3.67.44 \(\int \frac {-16 x^2+e^{576-48 x+x^2+(48-2 x) \log (x)+\log ^2(x)} (4 x^2+(-240+250 x-106 x^2+100 x^3-4 x^4+(-10+10 x-4 x^2+4 x^3) \log (x)) \log (5+2 x^2))}{(5 x+2 x^3) \log ^2(5+2 x^2)} \, dx\)

Optimal. Leaf size=26 \[ \frac {4-e^{(24-x+\log (x))^2}}{\log \left (5+2 x^2\right )} \]

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Rubi [F]  time = 38.27, 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 {-16 x^2+e^{576-48 x+x^2+(48-2 x) \log (x)+\log ^2(x)} \left (4 x^2+\left (-240+250 x-106 x^2+100 x^3-4 x^4+\left (-10+10 x-4 x^2+4 x^3\right ) \log (x)\right ) \log \left (5+2 x^2\right )\right )}{\left (5 x+2 x^3\right ) \log ^2\left (5+2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-16*x^2 + E^(576 - 48*x + x^2 + (48 - 2*x)*Log[x] + Log[x]^2)*(4*x^2 + (-240 + 250*x - 106*x^2 + 100*x^3
- 4*x^4 + (-10 + 10*x - 4*x^2 + 4*x^3)*Log[x])*Log[5 + 2*x^2]))/((5*x + 2*x^3)*Log[5 + 2*x^2]^2),x]

[Out]

