3.32.59 \(\int \frac {-6 x+e^{3+x^2} (-1-10 x^2)+(-x-2 e^{3+x^2} x^2) \log (x)+(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x)) \log (5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x))}{(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x)) \log ^2(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x))} \, dx\)

Optimal. Leaf size=21 \[ -5+\frac {x}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \]

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Rubi [F]  time = 3.85, 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 {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-6*x + E^(3 + x^2)*(-1 - 10*x^2) + (-x - 2*E^(3 + x^2)*x^2)*Log[x] + (5*E^(3 + x^2) + 5*x + (E^(3 + x^2)
+ x)*Log[x])*Log[5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Log[x]])/((5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Lo
g[x])*Log[5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Log[x]]^2),x]

[Out]

-Defer[Int][x/((E^(3 + x^2) + x)*Log[(E^(3 + x^2) + x)*(5 + Log[x])]^2), x] + 2*Defer[Int][x^3/((E^(3 + x^2) +
x)*Log[(E^(3 + x^2) + x)*(5 + Log[x])]^2), x] - Defer[Int][1/((5 + Log[x])*Log[(E^(3 + x^2) + x)*(5 + Log[x])
]^2), x] - 10*Defer[Int][x^2/((5 + Log[x])*Log[(E^(3 + x^2) + x)*(5 + Log[x])]^2), x] - 2*Defer[Int][(x^2*Log[
x])/((5 + Log[x])*Log[(E^(3 + x^2) + x)*(5 + Log[x])]^2), x] + Defer[Int][Log[(E^(3 + x^2) + x)*(5 + Log[x])]^
(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (e^{3+x^2}+x\right ) (5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=\int \frac {-6 x-e^{3+x^2} \left (1+10 x^2\right )-x \left (1+2 e^{3+x^2} x\right ) \log (x)+\left (e^{3+x^2}+x\right ) (5+\log (x)) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{\left (e^{3+x^2}+x\right ) (5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=\int \left (\frac {x \left (-1+2 x^2\right )}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {-1-10 x^2-2 x^2 \log (x)+5 \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )+\log (x) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx\\ &=\int \frac {x \left (-1+2 x^2\right )}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {-1-10 x^2-2 x^2 \log (x)+5 \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )+\log (x) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=\int \left (-\frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {2 x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx+\int \left (\frac {-1-10 x^2-2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx\\ &=2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {-1-10 x^2-2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \left (-\frac {1}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}-\frac {10 x^2}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}-\frac {2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ &=2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-2 \int \frac {x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-10 \int \frac {x^2}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {1}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 19, normalized size = 0.90 \begin {gather*} \frac {x}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x + E^(3 + x^2)*(-1 - 10*x^2) + (-x - 2*E^(3 + x^2)*x^2)*Log[x] + (5*E^(3 + x^2) + 5*x + (E^(3 +
x^2) + x)*Log[x])*Log[5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Log[x]])/((5*E^(3 + x^2) + 5*x + (E^(3 + x^2) +
x)*Log[x])*Log[5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Log[x]]^2),x]

[Out]

x/Log[(E^(3 + x^2) + x)*(5 + Log[x])]

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fricas [A]  time = 0.54, size = 28, normalized size = 1.33 \begin {gather*} \frac {x}{\log \left ({\left (x + e^{\left (x^{2} + 3\right )}\right )} \log \relax (x) + 5 \, x + 5 \, e^{\left (x^{2} + 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)*log((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)+(-2*x^2*exp(x^
2+3)-x)*log(x)+(-10*x^2-1)*exp(x^2+3)-6*x)/((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)/log((exp(x^2+3)+x)*log(x)+
5*exp(x^2+3)+5*x)^2,x, algorithm="fricas")

[Out]

x/log((x + e^(x^2 + 3))*log(x) + 5*x + 5*e^(x^2 + 3))

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giac [A]  time = 1.73, size = 30, normalized size = 1.43 \begin {gather*} \frac {x}{\log \left (x \log \relax (x) + e^{\left (x^{2} + 3\right )} \log \relax (x) + 5 \, x + 5 \, e^{\left (x^{2} + 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)*log((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)+(-2*x^2*exp(x^
2+3)-x)*log(x)+(-10*x^2-1)*exp(x^2+3)-6*x)/((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)/log((exp(x^2+3)+x)*log(x)+
5*exp(x^2+3)+5*x)^2,x, algorithm="giac")

