3.59.7 \(\int \frac {x^{x+x^2} (-x^2+(1+x-2 x^2) \log (x))}{\log (5)} \, dx\)

Optimal. Leaf size=17 \[ \frac {(1-x) x^{x+x^2}}{\log (5)} \]

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Rubi [F]  time = 0.27, 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 {x^{x+x^2} \left (-x^2+\left (1+x-2 x^2\right ) \log (x)\right )}{\log (5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 16, normalized size = 0.94 \begin {gather*} -\frac {(-1+x) x^{x (1+x)}}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(x + x^2)*(-x^2 + (1 + x - 2*x^2)*Log[x]))/Log[5],x]

[Out]

-(((-1 + x)*x^(x*(1 + x)))/Log[5])

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fricas [A]  time = 0.83, size = 16, normalized size = 0.94 \begin {gather*} -\frac {{\left (x - 1\right )} x^{x^{2} + x}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+x+1)*log(x)-x^2)*exp((x^2+x)*log(x))/log(5),x, algorithm="fricas")

[Out]

-(x - 1)*x^(x^2 + x)/log(5)

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giac [B]  time = 0.47, size = 35, normalized size = 2.06 \begin {gather*} -\frac {x e^{\left (x^{2} \log \relax (x) + x \log \relax (x)\right )} - e^{\left (x^{2} \log \relax (x) + x \log \relax (x)\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+x+1)*log(x)-x^2)*exp((x^2+x)*log(x))/log(5),x, algorithm="giac")

[Out]

-(x*e^(x^2*log(x) + x*log(x)) - e^(x^2*log(x) + x*log(x)))/log(5)

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maple [A]  time = 0.05, size = 18, normalized size = 1.06

method	result	size
risch	\(\frac {\left (1-x \right ) x^{\left (x +1\right ) x}}{\ln \relax (5)}\)	\(18\)
norman	\(\frac {{\mathrm e}^{\left (x^{2}+x \right ) \ln \relax (x )}}{\ln \relax (5)}-\frac {x \,{\mathrm e}^{\left (x^{2}+x \right ) \ln \relax (x )}}{\ln \relax (5)}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2+x+1)*ln(x)-x^2)*exp((x^2+x)*ln(x))/ln(5),x,method=_RETURNVERBOSE)

[Out]

(1-x)*x^((x+1)*x)/ln(5)

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maxima [A]  time = 0.41, size = 21, normalized size = 1.24 \begin {gather*} -\frac {{\left (x - 1\right )} e^{\left (x^{2} \log \relax (x) + x \log \relax (x)\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+x+1)*log(x)-x^2)*exp((x^2+x)*log(x))/log(5),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.16, size = 16, normalized size = 0.94 \begin {gather*} -\frac {x^{x^2+x}\,\left (x-1\right )}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(x)*(x + x^2))*(log(x)*(x - 2*x^2 + 1) - x^2))/log(5),x)

[Out]

-(x^(x + x^2)*(x - 1))/log(5)

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sympy [A]  time = 0.33, size = 15, normalized size = 0.88 \begin {gather*} \frac {\left (1 - x\right ) e^{\left (x^{2} + x\right ) \log {\relax (x )}}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2+x+1)*ln(x)-x**2)*exp((x**2+x)*ln(x))/ln(5),x)

[Out]

(1 - x)*exp((x**2 + x)*log(x))/log(5)

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