∫ 6 log(5)+ex(−27−18x log(5)−3x2 log2(5))_______________________________ (96+62x log(5)+10x2 log2(5)+ex(9+6x log(5)+x2 log2(5))) log(32+10x log(5)+ex(3+x log(5)) 3+x log(5) ) log2(log(32+10x log(5)+ex(3+x log(5)) 3+x log(5) )) dx

3.94.93 \(\int \frac {6 \log (5)+e^x (-27-18 x \log (5)-3 x^2 \log ^2(5))}{(96+62 x \log (5)+10 x^2 \log ^2(5)+e^x (9+6 x \log (5)+x^2 \log ^2(5))) \log (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}) \log ^2(\log (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}))} \, dx\)

Optimal. Leaf size=30 \[ e^4+\frac {3}{\log \left (\log \left (10+e^x+\frac {2}{x \left (\frac {3}{x}+\log (5)\right )}\right )\right )} \]

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Rubi [F] time = 7.77, 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 {6 \log (5)+e^x \left (-27-18 x \log (5)-3 x^2 \log ^2(5)\right )}{\left (96+62 x \log (5)+10 x^2 \log ^2(5)+e^x \left (9+6 x \log (5)+x^2 \log ^2(5)\right )\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Int[(6*Log[5] + E^x*(-27 - 18*x*Log[5] - 3*x^2*Log[5]^2))/((96 + 62*x*Log[5] + 10*x^2*Log[5]^2 + E^x*(9 + 6*x*

Log[5] + x^2*Log[5]^2))*Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*Log[5]

+ E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]^2),x]

[Out]

-3*Defer[Int][1/(Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*Log[5] + E^x*(

3 + x*Log[5]))/(3 + x*Log[5])]]^2), x] + 96*Defer[Int][1/((32 + 3*E^x + 10*x*Log[5] + E^x*x*Log[5])*Log[(32 +

10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5

])]]^2), x] + 30*Log[5]*Defer[Int][x/((32 + 3*E^x + 10*x*Log[5] + E^x*x*Log[5])*Log[(32 + 10*x*Log[5] + E^x*(3

+ x*Log[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]^2), x] + 6*Log[

5]*Defer[Int][1/((3 + x*Log[5])*(32 + 3*E^x + 10*x*Log[5] + E^x*x*Log[5])*Log[(32 + 10*x*Log[5] + E^x*(3 + x*L

og[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 \left (2 \log (5)-e^x (3+x \log (5))^2\right )}{(3+x \log (5)) \left (32+10 x \log (5)+e^x (3+x \log (5))\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx\\ &=3 \int \frac {2 \log (5)-e^x (3+x \log (5))^2}{(3+x \log (5)) \left (32+10 x \log (5)+e^x (3+x \log (5))\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx\\ &=3 \int \left (-\frac {1}{\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )}+\frac {2 \left (48+\log (5)+31 x \log (5)+5 x^2 \log ^2(5)\right )}{(3+x \log (5)) \left (32+3 e^x+10 x \log (5)+e^x x \log (5)\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )}\right ) \, dx\\ &=-\left (3 \int \frac {1}{\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx\right )+6 \int \frac {48+\log (5)+31 x \log (5)+5 x^2 \log ^2(5)}{(3+x \log (5)) \left (32+3 e^x+10 x \log (5)+e^x x \log (5)\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx\\ &=-\left (3 \int \frac {1}{\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx\right )+6 \int \left (\frac {16}{\left (32+3 e^x+10 x \log (5)+e^x x \log (5)\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )}+\frac {5 x \log (5)}{\left (32+3 e^x+10 x \log (5)+e^x x \log (5)\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )}+\frac {\log (5)}{(3+x \log (5)) \left (32+3 e^x+10 x \log (5)+e^x x \log (5)\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )}\right ) \, dx\\ &=-\left (3 \int \frac {1}{\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx\right )+96 \int \frac {1}{\left (32+3 e^x+10 x \log (5)+e^x x \log (5)\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx+(6 \log (5)) \int \frac {1}{(3+x \log (5)) \left (32+3 e^x+10 x \log (5)+e^x x \log (5)\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx+(30 \log (5)) \int \frac {x}{\left (32+3 e^x+10 x \log (5)+e^x x \log (5)\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.07, size = 32, normalized size = 1.07 \begin {gather*} \frac {3}{\log \left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(6*Log[5] + E^x*(-27 - 18*x*Log[5] - 3*x^2*Log[5]^2))/((96 + 62*x*Log[5] + 10*x^2*Log[5]^2 + E^x*(9

+ 6*x*Log[5] + x^2*Log[5]^2))*Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*L

og[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]^2),x]

[Out]

3/Log[Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]

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fricas [A] time = 0.69, size = 31, normalized size = 1.03 \begin {gather*} \frac {3}{\log \left (\log \left (\frac {{\left (x \log \relax (5) + 3\right )} e^{x} + 10 \, x \log \relax (5) + 32}{x \log \relax (5) + 3}\right )\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2*log(5)^2-18*x*log(5)-27)*exp(x)+6*log(5))/((x^2*log(5)^2+6*x*log(5)+9)*exp(x)+10*x^2*log(5)

^2+62*x*log(5)+96)/log(((x*log(5)+3)*exp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*lo

g(5)+32)/(x*log(5)+3)))^2,x, algorithm="fricas")

[Out]

3/log(log(((x*log(5) + 3)*e^x + 10*x*log(5) + 32)/(x*log(5) + 3)))