[Out]

x/log(x*log(x) + e^(x^2 + 3)*log(x) + 5*x + 5*e^(x^2 + 3))

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maple [C]  time = 0.11, size = 148, normalized size = 7.05

method	result	size
risch	\(\frac {2 i x}{\pi \,\mathrm {csgn}\left (i \left (5+\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \relax (x )\right )\right )-\pi \,\mathrm {csgn}\left (i \left (5+\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \relax (x )\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \relax (x )\right )\right )^{2}+\pi \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \relax (x )\right )\right )^{3}+2 i \ln \left (5+\ln \relax (x )\right )+2 i \ln \left ({\mathrm e}^{x^{2}+3}+x \right )}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+5*x)*ln((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+5*x)+(-2*x^2*exp(x^2+3)-x)*l
n(x)+(-10*x^2-1)*exp(x^2+3)-6*x)/((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+5*x)/ln((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+
5*x)^2,x,method=_RETURNVERBOSE)

[Out]

2*I*x/(Pi*csgn(I*(5+ln(x)))*csgn(I*(exp(x^2+3)+x))*csgn(I*(exp(x^2+3)+x)*(5+ln(x)))-Pi*csgn(I*(5+ln(x)))*csgn(
I*(exp(x^2+3)+x)*(5+ln(x)))^2-Pi*csgn(I*(exp(x^2+3)+x))*csgn(I*(exp(x^2+3)+x)*(5+ln(x)))^2+Pi*csgn(I*(exp(x^2+
3)+x)*(5+ln(x)))^3+2*I*ln(5+ln(x))+2*I*ln(exp(x^2+3)+x))

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maxima [A]  time = 0.80, size = 19, normalized size = 0.90 \begin {gather*} \frac {x}{\log \left (x + e^{\left (x^{2} + 3\right )}\right ) + \log \left (\log \relax (x) + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)*log((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)+(-2*x^2*exp(x^
2+3)-x)*log(x)+(-10*x^2-1)*exp(x^2+3)-6*x)/((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)/log((exp(x^2+3)+x)*log(x)+
5*exp(x^2+3)+5*x)^2,x, algorithm="maxima")

[Out]

x/(log(x + e^(x^2 + 3)) + log(log(x) + 5))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {6\,x+{\mathrm {e}}^{x^2+3}\,\left (10\,x^2+1\right )-\ln \left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \relax (x)\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )\,\left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \relax (x)\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )+\ln \relax (x)\,\left (x+2\,x^2\,{\mathrm {e}}^{x^2+3}\right )}{{\ln \left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \relax (x)\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )}^2\,\left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \relax (x)\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x + exp(x^2 + 3)*(10*x^2 + 1) - log(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))*(5*x + 5*exp(x^2
+ 3) + log(x)*(x + exp(x^2 + 3))) + log(x)*(x + 2*x^2*exp(x^2 + 3)))/(log(5*x + 5*exp(x^2 + 3) + log(x)*(x +
exp(x^2 + 3)))^2*(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))),x)

[Out]

int(-(6*x + exp(x^2 + 3)*(10*x^2 + 1) - log(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))*(5*x + 5*exp(x^2
+ 3) + log(x)*(x + exp(x^2 + 3))) + log(x)*(x + 2*x^2*exp(x^2 + 3)))/(log(5*x + 5*exp(x^2 + 3) + log(x)*(x +
exp(x^2 + 3)))^2*(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))), x)

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sympy [A]  time = 1.01, size = 26, normalized size = 1.24 \begin {gather*} \frac {x}{\log {\left (5 x + \left (x + e^{x^{2} + 3}\right ) \log {\relax (x )} + 5 e^{x^{2} + 3} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x**2+3)+x)*ln(x)+5*exp(x**2+3)+5*x)*ln((exp(x**2+3)+x)*ln(x)+5*exp(x**2+3)+5*x)+(-2*x**2*exp(
x**2+3)-x)*ln(x)+(-10*x**2-1)*exp(x**2+3)-6*x)/((exp(x**2+3)+x)*ln(x)+5*exp(x**2+3)+5*x)/ln((exp(x**2+3)+x)*ln
(x)+5*exp(x**2+3)+5*x)**2,x)

[Out]

x/log(5*x + (x + exp(x**2 + 3))*log(x) + 5*exp(x**2 + 3))

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