3.56.6 \(\int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)} (-48+30 x-12 x^2-36 x^3+(-48+12 x+36 x^2) \log (x))}{16+24 x+9 x^2} \, dx\)

Optimal. Leaf size=26 \[ 4+\frac {2 e^{(-x+\log (x))^2} x^2}{-\frac {4}{3}-x} \]

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Rubi [F]  time = 3.67, 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 {e^{x^2+(1-2 x) \log (x)+\log ^2(x)} \left (-48+30 x-12 x^2-36 x^3+\left (-48+12 x+36 x^2\right ) \log (x)\right )}{16+24 x+9 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x^2 + (1 - 2*x)*Log[x] + Log[x]^2)*(-48 + 30*x - 12*x^2 - 36*x^3 + (-48 + 12*x + 36*x^2)*Log[x]))/(16
+ 24*x + 9*x^2),x]

[Out]

(28*Defer[Int][E^(x^2 + (1 - 2*x)*Log[x] + Log[x]^2), x])/3 - 4*Defer[Int][E^(x^2 + (1 - 2*x)*Log[x] + Log[x]^
2)*x, x] - 24*Defer[Int][E^(x^2 + (1 - 2*x)*Log[x] + Log[x]^2)/(4 + 3*x)^2, x] - (130*Defer[Int][E^(x^2 + (1 -
2*x)*Log[x] + Log[x]^2)/(4 + 3*x), x])/3 + 4*Defer[Int][E^(x^2 + (1 - 2*x)*Log[x] + Log[x]^2)*Log[x], x] - 28
*Defer[Int][(E^(x^2 + (1 - 2*x)*Log[x] + Log[x]^2)*Log[x])/(4 + 3*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)} \left (-48+30 x-12 x^2-36 x^3+\left (-48+12 x+36 x^2\right ) \log (x)\right )}{(4+3 x)^2} \, dx\\ &=\int \left (-\frac {48 e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{(4+3 x)^2}+\frac {30 e^{x^2+(1-2 x) \log (x)+\log ^2(x)} x}{(4+3 x)^2}-\frac {12 e^{x^2+(1-2 x) \log (x)+\log ^2(x)} x^2}{(4+3 x)^2}-\frac {36 e^{x^2+(1-2 x) \log (x)+\log ^2(x)} x^3}{(4+3 x)^2}+\frac {12 e^{x^2+(1-2 x) \log (x)+\log ^2(x)} (-1+x) \log (x)}{4+3 x}\right ) \, dx\\ &=-\left (12 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)} x^2}{(4+3 x)^2} \, dx\right )+12 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)} (-1+x) \log (x)}{4+3 x} \, dx+30 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)} x}{(4+3 x)^2} \, dx-36 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)} x^3}{(4+3 x)^2} \, dx-48 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{(4+3 x)^2} \, dx\\ &=-\left (12 \int \left (\frac {1}{9} e^{x^2+(1-2 x) \log (x)+\log ^2(x)}+\frac {16 e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{9 (4+3 x)^2}-\frac {8 e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{9 (4+3 x)}\right ) \, dx\right )+12 \int \left (\frac {1}{3} e^{x^2+(1-2 x) \log (x)+\log ^2(x)} \log (x)-\frac {7 e^{x^2+(1-2 x) \log (x)+\log ^2(x)} \log (x)}{3 (4+3 x)}\right ) \, dx+30 \int \left (-\frac {4 e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{3 (4+3 x)^2}+\frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{3 (4+3 x)}\right ) \, dx-36 \int \left (-\frac {8}{27} e^{x^2+(1-2 x) \log (x)+\log ^2(x)}+\frac {1}{9} e^{x^2+(1-2 x) \log (x)+\log ^2(x)} x-\frac {64 e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{27 (4+3 x)^2}+\frac {16 e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{9 (4+3 x)}\right ) \, dx-48 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{(4+3 x)^2} \, dx\\ &=-\left (\frac {4}{3} \int e^{x^2+(1-2 x) \log (x)+\log ^2(x)} \, dx\right )-4 \int e^{x^2+(1-2 x) \log (x)+\log ^2(x)} x \, dx+4 \int e^{x^2+(1-2 x) \log (x)+\log ^2(x)} \log (x) \, dx+10 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{4+3 x} \, dx+\frac {32}{3} \int e^{x^2+(1-2 x) \log (x)+\log ^2(x)} \, dx+\frac {32}{3} \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{4+3 x} \, dx-\frac {64}{3} \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{(4+3 x)^2} \, dx-28 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)} \log (x)}{4+3 x} \, dx-40 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{(4+3 x)^2} \, dx-48 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{(4+3 x)^2} \, dx-64 \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{4+3 x} \, dx+\frac {256}{3} \int \frac {e^{x^2+(1-2 x) \log (x)+\log ^2(x)}}{(4+3 x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.96, size = 26, normalized size = 1.00 \begin {gather*} -\frac {6 e^{x^2+\log ^2(x)} x^{2-2 x}}{4+3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x^2 + (1 - 2*x)*Log[x] + Log[x]^2)*(-48 + 30*x - 12*x^2 - 36*x^3 + (-48 + 12*x + 36*x^2)*Log[x])
)/(16 + 24*x + 9*x^2),x]

