3.78.40 \(\int \frac {-5 e^7+e^6 (-5 e^4+20 e^2 x)+e^6 (20 e^2 x+10 e^4 x) \log (x)+(-e^4 x+e^3 (-2 e^4 x+8 e^2 x^2)+e^6 (4 e^4 x+8 e^2 x^2-16 x^3)) \log ^2(x)+(2 e^7 x^2+e^6 (2 e^4 x^2-8 e^2 x^3)) \log ^3(x)-e^{10} x^3 \log ^4(x)}{(e^4 x+e^3 (2 e^4 x-8 e^2 x^2)+e^6 (e^4 x-8 e^2 x^2+16 x^3)) \log ^2(x)+(-2 e^7 x^2+e^6 (-2 e^4 x^2+8 e^2 x^3)) \log ^3(x)+e^{10} x^3 \log ^4(x)} \, dx\)

Optimal. Leaf size=28 \[ -x+\frac {5}{\log (x) \left (1+\frac {1}{e^3}-x \left (\frac {4}{e^2}+\log (x)\right )\right )} \]

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Rubi [F]  time = 2.13, 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 {-5 e^7+e^6 \left (-5 e^4+20 e^2 x\right )+e^6 \left (20 e^2 x+10 e^4 x\right ) \log (x)+\left (-e^4 x+e^3 \left (-2 e^4 x+8 e^2 x^2\right )+e^6 \left (4 e^4 x+8 e^2 x^2-16 x^3\right )\right ) \log ^2(x)+\left (2 e^7 x^2+e^6 \left (2 e^4 x^2-8 e^2 x^3\right )\right ) \log ^3(x)-e^{10} x^3 \log ^4(x)}{\left (e^4 x+e^3 \left (2 e^4 x-8 e^2 x^2\right )+e^6 \left (e^4 x-8 e^2 x^2+16 x^3\right )\right ) \log ^2(x)+\left (-2 e^7 x^2+e^6 \left (-2 e^4 x^2+8 e^2 x^3\right )\right ) \log ^3(x)+e^{10} x^3 \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5*E^7 + E^6*(-5*E^4 + 20*E^2*x) + E^6*(20*E^2*x + 10*E^4*x)*Log[x] + (-(E^4*x) + E^3*(-2*E^4*x + 8*E^2*x
^2) + E^6*(4*E^4*x + 8*E^2*x^2 - 16*x^3))*Log[x]^2 + (2*E^7*x^2 + E^6*(2*E^4*x^2 - 8*E^2*x^3))*Log[x]^3 - E^10
*x^3*Log[x]^4)/((E^4*x + E^3*(2*E^4*x - 8*E^2*x^2) + E^6*(E^4*x - 8*E^2*x^2 + 16*x^3))*Log[x]^2 + (-2*E^7*x^2
+ E^6*(-2*E^4*x^2 + 8*E^2*x^3))*Log[x]^3 + E^10*x^3*Log[x]^4),x]

[Out]

