3.76.59 \(\int \frac {-1+2 x^2+2 x^3-2 e^{-4+e x} x^3+(1-x+e^{-4+e x} x) \log (x)}{-2 x^3+x \log (x)} \, dx\) [7559]

Optimal. Leaf size=30 \[ -x+\log \left (\frac {e^{-5+e^{-5+e x}} x}{2 x^2-\log (x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.98, antiderivative size = 26, normalized size of antiderivative = 0.87, number of steps used = 9, number of rules used = 5, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {2641, 6874, 2225, 45, 6816} \begin {gather*} -\log \left (2 x^2-\log (x)\right )-x+e^{e x-5}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2225

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 2641

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

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6874

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

Rubi steps

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

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Mathematica [A]
time = 0.28, size = 26, normalized size = 0.87 \begin {gather*} e^{-5+e x}-x+\log (x)-\log \left (2 x^2-\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 2.52, size = 25, normalized size = 0.83

method	result	size
default	\(-x +\ln \left (x \right )-\ln \left (-2 x^{2}+\ln \left (x \right )\right )+{\mathrm e}^{x \,{\mathrm e}-5}\)	\(25\)
risch	\(-x +\ln \left (x \right )-\ln \left (-2 x^{2}+\ln \left (x \right )\right )+{\mathrm e}^{x \,{\mathrm e}-5}\)	\(25\)
norman	\(\ln \left (x \right )-x +{\mathrm e}^{x \,{\mathrm e}-5}-\ln \left (2 x^{2}-\ln \left (x \right )\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 30, normalized size = 1.00 \begin {gather*} -{\left (x e^{5} - e^{\left (x e\right )}\right )} e^{\left (-5\right )} - \log \left (-2 \, x^{2} + \log \left (x\right )\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 36, normalized size = 1.20 \begin {gather*} -{\left (x e + e \log \left (-2 \, x^{2} + \log \left (x\right )\right ) - e \log \left (x\right ) - e^{\left (x e - 4\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.13, size = 22, normalized size = 0.73 \begin {gather*} - x + e^{e x - 5} + \log {\left (x \right )} - \log {\left (- 2 x^{2} + \log {\left (x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + exp(E*x - 5) + log(x) - log(-2*x**2 + log(x))

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Giac [A]
time = 0.41, size = 36, normalized size = 1.20 \begin {gather*} -{\left (x e^{5} + e^{5} \log \left (2 \, x^{2} - \log \left (x\right )\right ) - e^{5} \log \left (x\right ) - e^{\left (x e\right )}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.66, size = 24, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^{x\,\mathrm {e}-5}-x+\ln \left (x\right )-\ln \left (\ln \left (x\right )-2\,x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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