3.7.86 \(\int \frac {-3 x^2-6 x^3-9 \log ^2(5 e^{x+x^2})}{x^2+(6 x-6 x^2) \log (5 e^{x+x^2})+(9-18 x+9 x^2) \log ^2(5 e^{x+x^2})} \, dx\)

Optimal. Leaf size=24 \[ \frac {x}{-1+x-\frac {x}{3 \log \left (5 e^{x+x^2}\right )}} \]

________________________________________________________________________________________

Rubi [F]  time = 1.88, 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 {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*x^2 - 6*x^3 - 9*Log[5*E^(x + x^2)]^2)/(x^2 + (6*x - 6*x^2)*Log[5*E^(x + x^2)] + (9 - 18*x + 9*x^2)*Log
[5*E^(x + x^2)]^2),x]

[Out]

-(1 - x)^(-1) - Defer[Int][(x - 3*(-1 + x)*Log[5*E^(x + x^2)])^(-2), x] - Defer[Int][1/((1 - x)^2*(x - 3*(-1 +
x)*Log[5*E^(x + x^2)])^2), x] - 2*Defer[Int][1/((-1 + x)*(x - 3*(-1 + x)*Log[5*E^(x + x^2)])^2), x] - 3*Defer
[Int][x^2/(x - 3*(-1 + x)*Log[5*E^(x + x^2)])^2, x] - 6*Defer[Int][x^3/(x - 3*(-1 + x)*Log[5*E^(x + x^2)])^2,
x] - 2*Defer[Int][1/((1 - x)*(x - 3*(-1 + x)*Log[5*E^(x + x^2)])), x] - 2*Defer[Int][1/((-1 + x)^2*(-x - 3*Log
[5*E^(x + x^2)] + 3*x*Log[5*E^(x + x^2)])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 \left (-x^2-2 x^3-3 \log ^2\left (5 e^{x+x^2}\right )\right )}{\left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )^2} \, dx\\ &=3 \int \frac {-x^2-2 x^3-3 \log ^2\left (5 e^{x+x^2}\right )}{\left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )^2} \, dx\\ &=3 \int \left (-\frac {1}{3 (-1+x)^2}+\frac {x^2 \left (-4+9 x^2-6 x^3\right )}{3 (1-x)^2 \left (x+3 \log \left (5 e^{x+x^2}\right )-3 x \log \left (5 e^{x+x^2}\right )\right )^2}-\frac {2 x}{3 (-1+x)^2 \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )}\right ) \, dx\\ &=-\frac {1}{1-x}-2 \int \frac {x}{(-1+x)^2 \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )} \, dx+\int \frac {x^2 \left (-4+9 x^2-6 x^3\right )}{(1-x)^2 \left (x+3 \log \left (5 e^{x+x^2}\right )-3 x \log \left (5 e^{x+x^2}\right )\right )^2} \, dx\\ &=-\frac {1}{1-x}-2 \int \left (\frac {1}{(-1+x)^2 \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )}+\frac {1}{(-1+x) \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )}\right ) \, dx+\int \frac {x^2 \left (-4+9 x^2-6 x^3\right )}{(1-x)^2 \left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )^2} \, dx\\ &=-\frac {1}{1-x}-2 \int \frac {1}{(-1+x)^2 \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )} \, dx-2 \int \frac {1}{(-1+x) \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )} \, dx+\int \left (-\frac {1}{\left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2}-\frac {1}{(-1+x)^2 \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2}-\frac {2}{(-1+x) \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2}-\frac {3 x^2}{\left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2}-\frac {6 x^3}{\left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2}\right ) \, dx\\ &=-\frac {1}{1-x}-2 \int \frac {1}{(1-x) \left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )} \, dx-2 \int \frac {1}{(-1+x) \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2} \, dx-2 \int \frac {1}{(-1+x)^2 \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )} \, dx-3 \int \frac {x^2}{\left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2} \, dx-6 \int \frac {x^3}{\left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2} \, dx-\int \frac {1}{\left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2} \, dx-\int \frac {1}{(-1+x)^2 \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )^2} \, dx\\ &=-\frac {1}{1-x}-2 \int \frac {1}{(-1+x) \left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )^2} \, dx-2 \int \frac {1}{(1-x) \left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )} \, dx-2 \int \frac {1}{(-1+x)^2 \left (-x-3 \log \left (5 e^{x+x^2}\right )+3 x \log \left (5 e^{x+x^2}\right )\right )} \, dx-3 \int \frac {x^2}{\left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )^2} \, dx-6 \int \frac {x^3}{\left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )^2} \, dx-\int \frac {1}{\left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )^2} \, dx-\int \frac {1}{(1-x)^2 \left (x-3 (-1+x) \log \left (5 e^{x+x^2}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 1.22, size = 37, normalized size = 1.54 \begin {gather*} \frac {3 \left (x+3 \log \left (5 e^{x+x^2}\right )\right )}{-3 x+9 (-1+x) \log \left (5 e^{x+x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^2 - 6*x^3 - 9*Log[5*E^(x + x^2)]^2)/(x^2 + (6*x - 6*x^2)*Log[5*E^(x + x^2)] + (9 - 18*x + 9*x^
2)*Log[5*E^(x + x^2)]^2),x]

