3.82.73 \(\int \frac {(-x-16 e x^2-64 x^4) \log (x)+(-e^2-x-8 e x^2-16 x^4+(e^2+x+8 e x^2+16 x^4) \log (x)) \log (e^2+x+8 e x^2+16 x^4)+(e^2+x+8 e x^2+16 x^4+(-e^2-x-8 e x^2-16 x^4) \log (x)+(e^2+x+8 e x^2+16 x^4) \log ^2(x)) \log ^2(e^2+x+8 e x^2+16 x^4)}{(e^2+x+8 e x^2+16 x^4) \log ^2(x) \log ^2(e^2+x+8 e x^2+16 x^4)} \, dx\)

Optimal. Leaf size=27 \[ x+\frac {-x+\frac {x}{\log \left (x+\left (e+4 x^2\right )^2\right )}}{\log (x)} \]

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Rubi [F]  time = 6.99, 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 {\left (-x-16 e x^2-64 x^4\right ) \log (x)+\left (-e^2-x-8 e x^2-16 x^4+\left (e^2+x+8 e x^2+16 x^4\right ) \log (x)\right ) \log \left (e^2+x+8 e x^2+16 x^4\right )+\left (e^2+x+8 e x^2+16 x^4+\left (-e^2-x-8 e x^2-16 x^4\right ) \log (x)+\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x)\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2(x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-x - 16*E*x^2 - 64*x^4)*Log[x] + (-E^2 - x - 8*E*x^2 - 16*x^4 + (E^2 + x + 8*E*x^2 + 16*x^4)*Log[x])*Log
[E^2 + x + 8*E*x^2 + 16*x^4] + (E^2 + x + 8*E*x^2 + 16*x^4 + (-E^2 - x - 8*E*x^2 - 16*x^4)*Log[x] + (E^2 + x +
8*E*x^2 + 16*x^4)*Log[x]^2)*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2)/((E^2 + x + 8*E*x^2 + 16*x^4)*Log[x]^2*Log[E^2
+ x + 8*E*x^2 + 16*x^4]^2),x]

[Out]

x - x/Log[x] - 4*Defer[Int][1/(Log[x]*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2), x] + 4*E^2*Defer[Int][1/((E^2 + x +
8*E*x^2 + 16*x^4)*Log[x]*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2), x] + 3*Defer[Int][x/((E^2 + x + 8*E*x^2 + 16*x^4)
*Log[x]*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2), x] + 16*E*Defer[Int][x^2/((E^2 + x + 8*E*x^2 + 16*x^4)*Log[x]*Log[
E^2 + x + 8*E*x^2 + 16*x^4]^2), x] - Defer[Int][1/(Log[x]^2*Log[E^2 + x + 8*E*x^2 + 16*x^4]), x] + Defer[Int][
1/(Log[x]*Log[E^2 + x + 8*E*x^2 + 16*x^4]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-\log (x)+\log ^2(x)-\frac {x \left (1+16 e x+64 x^3\right ) \log (x)}{\left (e^2+x+8 e x^2+16 x^4\right ) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}+\frac {-1+\log (x)}{\log \left (e^2+x+8 e x^2+16 x^4\right )}}{\log ^2(x)} \, dx\\ &=\int \left (\frac {1-\log (x)+\log ^2(x)}{\log ^2(x)}-\frac {x \left (1+16 e x+64 x^3\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}+\frac {-1+\log (x)}{\log ^2(x) \log \left (e^2+x+8 e x^2+16 x^4\right )}\right ) \, dx\\ &=\int \frac {1-\log (x)+\log ^2(x)}{\log ^2(x)} \, dx-\int \frac {x \left (1+16 e x+64 x^3\right )}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx+\int \frac {-1+\log (x)}{\log ^2(x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \, dx\\ &=\int \left (1+\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \, dx-\int \left (\frac {4}{\log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}+\frac {-4 e^2-3 x-16 e x^2}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}\right ) \, dx+\int \left (-\frac {1}{\log ^2(x) \log \left (e^2+x+8 e x^2+16 x^4\right )}+\frac {1}{\log (x) \log \left (e^2+x+8 e x^2+16 x^4\right )}\right ) \, dx\\ &=x-4 \int \frac {1}{\log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx+\int \frac {1}{\log ^2(x)} \, dx-\int \frac {1}{\log (x)} \, dx-\int \frac {-4 e^2-3 x-16 e x^2}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx-\int \frac {1}{\log ^2(x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \, dx+\int \frac {1}{\log (x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \, dx\\ &=x-\frac {x}{\log (x)}-\text {li}(x)-4 \int \frac {1}{\log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx+\int \frac {1}{\log (x)} \, dx-\int \left (-\frac {4 e^2}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}-\frac {3 x}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}-\frac {16 e x^2}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )}\right ) \, dx-\int \frac {1}{\log ^2(x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \, dx+\int \frac {1}{\log (x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \, dx\\ &=x-\frac {x}{\log (x)}+3 \int \frac {x}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx-4 \int \frac {1}{\log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx+(16 e) \int \frac {x^2}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx+\left (4 e^2\right ) \int \frac {1}{\left (e^2+x+8 e x^2+16 x^4\right ) \log (x) \log ^2\left (e^2+x+8 e x^2+16 x^4\right )} \, dx-\int \frac {1}{\log ^2(x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \, dx+\int \frac {1}{\log (x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 34, normalized size = 1.26 \begin {gather*} x-\frac {x}{\log (x)}+\frac {x}{\log (x) \log \left (e^2+x+8 e x^2+16 x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-x - 16*E*x^2 - 64*x^4)*Log[x] + (-E^2 - x - 8*E*x^2 - 16*x^4 + (E^2 + x + 8*E*x^2 + 16*x^4)*Log[x
])*Log[E^2 + x + 8*E*x^2 + 16*x^4] + (E^2 + x + 8*E*x^2 + 16*x^4 + (-E^2 - x - 8*E*x^2 - 16*x^4)*Log[x] + (E^2
+ x + 8*E*x^2 + 16*x^4)*Log[x]^2)*Log[E^2 + x + 8*E*x^2 + 16*x^4]^2)/((E^2 + x + 8*E*x^2 + 16*x^4)*Log[x]^2*L
og[E^2 + x + 8*E*x^2 + 16*x^4]^2),x]

