3.53.10 \(\int \frac {e^{\frac {2 (e^4 x+3 \log (4))}{3 e^4}} (3 e^{2 x}+e^x (-15 x+25 x^2))-2 e^{2 x+\frac {2 (e^4 x+3 \log (4))}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^{\frac {2 x}{3}+\frac {2 \log (4)}{e^4}}}{5 e^{-x} x-\log (x)} \]

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Rubi [F]  time = 3.42, 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 {e^{\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} \left (3 e^{2 x}+e^x \left (-15 x+25 x^2\right )\right )-2 e^{2 x+\frac {2 \left (e^4 x+3 \log (4)\right )}{3 e^4}} x \log (x)}{75 x^3-30 e^x x^2 \log (x)+3 e^{2 x} x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(E^4*x + 3*Log[4]))/(3*E^4))*(3*E^(2*x) + E^x*(-15*x + 25*x^2)) - 2*E^(2*x + (2*(E^4*x + 3*Log[4]))
/(3*E^4))*x*Log[x])/(75*x^3 - 30*E^x*x^2*Log[x] + 3*E^(2*x)*x*Log[x]^2),x]

[Out]

5*16^E^(-4)*Defer[Int][(E^((5*x)/3)*x)/(5*x - E^x*Log[x])^2, x] - 16^E^(-4)*Defer[Int][E^((5*x)/3)/(x*Log[x]*(
5*x - E^x*Log[x])), x] - 15*16^E^(-4)*Defer[Subst][Defer[Int][E^(5*x)/(15*x - E^(3*x)*Log[3*x])^2, x], x, x/3]
+ 2^(1 + 4/E^4)*Defer[Subst][Defer[Int][E^(5*x)/(15*x - E^(3*x)*Log[3*x]), x], x, x/3] + 15*16^E^(-4)*Defer[S
ubst][Defer[Int][E^(5*x)/(Log[3*x]*(-15*x + E^(3*x)*Log[3*x])^2), x], x, x/3]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16^{\frac {1}{e^4}} e^{5 x/3} \left (3 e^x+5 x (-3+5 x)-2 e^x x \log (x)\right )}{3 x \left (5 x-e^x \log (x)\right )^2} \, dx\\ &=\frac {1}{3} 16^{\frac {1}{e^4}} \int \frac {e^{5 x/3} \left (3 e^x+5 x (-3+5 x)-2 e^x x \log (x)\right )}{x \left (5 x-e^x \log (x)\right )^2} \, dx\\ &=\frac {1}{3} 16^{\frac {1}{e^4}} \int \left (\frac {15 e^{5 x/3} (1-\log (x)+x \log (x))}{\log (x) \left (-5 x+e^x \log (x)\right )^2}+\frac {e^{5 x/3} (-3+2 x \log (x))}{x \log (x) \left (5 x-e^x \log (x)\right )}\right ) \, dx\\ &=\frac {1}{3} 16^{\frac {1}{e^4}} \int \frac {e^{5 x/3} (-3+2 x \log (x))}{x \log (x) \left (5 x-e^x \log (x)\right )} \, dx+\left (5\ 16^{\frac {1}{e^4}}\right ) \int \frac {e^{5 x/3} (1-\log (x)+x \log (x))}{\log (x) \left (-5 x+e^x \log (x)\right )^2} \, dx\\ &=\frac {1}{3} 16^{\frac {1}{e^4}} \int \left (\frac {2 e^{5 x/3}}{5 x-e^x \log (x)}-\frac {3 e^{5 x/3}}{x \log (x) \left (5 x-e^x \log (x)\right )}\right ) \, dx+\left (5\ 16^{\frac {1}{e^4}}\right ) \int \left (-\frac {e^{5 x/3}}{\left (5 x-e^x \log (x)\right )^2}+\frac {e^{5 x/3} x}{\left (5 x-e^x \log (x)\right )^2}+\frac {e^{5 x/3}}{\log (x) \left (-5 x+e^x \log (x)\right )^2}\right ) \, dx\\ &=\frac {1}{3} 2^{1+\frac {4}{e^4}} \int \frac {e^{5 x/3}}{5 x-e^x \log (x)} \, dx-16^{\frac {1}{e^4}} \int \frac {e^{5 x/3}}{x \log (x) \left (5 x-e^x \log (x)\right )} \, dx-\left (5\ 16^{\frac {1}{e^4}}\right ) \int \frac {e^{5 x/3}}{\left (5 x-e^x \log (x)\right )^2} \, dx+\left (5\ 16^{\frac {1}{e^4}}\right ) \int \frac {e^{5 x/3} x}{\left (5 x-e^x \log (x)\right )^2} \, dx+\left (5\ 16^{\frac {1}{e^4}}\right ) \int \frac {e^{5 x/3}}{\log (x) \left (-5 x+e^x \log (x)\right )^2} \, dx\\ &=2^{1+\frac {4}{e^4}} \operatorname {Subst}\left (\int \frac {e^{5 x}}{15 x-e^{3 x} \log (3 x)} \, dx,x,\frac {x}{3}\right )-16^{\frac {1}{e^4}} \int \frac {e^{5 x/3}}{x \log (x) \left (5 x-e^x \log (x)\right )} \, dx+\left (5\ 16^{\frac {1}{e^4}}\right ) \int \frac {e^{5 x/3} x}{\left (5 x-e^x \log (x)\right )^2} \, dx-\left (15\ 16^{\frac {1}{e^4}}\right ) \operatorname {Subst}\left (\int \frac {e^{5 x}}{\left (15 x-e^{3 x} \log (3 x)\right )^2} \, dx,x,\frac {x}{3}\right )+\left (15\ 16^{\frac {1}{e^4}}\right ) \operatorname {Subst}\left (\int \frac {e^{5 x}}{\log (3 x) \left (-15 x+e^{3 x} \log (3 x)\right )^2} \, dx,x,\frac {x}{3}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.58, size = 26, normalized size = 0.84 \begin {gather*} -\frac {16^{\frac {1}{e^4}} e^{5 x/3}}{-5 x+e^x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(E^4*x + 3*Log[4]))/(3*E^4))*(3*E^(2*x) + E^x*(-15*x + 25*x^2)) - 2*E^(2*x + (2*(E^4*x + 3*Lo
g[4]))/(3*E^4))*x*Log[x])/(75*x^3 - 30*E^x*x^2*Log[x] + 3*E^(2*x)*x*Log[x]^2),x]

