3.69.57 \(\int \frac {e^{\log ^2(\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4})} (-30 e x^3-e x \log (x)+(e (16-4 x)+e (-32+8 x) \log (x)) \log (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}))}{30 x^3+x \log (x)} \, dx\)

Optimal. Leaf size=30 \[ e^{1+\log ^2\left (\frac {1}{4} \left (6+\frac {\log (x)}{5 x^2}\right )^2\right )} (4-x) \]

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Rubi [B]  time = 0.12, antiderivative size = 129, normalized size of antiderivative = 4.30, number of steps used = 1, number of rules used = 1, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {2288} \begin {gather*} \frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} (e (4-x)-2 e (4-x) \log (x)) \left (900 x^4+60 x^2 \log (x)+\log ^2(x)\right )}{x^4 \left (30 x^3+x \log (x)\right ) \left (\frac {1800 x^3+30 x+60 x \log (x)+\frac {\log (x)}{x}}{x^4}-\frac {2 \left (900 x^4+60 x^2 \log (x)+\log ^2(x)\right )}{x^5}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]^2*(-30*E*x^3 - E*x*Log[x] + (E*(16 - 4*x) + E*(-32
+ 8*x)*Log[x])*Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]))/(30*x^3 + x*Log[x]),x]

[Out]

(E^Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]^2*(E*(4 - x) - 2*E*(4 - x)*Log[x])*(900*x^4 + 60*x^2*Lo
g[x] + Log[x]^2))/(x^4*(30*x^3 + x*Log[x])*((30*x + 1800*x^3 + Log[x]/x + 60*x*Log[x])/x^4 - (2*(900*x^4 + 60*
x^2*Log[x] + Log[x]^2))/x^5))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {e^{\log ^2\left (\frac {900 x^4+60 x^2 \log (x)+\log ^2(x)}{100 x^4}\right )} (e (4-x)-2 e (4-x) \log (x)) \left (900 x^4+60 x^2 \log (x)+\log ^2(x)\right )}{x^4 \left (30 x^3+x \log (x)\right ) \left (\frac {30 x+1800 x^3+\frac {\log (x)}{x}+60 x \log (x)}{x^4}-\frac {2 \left (900 x^4+60 x^2 \log (x)+\log ^2(x)\right )}{x^5}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 29, normalized size = 0.97 \begin {gather*} -e^{1+\log ^2\left (\frac {\left (30 x^2+\log (x)\right )^2}{100 x^4}\right )} (-4+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]^2*(-30*E*x^3 - E*x*Log[x] + (E*(16 - 4*x) + E
*(-32 + 8*x)*Log[x])*Log[(900*x^4 + 60*x^2*Log[x] + Log[x]^2)/(100*x^4)]))/(30*x^3 + x*Log[x]),x]

[Out]

-(E^(1 + Log[(30*x^2 + Log[x])^2/(100*x^4)]^2)*(-4 + x))

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fricas [A]  time = 0.53, size = 33, normalized size = 1.10 \begin {gather*} -{\left (x - 4\right )} e^{\left (\log \left (\frac {900 \, x^{4} + 60 \, x^{2} \log \relax (x) + \log \relax (x)^{2}}{100 \, x^{4}}\right )^{2} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-32)*exp(1)*log(x)+(-4*x+16)*exp(1))*log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)-x*exp(1)*
log(x)-30*x^3*exp(1))*exp(log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)^2)/(x*log(x)+30*x^3),x, algorithm="f
ricas")

[Out]

-(x - 4)*e^(log(1/100*(900*x^4 + 60*x^2*log(x) + log(x)^2)/x^4)^2 + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (30 \, x^{3} e + x e \log \relax (x) - 4 \, {\left (2 \, {\left (x - 4\right )} e \log \relax (x) - {\left (x - 4\right )} e\right )} \log \left (\frac {900 \, x^{4} + 60 \, x^{2} \log \relax (x) + \log \relax (x)^{2}}{100 \, x^{4}}\right )\right )} e^{\left (\log \left (\frac {900 \, x^{4} + 60 \, x^{2} \log \relax (x) + \log \relax (x)^{2}}{100 \, x^{4}}\right )^{2}\right )}}{30 \, x^{3} + x \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-32)*exp(1)*log(x)+(-4*x+16)*exp(1))*log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)-x*exp(1)*
log(x)-30*x^3*exp(1))*exp(log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)^2)/(x*log(x)+30*x^3),x, algorithm="g
iac")

[Out]

integrate(-(30*x^3*e + x*e*log(x) - 4*(2*(x - 4)*e*log(x) - (x - 4)*e)*log(1/100*(900*x^4 + 60*x^2*log(x) + lo
g(x)^2)/x^4))*e^(log(1/100*(900*x^4 + 60*x^2*log(x) + log(x)^2)/x^4)^2)/(30*x^3 + x*log(x)), x)

