3.85.82 \(\int \frac {-3 x^4+3 e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^3 \log (x)+e^{\frac {1}{3} (-5+\log (-x+e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} \log (x)))} (e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^2-x^3+e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} (-50-10 e^3 x) \log (x))}{-3 x^4+3 e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^3 \log (x)} \, dx\)

Optimal. Leaf size=31 \[ e^{\frac {1}{3} \left (-5+\log \left (-x+e^{\left (e^3+\frac {5}{x}\right )^2} \log (x)\right )\right )}+x \]

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Rubi [F]  time = 10.45, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-3 x^4+3 e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^3 \log (x)+\exp \left (\frac {1}{3} \left (-5+\log \left (-x+e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} \log (x)\right )\right )\right ) \left (e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^2-x^3+e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} \left (-50-10 e^3 x\right ) \log (x)\right )}{-3 x^4+3 e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^3 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*x^4 + 3*E^((25 + 10*E^3*x + E^6*x^2)/x^2)*x^3*Log[x] + E^((-5 + Log[-x + E^((25 + 10*E^3*x + E^6*x^2)/
x^2)*Log[x]])/3)*(E^((25 + 10*E^3*x + E^6*x^2)/x^2)*x^2 - x^3 + E^((25 + 10*E^3*x + E^6*x^2)/x^2)*(-50 - 10*E^
3*x)*Log[x]))/(-3*x^4 + 3*E^((25 + 10*E^3*x + E^6*x^2)/x^2)*x^3*Log[x]),x]

[Out]

x - Defer[Int][(-x + E^((5 + E^3*x)^2/x^2)*Log[x])^(-2/3), x]/(3*E^(5/3)) + Defer[Int][E^((5 + E^3*x)^2/x^2)/(
x*(-x + E^((5 + E^3*x)^2/x^2)*Log[x])^(2/3)), x]/(3*E^(5/3)) - (50*Defer[Int][(E^((5 + E^3*x)^2/x^2)*Log[x])/(
x^3*(-x + E^((5 + E^3*x)^2/x^2)*Log[x])^(2/3)), x])/(3*E^(5/3)) - (10*Defer[Int][(E^(3 + (5 + E^3*x)^2/x^2)*Lo
g[x])/(x^2*(-x + E^((5 + E^3*x)^2/x^2)*Log[x])^(2/3)), x])/(3*E^(5/3))

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 x^4-3 e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^3 \log (x)-\exp \left (\frac {1}{3} \left (-5+\log \left (-x+e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} \log (x)\right )\right )\right ) \left (e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^2-x^3+e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} \left (-50-10 e^3 x\right ) \log (x)\right )}{3 x^3 \left (x-e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )} \, dx\\ &=\frac {1}{3} \int \frac {3 x^4-3 e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^3 \log (x)-\exp \left (\frac {1}{3} \left (-5+\log \left (-x+e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} \log (x)\right )\right )\right ) \left (e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} x^2-x^3+e^{\frac {25+10 e^3 x+e^6 x^2}{x^2}} \left (-50-10 e^3 x\right ) \log (x)\right )}{x^3 \left (x-e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )} \, dx\\ &=\frac {1}{3} \int \left (3-\frac {-e^{\frac {\left (5+e^3 x\right )^2}{x^2}} x^2+x^3+50 e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)+10 e^{3+\frac {\left (5+e^3 x\right )^2}{x^2}} x \log (x)}{e^{5/3} x^3 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}}\right ) \, dx\\ &=x-\frac {\int \frac {-e^{\frac {\left (5+e^3 x\right )^2}{x^2}} x^2+x^3+50 e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)+10 e^{3+\frac {\left (5+e^3 x\right )^2}{x^2}} x \log (x)}{x^3 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}\\ &=x-\frac {\int \left (\frac {1}{\left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}}-\frac {e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \left (x^2-50 \log (x)-10 e^3 x \log (x)\right )}{x^3 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}}\right ) \, dx}{3 e^{5/3}}\\ &=x-\frac {\int \frac {1}{\left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}+\frac {\int \frac {e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \left (x^2-50 \log (x)-10 e^3 x \log (x)\right )}{x^3 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}\\ &=x-\frac {\int \frac {1}{\left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}+\frac {\int \left (\frac {e^{\frac {\left (5+e^3 x\right )^2}{x^2}}}{x \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}}-\frac {10 e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \left (5+e^3 x\right ) \log (x)}{x^3 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}}\right ) \, dx}{3 e^{5/3}}\\ &=x-\frac {\int \frac {1}{\left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}+\frac {\int \frac {e^{\frac {\left (5+e^3 x\right )^2}{x^2}}}{x \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}-\frac {10 \int \frac {e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \left (5+e^3 x\right ) \log (x)}{x^3 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}\\ &=x-\frac {\int \frac {1}{\left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}+\frac {\int \frac {e^{\frac {\left (5+e^3 x\right )^2}{x^2}}}{x \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}-\frac {10 \int \left (\frac {5 e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)}{x^3 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}}+\frac {e^{3+\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)}{x^2 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}}\right ) \, dx}{3 e^{5/3}}\\ &=x-\frac {\int \frac {1}{\left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}+\frac {\int \frac {e^{\frac {\left (5+e^3 x\right )^2}{x^2}}}{x \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}-\frac {10 \int \frac {e^{3+\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)}{x^2 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}-\frac {50 \int \frac {e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)}{x^3 \left (-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)\right )^{2/3}} \, dx}{3 e^{5/3}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.89, size = 41, normalized size = 1.32 \begin {gather*} \frac {1}{3} \left (3 x+\frac {3 \sqrt [3]{-x+e^{\frac {\left (5+e^3 x\right )^2}{x^2}} \log (x)}}{e^{5/3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^4 + 3*E^((25 + 10*E^3*x + E^6*x^2)/x^2)*x^3*Log[x] + E^((-5 + Log[-x + E^((25 + 10*E^3*x + E^6
*x^2)/x^2)*Log[x]])/3)*(E^((25 + 10*E^3*x + E^6*x^2)/x^2)*x^2 - x^3 + E^((25 + 10*E^3*x + E^6*x^2)/x^2)*(-50 -
10*E^3*x)*Log[x]))/(-3*x^4 + 3*E^((25 + 10*E^3*x + E^6*x^2)/x^2)*x^3*Log[x]),x]

