3.50.27 \(\int \frac {-1+2 e^x x^2 \log (5)+(-2 x^2+e^2 (8 x^2+4 x^3)) \log (5)-\log (\frac {4}{x})}{2 x^2 \log (5)} \, dx\)

Optimal. Leaf size=33 \[ e^x-x+e^2 (2+x)^2+\frac {\log \left (\frac {4}{x}\right )}{2 x \log (5)} \]

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Rubi [A]  time = 0.08, antiderivative size = 48, normalized size of antiderivative = 1.45, number of steps used = 9, number of rules used = 4, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {12, 14, 2194, 2304} \begin {gather*} e^2 x^2-\left (1-4 e^2\right ) x+\frac {e^x \log (25)}{2 \log (5)}+\frac {\log \left (\frac {4}{x}\right )}{2 x \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 2*E^x*x^2*Log[5] + (-2*x^2 + E^2*(8*x^2 + 4*x^3))*Log[5] - Log[4/x])/(2*x^2*Log[5]),x]

[Out]

-((1 - 4*E^2)*x) + E^2*x^2 + (E^x*Log[25])/(2*Log[5]) + Log[4/x]/(2*x*Log[5])

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 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
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{x^2} \, dx}{2 \log (5)}\\ &=\frac {\int \left (e^x \log (25)+\frac {-1-2 \left (1-4 e^2\right ) x^2 \log (5)+4 e^2 x^3 \log (5)-\log \left (\frac {4}{x}\right )}{x^2}\right ) \, dx}{2 \log (5)}\\ &=\frac {\int \frac {-1-2 \left (1-4 e^2\right ) x^2 \log (5)+4 e^2 x^3 \log (5)-\log \left (\frac {4}{x}\right )}{x^2} \, dx}{2 \log (5)}+\frac {\log (25) \int e^x \, dx}{2 \log (5)}\\ &=\frac {e^x \log (25)}{2 \log (5)}+\frac {\int \left (\frac {-1-2 \left (1-4 e^2\right ) x^2 \log (5)+4 e^2 x^3 \log (5)}{x^2}-\frac {\log \left (\frac {4}{x}\right )}{x^2}\right ) \, dx}{2 \log (5)}\\ &=\frac {e^x \log (25)}{2 \log (5)}+\frac {\int \frac {-1-2 \left (1-4 e^2\right ) x^2 \log (5)+4 e^2 x^3 \log (5)}{x^2} \, dx}{2 \log (5)}-\frac {\int \frac {\log \left (\frac {4}{x}\right )}{x^2} \, dx}{2 \log (5)}\\ &=-\frac {1}{2 x \log (5)}+\frac {e^x \log (25)}{2 \log (5)}+\frac {\log \left (\frac {4}{x}\right )}{2 x \log (5)}+\frac {\int \left (-\frac {1}{x^2}+2 \left (-1+4 e^2\right ) \log (5)+4 e^2 x \log (5)\right ) \, dx}{2 \log (5)}\\ &=-\left (\left (1-4 e^2\right ) x\right )+e^2 x^2+\frac {e^x \log (25)}{2 \log (5)}+\frac {\log \left (\frac {4}{x}\right )}{2 x \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 45, normalized size = 1.36 \begin {gather*} \frac {-2 x \log (5)+8 e^2 x \log (5)+2 e^2 x^2 \log (5)+e^x \log (25)+\frac {\log \left (\frac {4}{x}\right )}{x}}{\log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 2*E^x*x^2*Log[5] + (-2*x^2 + E^2*(8*x^2 + 4*x^3))*Log[5] - Log[4/x])/(2*x^2*Log[5]),x]

[Out]

(-2*x*Log[5] + 8*E^2*x*Log[5] + 2*E^2*x^2*Log[5] + E^x*Log[25] + Log[4/x]/x)/Log[25]

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fricas [A]  time = 0.70, size = 44, normalized size = 1.33 \begin {gather*} \frac {2 \, x e^{x} \log \relax (5) - 2 \, {\left (x^{2} - {\left (x^{3} + 4 \, x^{2}\right )} e^{2}\right )} \log \relax (5) + \log \left (\frac {4}{x}\right )}{2 \, x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-log(4/x)+2*x^2*log(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2)*log(5)-1)/x^2/log(5),x, algorithm=
"fricas")

