3.44.63 \(\int \frac {-10 x^2+e^x (-10+2 x-e^5 x^2-2 x^3)+(10 x^2+e^x (10+e^5 x^2+2 x^3)) \log (x)}{-2 e^x x^2+2 e^x x^2 \log (x)} \, dx\)

Optimal. Leaf size=29 \[ 3-5 e^{-x}-\frac {5}{x}+\frac {1}{2} x \left (e^5+x\right )+\log (-1+\log (x)) \]

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Rubi [A]  time = 1.30, antiderivative size = 35, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 7, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6741, 12, 6742, 2194, 14, 2302, 29} \begin {gather*} \frac {x^2}{2}+\frac {e^5 x}{2}-5 e^{-x}-\frac {5}{x}+\log (1-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x^2 + E^x*(-10 + 2*x - E^5*x^2 - 2*x^3) + (10*x^2 + E^x*(10 + E^5*x^2 + 2*x^3))*Log[x])/(-2*E^x*x^2 +
2*E^x*x^2*Log[x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 29

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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 6741

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

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 \frac {e^{-x} \left (10 x^2-e^x \left (-10+2 x-e^5 x^2-2 x^3\right )-\left (10 x^2+e^x \left (10+e^5 x^2+2 x^3\right )\right ) \log (x)\right )}{2 x^2 (1-\log (x))} \, dx\\ &=\frac {1}{2} \int \frac {e^{-x} \left (10 x^2-e^x \left (-10+2 x-e^5 x^2-2 x^3\right )-\left (10 x^2+e^x \left (10+e^5 x^2+2 x^3\right )\right ) \log (x)\right )}{x^2 (1-\log (x))} \, dx\\ &=\frac {1}{2} \int \left (10 e^{-x}+\frac {-10+2 x-e^5 x^2-2 x^3+10 \log (x)+e^5 x^2 \log (x)+2 x^3 \log (x)}{x^2 (-1+\log (x))}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-10+2 x-e^5 x^2-2 x^3+10 \log (x)+e^5 x^2 \log (x)+2 x^3 \log (x)}{x^2 (-1+\log (x))} \, dx+5 \int e^{-x} \, dx\\ &=-5 e^{-x}+\frac {1}{2} \int \left (\frac {10+e^5 x^2+2 x^3}{x^2}+\frac {2}{x (-1+\log (x))}\right ) \, dx\\ &=-5 e^{-x}+\frac {1}{2} \int \frac {10+e^5 x^2+2 x^3}{x^2} \, dx+\int \frac {1}{x (-1+\log (x))} \, dx\\ &=-5 e^{-x}+\frac {1}{2} \int \left (e^5+\frac {10}{x^2}+2 x\right ) \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-1+\log (x)\right )\\ &=-5 e^{-x}-\frac {5}{x}+\frac {e^5 x}{2}+\frac {x^2}{2}+\log (1-\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.53, size = 34, normalized size = 1.17 \begin {gather*} \frac {1}{2} \left (-10 e^{-x}-\frac {10}{x}+e^5 x+x^2+2 \log (1-\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x^2 + E^x*(-10 + 2*x - E^5*x^2 - 2*x^3) + (10*x^2 + E^x*(10 + E^5*x^2 + 2*x^3))*Log[x])/(-2*E^x
*x^2 + 2*E^x*x^2*Log[x]),x]

[Out]

(-10/E^x - 10/x + E^5*x + x^2 + 2*Log[1 - Log[x]])/2

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fricas [A]  time = 0.56, size = 37, normalized size = 1.28 \begin {gather*} \frac {{\left (2 \, x e^{x} \log \left (\log \relax (x) - 1\right ) + {\left (x^{3} + x^{2} e^{5} - 10\right )} e^{x} - 10 \, x\right )} e^{\left (-x\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2*exp(5)+2*x^3+10)*exp(x)+10*x^2)*log(x)+(-x^2*exp(5)-2*x^3+2*x-10)*exp(x)-10*x^2)/(2*x^2*exp(x
)*log(x)-2*exp(x)*x^2),x, algorithm="fricas")