[Out]

(3*(x + 3*Log[5*E^(x + x^2)]))/(-3*x + 9*(-1 + x)*Log[5*E^(x + x^2)])

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 32, normalized size = 1.33 \begin {gather*} \frac {3 \, x^{2} + 4 \, x + 3 \, \log \relax (5)}{3 \, x^{3} + 3 \, {\left (x - 1\right )} \log \relax (5) - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+
x))+x^2),x, algorithm="fricas")

[Out]

(3*x^2 + 4*x + 3*log(5))/(3*x^3 + 3*(x - 1)*log(5) - 4*x)

________________________________________________________________________________________

giac [A]  time = 0.26, size = 34, normalized size = 1.42 \begin {gather*} \frac {3 \, x^{2} + 4 \, x + 3 \, \log \relax (5)}{3 \, x^{3} + 3 \, x \log \relax (5) - 4 \, x - 3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+
x))+x^2),x, algorithm="giac")

[Out]

(3*x^2 + 4*x + 3*log(5))/(3*x^3 + 3*x*log(5) - 4*x - 3*log(5))

________________________________________________________________________________________

maple [C]  time = 0.22, size = 57, normalized size = 2.38

method	result	size
risch	\(\frac {1}{x -1}+\frac {2 i x^{2}}{\left (x -1\right ) \left (6 i x \ln \relax (5)+6 i x \ln \left ({\mathrm e}^{\left (x +1\right ) x}\right )-6 i \ln \relax (5)-2 i x -6 i \ln \left ({\mathrm e}^{\left (x +1\right ) x}\right )\right )}\)	\(57\)
default	\(-\frac {3 \left (-\frac {x}{9}-\frac {\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )}{3}\right )}{x^{3}+x \left (\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )-x^{2}-x \right )-\frac {x}{3}-\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )+x^{2}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9*ln(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*ln(5*exp(x^2+x))^2+(-6*x^2+6*x)*ln(5*exp(x^2+x))+x^2),
x,method=_RETURNVERBOSE)

[Out]

1/(x-1)+2*I*x^2/(x-1)/(6*I*x*ln(5)+6*I*x*ln(exp((x+1)*x))-6*I*ln(5)-2*I*x-6*I*ln(exp((x+1)*x)))

________________________________________________________________________________________

maxima [A]  time = 0.71, size = 34, normalized size = 1.42 \begin {gather*} \frac {3 \, x^{2} + 4 \, x + 3 \, \log \relax (5)}{3 \, x^{3} + x {\left (3 \, \log \relax (5) - 4\right )} - 3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+
x))+x^2),x, algorithm="maxima")

[Out]

(3*x^2 + 4*x + 3*log(5))/(3*x^3 + x*(3*log(5) - 4) - 3*log(5))

________________________________________________________________________________________

mupad [B]  time = 0.72, size = 30, normalized size = 1.25 \begin {gather*} \frac {3\,x^2+4\,x+\ln \left (125\right )}{3\,x^3+\left (\ln \left (125\right )-4\right )\,x-\ln \left (125\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*log(5*exp(x + x^2))^2 + 3*x^2 + 6*x^3)/(log(5*exp(x + x^2))^2*(9*x^2 - 18*x + 9) + log(5*exp(x + x^2))
*(6*x - 6*x^2) + x^2),x)

[Out]

(4*x + log(125) + 3*x^2)/(x*(log(125) - 4) - log(125) + 3*x^3)

________________________________________________________________________________________

sympy [A]  time = 1.20, size = 34, normalized size = 1.42 \begin {gather*} - \frac {- 3 x^{2} - 4 x - 3 \log {\relax (5 )}}{3 x^{3} + x \left (-4 + 3 \log {\relax (5 )}\right ) - 3 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-3*x**2 - 4*x - 3*log(5))/(3*x**3 + x*(-4 + 3*log(5)) - 3*log(5))

________________________________________________________________________________________