3.60.44 \(\int \frac {x-2 x^2+e^{2 e+2 e^5} (24-28 x+8 x^2)+(x-4 e^{2 e+2 e^5} x) \log (x)}{x} \, dx\)

Optimal. Leaf size=27 \[ \left (4 e^{2 e+2 e^5} (6-x)+x\right ) (-x+\log (x)) \]

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Rubi [B]  time = 0.11, antiderivative size = 80, normalized size of antiderivative = 2.96, number of steps used = 5, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {14, 2295} \begin {gather*} -\left (\left (1-4 e^{2 \left (e+e^5\right )}\right ) x^2\right )-\left (1-4 e^{2 \left (e+e^5\right )}\right ) x+\left (1-28 e^{2 \left (e+e^5\right )}\right ) x+\left (1-4 e^{2 \left (e+e^5\right )}\right ) x \log (x)+24 e^{2 \left (e+e^5\right )} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x - 2*x^2 + E^(2*E + 2*E^5)*(24 - 28*x + 8*x^2) + (x - 4*E^(2*E + 2*E^5)*x)*Log[x])/x,x]

[Out]

(1 - 28*E^(2*(E + E^5)))*x - (1 - 4*E^(2*(E + E^5)))*x - (1 - 4*E^(2*(E + E^5)))*x^2 + 24*E^(2*(E + E^5))*Log[
x] + (1 - 4*E^(2*(E + E^5)))*x*Log[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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {24 e^{2 \left (e+e^5\right )}+\left (1-28 e^{2 \left (e+e^5\right )}\right ) x-2 \left (1-4 e^{2 \left (e+e^5\right )}\right ) x^2}{x}+\left (1-4 e^{2 \left (e+e^5\right )}\right ) \log (x)\right ) \, dx\\ &=\left (1-4 e^{2 \left (e+e^5\right )}\right ) \int \log (x) \, dx+\int \frac {24 e^{2 \left (e+e^5\right )}+\left (1-28 e^{2 \left (e+e^5\right )}\right ) x-2 \left (1-4 e^{2 \left (e+e^5\right )}\right ) x^2}{x} \, dx\\ &=-\left (\left (1-4 e^{2 \left (e+e^5\right )}\right ) x\right )+\left (1-4 e^{2 \left (e+e^5\right )}\right ) x \log (x)+\int \left (1-28 e^{2 \left (e+e^5\right )}+\frac {24 e^{2 e+2 e^5}}{x}-2 \left (1-4 e^{2 \left (e+e^5\right )}\right ) x\right ) \, dx\\ &=\left (1-28 e^{2 \left (e+e^5\right )}\right ) x-\left (1-4 e^{2 \left (e+e^5\right )}\right ) x-\left (1-4 e^{2 \left (e+e^5\right )}\right ) x^2+24 e^{2 \left (e+e^5\right )} \log (x)+\left (1-4 e^{2 \left (e+e^5\right )}\right ) x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 25, normalized size = 0.93 \begin {gather*} \left (4 e^{2 \left (e+e^5\right )} (-6+x)-x\right ) (x-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - 2*x^2 + E^(2*E + 2*E^5)*(24 - 28*x + 8*x^2) + (x - 4*E^(2*E + 2*E^5)*x)*Log[x])/x,x]

[Out]

(4*E^(2*(E + E^5))*(-6 + x) - x)*(x - Log[x])

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fricas [A]  time = 0.77, size = 48, normalized size = 1.78 \begin {gather*} -x^{2} + 4 \, {\left (x^{2} - 6 \, x\right )} e^{\left (2 \, e^{5} + 2 \, e\right )} - {\left (4 \, {\left (x - 6\right )} e^{\left (2 \, e^{5} + 2 \, e\right )} - x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(exp(5)+exp(1))^2+x)*log(x)+(8*x^2-28*x+24)*exp(exp(5)+exp(1))^2-2*x^2+x)/x,x, algorithm="
fricas")

[Out]

-x^2 + 4*(x^2 - 6*x)*e^(2*e^5 + 2*e) - (4*(x - 6)*e^(2*e^5 + 2*e) - x)*log(x)

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giac [B]  time = 0.18, size = 67, normalized size = 2.48 \begin {gather*} 4 \, x^{2} e^{\left (2 \, e^{5} + 2 \, e\right )} - 4 \, x e^{\left (2 \, e^{5} + 2 \, e\right )} \log \relax (x) - x^{2} - 24 \, x e^{\left (2 \, e^{5} + 2 \, e\right )} + x \log \relax (x) + 24 \, e^{\left (2 \, e^{5} + 2 \, e\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(exp(5)+exp(1))^2+x)*log(x)+(8*x^2-28*x+24)*exp(exp(5)+exp(1))^2-2*x^2+x)/x,x, algorithm="
giac")

[Out]

4*x^2*e^(2*e^5 + 2*e) - 4*x*e^(2*e^5 + 2*e)*log(x) - x^2 - 24*x*e^(2*e^5 + 2*e) + x*log(x) + 24*e^(2*e^5 + 2*e
)*log(x)