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maple [C]  time = 0.86, size = 7474, normalized size = 249.13

method	result	size
risch	\(\text {Expression too large to display}\)	\(7474\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x-32)*exp(1)*ln(x)+(-4*x+16)*exp(1))*ln(1/100*(ln(x)^2+60*x^2*ln(x)+900*x^4)/x^4)-x*exp(1)*ln(x)-30*x
^3*exp(1))*exp(ln(1/100*(ln(x)^2+60*x^2*ln(x)+900*x^4)/x^4)^2)/(x*ln(x)+30*x^3),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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maxima [B]  time = 0.67, size = 123, normalized size = 4.10 \begin {gather*} -{\left (2^{8 \, \log \relax (5)} x e^{\left (4 \, \log \relax (5)^{2} + 4 \, \log \relax (2)^{2} + 1\right )} - 2^{8 \, \log \relax (5) + 2} e^{\left (4 \, \log \relax (5)^{2} + 4 \, \log \relax (2)^{2} + 1\right )}\right )} e^{\left (-8 \, \log \relax (5) \log \left (30 \, x^{2} + \log \relax (x)\right ) - 8 \, \log \relax (2) \log \left (30 \, x^{2} + \log \relax (x)\right ) + 4 \, \log \left (30 \, x^{2} + \log \relax (x)\right )^{2} + 16 \, \log \relax (5) \log \relax (x) + 16 \, \log \relax (2) \log \relax (x) - 16 \, \log \left (30 \, x^{2} + \log \relax (x)\right ) \log \relax (x) + 16 \, \log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-32)*exp(1)*log(x)+(-4*x+16)*exp(1))*log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)-x*exp(1)*
log(x)-30*x^3*exp(1))*exp(log(1/100*(log(x)^2+60*x^2*log(x)+900*x^4)/x^4)^2)/(x*log(x)+30*x^3),x, algorithm="m
axima")

[Out]

-(2^(8*log(5))*x*e^(4*log(5)^2 + 4*log(2)^2 + 1) - 2^(8*log(5) + 2)*e^(4*log(5)^2 + 4*log(2)^2 + 1))*e^(-8*log
(5)*log(30*x^2 + log(x)) - 8*log(2)*log(30*x^2 + log(x)) + 4*log(30*x^2 + log(x))^2 + 16*log(5)*log(x) + 16*lo
g(2)*log(x) - 16*log(30*x^2 + log(x))*log(x) + 16*log(x)^2)

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mupad [B]  time = 4.55, size = 103, normalized size = 3.43 \begin {gather*} \frac {{\mathrm {e}}^{{\ln \left (\frac {1}{x^4}\right )}^2+{\ln \left (900\,x^4+60\,x^2\,\ln \relax (x)+{\ln \relax (x)}^2\right )}^2+4\,{\ln \left (10\right )}^2}\,\left (4\,\mathrm {e}-x\,\mathrm {e}\right )\,{\left (\frac {1}{x^4}\right )}^{2\,\ln \left (900\,x^4+60\,x^2\,\ln \relax (x)+{\ln \relax (x)}^2\right )}}{{\left (\frac {1}{x^4}\right )}^{4\,\ln \left (10\right )}\,{\left (900\,x^4+60\,x^2\,\ln \relax (x)+{\ln \relax (x)}^2\right )}^{4\,\ln \left (10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(((3*x^2*log(x))/5 + log(x)^2/100 + 9*x^4)/x^4)^2)*(log(((3*x^2*log(x))/5 + log(x)^2/100 + 9*x^4)
/x^4)*(exp(1)*(4*x - 16) - exp(1)*log(x)*(8*x - 32)) + 30*x^3*exp(1) + x*exp(1)*log(x)))/(x*log(x) + 30*x^3),x
)

[Out]

(exp(log(1/x^4)^2 + log(60*x^2*log(x) + log(x)^2 + 900*x^4)^2 + 4*log(10)^2)*(4*exp(1) - x*exp(1))*(1/x^4)^(2*
log(60*x^2*log(x) + log(x)^2 + 900*x^4)))/((1/x^4)^(4*log(10))*(60*x^2*log(x) + log(x)^2 + 900*x^4)^(4*log(10)
))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x-32)*exp(1)*ln(x)+(-4*x+16)*exp(1))*ln(1/100*(ln(x)**2+60*x**2*ln(x)+900*x**4)/x**4)-x*exp(1)*
ln(x)-30*x**3*exp(1))*exp(ln(1/100*(ln(x)**2+60*x**2*ln(x)+900*x**4)/x**4)**2)/(x*ln(x)+30*x**3),x)

[Out]

Timed out

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