[Out]

-((16^E^(-4)*E^((5*x)/3))/(-5*x + E^x*Log[x]))

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fricas [A]  time = 0.60, size = 23, normalized size = 0.74 \begin {gather*} -\frac {2^{4 \, e^{\left (-4\right )}} e^{\left (\frac {5}{3} \, x\right )}}{e^{x} \log \relax (x) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(x)^2*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2*log(x)+(3*exp(x)^2+(25*x^2-15*x)*exp(x))*exp(1/
3*(6*log(2)+x*exp(4))/exp(4))^2)/(3*x*exp(x)^2*log(x)^2-30*x^2*exp(x)*log(x)+75*x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 3.36, size = 44, normalized size = 1.42 \begin {gather*} -\frac {5 \cdot 2^{4 \, e^{\left (-4\right )}} x e^{\left (\frac {2}{3} \, x\right )} + 2^{4 \, e^{\left (-4\right )}} e^{\left (\frac {5}{3} \, x\right )} \log \relax (x)}{e^{x} \log \relax (x)^{2} - 5 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(x)^2*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2*log(x)+(3*exp(x)^2+(25*x^2-15*x)*exp(x))*exp(1/
3*(6*log(2)+x*exp(4))/exp(4))^2)/(3*x*exp(x)^2*log(x)^2-30*x^2*exp(x)*log(x)+75*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 24, normalized size = 0.77

method	result	size
risch	\(\frac {2^{4 \,{\mathrm e}^{-4}} {\mathrm e}^{\frac {5 x}{3}}}{-{\mathrm e}^{x} \ln \relax (x )+5 x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*exp(x)^2*exp(1/3*(6*ln(2)+x*exp(4))/exp(4))^2*ln(x)+(3*exp(x)^2+(25*x^2-15*x)*exp(x))*exp(1/3*(6*ln(
2)+x*exp(4))/exp(4))^2)/(3*x*exp(x)^2*ln(x)^2-30*x^2*exp(x)*ln(x)+75*x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.55, size = 23, normalized size = 0.74 \begin {gather*} -\frac {2^{4 \, e^{\left (-4\right )}} e^{\left (\frac {5}{3} \, x\right )}}{e^{x} \log \relax (x) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(x)^2*exp(1/3*(6*log(2)+x*exp(4))/exp(4))^2*log(x)+(3*exp(x)^2+(25*x^2-15*x)*exp(x))*exp(1/
3*(6*log(2)+x*exp(4))/exp(4))^2)/(3*x*exp(x)^2*log(x)^2-30*x^2*exp(x)*log(x)+75*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.56, size = 21, normalized size = 0.68 \begin {gather*} \frac {{16}^{{\mathrm {e}}^{-4}}\,{\mathrm {e}}^{\frac {5\,x}{3}}}{5\,x-{\mathrm {e}}^x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp(-4)*(2*log(2) + (x*exp(4))/3))*(3*exp(2*x) - exp(x)*(15*x - 25*x^2)) - 2*x*exp(2*x)*exp(2*exp(-
4)*(2*log(2) + (x*exp(4))/3))*log(x))/(75*x^3 + 3*x*exp(2*x)*log(x)^2 - 30*x^2*exp(x)*log(x)),x)

[Out]

(16^exp(-4)*exp((5*x)/3))/(5*x - exp(x)*log(x))

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sympy [A]  time = 0.32, size = 24, normalized size = 0.77 \begin {gather*} - \frac {2^{\frac {4}{e^{4}}} e^{\frac {5 x}{3}}}{- 5 x + e^{x} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(x)**2*exp(1/3*(6*ln(2)+x*exp(4))/exp(4))**2*ln(x)+(3*exp(x)**2+(25*x**2-15*x)*exp(x))*exp(
1/3*(6*ln(2)+x*exp(4))/exp(4))**2)/(3*x*exp(x)**2*ln(x)**2-30*x**2*exp(x)*ln(x)+75*x**3),x)

[Out]

-2**(4*exp(-4))*exp(5*x/3)/(-5*x + exp(x)*log(x))

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