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x*exp(3)-50)*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)+x^2*exp((x^2*exp(3)^2+10*x*exp(3)+
25)/x^2)-x^3)*exp(1/3*log(exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)-x)-5/3)+3*x^3*exp((x^2*exp(3)^2+10*x*e
xp(3)+25)/x^2)*log(x)-3*x^4)/(3*x^3*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)-3*x^4),x, algorithm="fricas"
)

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 \, x^{3} e^{\left (\frac {x^{2} e^{6} + 10 \, x e^{3} + 25}{x^{2}}\right )} \log \relax (x) - 3 \, x^{4} - {\left (x^{3} - x^{2} e^{\left (\frac {x^{2} e^{6} + 10 \, x e^{3} + 25}{x^{2}}\right )} + 10 \, {\left (x e^{3} + 5\right )} e^{\left (\frac {x^{2} e^{6} + 10 \, x e^{3} + 25}{x^{2}}\right )} \log \relax (x)\right )} e^{\left (\frac {1}{3} \, \log \left (e^{\left (\frac {x^{2} e^{6} + 10 \, x e^{3} + 25}{x^{2}}\right )} \log \relax (x) - x\right ) - \frac {5}{3}\right )}}{3 \, {\left (x^{3} e^{\left (\frac {x^{2} e^{6} + 10 \, x e^{3} + 25}{x^{2}}\right )} \log \relax (x) - x^{4}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x*exp(3)-50)*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)+x^2*exp((x^2*exp(3)^2+10*x*exp(3)+
25)/x^2)-x^3)*exp(1/3*log(exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)-x)-5/3)+3*x^3*exp((x^2*exp(3)^2+10*x*e
xp(3)+25)/x^2)*log(x)-3*x^4)/(3*x^3*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)-3*x^4),x, algorithm="giac")

[Out]

integrate(1/3*(3*x^3*e^((x^2*e^6 + 10*x*e^3 + 25)/x^2)*log(x) - 3*x^4 - (x^3 - x^2*e^((x^2*e^6 + 10*x*e^3 + 25
)/x^2) + 10*(x*e^3 + 5)*e^((x^2*e^6 + 10*x*e^3 + 25)/x^2)*log(x))*e^(1/3*log(e^((x^2*e^6 + 10*x*e^3 + 25)/x^2)
*log(x) - x) - 5/3))/(x^3*e^((x^2*e^6 + 10*x*e^3 + 25)/x^2)*log(x) - x^4), x)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (-10 x \,{\mathrm e}^{3}-50\right ) {\mathrm e}^{\frac {x^{2} {\mathrm e}^{6}+10 x \,{\mathrm e}^{3}+25}{x^{2}}} \ln \relax (x )+x^{2} {\mathrm e}^{\frac {x^{2} {\mathrm e}^{6}+10 x \,{\mathrm e}^{3}+25}{x^{2}}}-x^{3}\right ) {\mathrm e}^{\frac {\ln \left ({\mathrm e}^{\frac {x^{2} {\mathrm e}^{6}+10 x \,{\mathrm e}^{3}+25}{x^{2}}} \ln \relax (x )-x \right )}{3}-\frac {5}{3}}+3 x^{3} {\mathrm e}^{\frac {x^{2} {\mathrm e}^{6}+10 x \,{\mathrm e}^{3}+25}{x^{2}}} \ln \relax (x )-3 x^{4}}{3 x^{3} {\mathrm e}^{\frac {x^{2} {\mathrm e}^{6}+10 x \,{\mathrm e}^{3}+25}{x^{2}}} \ln \relax (x )-3 x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-10*x*exp(3)-50)*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*ln(x)+x^2*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2
)-x^3)*exp(1/3*ln(exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*ln(x)-x)-5/3)+3*x^3*exp((x^2*exp(3)^2+10*x*exp(3)+25)
/x^2)*ln(x)-3*x^4)/(3*x^3*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*ln(x)-3*x^4),x)

