3.17.12 $\int \frac{e^{{\log }^2(2)}-16x}{e^{{\log }^2(2)}x-8x^2}dx$

Optimal. Leaf size=18 $$\log \left(\frac{1}{5}x(-\frac{1}{8}{e}^{{\log }^2(2)}+x)\right)$$

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.83, number of steps used = 1, number of rules used = 1, integrand size = 27, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.037, Rules used = {628} $$\begin{array}{c}\log (x{e}^{{\log }^2(2)}-8x^2)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^Log[2]^2 - 16*x)/(E^Log[2]^2*x - 8*x^2),x]

[Out]

Log[E^Log[2]^2*x - 8*x^2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\log (e^{{\log }^2(2)}x-8x^2)\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.78 $$\begin{array}{c}\log (e^{{\log }^2(2)}-8x)+\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^Log[2]^2 - 16*x)/(E^Log[2]^2*x - 8*x^2),x]

[Out]

Log[E^Log[2]^2 - 8*x] + Log[x]

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fricas [A]  time = 0.66, size = 15, normalized size = 0.83 $$\begin{array}{c}\log (8x^2-x{e}^{(\log (2{)}^2)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(log(2)^2)-16*x)/(x*exp(log(2)^2)-8*x^2),x, algorithm="fricas")

[Out]

log(8*x^2 - x*e^(log(2)^2))

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giac [A]  time = 0.22, size = 16, normalized size = 0.89 $$\begin{array}{c}\log \left(|8x^2-x{e}^{(\log (2{)}^2)}|\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(log(2)^2)-16*x)/(x*exp(log(2)^2)-8*x^2),x, algorithm="giac")

[Out]

log(abs(8*x^2 - x*e^(log(2)^2)))

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maple [A]  time = 0.31, size = 13, normalized size = 0.72

method	result	size
default	$\ln \left(x({\mathrm{e}}^{\ln (2{)}^2}-8x)\right)$	$13$
norman	$\ln (x)+\ln ({\mathrm{e}}^{\ln (2{)}^2}-8x)$	$14$
derivativedivides	$\ln (x{\mathrm{e}}^{\ln (2{)}^2}-8x^2)$	$15$
risch	$\ln (-x{\mathrm{e}}^{\ln (2{)}^2}+8x^2)$	$16$
meijerg	$\ln (1-8x{\mathrm{e}}^{-\ln (2{)}^2})+\ln (x)+3\ln (2)-\ln (2{)}^2+i\pi$	$31$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(ln(2)^2)-16*x)/(x*exp(ln(2)^2)-8*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x*(exp(ln(2)^2)-8*x))

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maxima [A]  time = 0.36, size = 15, normalized size = 0.83 $$\begin{array}{c}\log (8x^2-x{e}^{(\log (2{)}^2)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(log(2)^2)-16*x)/(x*exp(log(2)^2)-8*x^2),x, algorithm="maxima")

[Out]

log(8*x^2 - x*e^(log(2)^2))

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mupad [B]  time = 1.10, size = 14, normalized size = 0.78 $$\begin{array}{c}\ln \left(x(8x-{\mathrm{e}}^{{\ln (2)}^2})\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x - exp(log(2)^2))/(x*exp(log(2)^2) - 8*x^2),x)

[Out]

log(x*(8*x - exp(log(2)^2)))

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sympy [A]  time = 0.15, size = 14, normalized size = 0.78 $$\begin{array}{c}\log (8x^2-x{e}^{\log {(2)}^2})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(ln(2)**2)-16*x)/(x*exp(ln(2)**2)-8*x**2),x)

[Out]

log(8*x**2 - x*exp(log(2)**2))

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