4/Log[5 + 2*x^2] + ((2*I)*Defer[Int][(E^((-24 + x)^2 + Log[x]^2)*x^(49 - 2*x))/((I*Sqrt[5] - Sqrt[2]*x)*Log[5
+ 2*x^2]^2), x])/Sqrt[5] + ((2*I)*Defer[Int][(E^((-24 + x)^2 + Log[x]^2)*x^(49 - 2*x))/((I*Sqrt[5] + Sqrt[2]*x
)*Log[5 + 2*x^2]^2), x])/Sqrt[5] - 48*Defer[Int][(E^((-24 + x)^2 + Log[x]^2)*x^(47 - 2*x))/Log[5 + 2*x^2], x]
+ 50*Defer[Int][(E^((-24 + x)^2 + Log[x]^2)*x^(48 - 2*x))/Log[5 + 2*x^2], x] - 2*Defer[Int][(E^((-24 + x)^2 +
Log[x]^2)*x^(49 - 2*x))/Log[5 + 2*x^2], x] - 2*Defer[Int][(E^((-24 + x)^2 + Log[x]^2)*x^(47 - 2*x)*Log[x])/Log
[5 + 2*x^2], x] + 2*Defer[Int][(E^((-24 + x)^2 + Log[x]^2)*x^(48 - 2*x)*Log[x])/Log[5 + 2*x^2], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16 x^2+e^{576-48 x+x^2+(48-2 x) \log (x)+\log ^2(x)} \left (4 x^2+\left (-240+250 x-106 x^2+100 x^3-4 x^4+\left (-10+10 x-4 x^2+4 x^3\right ) \log (x)\right ) \log \left (5+2 x^2\right )\right )}{x \left (5+2 x^2\right ) \log ^2\left (5+2 x^2\right )} \, dx\\ &=\int \left (-\frac {16 x}{\left (5+2 x^2\right ) \log ^2\left (5+2 x^2\right )}-\frac {2 e^{(-24+x)^2+\log ^2(x)} x^{47-2 x} \left (-2 x^2+120 \log \left (5+2 x^2\right )-125 x \log \left (5+2 x^2\right )+53 x^2 \log \left (5+2 x^2\right )-50 x^3 \log \left (5+2 x^2\right )+2 x^4 \log \left (5+2 x^2\right )+5 \log (x) \log \left (5+2 x^2\right )-5 x \log (x) \log \left (5+2 x^2\right )+2 x^2 \log (x) \log \left (5+2 x^2\right )-2 x^3 \log (x) \log \left (5+2 x^2\right )\right )}{\left (5+2 x^2\right ) \log ^2\left (5+2 x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{47-2 x} \left (-2 x^2+120 \log \left (5+2 x^2\right )-125 x \log \left (5+2 x^2\right )+53 x^2 \log \left (5+2 x^2\right )-50 x^3 \log \left (5+2 x^2\right )+2 x^4 \log \left (5+2 x^2\right )+5 \log (x) \log \left (5+2 x^2\right )-5 x \log (x) \log \left (5+2 x^2\right )+2 x^2 \log (x) \log \left (5+2 x^2\right )-2 x^3 \log (x) \log \left (5+2 x^2\right )\right )}{\left (5+2 x^2\right ) \log ^2\left (5+2 x^2\right )} \, dx\right )-16 \int \frac {x}{\left (5+2 x^2\right ) \log ^2\left (5+2 x^2\right )} \, dx\\ &=-\left (2 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{47-2 x} \left (-2 x^2+\left (-5+5 x-2 x^2+2 x^3\right ) (-24+x-\log (x)) \log \left (5+2 x^2\right )\right )}{\left (5+2 x^2\right ) \log ^2\left (5+2 x^2\right )} \, dx\right )-8 \operatorname {Subst}\left (\int \frac {1}{(5+2 x) \log ^2(5+2 x)} \, dx,x,x^2\right )\\ &=-\left (2 \int \left (-\frac {2 e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\left (5+2 x^2\right ) \log ^2\left (5+2 x^2\right )}+\frac {e^{(-24+x)^2+\log ^2(x)} (-1+x) x^{47-2 x} (-24+x-\log (x))}{\log \left (5+2 x^2\right )}\right ) \, dx\right )-4 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,5+2 x^2\right )\\ &=-\left (2 \int \frac {e^{(-24+x)^2+\log ^2(x)} (-1+x) x^{47-2 x} (-24+x-\log (x))}{\log \left (5+2 x^2\right )} \, dx\right )+4 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\left (5+2 x^2\right ) \log ^2\left (5+2 x^2\right )} \, dx-4 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (5+2 x^2\right )\right )\\ &=\frac {4}{\log \left (5+2 x^2\right )}-2 \int \left (\frac {e^{(-24+x)^2+\log ^2(x)} x^{48-2 x} (-24+x-\log (x))}{\log \left (5+2 x^2\right )}+\frac {e^{(-24+x)^2+\log ^2(x)} x^{47-2 x} (24-x+\log (x))}{\log \left (5+2 x^2\right )}\right ) \, dx+4 \int \left (\frac {i e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{2 \sqrt {5} \left (i \sqrt {5}-\sqrt {2} x\right ) \log ^2\left (5+2 x^2\right )}+\frac {i e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{2 \sqrt {5} \left (i \sqrt {5}+\sqrt {2} x\right ) \log ^2\left (5+2 x^2\right )}\right ) \, dx\\ &=\frac {4}{\log \left (5+2 x^2\right )}-2 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{48-2 x} (-24+x-\log (x))}{\log \left (5+2 x^2\right )} \, dx-2 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{47-2 x} (24-x+\log (x))}{\log \left (5+2 x^2\right )} \, dx+\frac {(2 i) \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\left (i \sqrt {5}-\sqrt {2} x\right ) \log ^2\left (5+2 x^2\right )} \, dx}{\sqrt {5}}+\frac {(2 i) \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\left (i \sqrt {5}+\sqrt {2} x\right ) \log ^2\left (5+2 x^2\right )} \, dx}{\sqrt {5}}\\ &=\frac {4}{\log \left (5+2 x^2\right )}-2 \int \left (\frac {24 e^{(-24+x)^2+\log ^2(x)} x^{47-2 x}}{\log \left (5+2 x^2\right )}-\frac {e^{(-24+x)^2+\log ^2(x)} x^{48-2 x}}{\log \left (5+2 x^2\right )}+\frac {e^{(-24+x)^2+\log ^2(x)} x^{47-2 x} \log (x)}{\log \left (5+2 x^2\right )}\right ) \, dx-2 \int \left (-\frac {24 e^{(-24+x)^2+\log ^2(x)} x^{48-2 x}}{\log \left (5+2 x^2\right )}+\frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\log \left (5+2 x^2\right )}-\frac {e^{(-24+x)^2+\log ^2(x)} x^{48-2 x} \log (x)}{\log \left (5+2 x^2\right )}\right ) \, dx+\frac {(2 i) \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\left (i \sqrt {5}-\sqrt {2} x\right ) \log ^2\left (5+2 x^2\right )} \, dx}{\sqrt {5}}+\frac {(2 i) \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\left (i \sqrt {5}+\sqrt {2} x\right ) \log ^2\left (5+2 x^2\right )} \, dx}{\sqrt {5}}\\ &=\frac {4}{\log \left (5+2 x^2\right )}+2 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{48-2 x}}{\log \left (5+2 x^2\right )} \, dx-2 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\log \left (5+2 x^2\right )} \, dx-2 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{47-2 x} \log (x)}{\log \left (5+2 x^2\right )} \, dx+2 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{48-2 x} \log (x)}{\log \left (5+2 x^2\right )} \, dx-48 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{47-2 x}}{\log \left (5+2 x^2\right )} \, dx+48 \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{48-2 x}}{\log \left (5+2 x^2\right )} \, dx+\frac {(2 i) \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\left (i \sqrt {5}-\sqrt {2} x\right ) \log ^2\left (5+2 x^2\right )} \, dx}{\sqrt {5}}+\frac {(2 i) \int \frac {e^{(-24+x)^2+\log ^2(x)} x^{49-2 x}}{\left (i \sqrt {5}+\sqrt {2} x\right ) \log ^2\left (5+2 x^2\right )} \, dx}{\sqrt {5}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 34, normalized size = 1.31 \begin {gather*} \frac {4-e^{(-24+x)^2+\log ^2(x)} x^{48-2 x}}{\log \left (5+2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^2 + E^(576 - 48*x + x^2 + (48 - 2*x)*Log[x] + Log[x]^2)*(4*x^2 + (-240 + 250*x - 106*x^2 + 10
0*x^3 - 4*x^4 + (-10 + 10*x - 4*x^2 + 4*x^3)*Log[x])*Log[5 + 2*x^2]))/((5*x + 2*x^3)*Log[5 + 2*x^2]^2),x]

