3.89.31 \(\int \frac {-20 e^3 x-5 e^{3+\frac {-1-e^4 x+x^2}{x}} x+e^{3+\frac {-1-e^4 x+x^2}{x}} (5+5 x^2) \log (\frac {x}{2})+5 x^2 \log ^2(\frac {x}{2})}{x^2 \log ^2(\frac {x}{2})} \, dx\)

Optimal. Leaf size=34 \[ 5 \left (x+\frac {e^3 \left (4+e^{x-\frac {1+e^4 x}{x}}\right )}{\log \left (\frac {x}{2}\right )}\right ) \]

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Rubi [B]  time = 1.32, antiderivative size = 69, normalized size of antiderivative = 2.03, number of steps used = 7, number of rules used = 5, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6742, 2288, 6688, 2302, 30} \begin {gather*} \frac {5 e^{x-\frac {1}{x}-e^4+3} \left (x^2 \log \left (\frac {x}{2}\right )+\log \left (\frac {x}{2}\right )\right )}{\left (\frac {1}{x^2}+1\right ) x^2 \log ^2\left (\frac {x}{2}\right )}+5 x+\frac {20 e^3}{\log \left (\frac {x}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*E^3*x - 5*E^(3 + (-1 - E^4*x + x^2)/x)*x + E^(3 + (-1 - E^4*x + x^2)/x)*(5 + 5*x^2)*Log[x/2] + 5*x^2*
Log[x/2]^2)/(x^2*Log[x/2]^2),x]

[Out]

5*x + (20*E^3)/Log[x/2] + (5*E^(3 - E^4 - x^(-1) + x)*(Log[x/2] + x^2*Log[x/2]))/((1 + x^(-2))*x^2*Log[x/2]^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

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]]

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.45, size = 33, normalized size = 0.97 \begin {gather*} 5 x+\frac {5 e^3 \left (4+e^{-e^4-\frac {1}{x}+x}\right )}{\log \left (\frac {x}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*E^3*x - 5*E^(3 + (-1 - E^4*x + x^2)/x)*x + E^(3 + (-1 - E^4*x + x^2)/x)*(5 + 5*x^2)*Log[x/2] +
5*x^2*Log[x/2]^2)/(x^2*Log[x/2]^2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 37, normalized size = 1.09 \begin {gather*} \frac {5 \, {\left (x \log \left (\frac {1}{2} \, x\right ) + 4 \, e^{3} + e^{\left (\frac {x^{2} - x e^{4} + 3 \, x - 1}{x}\right )}\right )}}{\log \left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*log(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*log(1/2*x)-5*x*exp(3)*exp((-x*exp(4)+x
^2-1)/x)-20*x*exp(3))/x^2/log(1/2*x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 37, normalized size = 1.09 \begin {gather*} \frac {5 \, {\left (x \log \left (\frac {1}{2} \, x\right ) + 4 \, e^{3} + e^{\left (\frac {x^{2} - x e^{4} + 3 \, x - 1}{x}\right )}\right )}}{\log \left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*log(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*log(1/2*x)-5*x*exp(3)*exp((-x*exp(4)+x
^2-1)/x)-20*x*exp(3))/x^2/log(1/2*x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 34, normalized size = 1.00

method	result	size
risch	\(5 x +\frac {5 \,{\mathrm e}^{3} \left ({\mathrm e}^{-\frac {x \,{\mathrm e}^{4}-x^{2}+1}{x}}+4\right )}{\ln \left (\frac {x}{2}\right )}\)	\(34\)
default	\(5 x +\frac {5 \,{\mathrm e}^{3} {\mathrm e}^{\frac {-x \,{\mathrm e}^{4}+x^{2}-1}{x}}}{\ln \left (\frac {x}{2}\right )}+\frac {20 \,{\mathrm e}^{3}}{\ln \left (\frac {x}{2}\right )}\)	\(40\)
norman	\(\frac {20 x \,{\mathrm e}^{3}+5 x^{2} \ln \left (\frac {x}{2}\right )+5 x \,{\mathrm e}^{3} {\mathrm e}^{\frac {-x \,{\mathrm e}^{4}+x^{2}-1}{x}}}{x \ln \left (\frac {x}{2}\right )}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2*ln(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*ln(1/2*x)-5*x*exp(3)*exp((-x*exp(4)+x^2-1)/x)
-20*x*exp(3))/x^2/ln(1/2*x)^2,x,method=_RETURNVERBOSE)

[Out]

5*x+5*exp(3)*(exp(-(x*exp(4)-x^2+1)/x)+4)/ln(1/2*x)

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maxima [A]  time = 0.52, size = 41, normalized size = 1.21 \begin {gather*} 5 \, x - \frac {5 \, e^{\left (x - \frac {1}{x} + 3\right )}}{e^{\left (e^{4}\right )} \log \relax (2) - e^{\left (e^{4}\right )} \log \relax (x)} + \frac {20 \, e^{3}}{\log \left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*log(1/2*x)^2+(5*x^2+5)*exp(3)*exp((-x*exp(4)+x^2-1)/x)*log(1/2*x)-5*x*exp(3)*exp((-x*exp(4)+x
^2-1)/x)-20*x*exp(3))/x^2/log(1/2*x)^2,x, algorithm="maxima")

[Out]

5*x - 5*e^(x - 1/x + 3)/(e^(e^4)*log(2) - e^(e^4)*log(x)) + 20*e^3/log(1/2*x)

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mupad [B]  time = 5.87, size = 43, normalized size = 1.26 \begin {gather*} 5\,x-\frac {20\,{\mathrm {e}}^3}{\ln \relax (2)-\ln \relax (x)}-\frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-\frac {1}{x}}\,{\mathrm {e}}^x}{\ln \relax (2)-\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x*exp(3) - 5*x^2*log(x/2)^2 + 5*x*exp(3)*exp(-(x*exp(4) - x^2 + 1)/x) - log(x/2)*exp(3)*exp(-(x*exp(4
) - x^2 + 1)/x)*(5*x^2 + 5))/(x^2*log(x/2)^2),x)

[Out]

5*x - (20*exp(3))/(log(2) - log(x)) - (5*exp(-exp(4))*exp(3)*exp(-1/x)*exp(x))/(log(2) - log(x))

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sympy [A]  time = 0.31, size = 36, normalized size = 1.06 \begin {gather*} 5 x + \frac {5 e^{3} e^{\frac {x^{2} - x e^{4} - 1}{x}}}{\log {\left (\frac {x}{2} \right )}} + \frac {20 e^{3}}{\log {\left (\frac {x}{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2*ln(1/2*x)**2+(5*x**2+5)*exp(3)*exp((-x*exp(4)+x**2-1)/x)*ln(1/2*x)-5*x*exp(3)*exp((-x*exp(4)
+x**2-1)/x)-20*x*exp(3))/x**2/ln(1/2*x)**2,x)

[Out]

5*x + 5*exp(3)*exp((x**2 - x*exp(4) - 1)/x)/log(x/2) + 20*exp(3)/log(x/2)

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