3.62.30 \(\int \frac {x+(x+x^2+5\ 3^{5/x} \log (3)) \log (x)}{(64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} (-16 x^2-2 x^3)) \log (x)+(16 x^2-2\ 3^{5/x} x^2+2 x^3) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{-8+3^{5/x}-x-\log (x \log (x))} \]

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Rubi [A]  time = 0.43, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 2, number of rules used = 2, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 6686} \begin {gather*} -\frac {1}{x-243^{\frac {1}{x}}+\log (x \log (x))+8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + (x + x^2 + 5*3^(5/x)*Log[3])*Log[x])/((64*x^2 + 3^(10/x)*x^2 + 16*x^3 + x^4 + 3^(5/x)*(-16*x^2 - 2*x^
3))*Log[x] + (16*x^2 - 2*3^(5/x)*x^2 + 2*x^3)*Log[x]*Log[x*Log[x]] + x^2*Log[x]*Log[x*Log[x]]^2),x]

[Out]

-(8 - 243^x^(-1) + x + Log[x*Log[x]])^(-1)

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x+\left (x+x^2+5\ 243^{\frac {1}{x}} \log (3)\right ) \log (x)}{x^2 \log (x) \left (8-243^{\frac {1}{x}}+x+\log (x \log (x))\right )^2} \, dx\\ &=-\frac {1}{8-243^{\frac {1}{x}}+x+\log (x \log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 19, normalized size = 0.90 \begin {gather*} -\frac {1}{8-243^{\frac {1}{x}}+x+\log (x \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + (x + x^2 + 5*3^(5/x)*Log[3])*Log[x])/((64*x^2 + 3^(10/x)*x^2 + 16*x^3 + x^4 + 3^(5/x)*(-16*x^2
- 2*x^3))*Log[x] + (16*x^2 - 2*3^(5/x)*x^2 + 2*x^3)*Log[x]*Log[x*Log[x]] + x^2*Log[x]*Log[x*Log[x]]^2),x]

[Out]

-(8 - 243^x^(-1) + x + Log[x*Log[x]])^(-1)

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fricas [A]  time = 0.55, size = 21, normalized size = 1.00 \begin {gather*} \frac {1}{3^{\frac {5}{x}} - x - \log \left (x \log \relax (x)\right ) - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(3)*exp(5*log(3)/x)+x^2+x)*log(x)+x)/(x^2*log(x)*log(x*log(x))^2+(-2*x^2*exp(5*log(3)/x)+2*x^
3+16*x^2)*log(x)*log(x*log(x))+(x^2*exp(5*log(3)/x)^2+(-2*x^3-16*x^2)*exp(5*log(3)/x)+x^4+16*x^3+64*x^2)*log(x
)),x, algorithm="fricas")

[Out]

1/(3^(5/x) - x - log(x*log(x)) - 8)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} + 5 \cdot 3^{\frac {5}{x}} \log \relax (3) + x\right )} \log \relax (x) + x}{x^{2} \log \left (x \log \relax (x)\right )^{2} \log \relax (x) - 2 \, {\left (3^{\frac {5}{x}} x^{2} - x^{3} - 8 \, x^{2}\right )} \log \left (x \log \relax (x)\right ) \log \relax (x) + {\left (x^{4} + 3^{\frac {10}{x}} x^{2} + 16 \, x^{3} - 2 \, {\left (x^{3} + 8 \, x^{2}\right )} 3^{\frac {5}{x}} + 64 \, x^{2}\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(3)*exp(5*log(3)/x)+x^2+x)*log(x)+x)/(x^2*log(x)*log(x*log(x))^2+(-2*x^2*exp(5*log(3)/x)+2*x^
3+16*x^2)*log(x)*log(x*log(x))+(x^2*exp(5*log(3)/x)^2+(-2*x^3-16*x^2)*exp(5*log(3)/x)+x^4+16*x^3+64*x^2)*log(x
)),x, algorithm="giac")

