3.73.94 $\int \frac{-16-8\sqrt[4]{3}-16e^2+8x}{25+4\sqrt{3}+16e^4+e^2(32-16x)+\sqrt[4]{3}(16+16e^2-8x)-16x+4x^2}dx$

Optimal. Leaf size=22 $$\log (9+4{(2+\sqrt[4]{3}+2e^2-x)}^2)$$

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Rubi [B]  time = 0.05, antiderivative size = 51, normalized size of antiderivative = 2.32, number of steps used = 1, number of rules used = 1, integrand size = 67, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.015, Rules used = {1587} $$\begin{array}{c}\log (4x^2-16x+16e^2(2-x)+8\sqrt[4]{3}(2(1+e^2)-x)+16e^4+4\sqrt{3}+25)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-16 - 8*3^(1/4) - 16*E^2 + 8*x)/(25 + 4*Sqrt[3] + 16*E^4 + E^2*(32 - 16*x) + 3^(1/4)*(16 + 16*E^2 - 8*x)
- 16*x + 4*x^2),x]

[Out]

Log[25 + 4*Sqrt[3] + 16*E^4 + 16*E^2*(2 - x) + 8*3^(1/4)*(2*(1 + E^2) - x) - 16*x + 4*x^2]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\log (25+4\sqrt{3}+16e^4+16e^2(2-x)+8\sqrt[4]{3}(2(1+e^2)-x)-16x+4x^2)\end{array}\end{array}$$

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Mathematica [B]  time = 0.06, size = 59, normalized size = 2.68 $$\begin{array}{c}\log (25+16\sqrt[4]{3}+4\sqrt{3}+32e^2+16\sqrt[4]{3}{e}^2+16e^4-16x-8\sqrt[4]{3}x-16e^{2}x+4x^2)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 8*3^(1/4) - 16*E^2 + 8*x)/(25 + 4*Sqrt[3] + 16*E^4 + E^2*(32 - 16*x) + 3^(1/4)*(16 + 16*E^2 -
8*x) - 16*x + 4*x^2),x]

[Out]

Log[25 + 16*3^(1/4) + 4*Sqrt[3] + 32*E^2 + 16*3^(1/4)*E^2 + 16*E^4 - 16*x - 8*3^(1/4)*x - 16*E^2*x + 4*x^2]

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fricas [B]  time = 1.31, size = 39, normalized size = 1.77 $$\begin{array}{c}\log (4x^2-16(x-2)e^2-8\cdot 3^{\frac{1}{4}}(x-2e^2-2)-16x+4\sqrt{3}+16e^4+25)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*3^(1/4)-16*exp(2)+8*x-16)/(4*3^(1/2)+(16*exp(2)-8*x+16)*3^(1/4)+16*exp(2)^2+(-16*x+32)*exp(2)+4*
x^2-16*x+25),x, algorithm="fricas")

[Out]

log(4*x^2 - 16*(x - 2)*e^2 - 8*3^(1/4)*(x - 2*e^2 - 2) - 16*x + 4*sqrt(3) + 16*e^4 + 25)

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giac [B]  time = 0.18, size = 45, normalized size = 2.05 $$\begin{array}{c}\log (4x^2-16x{e}^2+16(3^{\frac{1}{4}}+2)e^2-8\cdot 3^{\frac{1}{4}}x-16x+4\sqrt{3}+16\cdot 3^{\frac{1}{4}}+16e^4+25)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*3^(1/4)-16*exp(2)+8*x-16)/(4*3^(1/2)+(16*exp(2)-8*x+16)*3^(1/4)+16*exp(2)^2+(-16*x+32)*exp(2)+4*
x^2-16*x+25),x, algorithm="giac")

[Out]

log(4*x^2 - 16*x*e^2 + 16*(3^(1/4) + 2)*e^2 - 8*3^(1/4)*x - 16*x + 4*sqrt(3) + 16*3^(1/4) + 16*e^4 + 25)