[Out]

x - x/Log[x] + x/(Log[x]*Log[E^2 + x + 8*E*x^2 + 16*x^4])

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fricas [A]  time = 1.37, size = 52, normalized size = 1.93 \begin {gather*} \frac {{\left (x \log \relax (x) - x\right )} \log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) + x}{\log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)^2+(-exp(1)^2-8*x^2*exp(1)-16*x^4-x)*log(x)+exp(1)^2+8*x^2*
exp(1)+16*x^4+x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)-exp(1)^2-8*x^2
*exp(1)-16*x^4-x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4-x)*log(x))/(exp(1)^2+8*x^2*exp(1)
+16*x^4+x)/log(x)^2/log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)^2,x, algorithm="fricas")

[Out]

((x*log(x) - x)*log(16*x^4 + 8*x^2*e + x + e^2) + x)/(log(16*x^4 + 8*x^2*e + x + e^2)*log(x))

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giac [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((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)^2+(-exp(1)^2-8*x^2*exp(1)-16*x^4-x)*log(x)+exp(1)^2+8*x^2*
exp(1)+16*x^4+x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)-exp(1)^2-8*x^2
*exp(1)-16*x^4-x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4-x)*log(x))/(exp(1)^2+8*x^2*exp(1)
+16*x^4+x)/log(x)^2/log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 4.79Not invertible Error: Bad Argument Value

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maple [A]  time = 0.09, size = 37, normalized size = 1.37

method	result	size
risch	\(\frac {x \left (\ln \relax (x )-1\right )}{\ln \relax (x )}+\frac {x}{\ln \relax (x ) \ln \left ({\mathrm e}^{2}+8 x^{2} {\mathrm e}+16 x^{4}+x \right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*ln(x)^2+(-exp(1)^2-8*x^2*exp(1)-16*x^4-x)*ln(x)+exp(1)^2+8*x^2*exp(1)+1
6*x^4+x)*ln(exp(1)^2+8*x^2*exp(1)+16*x^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*ln(x)-exp(1)^2-8*x^2*exp(1)-16
*x^4-x)*ln(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4-x)*ln(x))/(exp(1)^2+8*x^2*exp(1)+16*x^4+x)/l
n(x)^2/ln(exp(1)^2+8*x^2*exp(1)+16*x^4+x)^2,x,method=_RETURNVERBOSE)

[Out]

x*(ln(x)-1)/ln(x)+x/ln(x)/ln(exp(2)+8*x^2*exp(1)+16*x^4+x)

