3.26.39 \(\int \frac {(12 x^2+2 x^3) \log ^2(4 e^x)+(-15 x-3 x^2) \log (x)+(-18-3 x) \log (6+x)+(12 x^2+2 x^3) \log ^2(6+x)+\log (4 e^x) (18+3 x+(-24 x^2-4 x^3) \log (6+x))}{(18 x+3 x^2) \log ^2(4 e^x)+(-36 x-6 x^2) \log (4 e^x) \log (6+x)+(18 x+3 x^2) \log ^2(6+x)} \, dx\)

Optimal. Leaf size=31 \[ -e^6+\frac {x^2}{3}+\frac {\log (x)}{\log \left (4 e^x\right )-\log (6+x)} \]

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Rubi [F]  time = 1.21, 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 (12 x^2+2 x^3\right ) \log ^2\left (4 e^x\right )+\left (-15 x-3 x^2\right ) \log (x)+(-18-3 x) \log (6+x)+\left (12 x^2+2 x^3\right ) \log ^2(6+x)+\log \left (4 e^x\right ) \left (18+3 x+\left (-24 x^2-4 x^3\right ) \log (6+x)\right )}{\left (18 x+3 x^2\right ) \log ^2\left (4 e^x\right )+\left (-36 x-6 x^2\right ) \log \left (4 e^x\right ) \log (6+x)+\left (18 x+3 x^2\right ) \log ^2(6+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((12*x^2 + 2*x^3)*Log[4*E^x]^2 + (-15*x - 3*x^2)*Log[x] + (-18 - 3*x)*Log[6 + x] + (12*x^2 + 2*x^3)*Log[6
+ x]^2 + Log[4*E^x]*(18 + 3*x + (-24*x^2 - 4*x^3)*Log[6 + x]))/((18*x + 3*x^2)*Log[4*E^x]^2 + (-36*x - 6*x^2)*
Log[4*E^x]*Log[6 + x] + (18*x + 3*x^2)*Log[6 + x]^2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 27, normalized size = 0.87 \begin {gather*} \frac {1}{3} \left (x^2-\frac {3 \log (x)}{-\log \left (4 e^x\right )+\log (6+x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((12*x^2 + 2*x^3)*Log[4*E^x]^2 + (-15*x - 3*x^2)*Log[x] + (-18 - 3*x)*Log[6 + x] + (12*x^2 + 2*x^3)*
Log[6 + x]^2 + Log[4*E^x]*(18 + 3*x + (-24*x^2 - 4*x^3)*Log[6 + x]))/((18*x + 3*x^2)*Log[4*E^x]^2 + (-36*x - 6
*x^2)*Log[4*E^x]*Log[6 + x] + (18*x + 3*x^2)*Log[6 + x]^2),x]

[Out]

(x^2 - (3*Log[x])/(-Log[4*E^x] + Log[6 + x]))/3

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fricas [A]  time = 0.73, size = 40, normalized size = 1.29 \begin {gather*} \frac {x^{3} + 2 \, x^{2} \log \relax (2) - x^{2} \log \left (x + 6\right ) + 3 \, \log \relax (x)}{3 \, {\left (x + 2 \, \log \relax (2) - \log \left (x + 6\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+12*x^2)*log(4*exp(x))^2+((-4*x^3-24*x^2)*log(x+6)+3*x+18)*log(4*exp(x))+(2*x^3+12*x^2)*log(x
+6)^2+(-3*x-18)*log(x+6)+(-3*x^2-15*x)*log(x))/((3*x^2+18*x)*log(4*exp(x))^2+(-6*x^2-36*x)*log(x+6)*log(4*exp(
x))+(3*x^2+18*x)*log(x+6)^2),x, algorithm="fricas")

[Out]

1/3*(x^3 + 2*x^2*log(2) - x^2*log(x + 6) + 3*log(x))/(x + 2*log(2) - log(x + 6))

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giac [A]  time = 0.43, size = 23, normalized size = 0.74 \begin {gather*} \frac {1}{3} \, x^{2} + \frac {\log \relax (x)}{x + 2 \, \log \relax (2) - \log \left (x + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+12*x^2)*log(4*exp(x))^2+((-4*x^3-24*x^2)*log(x+6)+3*x+18)*log(4*exp(x))+(2*x^3+12*x^2)*log(x
+6)^2+(-3*x-18)*log(x+6)+(-3*x^2-15*x)*log(x))/((3*x^2+18*x)*log(4*exp(x))^2+(-6*x^2-36*x)*log(x+6)*log(4*exp(
x))+(3*x^2+18*x)*log(x+6)^2),x, algorithm="giac")