-x + 5*E^3*Defer[Int][1/(x*(-1 - E^3 + 4*E*x)*Log[x]^2), x] + 20*E^4*Defer[Int][1/((1 + E^3 - 4*E*x)^2*Log[x])
, x] - (5*E^8*Defer[Int][(-1 - E^3 + 4*E*x + E^3*x*Log[x])^(-2), x])/4 + (5*E^6*(4 + E^2 + 4*E^3 + E^5)*Defer[
Int][1/((1 + E^3 - 4*E*x)*(-1 - E^3 + 4*E*x + E^3*x*Log[x])^2), x])/4 - 5*E^6*(1 + E^3)*Defer[Int][1/((1 + E^3
- 4*E*x)^2*(-1 - E^3 + 4*E*x + E^3*x*Log[x])), x] + 5*E^6*Defer[Int][1/((1 + E^3 - 4*E*x)*(-1 - E^3 + 4*E*x +
E^3*x*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5 e^3 \left (1+e^3-4 e x\right )+10 e^4 \left (2+e^2\right ) x \log (x)-x \left (1+2 e^3-4 e^6-8 e x-8 e^4 x+16 e^2 x^2\right ) \log ^2(x)+2 e^3 x^2 \left (1+e^3-4 e x\right ) \log ^3(x)-e^6 x^3 \log ^4(x)}{x \log ^2(x) \left (1+e^3-4 e x-e^3 x \log (x)\right )^2} \, dx\\ &=\int \left (-1+\frac {5 e^3}{x \left (-1-e^3+4 e x\right ) \log ^2(x)}+\frac {20 e^4}{\left (1+e^3-4 e x\right )^2 \log (x)}+\frac {5 \left (e^6+e^9+e^9 x\right )}{\left (1+e^3-4 e x\right ) \left (-1-e^3+4 e x+e^3 x \log (x)\right )^2}-\frac {20 e^7 x}{\left (1+e^3-4 e x\right )^2 \left (-1-e^3+4 e x+e^3 x \log (x)\right )}\right ) \, dx\\ &=-x+5 \int \frac {e^6+e^9+e^9 x}{\left (1+e^3-4 e x\right ) \left (-1-e^3+4 e x+e^3 x \log (x)\right )^2} \, dx+\left (5 e^3\right ) \int \frac {1}{x \left (-1-e^3+4 e x\right ) \log ^2(x)} \, dx+\left (20 e^4\right ) \int \frac {1}{\left (1+e^3-4 e x\right )^2 \log (x)} \, dx-\left (20 e^7\right ) \int \frac {x}{\left (1+e^3-4 e x\right )^2 \left (-1-e^3+4 e x+e^3 x \log (x)\right )} \, dx\\ &=-x+5 \int \left (-\frac {e^8}{4 \left (-1-e^3+4 e x+e^3 x \log (x)\right )^2}+\frac {e^6 \left (4+e^2+4 e^3+e^5\right )}{4 \left (1+e^3-4 e x\right ) \left (-1-e^3+4 e x+e^3 x \log (x)\right )^2}\right ) \, dx+\left (5 e^3\right ) \int \frac {1}{x \left (-1-e^3+4 e x\right ) \log ^2(x)} \, dx+\left (20 e^4\right ) \int \frac {1}{\left (1+e^3-4 e x\right )^2 \log (x)} \, dx-\left (20 e^7\right ) \int \left (\frac {1+e^3}{4 e \left (1+e^3-4 e x\right )^2 \left (-1-e^3+4 e x+e^3 x \log (x)\right )}-\frac {1}{4 e \left (1+e^3-4 e x\right ) \left (-1-e^3+4 e x+e^3 x \log (x)\right )}\right ) \, dx\\ &=-x+\left (5 e^3\right ) \int \frac {1}{x \left (-1-e^3+4 e x\right ) \log ^2(x)} \, dx+\left (20 e^4\right ) \int \frac {1}{\left (1+e^3-4 e x\right )^2 \log (x)} \, dx+\left (5 e^6\right ) \int \frac {1}{\left (1+e^3-4 e x\right ) \left (-1-e^3+4 e x+e^3 x \log (x)\right )} \, dx-\frac {1}{4} \left (5 e^8\right ) \int \frac {1}{\left (-1-e^3+4 e x+e^3 x \log (x)\right )^2} \, dx-\left (5 e^6 \left (1+e^3\right )\right ) \int \frac {1}{\left (1+e^3-4 e x\right )^2 \left (-1-e^3+4 e x+e^3 x \log (x)\right )} \, dx+\frac {1}{4} \left (5 e^6 \left (4+e^2+4 e^3+e^5\right )\right ) \int \frac {1}{\left (1+e^3-4 e x\right ) \left (-1-e^3+4 e x+e^3 x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 61, normalized size = 2.18 \begin {gather*} -x+\frac {5 e^3}{\left (1+e^3-4 e x\right ) \log (x)}-\frac {5 e^6 x}{\left (1+e^3-4 e x\right ) \left (-1-e^3+4 e x+e^3 x \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*E^7 + E^6*(-5*E^4 + 20*E^2*x) + E^6*(20*E^2*x + 10*E^4*x)*Log[x] + (-(E^4*x) + E^3*(-2*E^4*x + 8
*E^2*x^2) + E^6*(4*E^4*x + 8*E^2*x^2 - 16*x^3))*Log[x]^2 + (2*E^7*x^2 + E^6*(2*E^4*x^2 - 8*E^2*x^3))*Log[x]^3
- E^10*x^3*Log[x]^4)/((E^4*x + E^3*(2*E^4*x - 8*E^2*x^2) + E^6*(E^4*x - 8*E^2*x^2 + 16*x^3))*Log[x]^2 + (-2*E^
7*x^2 + E^6*(-2*E^4*x^2 + 8*E^2*x^3))*Log[x]^3 + E^10*x^3*Log[x]^4),x]

[Out]

-x + (5*E^3)/((1 + E^3 - 4*E*x)*Log[x]) - (5*E^6*x)/((1 + E^3 - 4*E*x)*(-1 - E^3 + 4*E*x + E^3*x*Log[x]))

