3.44.20 \(\int \frac {-36 x-36 x^2+(-36 x-12 x^2) \log (\frac {1}{e^2 (81 x^2+108 x^3+54 x^4+12 x^5+x^6)})+(30+10 x) \log ^2(\frac {1}{e^2 (81 x^2+108 x^3+54 x^4+12 x^5+x^6)})}{(3+x) \log ^2(\frac {1}{e^2 (81 x^2+108 x^3+54 x^4+12 x^5+x^6)})} \, dx\)

Optimal. Leaf size=23 \[ 2 x \left (5-\frac {3 x}{\log \left (\frac {1}{e^2 x^2 (3+x)^4}\right )}\right ) \]

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Rubi [F]  time = 0.56, 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 {-36 x-36 x^2+\left (-36 x-12 x^2\right ) \log \left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )+(30+10 x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )}{(3+x) \log ^2\left (\frac {1}{e^2 \left (81 x^2+108 x^3+54 x^4+12 x^5+x^6\right )}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-36*x - 36*x^2 + (-36*x - 12*x^2)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))] + (30 + 10*x)*L
og[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))]^2)/((3 + x)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*
x^5 + x^6))]^2),x]

[Out]

10*x + 72*Defer[Int][(-2 + Log[1/(x^2*(3 + x)^4)])^(-2), x] - 36*Defer[Int][x/(-2 + Log[1/(x^2*(3 + x)^4)])^2,
x] - 216*Defer[Int][1/((3 + x)*(-2 + Log[1/(x^2*(3 + x)^4)])^2), x] - 12*Defer[Int][x/(-2 + Log[1/(x^2*(3 + x
)^4)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (60+38 x-6 x^2-\left (60+38 x+6 x^2\right ) \log \left (\frac {1}{x^2 (3+x)^4}\right )+5 (3+x) \log ^2\left (\frac {1}{x^2 (3+x)^4}\right )\right )}{(3+x) \left (2-\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx\\ &=2 \int \frac {60+38 x-6 x^2-\left (60+38 x+6 x^2\right ) \log \left (\frac {1}{x^2 (3+x)^4}\right )+5 (3+x) \log ^2\left (\frac {1}{x^2 (3+x)^4}\right )}{(3+x) \left (2-\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx\\ &=2 \int \left (5-\frac {18 x (1+x)}{(3+x) \left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2}-\frac {6 x}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )}\right ) \, dx\\ &=10 x-12 \int \frac {x}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )} \, dx-36 \int \frac {x (1+x)}{(3+x) \left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx\\ &=10 x-12 \int \frac {x}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )} \, dx-36 \int \left (-\frac {2}{\left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2}+\frac {x}{\left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2}+\frac {6}{(3+x) \left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2}\right ) \, dx\\ &=10 x-12 \int \frac {x}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )} \, dx-36 \int \frac {x}{\left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx+72 \int \frac {1}{\left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx-216 \int \frac {1}{(3+x) \left (-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 25, normalized size = 1.09 \begin {gather*} -2 \left (-5 x+\frac {3 x^2}{-2+\log \left (\frac {1}{x^2 (3+x)^4}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-36*x - 36*x^2 + (-36*x - 12*x^2)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))] + (30 + 1
0*x)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))]^2)/((3 + x)*Log[1/(E^2*(81*x^2 + 108*x^3 + 54*x^4
+ 12*x^5 + x^6))]^2),x]

[Out]

-2*(-5*x + (3*x^2)/(-2 + Log[1/(x^2*(3 + x)^4)]))

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fricas [B]  time = 0.90, size = 73, normalized size = 3.17 \begin {gather*} -\frac {2 \, {\left (3 \, x^{2} - 5 \, x \log \left (\frac {e^{\left (-2\right )}}{x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}}\right )\right )}}{\log \left (\frac {e^{\left (-2\right )}}{x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x+30)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2+(-12*x^2-36*x)*log(1/(x^6+12*x^5+54*x^
4+108*x^3+81*x^2)/exp(2))-36*x^2-36*x)/(3+x)/log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2,x, algorithm="
fricas")

[Out]

-2*(3*x^2 - 5*x*log(e^(-2)/(x^6 + 12*x^5 + 54*x^4 + 108*x^3 + 81*x^2)))/log(e^(-2)/(x^6 + 12*x^5 + 54*x^4 + 10
8*x^3 + 81*x^2))

