3.94.29 \(\int \frac {(15 x+4 x^2+e^x (-6 x-x^2)) \log (x^2)+(20+4 x) \log (2 \log (x^2))+x \log (x^2) \log ^2(2 \log (x^2))}{x \log (x^2)} \, dx\) [9329]

Optimal. Leaf size=25 \[ 1+(5+x) \left (5-e^x+2 x+\log ^2\left (2 \log \left (x^2\right )\right )\right ) \]

[Out]

1+(5+x)*(ln(2*ln(x^2))^2+5-exp(x)+2*x)

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Rubi [F]
time = 0.59, 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 {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((15*x + 4*x^2 + E^x*(-6*x - x^2))*Log[x^2] + (20 + 4*x)*Log[2*Log[x^2]] + x*Log[x^2]*Log[2*Log[x^2]]^2)/(
x*Log[x^2]),x]

[Out]

E^x + 15*x + 2*x^2 - E^x*(6 + x) + 5*Log[2*Log[x^2]]^2 + 4*Defer[Int][Log[2*Log[x^2]]/Log[x^2], x] + Defer[Int
][Log[2*Log[x^2]]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (15+4 x-e^x (6+x)+\frac {4 (5+x) \log \left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )}+\log ^2\left (2 \log \left (x^2\right )\right )\right ) \, dx\\ &=15 x+2 x^2+4 \int \frac {(5+x) \log \left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx-\int e^x (6+x) \, dx+\int \log ^2\left (2 \log \left (x^2\right )\right ) \, dx\\ &=15 x+2 x^2-e^x (6+x)+4 \int \left (\frac {\log \left (2 \log \left (x^2\right )\right )}{\log \left (x^2\right )}+\frac {5 \log \left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )}\right ) \, dx+\int e^x \, dx+\int \log ^2\left (2 \log \left (x^2\right )\right ) \, dx\\ &=e^x+15 x+2 x^2-e^x (6+x)+4 \int \frac {\log \left (2 \log \left (x^2\right )\right )}{\log \left (x^2\right )} \, dx+20 \int \frac {\log \left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx+\int \log ^2\left (2 \log \left (x^2\right )\right ) \, dx\\ &=e^x+15 x+2 x^2-e^x (6+x)+4 \int \frac {\log \left (2 \log \left (x^2\right )\right )}{\log \left (x^2\right )} \, dx+10 \text {Subst}\left (\int \frac {\log (2 x)}{x} \, dx,x,\log \left (x^2\right )\right )+\int \log ^2\left (2 \log \left (x^2\right )\right ) \, dx\\ &=e^x+15 x+2 x^2-e^x (6+x)+5 \log ^2\left (2 \log \left (x^2\right )\right )+4 \int \frac {\log \left (2 \log \left (x^2\right )\right )}{\log \left (x^2\right )} \, dx+\int \log ^2\left (2 \log \left (x^2\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.12, size = 29, normalized size = 1.16 \begin {gather*} -e^x (5+x)+x (15+2 x)+(5+x) \log ^2\left (2 \log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((15*x + 4*x^2 + E^x*(-6*x - x^2))*Log[x^2] + (20 + 4*x)*Log[2*Log[x^2]] + x*Log[x^2]*Log[2*Log[x^2]
]^2)/(x*Log[x^2]),x]

[Out]

-(E^x*(5 + x)) + x*(15 + 2*x) + (5 + x)*Log[2*Log[x^2]]^2

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 22.04, size = 59, normalized size = 2.36

method	result	size
risch	\(\left (5+x \right ) \ln \left (4 \ln \left (x \right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}\right )^{2}+2 x^{2}-{\mathrm e}^{x} x +15 x -5 \,{\mathrm e}^{x}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(x^2)*ln(2*ln(x^2))^2+(20+4*x)*ln(2*ln(x^2))+((-x^2-6*x)*exp(x)+4*x^2+15*x)*ln(x^2))/x/ln(x^2),x,meth
od=_RETURNVERBOSE)

[Out]

(5+x)*ln(4*ln(x)-I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)^2+2*x^2-exp(x)*x+15*x-5*exp(x)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
time = 0.51, size = 50, normalized size = 2.00 \begin {gather*} 4 \, x \log \left (2\right )^{2} + {\left (x + 5\right )} \log \left (\log \left (x\right )\right )^{2} + 2 \, x^{2} - {\left (x - 1\right )} e^{x} + 4 \, {\left (x \log \left (2\right ) + 5 \, \log \left (2\right )\right )} \log \left (\log \left (x\right )\right ) + 15 \, x - 6 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x^2)*log(2*log(x^2))^2+(20+4*x)*log(2*log(x^2))+((-x^2-6*x)*exp(x)+4*x^2+15*x)*log(x^2))/x/lo
g(x^2),x, algorithm="maxima")

[Out]

4*x*log(2)^2 + (x + 5)*log(log(x))^2 + 2*x^2 - (x - 1)*e^x + 4*(x*log(2) + 5*log(2))*log(log(x)) + 15*x - 6*e^
x

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Fricas [A]
time = 0.36, size = 29, normalized size = 1.16 \begin {gather*} {\left (x + 5\right )} \log \left (2 \, \log \left (x^{2}\right )\right )^{2} + 2 \, x^{2} - {\left (x + 5\right )} e^{x} + 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x^2)*log(2*log(x^2))^2+(20+4*x)*log(2*log(x^2))+((-x^2-6*x)*exp(x)+4*x^2+15*x)*log(x^2))/x/lo
g(x^2),x, algorithm="fricas")

[Out]

(x + 5)*log(2*log(x^2))^2 + 2*x^2 - (x + 5)*e^x + 15*x

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Sympy [A]
time = 0.21, size = 29, normalized size = 1.16 \begin {gather*} 2 x^{2} + 15 x + \left (- x - 5\right ) e^{x} + \left (x + 5\right ) \log {\left (2 \log {\left (x^{2} \right )} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(x**2)*ln(2*ln(x**2))**2+(20+4*x)*ln(2*ln(x**2))+((-x**2-6*x)*exp(x)+4*x**2+15*x)*ln(x**2))/x/l
n(x**2),x)

[Out]

2*x**2 + 15*x + (-x - 5)*exp(x) + (x + 5)*log(2*log(x**2))**2

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Giac [A]
time = 0.43, size = 40, normalized size = 1.60 \begin {gather*} x \log \left (2 \, \log \left (x^{2}\right )\right )^{2} + 2 \, x^{2} - x e^{x} + 5 \, \log \left (2 \, \log \left (x^{2}\right )\right )^{2} + 15 \, x - 5 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x^2)*log(2*log(x^2))^2+(20+4*x)*log(2*log(x^2))+((-x^2-6*x)*exp(x)+4*x^2+15*x)*log(x^2))/x/lo
g(x^2),x, algorithm="giac")

[Out]

x*log(2*log(x^2))^2 + 2*x^2 - x*e^x + 5*log(2*log(x^2))^2 + 15*x - 5*e^x

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Mupad [B]
time = 7.54, size = 29, normalized size = 1.16 \begin {gather*} 15\,x-{\mathrm {e}}^x\,\left (x+5\right )+{\ln \left (2\,\ln \left (x^2\right )\right )}^2\,\left (x+5\right )+2\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)*(15*x - exp(x)*(6*x + x^2) + 4*x^2) + log(2*log(x^2))*(4*x + 20) + x*log(2*log(x^2))^2*log(x^2))
/(x*log(x^2)),x)

[Out]

15*x - exp(x)*(x + 5) + log(2*log(x^2))^2*(x + 5) + 2*x^2

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