[Out]

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

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giac [A]  time = 0.12, size = 50, normalized size = 1.52 \begin {gather*} \frac {2 \, x^{3} e^{2} \log \relax (5) + 8 \, x^{2} e^{2} \log \relax (5) - 2 \, x^{2} \log \relax (5) + 2 \, x e^{x} \log \relax (5) + 2 \, \log \relax (2) - \log \relax (x)}{2 \, x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-log(4/x)+2*x^2*log(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2)*log(5)-1)/x^2/log(5),x, algorithm=
"giac")

[Out]

1/2*(2*x^3*e^2*log(5) + 8*x^2*e^2*log(5) - 2*x^2*log(5) + 2*x*e^x*log(5) + 2*log(2) - log(x))/(x*log(5))

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maple [A]  time = 0.13, size = 42, normalized size = 1.27

method	result	size
norman	\(\frac {x^{3} {\mathrm e}^{2}+\left (4 \,{\mathrm e}^{2}-1\right ) x^{2}+{\mathrm e}^{x} x +\frac {\ln \left (\frac {4}{x}\right )}{2 \ln \relax (5)}}{x}\)	\(42\)
default	\(\frac {2 \ln \relax (5) {\mathrm e}^{2} x^{2}+8 x \,{\mathrm e}^{2} \ln \relax (5)-2 x \ln \relax (5)+\frac {\ln \left (\frac {4}{x}\right )}{x}+2 \,{\mathrm e}^{x} \ln \relax (5)}{2 \ln \relax (5)}\)	\(45\)
risch	\(-\frac {\ln \relax (x )}{2 \ln \relax (5) x}+\frac {4 \ln \relax (5) {\mathrm e}^{2} x^{3}+16 \ln \relax (5) {\mathrm e}^{2} x^{2}-4 x^{2} \ln \relax (5)+4 x \,{\mathrm e}^{x} \ln \relax (5)+4 \ln \relax (2)}{4 \ln \relax (5) x}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-ln(4/x)+2*x^2*ln(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2)*ln(5)-1)/x^2/ln(5),x,method=_RETURNVERBOSE
)

[Out]

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

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maxima [A]  time = 0.36, size = 44, normalized size = 1.33 \begin {gather*} \frac {2 \, x^{2} e^{2} \log \relax (5) + 8 \, x e^{2} \log \relax (5) - 2 \, x \log \relax (5) + 2 \, e^{x} \log \relax (5) + \frac {\log \left (\frac {4}{x}\right )}{x}}{2 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-log(4/x)+2*x^2*log(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2)*log(5)-1)/x^2/log(5),x, algorithm=
"maxima")

[Out]

1/2*(2*x^2*e^2*log(5) + 8*x*e^2*log(5) - 2*x*log(5) + 2*e^x*log(5) + log(4/x)/x)/log(5)

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mupad [B]  time = 4.49, size = 40, normalized size = 1.21 \begin {gather*} x^2\,{\mathrm {e}}^2+x\,\left (4\,{\mathrm {e}}^2-1\right )+\frac {{\mathrm {e}}^x\,\ln \left (25\right )}{2\,\ln \relax (5)}+\frac {\ln \left (\frac {4}{x}\right )}{2\,x\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(4/x)/2 - (log(5)*(exp(2)*(8*x^2 + 4*x^3) - 2*x^2))/2 - x^2*exp(x)*log(5) + 1/2)/(x^2*log(5)),x)

[Out]

x^2*exp(2) + x*(4*exp(2) - 1) + (exp(x)*log(25))/(2*log(5)) + log(4/x)/(2*x*log(5))

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sympy [A]  time = 0.37, size = 29, normalized size = 0.88 \begin {gather*} x^{2} e^{2} + x \left (-1 + 4 e^{2}\right ) + e^{x} + \frac {\log {\left (\frac {4}{x} \right )}}{2 x \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-ln(4/x)+2*x**2*ln(5)*exp(x)+((4*x**3+8*x**2)*exp(1)**2-2*x**2)*ln(5)-1)/x**2/ln(5),x)

[Out]

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

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