3.41.25 $\int \frac{{\log }^2(e^{-10+2x+2{\log }^2(x)})+\log (e^{-10+2x+2{\log }^2(x)})(4x+8\log (x))}{{\log }^4(2)}dx$

Optimal. Leaf size=22 $$\frac{x{\log }^2\left(e^{-10+2x+2{\log }^2(x)}\right)}{{\log }^4(2)}$$

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Rubi [F]  time = 0.22, 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 = {} $$\begin{array}{c}\int \frac{{\log }^2\left(e^{-10+2x+2{\log }^2(x)}\right)+\log \left(e^{-10+2x+2{\log }^2(x)}\right)(4x+8\log (x))}{{\log }^4(2)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(Log[E^(-10 + 2*x + 2*Log[x]^2)]^2 + Log[E^(-10 + 2*x + 2*Log[x]^2)]*(4*x + 8*Log[x]))/Log[2]^4,x]

[Out]

(4*Defer[Int][x*Log[E^(2*(-5 + x + Log[x]^2))], x])/Log[2]^4 + Defer[Int][Log[E^(2*(-5 + x + Log[x]^2))]^2, x]
/Log[2]^4 + (8*Defer[Int][Log[E^(2*(-5 + x + Log[x]^2))]*Log[x], x])/Log[2]^4

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int ({\log }^2\left(e^{-10+2x+2{\log }^2(x)}\right)+\log \left(e^{-10+2x+2{\log }^2(x)}\right)(4x+8\log (x)))dx}{{\log }^4(2)}\\ & =\frac{\int {\log }^2\left(e^{-10+2x+2{\log }^2(x)}\right)dx}{{\log }^4(2)}+\frac{\int \log \left(e^{-10+2x+2{\log }^2(x)}\right)(4x+8\log (x))dx}{{\log }^4(2)}\\ & =\frac{\int {\log }^2\left(e^{2(-5+x+{\log }^2(x))}\right)dx}{{\log }^4(2)}+\frac{\int 4\log \left(e^{2(-5+x+{\log }^2(x))}\right)(x+2\log (x))dx}{{\log }^4(2)}\\ & =\frac{\int {\log }^2\left(e^{2(-5+x+{\log }^2(x))}\right)dx}{{\log }^4(2)}+\frac{4\int \log \left(e^{2(-5+x+{\log }^2(x))}\right)(x+2\log (x))dx}{{\log }^4(2)}\\ & =\frac{\int {\log }^2\left(e^{2(-5+x+{\log }^2(x))}\right)dx}{{\log }^4(2)}+\frac{4\int (x\log \left(e^{2(-5+x+{\log }^2(x))}\right)+2\log \left(e^{2(-5+x+{\log }^2(x))}\right)\log (x))dx}{{\log }^4(2)}\\ & =\frac{\int {\log }^2\left(e^{2(-5+x+{\log }^2(x))}\right)dx}{{\log }^4(2)}+\frac{4\int x\log \left(e^{2(-5+x+{\log }^2(x))}\right)dx}{{\log }^4(2)}+\frac{8\int \log \left(e^{2(-5+x+{\log }^2(x))}\right)\log (x)dx}{{\log }^4(2)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.08, size = 20, normalized size = 0.91 $$\begin{array}{c}\frac{x{\log }^2\left(e^{2(-5+x+{\log }^2(x))}\right)}{{\log }^4(2)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(Log[E^(-10 + 2*x + 2*Log[x]^2)]^2 + Log[E^(-10 + 2*x + 2*Log[x]^2)]*(4*x + 8*Log[x]))/Log[2]^4,x]

[Out]

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

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fricas [A]  time = 0.93, size = 37, normalized size = 1.68 $$\begin{array}{c}\frac{4(x\log (x{)}^4+x^3+2(x^2-5x)\log (x{)}^2-10x^2+25x)}{\log (2{)}^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(x)^2*exp(log(x)^2-5)^2)^2+(8*log(x)+4*x)*log(exp(x)^2*exp(log(x)^2-5)^2))/log(2)^4,x, algor
ithm="fricas")

[Out]

4*(x*log(x)^4 + x^3 + 2*(x^2 - 5*x)*log(x)^2 - 10*x^2 + 25*x)/log(2)^4

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giac [A]  time = 0.23, size = 40, normalized size = 1.82 $$\begin{array}{c}\frac{4(x\log (x{)}^4+2x^2\log (x{)}^2+x^3-10x\log (x{)}^2-10x^2+25x)}{\log (2{)}^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(x)^2*exp(log(x)^2-5)^2)^2+(8*log(x)+4*x)*log(exp(x)^2*exp(log(x)^2-5)^2))/log(2)^4,x, algor
ithm="giac")

[Out]

4*(x*log(x)^4 + 2*x^2*log(x)^2 + x^3 - 10*x*log(x)^2 - 10*x^2 + 25*x)/log(2)^4

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maple [B]  time = 0.44, size = 346, normalized size = 15.73