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maxima [A]  time = 0.43, size = 52, normalized size = 1.93 \begin {gather*} \frac {{\left (x \log \relax (x) - x\right )} \log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) + x}{\log \left (16 \, x^{4} + 8 \, x^{2} e + x + e^{2}\right ) \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)^2+(-exp(1)^2-8*x^2*exp(1)-16*x^4-x)*log(x)+exp(1)^2+8*x^2*
exp(1)+16*x^4+x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)^2+((exp(1)^2+8*x^2*exp(1)+16*x^4+x)*log(x)-exp(1)^2-8*x^2
*exp(1)-16*x^4-x)*log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)+(-16*x^2*exp(1)-64*x^4-x)*log(x))/(exp(1)^2+8*x^2*exp(1)
+16*x^4+x)/log(x)^2/log(exp(1)^2+8*x^2*exp(1)+16*x^4+x)^2,x, algorithm="maxima")

[Out]

((x*log(x) - x)*log(16*x^4 + 8*x^2*e + x + e^2) + x)/(log(16*x^4 + 8*x^2*e + x + e^2)*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {\left (\ln \relax (x)\,\left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )-{\mathrm {e}}^2-x-8\,x^2\,\mathrm {e}-{\ln \relax (x)}^2\,\left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )-16\,x^4\right )\,{\ln \left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )}^2+\left (x+{\mathrm {e}}^2-\ln \relax (x)\,\left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )+8\,x^2\,\mathrm {e}+16\,x^4\right )\,\ln \left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )+\ln \relax (x)\,\left (64\,x^4+16\,\mathrm {e}\,x^2+x\right )}{{\ln \left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )}^2\,{\ln \relax (x)}^2\,\left (16\,x^4+8\,\mathrm {e}\,x^2+x+{\mathrm {e}}^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x + exp(2) + 8*x^2*exp(1) + 16*x^4)*(x + exp(2) - log(x)*(x + exp(2) + 8*x^2*exp(1) + 16*x^4) + 8*x^
2*exp(1) + 16*x^4) - log(x + exp(2) + 8*x^2*exp(1) + 16*x^4)^2*(x + exp(2) - log(x)*(x + exp(2) + 8*x^2*exp(1)
+ 16*x^4) + 8*x^2*exp(1) + log(x)^2*(x + exp(2) + 8*x^2*exp(1) + 16*x^4) + 16*x^4) + log(x)*(x + 16*x^2*exp(1
) + 64*x^4))/(log(x + exp(2) + 8*x^2*exp(1) + 16*x^4)^2*log(x)^2*(x + exp(2) + 8*x^2*exp(1) + 16*x^4)),x)

[Out]

-int((log(x + exp(2) + 8*x^2*exp(1) + 16*x^4)*(x + exp(2) - log(x)*(x + exp(2) + 8*x^2*exp(1) + 16*x^4) + 8*x^
2*exp(1) + 16*x^4) - log(x + exp(2) + 8*x^2*exp(1) + 16*x^4)^2*(x + exp(2) - log(x)*(x + exp(2) + 8*x^2*exp(1)
+ 16*x^4) + 8*x^2*exp(1) + log(x)^2*(x + exp(2) + 8*x^2*exp(1) + 16*x^4) + 16*x^4) + log(x)*(x + 16*x^2*exp(1
) + 64*x^4))/(log(x + exp(2) + 8*x^2*exp(1) + 16*x^4)^2*log(x)^2*(x + exp(2) + 8*x^2*exp(1) + 16*x^4)), x)

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sympy [A]  time = 0.41, size = 31, normalized size = 1.15 \begin {gather*} x - \frac {x}{\log {\relax (x )}} + \frac {x}{\log {\relax (x )} \log {\left (16 x^{4} + 8 e x^{2} + x + e^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(1)**2+8*x**2*exp(1)+16*x**4+x)*ln(x)**2+(-exp(1)**2-8*x**2*exp(1)-16*x**4-x)*ln(x)+exp(1)**2+
8*x**2*exp(1)+16*x**4+x)*ln(exp(1)**2+8*x**2*exp(1)+16*x**4+x)**2+((exp(1)**2+8*x**2*exp(1)+16*x**4+x)*ln(x)-e
xp(1)**2-8*x**2*exp(1)-16*x**4-x)*ln(exp(1)**2+8*x**2*exp(1)+16*x**4+x)+(-16*x**2*exp(1)-64*x**4-x)*ln(x))/(ex
p(1)**2+8*x**2*exp(1)+16*x**4+x)/ln(x)**2/ln(exp(1)**2+8*x**2*exp(1)+16*x**4+x)**2,x)

[Out]

x - x/log(x) + x/(log(x)*log(16*x**4 + 8*E*x**2 + x + exp(2)))

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