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

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

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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 = {} $$\begin{array}{c}\int \frac{-3x^4+3e^{\frac{25+10e^{3}x+e^6{x}^2}{x^2}}{x}^3\log (x)+\exp \left(\frac{1}{3}(-5+\log (-x+e^{\frac{25+10e^{3}x+e^6{x}^2}{x^2}}\log (x)))\right)(e^{\frac{25+10e^{3}x+e^6{x}^2}{x^2}}{x}^2-x^3+e^{\frac{25+10e^{3}x+e^6{x}^2}{x^2}}(-50-10e^{3}x)\log (x))}{-3x^4+3e^{\frac{25+10e^{3}x+e^6{x}^2}{x^2}}{x}^3\log (x)}dx\end{array}$$

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

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

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 $$\begin{array}{c}\text{Exception raised: TypeError}\end{array}$$

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 $$\begin{array}{c}\int \frac{3x^3{e}^{\left(\frac{x^2{e}^6+10x{e}^3+25}{x^2}\right)}\log (x)-3x^4-(x^3-x^2{e}^{\left(\frac{x^2{e}^6+10x{e}^3+25}{x^2}\right)}+10(x{e}^3+5)e^{\left(\frac{x^2{e}^6+10x{e}^3+25}{x^2}\right)}\log (x))e^{(\frac{1}{3}\log (e^{\left(\frac{x^2{e}^6+10x{e}^3+25}{x^2}\right)}\log (x)-x)-\frac{5}{3})}}{3(x^3{e}^{\left(\frac{x^2{e}^6+10x{e}^3+25}{x^2}\right)}\log (x)-x^4)}dx\end{array}$$

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 $$\int \frac{((-10x{\mathrm{e}}^3-50){\mathrm{e}}^{\frac{x^2{\mathrm{e}}^6+10x{\mathrm{e}}^3+25}{x^2}}\ln (x)+x^2{\mathrm{e}}^{\frac{x^2{\mathrm{e}}^6+10x{\mathrm{e}}^3+25}{x^2}}-x^3){\mathrm{e}}^{\frac{\ln ({\mathrm{e}}^{\frac{x^2{\mathrm{e}}^6+10x{\mathrm{e}}^3+25}{x^2}}\ln (x)-x)}{3}-\frac{5}{3}}+3x^3{\mathrm{e}}^{\frac{x^2{\mathrm{e}}^6+10x{\mathrm{e}}^3+25}{x^2}}\ln (x)-3x^4}{3x^3{\mathrm{e}}^{\frac{x^2{\mathrm{e}}^6+10x{\mathrm{e}}^3+25}{x^2}}\ln (x)-3x^4}dx$$

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 $$\begin{array}{c}(x{e}^{\frac{5}{3}}+{(e^{(\frac{10e^3}{x}+\frac{25}{x^2}+e^6)}\log (x)-x)}^{\frac{1}{3}})e^{(-\frac{5}{3})}\end{array}$$

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 $$\begin{array}{c}\int \frac{{\mathrm{e}}^{\frac{\ln ({\mathrm{e}}^{\frac{{\mathrm{e}}^6{x}^2+10{\mathrm{e}}^{3}x+25}{x^2}}\ln (x)-x)}{3}-\frac{5}{3}}(x^3-x^2{\mathrm{e}}^{\frac{{\mathrm{e}}^6{x}^2+10{\mathrm{e}}^{3}x+25}{x^2}}+{\mathrm{e}}^{\frac{{\mathrm{e}}^6{x}^2+10{\mathrm{e}}^{3}x+25}{x^2}}\ln (x)(10x{\mathrm{e}}^3+50))+3x^4-3x^3{\mathrm{e}}^{\frac{{\mathrm{e}}^6{x}^2+10{\mathrm{e}}^{3}x+25}{x^2}}\ln (x)}{3x^4-3x^3{\mathrm{e}}^{\frac{{\mathrm{e}}^6{x}^2+10{\mathrm{e}}^{3}x+25}{x^2}}\ln (x)}dx\end{array}$$

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 $$\begin{array}{c}\text{Timed out}\end{array}$$

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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