method	result	size
default	$\frac{100x+8x^2\ln (x{)}^2-\frac{4x^3}{3}-40x^2+4x\ln (x{)}^4-40x\ln (x{)}^2+4x^2\ln \left({\mathrm{e}}^x\right)+{(\ln \left({\mathrm{e}}^{2x}{\mathrm{e}}^{2\ln (x{)}^2-10}\right)-2\ln \left({\mathrm{e}}^x\right)-2\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right))}^{2}x-40(\ln \left({\mathrm{e}}^x\right)-x)x+8(\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right)-\ln (x{)}^2+5)x^2+4(\ln \left({\mathrm{e}}^{2x}{\mathrm{e}}^{2\ln (x{)}^2-10}\right)-2\ln \left({\mathrm{e}}^x\right)-2\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right))x^2-40(\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right)-\ln (x{)}^2+5)x+4{(\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right)-\ln (x{)}^2+5)}^{2}x+\frac{4\ln {\left({\mathrm{e}}^x\right)}^3}{3}+8\ln (x{)}^{2}x(\ln \left({\mathrm{e}}^x\right)-x)+8(\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right)-\ln (x{)}^2+5)(\ln \left({\mathrm{e}}^x\right)-x)x+4(\ln \left({\mathrm{e}}^{2x}{\mathrm{e}}^{2\ln (x{)}^2-10}\right)-2\ln \left({\mathrm{e}}^x\right)-2\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right))(\ln \left({\mathrm{e}}^x\right)-x)x+8\ln (x{)}^{2}x(\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right)-\ln (x{)}^2+5)+4\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right)x(\ln \left({\mathrm{e}}^{2x}{\mathrm{e}}^{2\ln (x{)}^2-10}\right)-2\ln \left({\mathrm{e}}^x\right)-2\ln \left({\mathrm{e}}^{\ln (x{)}^2-5}\right))}{\ln (2{)}^4}$	$346$
risch	$\text{Expression too large to display}$	$2993$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(exp(x)^2*exp(ln(x)^2-5)^2)^2+(8*ln(x)+4*x)*ln(exp(x)^2*exp(ln(x)^2-5)^2))/ln(2)^4,x,method=_RETURNVERB
OSE)

[Out]

1/ln(2)^4*(100*x+8*x^2*ln(x)^2-4/3*x^3-40*x^2+4*x*ln(x)^4-40*x*ln(x)^2+4*x^2*ln(exp(x))+(ln(exp(x)^2*exp(ln(x)
^2-5)^2)-2*ln(exp(x))-2*ln(exp(ln(x)^2-5)))^2*x-40*(ln(exp(x))-x)*x+8*(ln(exp(ln(x)^2-5))-ln(x)^2+5)*x^2+4*(ln
(exp(x)^2*exp(ln(x)^2-5)^2)-2*ln(exp(x))-2*ln(exp(ln(x)^2-5)))*x^2-40*(ln(exp(ln(x)^2-5))-ln(x)^2+5)*x+4*(ln(e
xp(ln(x)^2-5))-ln(x)^2+5)^2*x+4/3*ln(exp(x))^3+8*ln(x)^2*x*(ln(exp(x))-x)+8*(ln(exp(ln(x)^2-5))-ln(x)^2+5)*(ln
(exp(x))-x)*x+4*(ln(exp(x)^2*exp(ln(x)^2-5)^2)-2*ln(exp(x))-2*ln(exp(ln(x)^2-5)))*(ln(exp(x))-x)*x+8*ln(x)^2*x
*(ln(exp(ln(x)^2-5))-ln(x)^2+5)+4*ln(exp(ln(x)^2-5))*x*(ln(exp(x)^2*exp(ln(x)^2-5)^2)-2*ln(exp(x))-2*ln(exp(ln
(x)^2-5))))

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maxima [A]  time = 0.35, size = 21, normalized size = 0.95 $$\begin{array}{c}\frac{x\log {\left(e^{(2\log (x{)}^2+2x-10)}\right)}^2}{\log (2{)}^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(exp(x)^2*exp(log(x)^2-5)^2)^2+(8*log(x)+4*x)*log(exp(x)^2*exp(log(x)^2-5)^2))/log(2)^4,x, algor
ithm="maxima")

[Out]

x*log(e^(2*log(x)^2 + 2*x - 10))^2/log(2)^4

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mupad [B]  time = 3.63, size = 74, normalized size = 3.36 $$\begin{array}{c}\frac{4x^3}{{\ln (2)}^4}+\frac{4x^2\ln \left({\mathrm{e}}^{2{\ln (x)}^2}\right)}{{\ln (2)}^4}-\frac{40x^2}{{\ln (2)}^4}+\frac{x{\ln \left({\mathrm{e}}^{2{\ln (x)}^2}\right)}^2}{{\ln (2)}^4}-\frac{20x\ln \left({\mathrm{e}}^{2{\ln (x)}^2}\right)}{{\ln (2)}^4}+\frac{100x}{{\ln (2)}^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(2*x)*exp(2*log(x)^2 - 10))*(4*x + 8*log(x)) + log(exp(2*x)*exp(2*log(x)^2 - 10))^2)/log(2)^4,x)

[Out]

(4*x^3)/log(2)^4 - (40*x^2)/log(2)^4 + (100*x)/log(2)^4 - (20*x*log(exp(2*log(x)^2)))/log(2)^4 + (x*log(exp(2*
log(x)^2))^2)/log(2)^4 + (4*x^2*log(exp(2*log(x)^2)))/log(2)^4

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sympy [B]  time = 0.42, size = 60, normalized size = 2.73 $$\begin{array}{c}\frac{4x^3}{\log {(2)}^4}-\frac{40x^2}{\log {(2)}^4}+\frac{4x\log {(x)}^4}{\log {(2)}^4}+\frac{100x}{\log {(2)}^4}+\frac{(8x^2-40x)\log {(x)}^2}{\log {(2)}^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(exp(x)**2*exp(ln(x)**2-5)**2)**2+(8*ln(x)+4*x)*ln(exp(x)**2*exp(ln(x)**2-5)**2))/ln(2)**4,x)

[Out]

4*x**3/log(2)**4 - 40*x**2/log(2)**4 + 4*x*log(x)**4/log(2)**4 + 100*x/log(2)**4 + (8*x**2 - 40*x)*log(x)**2/l
og(2)**4

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