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giac [A] time = 0.58, size = 33, normalized size = 1.10 \begin {gather*} \frac {3}{\log \left (\log \left (x e^{x} \log \relax (5) + 10 \, x \log \relax (5) + 3 \, e^{x} + 32\right ) - \log \left (x \log \relax (5) + 3\right )\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2*log(5)^2-18*x*log(5)-27)*exp(x)+6*log(5))/((x^2*log(5)^2+6*x*log(5)+9)*exp(x)+10*x^2*log(5)

^2+62*x*log(5)+96)/log(((x*log(5)+3)*exp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*lo

g(5)+32)/(x*log(5)+3)))^2,x, algorithm="giac")

[Out]

3/log(log(x*e^x*log(5) + 10*x*log(5) + 3*e^x + 32) - log(x*log(5) + 3))

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maple [C] time = 0.22, size = 149, normalized size = 4.97


method	result	size

risch	\(\frac {3}{\ln \left (-\ln \left (x \ln \relax (5)+3\right )+\ln \left (\ln \relax (5) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (\ln \relax (5) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )}{x \ln \relax (5)+3}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\ln \relax (5) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )}{x \ln \relax (5)+3}\right )+\mathrm {csgn}\left (\frac {i}{x \ln \relax (5)+3}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\ln \relax (5) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )}{x \ln \relax (5)+3}\right )+\mathrm {csgn}\left (i \left (\ln \relax (5) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )\right )\right )}{2}\right )}\)	\(149\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^2*ln(5)^2-18*x*ln(5)-27)*exp(x)+6*ln(5))/((x^2*ln(5)^2+6*x*ln(5)+9)*exp(x)+10*x^2*ln(5)^2+62*x*ln(5

)+96)/ln(((x*ln(5)+3)*exp(x)+10*x*ln(5)+32)/(x*ln(5)+3))/ln(ln(((x*ln(5)+3)*exp(x)+10*x*ln(5)+32)/(x*ln(5)+3))

)^2,x,method=_RETURNVERBOSE)

[Out]

3/ln(-ln(x*ln(5)+3)+ln(ln(5)*(exp(x)+10)*x+3*exp(x)+32)-1/2*I*Pi*csgn(I/(x*ln(5)+3)*(ln(5)*(exp(x)+10)*x+3*exp

(x)+32))*(-csgn(I/(x*ln(5)+3)*(ln(5)*(exp(x)+10)*x+3*exp(x)+32))+csgn(I/(x*ln(5)+3)))*(-csgn(I/(x*ln(5)+3)*(ln

(5)*(exp(x)+10)*x+3*exp(x)+32))+csgn(I*(ln(5)*(exp(x)+10)*x+3*exp(x)+32))))

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maxima [A] time = 0.79, size = 32, normalized size = 1.07 \begin {gather*} \frac {3}{\log \left (\log \left ({\left (x \log \relax (5) + 3\right )} e^{x} + 10 \, x \log \relax (5) + 32\right ) - \log \left (x \log \relax (5) + 3\right )\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2*log(5)^2-18*x*log(5)-27)*exp(x)+6*log(5))/((x^2*log(5)^2+6*x*log(5)+9)*exp(x)+10*x^2*log(5)

^2+62*x*log(5)+96)/log(((x*log(5)+3)*exp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*lo

g(5)+32)/(x*log(5)+3)))^2,x, algorithm="maxima")

[Out]

3/log(log((x*log(5) + 3)*e^x + 10*x*log(5) + 32) - log(x*log(5) + 3))

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mupad [B] time = 8.80, size = 31, normalized size = 1.03 \begin {gather*} \frac {3}{\ln \left (\ln \left (\frac {10\,x\,\ln \relax (5)+{\mathrm {e}}^x\,\left (x\,\ln \relax (5)+3\right )+32}{x\,\ln \relax (5)+3}\right )\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((6*log(5) - exp(x)*(3*x^2*log(5)^2 + 18*x*log(5) + 27))/(log((10*x*log(5) + exp(x)*(x*log(5) + 3) + 32)/(x

*log(5) + 3))*log(log((10*x*log(5) + exp(x)*(x*log(5) + 3) + 32)/(x*log(5) + 3)))^2*(10*x^2*log(5)^2 + 62*x*lo

g(5) + exp(x)*(x^2*log(5)^2 + 6*x*log(5) + 9) + 96)),x)

[Out]

3/log(log((10*x*log(5) + exp(x)*(x*log(5) + 3) + 32)/(x*log(5) + 3)))

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sympy [A] time = 3.90, size = 29, normalized size = 0.97 \begin {gather*} \frac {3}{\log {\left (\log {\left (\frac {10 x \log {\relax (5 )} + \left (x \log {\relax (5 )} + 3\right ) e^{x} + 32}{x \log {\relax (5 )} + 3} \right )} \right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**2*ln(5)**2-18*x*ln(5)-27)*exp(x)+6*ln(5))/((x**2*ln(5)**2+6*x*ln(5)+9)*exp(x)+10*x**2*ln(5)*

*2+62*x*ln(5)+96)/ln(((x*ln(5)+3)*exp(x)+10*x*ln(5)+32)/(x*ln(5)+3))/ln(ln(((x*ln(5)+3)*exp(x)+10*x*ln(5)+32)/

(x*ln(5)+3)))**2,x)

[Out]

3/log(log((10*x*log(5) + (x*log(5) + 3)*exp(x) + 32)/(x*log(5) + 3)))

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