[Out]

1/3*x^2 + log(x)/(x + 2*log(2) - log(x + 6))

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maple [C]  time = 1.00, size = 33, normalized size = 1.06

method	result	size
risch	\(\frac {x^{2}}{3}+\frac {2 i \ln \relax (x )}{4 i \ln \relax (2)-2 i \ln \left (x +6\right )+2 i \ln \left ({\mathrm e}^{x}\right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3+12*x^2)*ln(4*exp(x))^2+((-4*x^3-24*x^2)*ln(x+6)+3*x+18)*ln(4*exp(x))+(2*x^3+12*x^2)*ln(x+6)^2+(-3*
x-18)*ln(x+6)+(-3*x^2-15*x)*ln(x))/((3*x^2+18*x)*ln(4*exp(x))^2+(-6*x^2-36*x)*ln(x+6)*ln(4*exp(x))+(3*x^2+18*x
)*ln(x+6)^2),x,method=_RETURNVERBOSE)

[Out]

1/3*x^2+2*I*ln(x)/(4*I*ln(2)-2*I*ln(x+6)+2*I*ln(exp(x)))

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maxima [A]  time = 0.58, size = 40, normalized size = 1.29 \begin {gather*} \frac {x^{3} + 2 \, x^{2} \log \relax (2) - x^{2} \log \left (x + 6\right ) + 3 \, \log \relax (x)}{3 \, {\left (x + 2 \, \log \relax (2) - \log \left (x + 6\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+12*x^2)*log(4*exp(x))^2+((-4*x^3-24*x^2)*log(x+6)+3*x+18)*log(4*exp(x))+(2*x^3+12*x^2)*log(x
+6)^2+(-3*x-18)*log(x+6)+(-3*x^2-15*x)*log(x))/((3*x^2+18*x)*log(4*exp(x))^2+(-6*x^2-36*x)*log(x+6)*log(4*exp(
x))+(3*x^2+18*x)*log(x+6)^2),x, algorithm="maxima")

[Out]

1/3*(x^3 + 2*x^2*log(2) - x^2*log(x + 6) + 3*log(x))/(x + 2*log(2) - log(x + 6))

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mupad [B]  time = 1.64, size = 40, normalized size = 1.29 \begin {gather*} \frac {3\,\ln \relax (x)-x^2\,\ln \left (x+6\right )+x^2\,\ln \relax (4)+x^3}{3\,\left (x-\ln \left (x+6\right )+\ln \relax (4)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4*exp(x))^2*(12*x^2 + 2*x^3) + log(4*exp(x))*(3*x - log(x + 6)*(24*x^2 + 4*x^3) + 18) - log(x)*(15*x
+ 3*x^2) + log(x + 6)^2*(12*x^2 + 2*x^3) - log(x + 6)*(3*x + 18))/(log(x + 6)^2*(18*x + 3*x^2) + log(4*exp(x))
^2*(18*x + 3*x^2) - log(x + 6)*log(4*exp(x))*(36*x + 6*x^2)),x)

[Out]

(3*log(x) - x^2*log(x + 6) + x^2*log(4) + x^3)/(3*(x - log(x + 6) + log(4)))

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sympy [A]  time = 0.47, size = 19, normalized size = 0.61 \begin {gather*} \frac {x^{2}}{3} - \frac {\log {\relax (x )}}{- x + \log {\left (x + 6 \right )} - 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3+12*x**2)*ln(4*exp(x))**2+((-4*x**3-24*x**2)*ln(x+6)+3*x+18)*ln(4*exp(x))+(2*x**3+12*x**2)*l
n(x+6)**2+(-3*x-18)*ln(x+6)+(-3*x**2-15*x)*ln(x))/((3*x**2+18*x)*ln(4*exp(x))**2+(-6*x**2-36*x)*ln(x+6)*ln(4*e
xp(x))+(3*x**2+18*x)*ln(x+6)**2),x)

[Out]

x**2/3 - log(x)/(-x + log(x + 6) - 2*log(2))

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