^{2 a} x^4}-\frac {1}{2} e^{a/2} \text {ArcTan}\left (e^{a/2} x\right )+\frac {1}{2} e^{a/2} \tanh ^{-1}\left (e^{a/2} x\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {5657, 473, 468, 304, 209, 212} \begin {gather*} -\frac {1}{2} e^{a/2} \text {ArcTan}\left (e^{a/2} x\right )-\frac {1}{x \left (1-e^{2 a} x^4\right )}+\frac {2 e^{2 a} x^3}{1-e^{2 a} x^4}+\frac {1}{2} e^{a/2} \tanh ^{-1}\left (e^{a/2} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Coth[a + 2*Log[x]]^2/x^2,x]

[Out]

-(1/(x*(1 - E^(2*a)*x^4))) + (2*E^(2*a)*x^3)/(1 - E^(2*a)*x^4) - (E^(a/2)*ArcTan[E^(a/2)*x])/2 + (E^(a/2)*ArcT
anh[E^(a/2)*x])/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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 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 \frac {\coth ^2(a+2 \log (x))}{x^2} \, dx &=\int \frac {\coth ^2(a+2 \log (x))}{x^2} \, dx\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 1.93, size = 153, normalized size = 1.78 \begin {gather*} \frac {e^{-2 a} \left (-343-1163 e^{2 a} x^4-241 e^{4 a} x^8+3 e^{6 a} x^{12}+\left (343+632 e^{2 a} x^4+362 e^{4 a} x^8-56 e^{6 a} x^{12}-e^{8 a} x^{16}\right ) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};e^{2 a} x^4\right )\right )}{384 x^5}+\frac {16}{231} e^{2 a} x^3 \left (1+e^{2 a} x^4\right )^2 \, _4F_3\left (\frac {3}{4},2,2,2;1,1,\frac {15}{4};e^{2 a} x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Coth[a + 2*Log[x]]^2/x^2,x]

[Out]

(-343 - 1163*E^(2*a)*x^4 - 241*E^(4*a)*x^8 + 3*E^(6*a)*x^12 + (343 + 632*E^(2*a)*x^4 + 362*E^(4*a)*x^8 - 56*E^
(6*a)*x^12 - E^(8*a)*x^16)*Hypergeometric2F1[3/4, 1, 7/4, E^(2*a)*x^4])/(384*E^(2*a)*x^5) + (16*E^(2*a)*x^3*(1
+ E^(2*a)*x^4)^2*HypergeometricPFQ[{3/4, 2, 2, 2}, {1, 1, 15/4}, E^(2*a)*x^4])/231

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.61, size = 104, normalized size = 1.21

method	result	size
risch	\(\frac {-2 \,{\mathrm e}^{2 a} x^{4}+1}{x \left (-1+{\mathrm e}^{2 a} x^{4}\right )}+\frac {\sqrt {{\mathrm e}^{a}}\, \ln \left (-\left ({\mathrm e}^{a}\right )^{\frac {3}{2}}-{\mathrm e}^{2 a} x \right )}{4}-\frac {\sqrt {{\mathrm e}^{a}}\, \ln \left (\left ({\mathrm e}^{a}\right )^{\frac {3}{2}}-{\mathrm e}^{2 a} x \right )}{4}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{2}+{\mathrm e}^{a}\right )}{\sum }\textit {\_R} \ln \left (\left (-5 \textit {\_R}^{4}+4 \,{\mathrm e}^{2 a}\right ) x -\textit {\_R}^{3}\right )\right )}{4}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*exp(2*a)*x^4+1)/x/(-1+exp(2*a)*x^4)+1/4*exp(a)^(1/2)*ln(-exp(a)^(3/2)-exp(2*a)*x)-1/4*exp(a)^(1/2)*ln(exp(
a)^(3/2)-exp(2*a)*x)+1/4*sum(_R*ln((-5*_R^4+4*exp(2*a))*x-_R^3),_R=RootOf(_Z^2+exp(a)))

________________________________________________________________________________________

Maxima [A]
time = 0.48, size = 69, normalized size = 0.80 \begin {gather*} \frac {1}{2} \, \arctan \left (\frac {e^{\left (-\frac {1}{2} \, a\right )}}{x}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, e^{\left (\frac {1}{2} \, a\right )} \log \left (\frac {\frac {1}{x} - e^{\left (\frac {1}{2} \, a\right )}}{\frac {1}{x} + e^{\left (\frac {1}{2} \, a\right )}}\right ) - \frac {1}{x} + \frac {e^{\left (2 \, a\right )}}{x {\left (\frac {1}{x^{4}} - e^{\left (2 \, a\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 97, normalized size = 1.13 \begin {gather*} -\frac {8 \, x^{4} e^{\left (2 \, a\right )} + 2 \, {\left (x^{5} e^{\left (2 \, a\right )} - x\right )} \arctan \left (x e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - {\left (x^{5} e^{\left (2 \, a\right )} - x\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 ) - 4}{4 \, {\left (x^{5} e^{\left (2 \, a\right )} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*x^4*e^(2*a) + 2*(x^5*e^(2*a) - x)*arctan(x*e^(1/2*a))*e^(1/2*a) - (x^5*e^(2*a) - x)*e^(1/2*a)*log((x^2
*e^a + 2*x*e^(1/2*a) + 1)/(x^2*e^a - 1)) - 4)/(x^5*e^(2*a) - x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.40, size = 77, normalized size = 0.90 \begin {gather*} -\frac {1}{2} \, \arctan \left (x e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, e^{\left (\frac {1}{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 ) - \frac {2 \, x^{4} e^{\left (2 \, a\right )} - 1}{x^{5} e^{\left (2 \, a\right )} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctan(x*e^(1/2*a))*e^(1/2*a) - 1/4*e^(1/2*a)*log(abs(2*x*e^a - 2*e^(1/2*a))/abs(2*x*e^a + 2*e^(1/2*a)))
- (2*x^4*e^(2*a) - 1)/(x^5*e^(2*a) - x)

________________________________________________________________________________________

Mupad [B]
time = 1.21, size = 60, normalized size = 0.70 \begin {gather*} \frac {{\left ({\mathrm {e}}^{2\,a}\right )}^{1/4}\,\mathrm {atanh}\left (x\,{\left ({\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2}-\frac {{\left ({\mathrm {e}}^{2\,a}\right )}^{1/4}\,\mathrm {atan}\left (x\,{\left ({\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2}+\frac {2\,x^4\,{\mathrm {e}}^{2\,a}-1}{x-x^5\,{\mathrm {e}}^{2\,a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + 2*log(x))^2/x^2,x)

[Out]

(exp(2*a)^(1/4)*atanh(x*exp(2*a)^(1/4)))/2 - (exp(2*a)^(1/4)*atan(x*exp(2*a)^(1/4)))/2 + (2*x^4*exp(2*a) - 1)/
(x - x^5*exp(2*a))

________________________________________________________________________________________