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fricas [B]  time = 0.86, size = 61, normalized size = 2.18 \begin {gather*} -\frac {x^{2} e^{3} \log \relax (x)^{2} + {\left (4 \, x^{2} e - x e^{3} - x\right )} \log \relax (x) + 5 \, e^{3}}{x e^{3} \log \relax (x)^{2} + {\left (4 \, x e - e^{3} - 1\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(2)^2*exp(3)^2*log(x)^4+((2*x^2*exp(2)^2-8*x^3*exp(2))*exp(3)^2+2*x^2*exp(2)^2*exp(3))*log(
x)^3+((4*x*exp(2)^2+8*x^2*exp(2)-16*x^3)*exp(3)^2+(-2*x*exp(2)^2+8*x^2*exp(2))*exp(3)-x*exp(2)^2)*log(x)^2+(10
*x*exp(2)^2+20*exp(2)*x)*exp(3)^2*log(x)+(-5*exp(2)^2+20*exp(2)*x)*exp(3)^2-5*exp(2)^2*exp(3))/(x^3*exp(2)^2*e
xp(3)^2*log(x)^4+((-2*x^2*exp(2)^2+8*x^3*exp(2))*exp(3)^2-2*x^2*exp(2)^2*exp(3))*log(x)^3+((x*exp(2)^2-8*x^2*e
xp(2)+16*x^3)*exp(3)^2+(2*x*exp(2)^2-8*x^2*exp(2))*exp(3)+x*exp(2)^2)*log(x)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.94, size = 66, normalized size = 2.36 \begin {gather*} -\frac {x^{2} e^{3} \log \relax (x)^{2} + 4 \, x^{2} e \log \relax (x) - x e^{3} \log \relax (x) - x \log \relax (x) + 5 \, e^{3}}{x e^{3} \log \relax (x)^{2} + 4 \, x e \log \relax (x) - e^{3} \log \relax (x) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(2)^2*exp(3)^2*log(x)^4+((2*x^2*exp(2)^2-8*x^3*exp(2))*exp(3)^2+2*x^2*exp(2)^2*exp(3))*log(
x)^3+((4*x*exp(2)^2+8*x^2*exp(2)-16*x^3)*exp(3)^2+(-2*x*exp(2)^2+8*x^2*exp(2))*exp(3)-x*exp(2)^2)*log(x)^2+(10
*x*exp(2)^2+20*exp(2)*x)*exp(3)^2*log(x)+(-5*exp(2)^2+20*exp(2)*x)*exp(3)^2-5*exp(2)^2*exp(3))/(x^3*exp(2)^2*e
xp(3)^2*log(x)^4+((-2*x^2*exp(2)^2+8*x^3*exp(2))*exp(3)^2-2*x^2*exp(2)^2*exp(3))*log(x)^3+((x*exp(2)^2-8*x^2*e
xp(2)+16*x^3)*exp(3)^2+(2*x*exp(2)^2-8*x^2*exp(2))*exp(3)+x*exp(2)^2)*log(x)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.17, size = 32, normalized size = 1.14

method	result	size
risch	\(-x -\frac {5 \,{\mathrm e}^{3}}{\left (x \,{\mathrm e}^{3} \ln \relax (x )-{\mathrm e}^{3}+4 x \,{\mathrm e}-1\right ) \ln \relax (x )}\)	\(32\)
norman	\(\frac {\frac {{\mathrm e}^{4} {\mathrm e}^{-3} \left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) \ln \relax (x )}{4}+\left (-\frac {{\mathrm e}^{3} {\mathrm e}^{4}}{4}-\frac {{\mathrm e}^{4}}{4}\right ) x \ln \relax (x )^{2}-5 \,{\mathrm e}^{2} {\mathrm e}^{3}-4 x^{2} {\mathrm e}^{3} \ln \relax (x )-{\mathrm e}^{2} \ln \relax (x )^{2} {\mathrm e}^{3} x^{2}}{\ln \relax (x ) \left ({\mathrm e}^{3} {\mathrm e}^{2} x \ln \relax (x )-{\mathrm e}^{2} {\mathrm e}^{3}+4 x \,{\mathrm e}^{3}-{\mathrm e}^{2}\right )}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^3*exp(2)^2*exp(3)^2*ln(x)^4+((2*x^2*exp(2)^2-8*x^3*exp(2))*exp(3)^2+2*x^2*exp(2)^2*exp(3))*ln(x)^3+((4
*x*exp(2)^2+8*x^2*exp(2)-16*x^3)*exp(3)^2+(-2*x*exp(2)^2+8*x^2*exp(2))*exp(3)-x*exp(2)^2)*ln(x)^2+(10*x*exp(2)
^2+20*exp(2)*x)*exp(3)^2*ln(x)+(-5*exp(2)^2+20*exp(2)*x)*exp(3)^2-5*exp(2)^2*exp(3))/(x^3*exp(2)^2*exp(3)^2*ln
(x)^4+((-2*x^2*exp(2)^2+8*x^3*exp(2))*exp(3)^2-2*x^2*exp(2)^2*exp(3))*ln(x)^3+((x*exp(2)^2-8*x^2*exp(2)+16*x^3
)*exp(3)^2+(2*x*exp(2)^2-8*x^2*exp(2))*exp(3)+x*exp(2)^2)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