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maple [A]  time = 0.42, size = 44, normalized size = 2.00

method	result	size
derivativedivides	$\ln (4\sqrt{3}+(16{\mathrm{e}}^2-8x+16)3^{\frac{1}{4}}+16{\mathrm{e}}^4+(-16x+32){\mathrm{e}}^2+4x^2-16x+25)$	$44$
risch	$\ln (4x^2+(-83^{\frac{1}{4}}-16{\mathrm{e}}^2-16)x+16{\mathrm{e}}^4+163^{\frac{1}{4}}{\mathrm{e}}^2+32{\mathrm{e}}^2+4\sqrt{3}+163^{\frac{1}{4}}+25)$	$47$
default	$\ln (16{\mathrm{e}}^4+163^{\frac{1}{4}}{\mathrm{e}}^2-16{\mathrm{e}}^{2}x-83^{\frac{1}{4}}x+4x^2+32{\mathrm{e}}^2+4\sqrt{3}+163^{\frac{1}{4}}-16x+25)$	$50$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*3^(1/4)-16*exp(2)+8*x-16)/(4*3^(1/2)+(16*exp(2)-8*x+16)*3^(1/4)+16*exp(2)^2+(-16*x+32)*exp(2)+4*x^2-16
*x+25),x,method=_RETURNVERBOSE)

[Out]

ln(4*3^(1/2)+(16*exp(2)-8*x+16)*3^(1/4)+16*exp(2)^2+(-16*x+32)*exp(2)+4*x^2-16*x+25)

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maxima [B]  time = 0.44, size = 39, normalized size = 1.77 $$\begin{array}{c}\log (4x^2-16(x-2)e^2-8\cdot 3^{\frac{1}{4}}(x-2e^2-2)-16x+4\sqrt{3}+16e^4+25)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*3^(1/4)-16*exp(2)+8*x-16)/(4*3^(1/2)+(16*exp(2)-8*x+16)*3^(1/4)+16*exp(2)^2+(-16*x+32)*exp(2)+4*
x^2-16*x+25),x, algorithm="maxima")

[Out]

log(4*x^2 - 16*(x - 2)*e^2 - 8*3^(1/4)*(x - 2*e^2 - 2) - 16*x + 4*sqrt(3) + 16*e^4 + 25)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 $$\begin{array}{c}\int -\frac{16{\mathrm{e}}^2-8x+83^{1/4}+16}{16{\mathrm{e}}^4-16x+4\sqrt{3}+3^{1/4}(16{\mathrm{e}}^2-8x+16)+4x^2-{\mathrm{e}}^2(16x-32)+25}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(2) - 8*x + 8*3^(1/4) + 16)/(16*exp(4) - 16*x + 4*3^(1/2) + 3^(1/4)*(16*exp(2) - 8*x + 16) + 4*x^2
- exp(2)*(16*x - 32) + 25),x)

[Out]

int(-(16*exp(2) - 8*x + 8*3^(1/4) + 16)/(16*exp(4) - 16*x + 4*3^(1/2) + 3^(1/4)*(16*exp(2) - 8*x + 16) + 4*x^2
- exp(2)*(16*x - 32) + 25), x)

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sympy [B]  time = 0.35, size = 58, normalized size = 2.64 $$\begin{array}{c}\log (4x^2+x(-16e^2-16-8\sqrt[4]{3})+4\sqrt{3}+16\sqrt[4]{3}+25+16\sqrt[4]{3}{e}^2+32e^2+16e^4)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*3**(1/4)-16*exp(2)+8*x-16)/(4*3**(1/2)+(16*exp(2)-8*x+16)*3**(1/4)+16*exp(2)**2+(-16*x+32)*exp(2
)+4*x**2-16*x+25),x)

[Out]

log(4*x**2 + x*(-16*exp(2) - 16 - 8*3**(1/4)) + 4*sqrt(3) + 16*3**(1/4) + 25 + 16*3**(1/4)*exp(2) + 32*exp(2)
+ 16*exp(4))

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