[Out]

(4 - E^((-24 + x)^2 + Log[x]^2)*x^(48 - 2*x))/Log[5 + 2*x^2]

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fricas [A]  time = 0.53, size = 34, normalized size = 1.31 \begin {gather*} -\frac {e^{\left (x^{2} - 2 \, {\left (x - 24\right )} \log \relax (x) + \log \relax (x)^{2} - 48 \, x + 576\right )} - 4}{\log \left (2 \, x^{2} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x^3-4*x^2+10*x-10)*log(x)-4*x^4+100*x^3-106*x^2+250*x-240)*log(2*x^2+5)+4*x^2)*exp(log(x)^2+(-
2*x+48)*log(x)+x^2-48*x+576)-16*x^2)/(2*x^3+5*x)/log(2*x^2+5)^2,x, algorithm="fricas")

[Out]

-(e^(x^2 - 2*(x - 24)*log(x) + log(x)^2 - 48*x + 576) - 4)/log(2*x^2 + 5)

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giac [A]  time = 0.47, size = 36, normalized size = 1.38 \begin {gather*} -\frac {e^{\left (x^{2} - 2 \, x \log \relax (x) + \log \relax (x)^{2} - 48 \, x + 48 \, \log \relax (x) + 576\right )} - 4}{\log \left (2 \, x^{2} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x^3-4*x^2+10*x-10)*log(x)-4*x^4+100*x^3-106*x^2+250*x-240)*log(2*x^2+5)+4*x^2)*exp(log(x)^2+(-
2*x+48)*log(x)+x^2-48*x+576)-16*x^2)/(2*x^3+5*x)/log(2*x^2+5)^2,x, algorithm="giac")