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maple [B]  time = 0.07, size = 65, normalized size = 2.41

method	result	size
norman	\(24 \,{\mathrm e}^{2 \,{\mathrm e}^{5}} {\mathrm e}^{2 \,{\mathrm e}} \ln \relax (x )+\left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{5}} {\mathrm e}^{2 \,{\mathrm e}}-1\right ) x^{2}+\left (-4 \,{\mathrm e}^{2 \,{\mathrm e}^{5}} {\mathrm e}^{2 \,{\mathrm e}}+1\right ) x \ln \relax (x )-24 x \,{\mathrm e}^{2 \,{\mathrm e}^{5}} {\mathrm e}^{2 \,{\mathrm e}}\)	\(65\)
risch	\(-x \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{5}+2 \,{\mathrm e}}-1\right ) \ln \relax (x )+4 x^{2} {\mathrm e}^{2 \,{\mathrm e}^{5}+2 \,{\mathrm e}}-24 x \,{\mathrm e}^{2 \,{\mathrm e}^{5}+2 \,{\mathrm e}}-x^{2}+24 \ln \relax (x ) {\mathrm e}^{2 \,{\mathrm e}^{5}+2 \,{\mathrm e}}\)	\(68\)
default	\(-4 \,{\mathrm e}^{2 \,{\mathrm e}^{5}} {\mathrm e}^{2 \,{\mathrm e}} \left (x \ln \relax (x )-x \right )+4 \,{\mathrm e}^{2 \,{\mathrm e}^{5}} {\mathrm e}^{2 \,{\mathrm e}} x^{2}-28 x \,{\mathrm e}^{2 \,{\mathrm e}^{5}} {\mathrm e}^{2 \,{\mathrm e}}+x \ln \relax (x )+24 \,{\mathrm e}^{2 \,{\mathrm e}^{5}} {\mathrm e}^{2 \,{\mathrm e}} \ln \relax (x )-x^{2}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*exp(exp(5)+exp(1))^2+x)*ln(x)+(8*x^2-28*x+24)*exp(exp(5)+exp(1))^2-2*x^2+x)/x,x,method=_RETURNVERBO
SE)

[Out]

24*ln(x)*exp(exp(5))^2*exp(exp(1))^2+(4*exp(exp(5))^2*exp(exp(1))^2-1)*x^2+(-4*exp(exp(5))^2*exp(exp(1))^2+1)*
x*ln(x)-24*x*exp(exp(5))^2*exp(exp(1))^2

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maxima [B]  time = 0.44, size = 72, normalized size = 2.67 \begin {gather*} 4 \, x^{2} e^{\left (2 \, e^{5} + 2 \, e\right )} - x^{2} - 4 \, {\left (x \log \relax (x) - x\right )} e^{\left (2 \, e^{5} + 2 \, e\right )} - 28 \, x e^{\left (2 \, e^{5} + 2 \, e\right )} + x \log \relax (x) + 24 \, e^{\left (2 \, e^{5} + 2 \, e\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(exp(5)+exp(1))^2+x)*log(x)+(8*x^2-28*x+24)*exp(exp(5)+exp(1))^2-2*x^2+x)/x,x, algorithm="
maxima")

[Out]

4*x^2*e^(2*e^5 + 2*e) - x^2 - 4*(x*log(x) - x)*e^(2*e^5 + 2*e) - 28*x*e^(2*e^5 + 2*e) + x*log(x) + 24*e^(2*e^5
+ 2*e)*log(x)

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mupad [B]  time = 4.34, size = 35, normalized size = 1.30 \begin {gather*} -\left (x-\ln \relax (x)\right )\,\left (x+24\,{\mathrm {e}}^{2\,\mathrm {e}+2\,{\mathrm {e}}^5}-4\,x\,{\mathrm {e}}^{2\,\mathrm {e}+2\,{\mathrm {e}}^5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + exp(2*exp(1) + 2*exp(5))*(8*x^2 - 28*x + 24) + log(x)*(x - 4*x*exp(2*exp(1) + 2*exp(5))) - 2*x^2)/x,x
)

[Out]

-(x - log(x))*(x + 24*exp(2*exp(1) + 2*exp(5)) - 4*x*exp(2*exp(1) + 2*exp(5)))

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sympy [B]  time = 0.21, size = 76, normalized size = 2.81 \begin {gather*} - x^{2} \left (- 4 e^{2 e} e^{2 e^{5}} + 1\right ) - 24 x e^{2 e} e^{2 e^{5}} + \left (- 4 x e^{2 e} e^{2 e^{5}} + x\right ) \log {\relax (x )} + 24 e^{2 e} e^{2 e^{5}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(exp(5)+exp(1))**2+x)*ln(x)+(8*x**2-28*x+24)*exp(exp(5)+exp(1))**2-2*x**2+x)/x,x)

[Out]

-x**2*(-4*exp(2*E)*exp(2*exp(5)) + 1) - 24*x*exp(2*E)*exp(2*exp(5)) + (-4*x*exp(2*E)*exp(2*exp(5)) + x)*log(x)
+ 24*exp(2*E)*exp(2*exp(5))*log(x)

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