[Out]

int((((-10*x*exp(3)-50)*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*ln(x)+x^2*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2
)-x^3)*exp(1/3*ln(exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*ln(x)-x)-5/3)+3*x^3*exp((x^2*exp(3)^2+10*x*exp(3)+25)
/x^2)*ln(x)-3*x^4)/(3*x^3*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*ln(x)-3*x^4),x)

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maxima [A]  time = 0.53, size = 33, normalized size = 1.06 \begin {gather*} {\left (x e^{\frac {5}{3}} + {\left (e^{\left (\frac {10 \, e^{3}}{x} + \frac {25}{x^{2}} + e^{6}\right )} \log \relax (x) - x\right )}^{\frac {1}{3}}\right )} e^{\left (-\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x*exp(3)-50)*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)+x^2*exp((x^2*exp(3)^2+10*x*exp(3)+
25)/x^2)-x^3)*exp(1/3*log(exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)-x)-5/3)+3*x^3*exp((x^2*exp(3)^2+10*x*e
xp(3)+25)/x^2)*log(x)-3*x^4)/(3*x^3*exp((x^2*exp(3)^2+10*x*exp(3)+25)/x^2)*log(x)-3*x^4),x, algorithm="maxima"
)

[Out]

(x*e^(5/3) + (e^(10*e^3/x + 25/x^2 + e^6)*log(x) - x)^(1/3))*e^(-5/3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {\ln \left ({\mathrm {e}}^{\frac {{\mathrm {e}}^6\,x^2+10\,{\mathrm {e}}^3\,x+25}{x^2}}\,\ln \relax (x)-x\right )}{3}-\frac {5}{3}}\,\left (x^3-x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^6\,x^2+10\,{\mathrm {e}}^3\,x+25}{x^2}}+{\mathrm {e}}^{\frac {{\mathrm {e}}^6\,x^2+10\,{\mathrm {e}}^3\,x+25}{x^2}}\,\ln \relax (x)\,\left (10\,x\,{\mathrm {e}}^3+50\right )\right )+3\,x^4-3\,x^3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^6\,x^2+10\,{\mathrm {e}}^3\,x+25}{x^2}}\,\ln \relax (x)}{3\,x^4-3\,x^3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^6\,x^2+10\,{\mathrm {e}}^3\,x+25}{x^2}}\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(exp((10*x*exp(3) + x^2*exp(6) + 25)/x^2)*log(x) - x)/3 - 5/3)*(x^3 - x^2*exp((10*x*exp(3) + x^2*e
xp(6) + 25)/x^2) + exp((10*x*exp(3) + x^2*exp(6) + 25)/x^2)*log(x)*(10*x*exp(3) + 50)) + 3*x^4 - 3*x^3*exp((10
*x*exp(3) + x^2*exp(6) + 25)/x^2)*log(x))/(3*x^4 - 3*x^3*exp((10*x*exp(3) + x^2*exp(6) + 25)/x^2)*log(x)),x)

[Out]

int((exp(log(exp((10*x*exp(3) + x^2*exp(6) + 25)/x^2)*log(x) - x)/3 - 5/3)*(x^3 - x^2*exp((10*x*exp(3) + x^2*e
xp(6) + 25)/x^2) + exp((10*x*exp(3) + x^2*exp(6) + 25)/x^2)*log(x)*(10*x*exp(3) + 50)) + 3*x^4 - 3*x^3*exp((10
*x*exp(3) + x^2*exp(6) + 25)/x^2)*log(x))/(3*x^4 - 3*x^3*exp((10*x*exp(3) + x^2*exp(6) + 25)/x^2)*log(x)), 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((((-10*x*exp(3)-50)*exp((x**2*exp(3)**2+10*x*exp(3)+25)/x**2)*ln(x)+x**2*exp((x**2*exp(3)**2+10*x*ex
p(3)+25)/x**2)-x**3)*exp(1/3*ln(exp((x**2*exp(3)**2+10*x*exp(3)+25)/x**2)*ln(x)-x)-5/3)+3*x**3*exp((x**2*exp(3
)**2+10*x*exp(3)+25)/x**2)*ln(x)-3*x**4)/(3*x**3*exp((x**2*exp(3)**2+10*x*exp(3)+25)/x**2)*ln(x)-3*x**4),x)

[Out]

Timed out

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