3.93.100 \(\int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} (-10 x^3+e^{2/x} (-10+10 x^2)+(-20+20 x^2) \log (15))}{x^2-2 x^4+x^6} \, dx\)

Optimal. Leaf size=34 \[ \frac {5 e^{1-e^{2/x}-\frac {4 \log (15)}{x}}}{x \left (-\frac {1}{x}+x\right )} \]

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Rubi [B]  time = 0.63, antiderivative size = 126, normalized size of antiderivative = 3.71, number of steps used = 3, number of rules used = 3, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {1594, 28, 2288} \begin {gather*} -\frac {2\ 3^{-4/x} 5^{1-\frac {4}{x}} e^{\frac {x-e^{2/x} x}{x}} \left (e^{2/x} \left (1-x^2\right )+2 \left (1-x^2\right ) \log (15)\right )}{x^2 \left (1-x^2\right )^2 \left (\frac {-e^{2/x}+\frac {2 e^{2/x}}{x}+1}{x}-\frac {-e^{2/x} x+x-4 \log (15)}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x - E^(2/x)*x - 4*Log[15])/x)*(-10*x^3 + E^(2/x)*(-10 + 10*x^2) + (-20 + 20*x^2)*Log[15]))/(x^2 - 2*x
^4 + x^6),x]

[Out]

(-2*5^(1 - 4/x)*E^((x - E^(2/x)*x)/x)*(E^(2/x)*(1 - x^2) + 2*(1 - x^2)*Log[15]))/(3^(4/x)*x^2*(1 - x^2)^2*((1
- E^(2/x) + (2*E^(2/x))/x)/x - (x - E^(2/x)*x - 4*Log[15])/x^2))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2 \left (1-2 x^2+x^4\right )} \, dx\\ &=\int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2 \left (-1+x^2\right )^2} \, dx\\ &=-\frac {2\ 3^{-4/x} 5^{1-\frac {4}{x}} e^{\frac {x-e^{2/x} x}{x}} \left (e^{2/x} \left (1-x^2\right )+2 \left (1-x^2\right ) \log (15)\right )}{x^2 \left (1-x^2\right )^2 \left (\frac {1-e^{2/x}+\frac {2 e^{2/x}}{x}}{x}-\frac {x-e^{2/x} x-4 \log (15)}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 37, normalized size = 1.09 \begin {gather*} \frac {5^{\frac {-4+x}{x}} 81^{-1/x} e^{1-e^{2/x}}}{-1+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x - E^(2/x)*x - 4*Log[15])/x)*(-10*x^3 + E^(2/x)*(-10 + 10*x^2) + (-20 + 20*x^2)*Log[15]))/(x^2
- 2*x^4 + x^6),x]

[Out]

(5^((-4 + x)/x)*E^(1 - E^(2/x)))/(81^x^(-1)*(-1 + x^2))

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fricas [A]  time = 1.21, size = 31, normalized size = 0.91 \begin {gather*} \frac {5 \, e^{\left (-\frac {x e^{\frac {2}{x}} - x + 4 \, \log \left (15\right )}{x}\right )}}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-10)*exp(2/x)+(20*x^2-20)*log(15)-10*x^3)*exp((-x*exp(2/x)-4*log(15)+x)/x)/(x^6-2*x^4+x^2),x
, algorithm="fricas")

[Out]

5*e^(-(x*e^(2/x) - x + 4*log(15))/x)/(x^2 - 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {10 \, {\left (x^{3} - {\left (x^{2} - 1\right )} e^{\frac {2}{x}} - 2 \, {\left (x^{2} - 1\right )} \log \left (15\right )\right )} e^{\left (-\frac {x e^{\frac {2}{x}} - x + 4 \, \log \left (15\right )}{x}\right )}}{x^{6} - 2 \, x^{4} + x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-10)*exp(2/x)+(20*x^2-20)*log(15)-10*x^3)*exp((-x*exp(2/x)-4*log(15)+x)/x)/(x^6-2*x^4+x^2),x
, algorithm="giac")

[Out]

integrate(-10*(x^3 - (x^2 - 1)*e^(2/x) - 2*(x^2 - 1)*log(15))*e^(-(x*e^(2/x) - x + 4*log(15))/x)/(x^6 - 2*x^4
+ x^2), x)

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maple [A]  time = 0.24, size = 30, normalized size = 0.88

method	result	size
norman	\(\frac {5 \,{\mathrm e}^{\frac {-x \,{\mathrm e}^{\frac {2}{x}}-4 \ln \left (15\right )+x}{x}}}{x^{2}-1}\)	\(30\)
risch	\(\frac {5 \left (\frac {1}{625}\right )^{\frac {1}{x}} \left (\frac {1}{81}\right )^{\frac {1}{x}} {\mathrm e}^{-{\mathrm e}^{\frac {2}{x}}+1}}{x^{2}-1}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2-10)*exp(2/x)+(20*x^2-20)*ln(15)-10*x^3)*exp((-x*exp(2/x)-4*ln(15)+x)/x)/(x^6-2*x^4+x^2),x,method=
_RETURNVERBOSE)

[Out]

5*exp((-x*exp(2/x)-4*ln(15)+x)/x)/(x^2-1)

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maxima [A]  time = 0.54, size = 34, normalized size = 1.00 \begin {gather*} \frac {5 \, e^{\left (-\frac {4 \, \log \relax (5)}{x} - \frac {4 \, \log \relax (3)}{x} - e^{\frac {2}{x}} + 1\right )}}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2-10)*exp(2/x)+(20*x^2-20)*log(15)-10*x^3)*exp((-x*exp(2/x)-4*log(15)+x)/x)/(x^6-2*x^4+x^2),x
, algorithm="maxima")

[Out]

5*e^(-4*log(5)/x - 4*log(3)/x - e^(2/x) + 1)/(x^2 - 1)

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mupad [B]  time = 8.27, size = 29, normalized size = 0.85 \begin {gather*} \frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^{2/x}}\,\mathrm {e}}{{15}^{4/x}\,\left (x^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(4*log(15) - x + x*exp(2/x))/x)*(log(15)*(20*x^2 - 20) + exp(2/x)*(10*x^2 - 10) - 10*x^3))/(x^2 - 2*
x^4 + x^6),x)

[Out]

(5*exp(-exp(2/x))*exp(1))/(15^(4/x)*(x^2 - 1))

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sympy [A]  time = 0.33, size = 22, normalized size = 0.65 \begin {gather*} \frac {5 e^{\frac {- x e^{\frac {2}{x}} + x - 4 \log {\left (15 \right )}}{x}}}{x^{2} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2-10)*exp(2/x)+(20*x**2-20)*ln(15)-10*x**3)*exp((-x*exp(2/x)-4*ln(15)+x)/x)/(x**6-2*x**4+x**
2),x)

[Out]

5*exp((-x*exp(2/x) + x - 4*log(15))/x)/(x**2 - 1)

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