-x-5*exp(3)/(x*exp(3)*ln(x)-exp(3)+4*x*exp(1)-1)/ln(x)

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maxima [B]  time = 0.44, size = 60, normalized size = 2.14 \begin {gather*} -\frac {x^{2} e^{3} \log \relax (x)^{2} + {\left (4 \, x^{2} e - x {\left (e^{3} + 1\right )}\right )} \log \relax (x) + 5 \, e^{3}}{x e^{3} \log \relax (x)^{2} + {\left (4 \, x e - e^{3} - 1\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(2)^2*exp(3)^2*log(x)^4+((2*x^2*exp(2)^2-8*x^3*exp(2))*exp(3)^2+2*x^2*exp(2)^2*exp(3))*log(
x)^3+((4*x*exp(2)^2+8*x^2*exp(2)-16*x^3)*exp(3)^2+(-2*x*exp(2)^2+8*x^2*exp(2))*exp(3)-x*exp(2)^2)*log(x)^2+(10
*x*exp(2)^2+20*exp(2)*x)*exp(3)^2*log(x)+(-5*exp(2)^2+20*exp(2)*x)*exp(3)^2-5*exp(2)^2*exp(3))/(x^3*exp(2)^2*e
xp(3)^2*log(x)^4+((-2*x^2*exp(2)^2+8*x^3*exp(2))*exp(3)^2-2*x^2*exp(2)^2*exp(3))*log(x)^3+((x*exp(2)^2-8*x^2*e
xp(2)+16*x^3)*exp(3)^2+(2*x*exp(2)^2-8*x^2*exp(2))*exp(3)+x*exp(2)^2)*log(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.76, size = 34, normalized size = 1.21 \begin {gather*} -x-\frac {5}{x\,\left ({\ln \relax (x)}^2-\frac {{\mathrm {e}}^{-3}\,\ln \relax (x)\,\left ({\mathrm {e}}^3-4\,x\,\mathrm {e}+1\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(7) + exp(6)*(5*exp(4) - 20*x*exp(2)) + log(x)^2*(x*exp(4) - exp(6)*(4*x*exp(4) + 8*x^2*exp(2) - 16
*x^3) + exp(3)*(2*x*exp(4) - 8*x^2*exp(2))) + log(x)^3*(exp(6)*(8*x^3*exp(2) - 2*x^2*exp(4)) - 2*x^2*exp(7)) -
exp(6)*log(x)*(20*x*exp(2) + 10*x*exp(4)) + x^3*exp(10)*log(x)^4)/(log(x)^2*(exp(6)*(x*exp(4) - 8*x^2*exp(2)
+ 16*x^3) + x*exp(4) + exp(3)*(2*x*exp(4) - 8*x^2*exp(2))) + log(x)^3*(exp(6)*(8*x^3*exp(2) - 2*x^2*exp(4)) -
2*x^2*exp(7)) + x^3*exp(10)*log(x)^4),x)

[Out]

- x - 5/(x*(log(x)^2 - (exp(-3)*log(x)*(exp(3) - 4*x*exp(1) + 1))/x))

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sympy [A]  time = 0.33, size = 32, normalized size = 1.14 \begin {gather*} - x - \frac {5 e^{3}}{x e^{3} \log {\relax (x )}^{2} + \left (4 e x - e^{3} - 1\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**3*exp(2)**2*exp(3)**2*ln(x)**4+((2*x**2*exp(2)**2-8*x**3*exp(2))*exp(3)**2+2*x**2*exp(2)**2*exp
(3))*ln(x)**3+((4*x*exp(2)**2+8*x**2*exp(2)-16*x**3)*exp(3)**2+(-2*x*exp(2)**2+8*x**2*exp(2))*exp(3)-x*exp(2)*
*2)*ln(x)**2+(10*x*exp(2)**2+20*exp(2)*x)*exp(3)**2*ln(x)+(-5*exp(2)**2+20*exp(2)*x)*exp(3)**2-5*exp(2)**2*exp
(3))/(x**3*exp(2)**2*exp(3)**2*ln(x)**4+((-2*x**2*exp(2)**2+8*x**3*exp(2))*exp(3)**2-2*x**2*exp(2)**2*exp(3))*
ln(x)**3+((x*exp(2)**2-8*x**2*exp(2)+16*x**3)*exp(3)**2+(2*x*exp(2)**2-8*x**2*exp(2))*exp(3)+x*exp(2)**2)*ln(x
)**2),x)

[Out]

-x - 5*exp(3)/(x*exp(3)*log(x)**2 + (4*E*x - exp(3) - 1)*log(x))

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