3.36.5 \(\int \frac {e^{x+\log ^4(\frac {e^x}{10})} (-1+x+4 x \log ^3(\frac {e^x}{10}))}{x^2} \, dx\)

Optimal. Leaf size=18 \[ \frac {e^{x+\log ^4\left (\frac {e^x}{10}\right )}}{x} \]

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Rubi [B]  time = 0.11, antiderivative size = 49, normalized size of antiderivative = 2.72, number of steps used = 1, number of rules used = 1, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2288} \begin {gather*} \frac {e^{x+\log ^4\left (\frac {e^x}{10}\right )} \left (x+4 x \log ^3\left (\frac {e^x}{10}\right )\right )}{x^2 \left (4 \log ^3\left (\frac {e^x}{10}\right )+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x + Log[E^x/10]^4)*(-1 + x + 4*x*Log[E^x/10]^3))/x^2,x]

[Out]

(E^(x + Log[E^x/10]^4)*(x + 4*x*Log[E^x/10]^3))/(x^2*(1 + 4*Log[E^x/10]^3))

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} &=\frac {e^{x+\log ^4\left (\frac {e^x}{10}\right )} \left (x+4 x \log ^3\left (\frac {e^x}{10}\right )\right )}{x^2 \left (1+4 \log ^3\left (\frac {e^x}{10}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 18, normalized size = 1.00 \begin {gather*} \frac {e^{x+\log ^4\left (\frac {e^x}{10}\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x + Log[E^x/10]^4)*(-1 + x + 4*x*Log[E^x/10]^3))/x^2,x]

[Out]

E^(x + Log[E^x/10]^4)/x

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fricas [B]  time = 0.52, size = 37, normalized size = 2.06 \begin {gather*} \frac {e^{\left (x^{4} - 4 \, x^{3} \log \left (10\right ) + 6 \, x^{2} \log \left (10\right )^{2} - 4 \, x \log \left (10\right )^{3} + \log \left (10\right )^{4} + x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(1/10*exp(x))^3+x-1)*exp(log(1/10*exp(x))^4+x)/x^2,x, algorithm="fricas")

[Out]

e^(x^4 - 4*x^3*log(10) + 6*x^2*log(10)^2 - 4*x*log(10)^3 + log(10)^4 + x)/x

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giac [B]  time = 0.28, size = 37, normalized size = 2.06 \begin {gather*} \frac {e^{\left (x^{4} - 4 \, x^{3} \log \left (10\right ) + 6 \, x^{2} \log \left (10\right )^{2} - 4 \, x \log \left (10\right )^{3} + \log \left (10\right )^{4} + x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(1/10*exp(x))^3+x-1)*exp(log(1/10*exp(x))^4+x)/x^2,x, algorithm="giac")

[Out]

e^(x^4 - 4*x^3*log(10) + 6*x^2*log(10)^2 - 4*x*log(10)^3 + log(10)^4 + x)/x

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maple [A]  time = 0.35, size = 15, normalized size = 0.83

method	result	size
default	\(\frac {{\mathrm e}^{\ln \left (\frac {{\mathrm e}^{x}}{10}\right )^{4}+x}}{x}\)	\(15\)
risch	\(\frac {625^{\ln \relax (2)^{3}} \left ({\mathrm e}^{x}\right )^{-4 \ln \relax (2)^{3}} \left ({\mathrm e}^{x}\right )^{-12 \ln \relax (2)^{2} \ln \relax (5)} 16^{\ln \relax (5)^{3}} \left ({\mathrm e}^{x}\right )^{-12 \ln \relax (5)^{2} \ln \relax (2)} 4096^{\ln \relax (5) \ln \left ({\mathrm e}^{x}\right )^{2}} \left (\frac {1}{16}\right )^{\ln \left ({\mathrm e}^{x}\right )^{3}} \left ({\mathrm e}^{x}\right )^{-4 \ln \relax (5)^{3}} \left (\frac {1}{625}\right )^{\ln \left ({\mathrm e}^{x}\right )^{3}} {\mathrm e}^{\ln \relax (2)^{4}+6 \ln \relax (2)^{2} \ln \relax (5)^{2}+6 \ln \relax (2)^{2} \ln \left ({\mathrm e}^{x}\right )^{2}+\ln \relax (5)^{4}+6 \ln \relax (5)^{2} \ln \left ({\mathrm e}^{x}\right )^{2}+\ln \left ({\mathrm e}^{x}\right )^{4}+x}}{x}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(1/10*exp(x))^3+x-1)*exp(ln(1/10*exp(x))^4+x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(ln(1/10*exp(x))^4+x)/x

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maxima [B]  time = 0.83, size = 117, normalized size = 6.50 \begin {gather*} \frac {5^{4 \, \log \relax (2)^{3}} 2^{4 \, \log \relax (5)^{3}} e^{\left (x^{4} - 4 \, x^{3} \log \relax (5) + 6 \, x^{2} \log \relax (5)^{2} - 4 \, x \log \relax (5)^{3} + \log \relax (5)^{4} - 4 \, x^{3} \log \relax (2) + 12 \, x^{2} \log \relax (5) \log \relax (2) - 12 \, x \log \relax (5)^{2} \log \relax (2) + 6 \, x^{2} \log \relax (2)^{2} - 12 \, x \log \relax (5) \log \relax (2)^{2} + 6 \, \log \relax (5)^{2} \log \relax (2)^{2} - 4 \, x \log \relax (2)^{3} + \log \relax (2)^{4} + x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(1/10*exp(x))^3+x-1)*exp(log(1/10*exp(x))^4+x)/x^2,x, algorithm="maxima")

[Out]

5^(4*log(2)^3)*2^(4*log(5)^3)*e^(x^4 - 4*x^3*log(5) + 6*x^2*log(5)^2 - 4*x*log(5)^3 + log(5)^4 - 4*x^3*log(2)
+ 12*x^2*log(5)*log(2) - 12*x*log(5)^2*log(2) + 6*x^2*log(2)^2 - 12*x*log(5)*log(2)^2 + 6*log(5)^2*log(2)^2 -
4*x*log(2)^3 + log(2)^4 + x)/x

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mupad [B]  time = 2.25, size = 42, normalized size = 2.33 \begin {gather*} \frac {{\mathrm {e}}^{-4\,x\,{\ln \left (10\right )}^3}\,{\mathrm {e}}^{{\ln \left (10\right )}^4}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{6\,x^2\,{\ln \left (10\right )}^2}\,{\mathrm {e}}^x}{{10}^{4\,x^3}\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(exp(x)/10)^4)*(x + 4*x*log(exp(x)/10)^3 - 1))/x^2,x)

[Out]

(exp(-4*x*log(10)^3)*exp(log(10)^4)*exp(x^4)*exp(6*x^2*log(10)^2)*exp(x))/(10^(4*x^3)*x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(1/10*exp(x))**3+x-1)*exp(ln(1/10*exp(x))**4+x)/x**2,x)

[Out]

Timed out

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