[Out]

integrate(((x^2 + 5*3^(5/x)*log(3) + x)*log(x) + x)/(x^2*log(x*log(x))^2*log(x) - 2*(3^(5/x)*x^2 - x^3 - 8*x^2
)*log(x*log(x))*log(x) + (x^4 + 3^(10/x)*x^2 + 16*x^3 - 2*(x^3 + 8*x^2)*3^(5/x) + 64*x^2)*log(x)), x)

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maple [C]  time = 0.13, size = 98, normalized size = 4.67

method	result	size
risch	\(-\frac {2 i}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2}-\pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2}+\pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3}+2 i x -2 i 243^{\frac {1}{x}}+2 i \ln \relax (x )+2 i \ln \left (\ln \relax (x )\right )+16 i}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*ln(3)*exp(5*ln(3)/x)+x^2+x)*ln(x)+x)/(x^2*ln(x)*ln(x*ln(x))^2+(-2*x^2*exp(5*ln(3)/x)+2*x^3+16*x^2)*ln(
x)*ln(x*ln(x))+(x^2*exp(5*ln(3)/x)^2+(-2*x^3-16*x^2)*exp(5*ln(3)/x)+x^4+16*x^3+64*x^2)*ln(x)),x,method=_RETURN
VERBOSE)

[Out]

-2*I/(Pi*csgn(I*x)*csgn(I*ln(x))*csgn(I*x*ln(x))-Pi*csgn(I*x)*csgn(I*x*ln(x))^2-Pi*csgn(I*ln(x))*csgn(I*x*ln(x
))^2+Pi*csgn(I*x*ln(x))^3+2*I*x-2*I*243^(1/x)+2*I*ln(x)+2*I*ln(ln(x))+16*I)

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maxima [A]  time = 0.48, size = 23, normalized size = 1.10 \begin {gather*} \frac {1}{3^{\frac {5}{x}} - x - \log \relax (x) - \log \left (\log \relax (x)\right ) - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(3)*exp(5*log(3)/x)+x^2+x)*log(x)+x)/(x^2*log(x)*log(x*log(x))^2+(-2*x^2*exp(5*log(3)/x)+2*x^
3+16*x^2)*log(x)*log(x*log(x))+(x^2*exp(5*log(3)/x)^2+(-2*x^3-16*x^2)*exp(5*log(3)/x)+x^4+16*x^3+64*x^2)*log(x
)),x, algorithm="maxima")

[Out]

1/(3^(5/x) - x - log(x) - log(log(x)) - 8)

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mupad [B]  time = 4.68, size = 21, normalized size = 1.00 \begin {gather*} -\frac {1}{x+\ln \left (x\,\ln \relax (x)\right )-3^{5/x}+8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(x)*(x + 5*exp((5*log(3))/x)*log(3) + x^2))/(log(x)*(x^2*exp((10*log(3))/x) - exp((5*log(3))/x)*(1
6*x^2 + 2*x^3) + 64*x^2 + 16*x^3 + x^4) + log(x*log(x))*log(x)*(16*x^2 - 2*x^2*exp((5*log(3))/x) + 2*x^3) + x^
2*log(x*log(x))^2*log(x)),x)

[Out]

-1/(x + log(x*log(x)) - 3^(5/x) + 8)

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sympy [A]  time = 0.37, size = 19, normalized size = 0.90 \begin {gather*} \frac {1}{- x + e^{\frac {5 \log {\relax (3 )}}{x}} - \log {\left (x \log {\relax (x )} \right )} - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*ln(3)*exp(5*ln(3)/x)+x**2+x)*ln(x)+x)/(x**2*ln(x)*ln(x*ln(x))**2+(-2*x**2*exp(5*ln(3)/x)+2*x**3+
16*x**2)*ln(x)*ln(x*ln(x))+(x**2*exp(5*ln(3)/x)**2+(-2*x**3-16*x**2)*exp(5*ln(3)/x)+x**4+16*x**3+64*x**2)*ln(x
)),x)

[Out]

1/(-x + exp(5*log(3)/x) - log(x*log(x)) - 8)

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