3.83.9 \(\int \frac {e^5 (-2-4 x^2-2 x^4)-4 e^5 x^2 \log (x)}{(x+2 x^3+x^5) \log (x)} \, dx\)

Optimal. Leaf size=21 \[ e^5 \log \left (\frac {e^{\frac {2}{1+x^2}}}{\log ^2(x)}\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1594, 28, 6688, 261, 2302, 29} \begin {gather*} \frac {2 e^5}{x^2+1}-2 e^5 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^5)/(1 + x^2) - 2*E^5*Log[Log[x]]

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.00 \begin {gather*} \frac {2 e^5}{1+x^2}-2 e^5 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^5)/(1 + x^2) - 2*E^5*Log[Log[x]]

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fricas [A]  time = 0.70, size = 25, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left ({\left (x^{2} + 1\right )} e^{5} \log \left (\log \relax (x)\right ) - e^{5}\right )}}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2*exp(5)*log(x)+(-2*x^4-4*x^2-2)*exp(5))/(x^5+2*x^3+x)/log(x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 29, normalized size = 1.38 \begin {gather*} -\frac {2 \, {\left (x^{2} e^{5} \log \left (\log \relax (x)\right ) + e^{5} \log \left (\log \relax (x)\right ) - e^{5}\right )}}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2*exp(5)*log(x)+(-2*x^4-4*x^2-2)*exp(5))/(x^5+2*x^3+x)/log(x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 20, normalized size = 0.95

method	result	size
default	\(-2 \,{\mathrm e}^{5} \ln \left (\ln \relax (x )\right )+\frac {2 \,{\mathrm e}^{5}}{x^{2}+1}\)	\(20\)
norman	\(-2 \,{\mathrm e}^{5} \ln \left (\ln \relax (x )\right )+\frac {2 \,{\mathrm e}^{5}}{x^{2}+1}\)	\(20\)
risch	\(-2 \,{\mathrm e}^{5} \ln \left (\ln \relax (x )\right )+\frac {2 \,{\mathrm e}^{5}}{x^{2}+1}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^2*exp(5)*ln(x)+(-2*x^4-4*x^2-2)*exp(5))/(x^5+2*x^3+x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.49, size = 19, normalized size = 0.90 \begin {gather*} -2 \, e^{5} \log \left (\log \relax (x)\right ) + \frac {2 \, e^{5}}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2*exp(5)*log(x)+(-2*x^4-4*x^2-2)*exp(5))/(x^5+2*x^3+x)/log(x),x, algorithm="maxima")

[Out]

-2*e^5*log(log(x)) + 2*e^5/(x^2 + 1)

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mupad [B]  time = 5.14, size = 19, normalized size = 0.90 \begin {gather*} \frac {2\,{\mathrm {e}}^5}{x^2+1}-2\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5)*(4*x^2 + 2*x^4 + 2) + 4*x^2*exp(5)*log(x))/(log(x)*(x + 2*x^3 + x^5)),x)

[Out]

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

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sympy [A]  time = 0.14, size = 20, normalized size = 0.95 \begin {gather*} - 2 e^{5} \log {\left (\log {\relax (x )} \right )} + \frac {4 e^{5}}{2 x^{2} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**2*exp(5)*ln(x)+(-2*x**4-4*x**2-2)*exp(5))/(x**5+2*x**3+x)/ln(x),x)

[Out]

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

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