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giac [A]  time = 0.43, size = 38, normalized size = 1.65 \begin {gather*} 10 \, x + \frac {6 \, x^{2}}{\log \left (x^{6} + 12 \, x^{5} + 54 \, x^{4} + 108 \, x^{3} + 81 \, x^{2}\right ) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x+30)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2+(-12*x^2-36*x)*log(1/(x^6+12*x^5+54*x^
4+108*x^3+81*x^2)/exp(2))-36*x^2-36*x)/(3+x)/log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2,x, algorithm="
giac")

[Out]

10*x + 6*x^2/(log(x^6 + 12*x^5 + 54*x^4 + 108*x^3 + 81*x^2) + 2)

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maple [A]  time = 0.09, size = 42, normalized size = 1.83

method	result	size
risch	\(10 x -\frac {6 x^{2}}{\ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}\)	\(42\)
norman	\(\frac {-6 x^{2}+10 x \ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}{\ln \left (\frac {{\mathrm e}^{-2}}{x^{6}+12 x^{5}+54 x^{4}+108 x^{3}+81 x^{2}}\right )}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x+30)*ln(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2+(-12*x^2-36*x)*ln(1/(x^6+12*x^5+54*x^4+108*x^
3+81*x^2)/exp(2))-36*x^2-36*x)/(3+x)/ln(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2,x,method=_RETURNVERBOSE
)

[Out]

10*x-6*x^2/ln(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)*exp(-2))

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maxima [A]  time = 0.39, size = 34, normalized size = 1.48 \begin {gather*} \frac {3 \, x^{2} + 20 \, x \log \left (x + 3\right ) + 10 \, x \log \relax (x) + 10 \, x}{2 \, \log \left (x + 3\right ) + \log \relax (x) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x+30)*log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2+(-12*x^2-36*x)*log(1/(x^6+12*x^5+54*x^
4+108*x^3+81*x^2)/exp(2))-36*x^2-36*x)/(3+x)/log(1/(x^6+12*x^5+54*x^4+108*x^3+81*x^2)/exp(2))^2,x, algorithm="
maxima")

[Out]

(3*x^2 + 20*x*log(x + 3) + 10*x*log(x) + 10*x)/(2*log(x + 3) + log(x) + 1)

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mupad [B]  time = 3.42, size = 99, normalized size = 4.30 \begin {gather*} 14\,x+\frac {4}{x+1}+2\,x^2-\frac {6\,x^2+\frac {2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{-2}}{x^6+12\,x^5+54\,x^4+108\,x^3+81\,x^2}\right )\,\left (x+3\right )}{x+1}}{\ln \left (\frac {{\mathrm {e}}^{-2}}{x^6+12\,x^5+54\,x^4+108\,x^3+81\,x^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(36*x + log(exp(-2)/(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))*(36*x + 12*x^2) - log(exp(-2)/(81*x^2 + 1
08*x^3 + 54*x^4 + 12*x^5 + x^6))^2*(10*x + 30) + 36*x^2)/(log(exp(-2)/(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^
6))^2*(x + 3)),x)

[Out]

14*x + 4/(x + 1) + 2*x^2 - (6*x^2 + (2*x^2*log(exp(-2)/(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))*(x + 3))/(x
+ 1))/log(exp(-2)/(81*x^2 + 108*x^3 + 54*x^4 + 12*x^5 + x^6))

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sympy [A]  time = 0.20, size = 37, normalized size = 1.61 \begin {gather*} - \frac {6 x^{2}}{\log {\left (\frac {1}{\left (x^{6} + 12 x^{5} + 54 x^{4} + 108 x^{3} + 81 x^{2}\right ) e^{2}} \right )}} + 10 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x+30)*ln(1/(x**6+12*x**5+54*x**4+108*x**3+81*x**2)/exp(2))**2+(-12*x**2-36*x)*ln(1/(x**6+12*x**
5+54*x**4+108*x**3+81*x**2)/exp(2))-36*x**2-36*x)/(3+x)/ln(1/(x**6+12*x**5+54*x**4+108*x**3+81*x**2)/exp(2))**
2,x)

[Out]

-6*x**2/log(exp(-2)/(x**6 + 12*x**5 + 54*x**4 + 108*x**3 + 81*x**2)) + 10*x

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