3.98.47 \(\int \frac {90 x+45 x^2-60 x^4+e^x (-15+20 x^2)+(20 x+5 e^x x^2-15 x^4) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {5}{4} \left (e^x-x^3+\log \left (\left (3-x^2 (4+\log (x))\right )^2\right )\right ) \]

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Rubi [F]  time = 0.74, 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 {90 x+45 x^2-60 x^4+e^x \left (-15+20 x^2\right )+\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{-12+16 x^2+4 x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(90*x + 45*x^2 - 60*x^4 + E^x*(-15 + 20*x^2) + (20*x + 5*E^x*x^2 - 15*x^4)*Log[x])/(-12 + 16*x^2 + 4*x^2*L
og[x]),x]

[Out]

(5*E^x)/4 - (5*x^3)/4 + 5*Log[x] + 15*Defer[Int][1/(x*(-3 + 4*x^2 + x^2*Log[x])), x] + (5*Defer[Int][x/(-3 + 4
*x^2 + x^2*Log[x]), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-90 x-45 x^2+60 x^4-e^x \left (-15+20 x^2\right )-\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{4 \left (3-4 x^2-x^2 \log (x)\right )} \, dx\\ &=\frac {1}{4} \int \frac {-90 x-45 x^2+60 x^4-e^x \left (-15+20 x^2\right )-\left (20 x+5 e^x x^2-15 x^4\right ) \log (x)}{3-4 x^2-x^2 \log (x)} \, dx\\ &=\frac {1}{4} \int \left (5 e^x-\frac {5 x \left (-18-9 x+12 x^3-4 \log (x)+3 x^3 \log (x)\right )}{-3+4 x^2+x^2 \log (x)}\right ) \, dx\\ &=\frac {5 \int e^x \, dx}{4}-\frac {5}{4} \int \frac {x \left (-18-9 x+12 x^3-4 \log (x)+3 x^3 \log (x)\right )}{-3+4 x^2+x^2 \log (x)} \, dx\\ &=\frac {5 e^x}{4}-\frac {5}{4} \int \left (\frac {-4+3 x^3}{x}-\frac {2 \left (6+x^2\right )}{x \left (-3+4 x^2+x^2 \log (x)\right )}\right ) \, dx\\ &=\frac {5 e^x}{4}-\frac {5}{4} \int \frac {-4+3 x^3}{x} \, dx+\frac {5}{2} \int \frac {6+x^2}{x \left (-3+4 x^2+x^2 \log (x)\right )} \, dx\\ &=\frac {5 e^x}{4}-\frac {5}{4} \int \left (-\frac {4}{x}+3 x^2\right ) \, dx+\frac {5}{2} \int \left (\frac {6}{x \left (-3+4 x^2+x^2 \log (x)\right )}+\frac {x}{-3+4 x^2+x^2 \log (x)}\right ) \, dx\\ &=\frac {5 e^x}{4}-\frac {5 x^3}{4}+5 \log (x)+\frac {5}{2} \int \frac {x}{-3+4 x^2+x^2 \log (x)} \, dx+15 \int \frac {1}{x \left (-3+4 x^2+x^2 \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 30, normalized size = 1.11 \begin {gather*} \frac {5}{4} \left (e^x-x^3+2 \log \left (3-4 x^2-x^2 \log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(90*x + 45*x^2 - 60*x^4 + E^x*(-15 + 20*x^2) + (20*x + 5*E^x*x^2 - 15*x^4)*Log[x])/(-12 + 16*x^2 + 4
*x^2*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.57, size = 34, normalized size = 1.26 \begin {gather*} -\frac {5}{4} \, x^{3} + \frac {5}{4} \, e^{x} + 5 \, \log \relax (x) + \frac {5}{2} \, \log \left (\frac {x^{2} \log \relax (x) + 4 \, x^{2} - 3}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)*x^2-15*x^4+20*x)*log(x)+(20*x^2-15)*exp(x)-60*x^4+45*x^2+90*x)/(4*x^2*log(x)+16*x^2-12),x
, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 27, normalized size = 1.00 \begin {gather*} -\frac {5}{4} \, x^{3} + \frac {5}{4} \, e^{x} + \frac {5}{2} \, \log \left (-x^{2} \log \relax (x) - 4 \, x^{2} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)*x^2-15*x^4+20*x)*log(x)+(20*x^2-15)*exp(x)-60*x^4+45*x^2+90*x)/(4*x^2*log(x)+16*x^2-12),x
, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 27, normalized size = 1.00

method	result	size
default	\(-\frac {5 x^{3}}{4}+\frac {5 \ln \left (x^{2} \ln \relax (x )+4 x^{2}-3\right )}{2}+\frac {5 \,{\mathrm e}^{x}}{4}\)	\(27\)
norman	\(-\frac {5 x^{3}}{4}+\frac {5 \,{\mathrm e}^{x}}{4}+\frac {5 \ln \left (4 x^{2} \ln \relax (x )+16 x^{2}-12\right )}{2}\)	\(28\)
risch	\(-\frac {5 x^{3}}{4}+5 \ln \relax (x )+\frac {5 \,{\mathrm e}^{x}}{4}+\frac {5 \ln \left (\ln \relax (x )+\frac {4 x^{2}-3}{x^{2}}\right )}{2}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(x)*x^2-15*x^4+20*x)*ln(x)+(20*x^2-15)*exp(x)-60*x^4+45*x^2+90*x)/(4*x^2*ln(x)+16*x^2-12),x,method=
_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 34, normalized size = 1.26 \begin {gather*} -\frac {5}{4} \, x^{3} + \frac {5}{4} \, e^{x} + 5 \, \log \relax (x) + \frac {5}{2} \, \log \left (\frac {x^{2} \log \relax (x) + 4 \, x^{2} - 3}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)*x^2-15*x^4+20*x)*log(x)+(20*x^2-15)*exp(x)-60*x^4+45*x^2+90*x)/(4*x^2*log(x)+16*x^2-12),x
, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.08, size = 34, normalized size = 1.26 \begin {gather*} \frac {5\,\ln \left (\frac {x^2\,\ln \relax (x)+4\,x^2-3}{x^2}\right )}{2}+\frac {5\,{\mathrm {e}}^x}{4}+5\,\ln \relax (x)-\frac {5\,x^3}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*x + log(x)*(20*x + 5*x^2*exp(x) - 15*x^4) + exp(x)*(20*x^2 - 15) + 45*x^2 - 60*x^4)/(4*x^2*log(x) + 16
*x^2 - 12),x)

[Out]

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

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sympy [A]  time = 0.45, size = 36, normalized size = 1.33 \begin {gather*} - \frac {5 x^{3}}{4} + \frac {5 e^{x}}{4} + 5 \log {\relax (x )} + \frac {5 \log {\left (\log {\relax (x )} + \frac {4 x^{2} - 3}{x^{2}} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)*x**2-15*x**4+20*x)*ln(x)+(20*x**2-15)*exp(x)-60*x**4+45*x**2+90*x)/(4*x**2*ln(x)+16*x**2-
12),x)

[Out]

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

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