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

Optimal. Leaf size=19 \[ e^{3+\log ^2(x)} \log (5) \log \left (-\frac {5}{2 x^2}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {2288} \begin {gather*} \log (5) e^{\log ^2(x)+3} \log \left (-\frac {5}{2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \begin {gather*} e^{3+\log ^2(x)} \log (5) \log \left (-\frac {5}{2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.64, size = 57, normalized size = 3.00 \begin {gather*} e^{\left (-\frac {1}{4} \, \pi ^{2} + \frac {1}{2} i \, \pi \log \left (\frac {5}{2}\right ) + \frac {1}{4} \, \log \left (\frac {5}{2}\right )^{2} - \frac {1}{2} i \, \pi \log \left (-\frac {5}{2 \, x^{2}}\right ) - \frac {1}{2} \, \log \left (\frac {5}{2}\right ) \log \left (-\frac {5}{2 \, x^{2}}\right ) + \frac {1}{4} \, \log \left (-\frac {5}{2 \, x^{2}}\right )^{2} + 3\right )} \log \relax (5) \log \left (-\frac {5}{2 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (\log \relax (5) \log \relax (x) \log \left (-\frac {5}{2 \, x^{2}}\right ) - \log \relax (5)\right )} e^{\left (\log \relax (x)^{2} + 3\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(2*(log(5)*log(x)*log(-5/2/x^2) - log(5))*e^(log(x)^2 + 3)/x, x)

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maple [B]  time = 0.06, size = 36, normalized size = 1.89

method	result	size
default	\({\mathrm e}^{3} \ln \relax (5) \left (\ln \left (-\frac {5}{2 x^{2}}\right )+2 \ln \relax (x )\right ) {\mathrm e}^{\ln \relax (x )^{2}}-2 \,{\mathrm e}^{3} \ln \relax (5) {\mathrm e}^{\ln \relax (x )^{2}} \ln \relax (x )\)	\(36\)
risch	\(\left (\frac {i \ln \relax (5) \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-i \ln \relax (5) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\frac {i \ln \relax (5) \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}-i \ln \relax (5) \pi \mathrm {csgn}\left (\frac {i}{x^{2}}\right )^{2}+i \ln \relax (5) \pi \mathrm {csgn}\left (\frac {i}{x^{2}}\right )^{3}+i \ln \relax (5) \pi +\ln \relax (5)^{2}-\ln \relax (2) \ln \relax (5)-2 \ln \relax (5) \ln \relax (x )\right ) {\mathrm e}^{\ln \relax (x )^{2}+3}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 16, normalized size = 0.84 \begin {gather*} e^{\left (\log \relax (x)^{2} + 3\right )} \log \relax (5) \log \left (-\frac {5}{2 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(log(x)^2 + 3)*log(5)*log(-5/2/x^2)

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mupad [B]  time = 2.47, size = 21, normalized size = 1.11 \begin {gather*} {\mathrm {e}}^{{\ln \relax (x)}^2+3}\,\ln \relax (5)\,\left (\ln \left (\frac {5}{2\,x^2}\right )+\pi \,1{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(log(x)^2 + 3)*log(5)*(pi*1i + log(5/(2*x^2)))

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sympy [C]  time = 0.52, size = 36, normalized size = 1.89 \begin {gather*} - \left (2 \log {\relax (5 )} \log {\relax (x )} - \log {\relax (5 )}^{2} + \log {\relax (2 )} \log {\relax (5 )} - i \pi \log {\relax (5 )}\right ) e^{\log {\relax (x )}^{2} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*log(5)*log(x) - log(5)**2 + log(2)*log(5) - I*pi*log(5))*exp(log(x)**2 + 3)

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