[Out]

1/2*(2*x*e^x*log(log(x) - 1) + (x^3 + x^2*e^5 - 10)*e^x - 10*x)*e^(-x)/x

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giac [A]  time = 0.16, size = 31, normalized size = 1.07 \begin {gather*} \frac {x^{3} + x^{2} e^{5} - 10 \, x e^{\left (-x\right )} + 2 \, x \log \left (\log \relax (x) - 1\right ) - 10}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2*exp(5)+2*x^3+10)*exp(x)+10*x^2)*log(x)+(-x^2*exp(5)-2*x^3+2*x-10)*exp(x)-10*x^2)/(2*x^2*exp(x
)*log(x)-2*exp(x)*x^2),x, algorithm="giac")

[Out]

1/2*(x^3 + x^2*e^5 - 10*x*e^(-x) + 2*x*log(log(x) - 1) - 10)/x

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maple [A]  time = 0.08, size = 29, normalized size = 1.00

method	result	size
default	\(\frac {-10+x^{3}+x^{2} {\mathrm e}^{5}}{2 x}+\ln \left (\ln \relax (x )-1\right )-5 \,{\mathrm e}^{-x}\)	\(29\)
risch	\(\frac {\left (x^{2} {\mathrm e}^{5+x}+{\mathrm e}^{x} x^{3}-10 x -10 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{2 x}+\ln \left (\ln \relax (x )-1\right )\)	\(38\)
norman	\(\frac {\left (-5 x +\frac {{\mathrm e}^{x} x^{3}}{2}+\frac {x^{2} {\mathrm e}^{5} {\mathrm e}^{x}}{2}-5 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x}+\ln \left (\ln \relax (x )-1\right )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2*exp(5)+2*x^3+10)*exp(x)+10*x^2)*ln(x)+(-x^2*exp(5)-2*x^3+2*x-10)*exp(x)-10*x^2)/(2*x^2*exp(x)*ln(x)
-2*exp(x)*x^2),x,method=_RETURNVERBOSE)

[Out]

1/2*(-10+x^3+x^2*exp(5))/x+ln(ln(x)-1)-5/exp(x)

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maxima [A]  time = 0.40, size = 33, normalized size = 1.14 \begin {gather*} \frac {{\left ({\left (x^{3} + x^{2} e^{5} - 10\right )} e^{x} - 10 \, x\right )} e^{\left (-x\right )}}{2 \, x} + \log \left (\log \relax (x) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2*exp(5)+2*x^3+10)*exp(x)+10*x^2)*log(x)+(-x^2*exp(5)-2*x^3+2*x-10)*exp(x)-10*x^2)/(2*x^2*exp(x
)*log(x)-2*exp(x)*x^2),x, algorithm="maxima")

[Out]

1/2*((x^3 + x^2*e^5 - 10)*e^x - 10*x)*e^(-x)/x + log(log(x) - 1)

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mupad [B]  time = 3.18, size = 27, normalized size = 0.93 \begin {gather*} \ln \left (\ln \relax (x)-1\right )-5\,{\mathrm {e}}^{-x}+\frac {x\,{\mathrm {e}}^5}{2}-\frac {5}{x}+\frac {x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x^2*exp(5) - 2*x + 2*x^3 + 10) - log(x)*(exp(x)*(x^2*exp(5) + 2*x^3 + 10) + 10*x^2) + 10*x^2)/(2*
x^2*exp(x) - 2*x^2*exp(x)*log(x)),x)

[Out]

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

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sympy [A]  time = 0.36, size = 26, normalized size = 0.90 \begin {gather*} \frac {x^{2}}{2} + \frac {x e^{5}}{2} + \log {\left (\log {\relax (x )} - 1 \right )} - 5 e^{- x} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2*exp(5)+2*x**3+10)*exp(x)+10*x**2)*ln(x)+(-x**2*exp(5)-2*x**3+2*x-10)*exp(x)-10*x**2)/(2*x**2
*exp(x)*ln(x)-2*exp(x)*x**2),x)

[Out]

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

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