[Out]

-(e^(x^2 - 2*x*log(x) + log(x)^2 - 48*x + 48*log(x) + 576) - 4)/log(2*x^2 + 5)

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maple [A]  time = 0.07, size = 46, normalized size = 1.77

method	result	size
risch	\(\frac {4}{\ln \left (2 x^{2}+5\right )}-\frac {x^{-2 x +48} {\mathrm e}^{\ln \relax (x )^{2}+576+x^{2}-48 x}}{\ln \left (2 x^{2}+5\right )}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((4*x^3-4*x^2+10*x-10)*ln(x)-4*x^4+100*x^3-106*x^2+250*x-240)*ln(2*x^2+5)+4*x^2)*exp(ln(x)^2+(-2*x+48)*l
n(x)+x^2-48*x+576)-16*x^2)/(2*x^3+5*x)/ln(2*x^2+5)^2,x,method=_RETURNVERBOSE)

[Out]

4/ln(2*x^2+5)-1/ln(2*x^2+5)*x^(-2*x+48)*exp(ln(x)^2+576+x^2-48*x)

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maxima [A]  time = 0.48, size = 46, normalized size = 1.77 \begin {gather*} -\frac {x^{48} e^{\left (x^{2} - 2 \, x \log \relax (x) + \log \relax (x)^{2} - 48 \, x + 576\right )}}{\log \left (2 \, x^{2} + 5\right )} + \frac {4}{\log \left (2 \, x^{2} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x^3-4*x^2+10*x-10)*log(x)-4*x^4+100*x^3-106*x^2+250*x-240)*log(2*x^2+5)+4*x^2)*exp(log(x)^2+(-
2*x+48)*log(x)+x^2-48*x+576)-16*x^2)/(2*x^3+5*x)/log(2*x^2+5)^2,x, algorithm="maxima")

[Out]

-x^48*e^(x^2 - 2*x*log(x) + log(x)^2 - 48*x + 576)/log(2*x^2 + 5) + 4/log(2*x^2 + 5)

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mupad [B]  time = 4.42, size = 50, normalized size = 1.92 \begin {gather*} \frac {4}{\ln \left (2\,x^2+5\right )}-\frac {x^{48}\,{\mathrm {e}}^{-48\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{576}\,{\mathrm {e}}^{{\ln \relax (x)}^2}}{x^{2\,x}\,\ln \left (2\,x^2+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(x)^2 - 48*x - log(x)*(2*x - 48) + x^2 + 576)*(log(2*x^2 + 5)*(250*x - 106*x^2 + 100*x^3 - 4*x^4 +
log(x)*(10*x - 4*x^2 + 4*x^3 - 10) - 240) + 4*x^2) - 16*x^2)/(log(2*x^2 + 5)^2*(5*x + 2*x^3)),x)

[Out]

4/log(2*x^2 + 5) - (x^48*exp(-48*x)*exp(x^2)*exp(576)*exp(log(x)^2))/(x^(2*x)*log(2*x^2 + 5))

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sympy [B]  time = 0.55, size = 41, normalized size = 1.58 \begin {gather*} - \frac {e^{x^{2} - 48 x + \left (48 - 2 x\right ) \log {\relax (x )} + \log {\relax (x )}^{2} + 576}}{\log {\left (2 x^{2} + 5 \right )}} + \frac {4}{\log {\left (2 x^{2} + 5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x**3-4*x**2+10*x-10)*ln(x)-4*x**4+100*x**3-106*x**2+250*x-240)*ln(2*x**2+5)+4*x**2)*exp(ln(x)*
*2+(-2*x+48)*ln(x)+x**2-48*x+576)-16*x**2)/(2*x**3+5*x)/ln(2*x**2+5)**2,x)

[Out]

-exp(x**2 - 48*x + (48 - 2*x)*log(x) + log(x)**2 + 576)/log(2*x**2 + 5) + 4/log(2*x**2 + 5)

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