[Out]

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

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fricas [A]  time = 0.75, size = 28, normalized size = 1.08 \begin {gather*} -\frac {6 \, x e^{\left (x^{2} - {\left (2 \, x - 1\right )} \log \relax (x) + \log \relax (x)^{2}\right )}}{3 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^2+12*x-48)*log(x)-36*x^3-12*x^2+30*x-48)*exp(log(x)^2+(1-2*x)*log(x)+x^2)/(9*x^2+24*x+16),x,
algorithm="fricas")

[Out]

-6*x*e^(x^2 - (2*x - 1)*log(x) + log(x)^2)/(3*x + 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {6 \, {\left (6 \, x^{3} + 2 \, x^{2} - 2 \, {\left (3 \, x^{2} + x - 4\right )} \log \relax (x) - 5 \, x + 8\right )} e^{\left (x^{2} - {\left (2 \, x - 1\right )} \log \relax (x) + \log \relax (x)^{2}\right )}}{9 \, x^{2} + 24 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^2+12*x-48)*log(x)-36*x^3-12*x^2+30*x-48)*exp(log(x)^2+(1-2*x)*log(x)+x^2)/(9*x^2+24*x+16),x,
algorithm="giac")

[Out]

integrate(-6*(6*x^3 + 2*x^2 - 2*(3*x^2 + x - 4)*log(x) - 5*x + 8)*e^(x^2 - (2*x - 1)*log(x) + log(x)^2)/(9*x^2
+ 24*x + 16), x)

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maple [A]  time = 0.28, size = 27, normalized size = 1.04

method	result	size
risch	\(-\frac {6 x \,x^{1-2 x} {\mathrm e}^{\ln \relax (x )^{2}+x^{2}}}{4+3 x}\)	\(27\)
norman	\(-\frac {6 x \,{\mathrm e}^{\ln \relax (x )^{2}+\left (1-2 x \right ) \ln \relax (x )+x^{2}}}{4+3 x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((36*x^2+12*x-48)*ln(x)-36*x^3-12*x^2+30*x-48)*exp(ln(x)^2+(1-2*x)*ln(x)+x^2)/(9*x^2+24*x+16),x,method=_RE
TURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.46, size = 26, normalized size = 1.00 \begin {gather*} -\frac {6 \, x^{2} e^{\left (x^{2} - 2 \, x \log \relax (x) + \log \relax (x)^{2}\right )}}{3 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^2+12*x-48)*log(x)-36*x^3-12*x^2+30*x-48)*exp(log(x)^2+(1-2*x)*log(x)+x^2)/(9*x^2+24*x+16),x,
algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.87, size = 26, normalized size = 1.00 \begin {gather*} -\frac {2\,x^2\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{{\ln \relax (x)}^2}}{x^{2\,x}\,\left (x+\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(x)^2 - log(x)*(2*x - 1) + x^2)*(12*x^2 - log(x)*(12*x + 36*x^2 - 48) - 30*x + 36*x^3 + 48))/(24*
x + 9*x^2 + 16),x)

[Out]

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

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sympy [A]  time = 0.32, size = 27, normalized size = 1.04 \begin {gather*} - \frac {6 x e^{x^{2} + \left (1 - 2 x\right ) \log {\relax (x )} + \log {\relax (x )}^{2}}}{3 x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x**2+12*x-48)*ln(x)-36*x**3-12*x**2+30*x-48)*exp(ln(x)**2+(1-2*x)*ln(x)+x**2)/(9*x**2+24*x+16),
x)

[Out]

-6*x*exp(x**2 + (1 - 2*x)*log(x) + log(x)**2)/(3*x + 4)

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