3.75.48 \(\int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+(-2 \log (3)+e^{1+x} (-2 x+2 x^2) \log (3)) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx\)

Optimal. Leaf size=23 \[ \log (3) \left (x-\left (e^x-\frac {\log (x)}{e x}\right )^2\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 44, normalized size of antiderivative = 1.91, number of steps used = 10, number of rules used = 6, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 14, 2194, 2288, 2304, 2305} \begin {gather*} -\frac {\log (3) \log ^2(x)}{e^2 x^2}+\frac {2 e^{x-1} \log (3) \log (x)}{x}-\frac {1}{2} e^{2 x} \log (9)+x \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*Log[3] - (E^(2*x)*Log[9])/2 + (2*E^(-1 + x)*Log[3]*Log[x])/x - (Log[3]*Log[x]^2)/(E^2*x^2)

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 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 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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{x^3} \, dx}{e^2}\\ &=\frac {\int \left (-e^{2+2 x} \log (9)+\frac {2 e^{1+x} \log (3) (1-\log (x)+x \log (x))}{x^2}+\frac {\log (3) \left (e^2 x^3-2 \log (x)+2 \log ^2(x)\right )}{x^3}\right ) \, dx}{e^2}\\ &=\frac {\log (3) \int \frac {e^2 x^3-2 \log (x)+2 \log ^2(x)}{x^3} \, dx}{e^2}+\frac {(2 \log (3)) \int \frac {e^{1+x} (1-\log (x)+x \log (x))}{x^2} \, dx}{e^2}-\frac {\log (9) \int e^{2+2 x} \, dx}{e^2}\\ &=-\frac {1}{2} e^{2 x} \log (9)+\frac {2 e^{-1+x} \log (3) \log (x)}{x}+\frac {\log (3) \int \left (e^2-\frac {2 \log (x)}{x^3}+\frac {2 \log ^2(x)}{x^3}\right ) \, dx}{e^2}\\ &=x \log (3)-\frac {1}{2} e^{2 x} \log (9)+\frac {2 e^{-1+x} \log (3) \log (x)}{x}-\frac {(2 \log (3)) \int \frac {\log (x)}{x^3} \, dx}{e^2}+\frac {(2 \log (3)) \int \frac {\log ^2(x)}{x^3} \, dx}{e^2}\\ &=\frac {\log (3)}{2 e^2 x^2}+x \log (3)-\frac {1}{2} e^{2 x} \log (9)+\frac {\log (3) \log (x)}{e^2 x^2}+\frac {2 e^{-1+x} \log (3) \log (x)}{x}-\frac {\log (3) \log ^2(x)}{e^2 x^2}+\frac {(2 \log (3)) \int \frac {\log (x)}{x^3} \, dx}{e^2}\\ &=x \log (3)-\frac {1}{2} e^{2 x} \log (9)+\frac {2 e^{-1+x} \log (3) \log (x)}{x}-\frac {\log (3) \log ^2(x)}{e^2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 41, normalized size = 1.78 \begin {gather*} -\frac {\log (3) \left (e^{2+2 x}-e^2 x-\frac {2 e^{1+x} \log (x)}{x}+\frac {\log ^2(x)}{x^2}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.92, size = 47, normalized size = 2.04 \begin {gather*} \frac {{\left (x^{3} e^{2} \log \relax (3) - x^{2} e^{\left (2 \, x + 2\right )} \log \relax (3) + 2 \, x e^{\left (x + 1\right )} \log \relax (3) \log \relax (x) - \log \relax (3) \log \relax (x)^{2}\right )} e^{\left (-2\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.31, size = 47, normalized size = 2.04 \begin {gather*} \frac {{\left (x^{3} e^{2} \log \relax (3) - x^{2} e^{\left (2 \, x + 2\right )} \log \relax (3) + 2 \, x e^{\left (x + 1\right )} \log \relax (3) \log \relax (x) - \log \relax (3) \log \relax (x)^{2}\right )} e^{\left (-2\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 40, normalized size = 1.74

method	result	size
risch	\(\frac {2 \ln \relax (3) \ln \relax (x ) {\mathrm e}^{x -1}}{x}+x \ln \relax (3)-\ln \relax (3) {\mathrm e}^{2 x}-\frac {{\mathrm e}^{-2} \ln \relax (3) \ln \relax (x )^{2}}{x^{2}}\)	\(40\)
default	\({\mathrm e}^{-2} \left (\frac {2 \,{\mathrm e} \ln \relax (3) {\mathrm e}^{x} \ln \relax (x )}{x}+x \,{\mathrm e}^{2} \ln \relax (3)-{\mathrm e}^{2} \ln \relax (3) {\mathrm e}^{2 x}-\frac {\ln \relax (3) \ln \relax (x )^{2}}{x^{2}}\right )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{2} \, {\left (2 \, x e^{2} \log \relax (3) + 4 \, e \Gamma \left (-1, -x\right ) \log \relax (3) + {\left (\frac {2 \, \log \relax (x)}{x^{2}} + \frac {1}{x^{2}}\right )} \log \relax (3) - 2 \, e^{\left (2 \, x + 2\right )} \log \relax (3) - 4 \, \int \frac {e^{\left (x + 1\right )}}{x^{2}}\,{d x} \log \relax (3) + \frac {4 \, x e^{\left (x + 1\right )} \log \relax (3) \log \relax (x) - 2 \, \log \relax (3) \log \relax (x)^{2} - 2 \, \log \relax (3) \log \relax (x) - \log \relax (3)}{x^{2}}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*x*e^2*log(3) + 4*e*gamma(-1, -x)*log(3) + (2*log(x)/x^2 + 1/x^2)*log(3) - 2*e^(2*x + 2)*log(3) - 4*inte
grate(e^(x + 1)/x^2, x)*log(3) + (4*x*e^(x + 1)*log(3)*log(x) - 2*log(3)*log(x)^2 - 2*log(3)*log(x) - log(3))/
x^2)*e^(-2)

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mupad [B]  time = 4.63, size = 40, normalized size = 1.74 \begin {gather*} -\frac {{\mathrm {e}}^{-2}\,\ln \relax (3)\,\left ({\ln \relax (x)}^2-x^3\,{\mathrm {e}}^2+x^2\,{\mathrm {e}}^{2\,x+2}-2\,x\,{\mathrm {e}}^{x+1}\,\ln \relax (x)\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.43, size = 49, normalized size = 2.13 \begin {gather*} x \log {\relax (3 )} + \frac {- e x e^{2 x} \log {\relax (3 )} + 2 e^{x} \log {\relax (3 )} \log {\relax (x )}}{e x} - \frac {\log {\relax (3 )} \log {\relax (x )}^{2}}{x^{2} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(3) + (-E*x*exp(2*x)*log(3) + 2*exp(x)*log(3)*log(x))*exp(-1)/x - exp(-2)*log(3)*log(x)**2/x**2

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