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

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

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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 {gather*} \log \left (4 x^2-16 x+16 e^2 (2-x)+8 \sqrt [4]{3} \left (2 \left (1+e^2\right )-x\right )+16 e^4+4 \sqrt {3}+25\right ) \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\log \left (25+4 \sqrt {3}+16 e^4+16 e^2 (2-x)+8 \sqrt [4]{3} \left (2 \left (1+e^2\right )-x\right )-16 x+4 x^2\right )\\ \end {aligned} \end {gather*}

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

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 {gather*} \log \left (4 \, x^{2} - 16 \, {\left (x - 2\right )} e^{2} - 8 \cdot 3^{\frac {1}{4}} {\left (x - 2 \, e^{2} - 2\right )} - 16 \, x + 4 \, \sqrt {3} + 16 \, e^{4} + 25\right ) \end {gather*}

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 {gather*} \log \left (4 \, x^{2} - 16 \, x e^{2} + 16 \, {\left (3^{\frac {1}{4}} + 2\right )} e^{2} - 8 \cdot 3^{\frac {1}{4}} x - 16 \, x + 4 \, \sqrt {3} + 16 \cdot 3^{\frac {1}{4}} + 16 \, e^{4} + 25\right ) \end {gather*}

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	\(\ln \left (4 \sqrt {3}+\left (16 \,{\mathrm e}^{2}-8 x +16\right ) 3^{\frac {1}{4}}+16 \,{\mathrm e}^{4}+\left (-16 x +32\right ) {\mathrm e}^{2}+4 x^{2}-16 x +25\right )\)	\(44\)
risch	\(\ln \left (4 x^{2}+\left (-8 \,3^{\frac {1}{4}}-16 \,{\mathrm e}^{2}-16\right ) x +16 \,{\mathrm e}^{4}+16 \,3^{\frac {1}{4}} {\mathrm e}^{2}+32 \,{\mathrm e}^{2}+4 \sqrt {3}+16 \,3^{\frac {1}{4}}+25\right )\)	\(47\)
default	\(\ln \left (16 \,{\mathrm e}^{4}+16 \,3^{\frac {1}{4}} {\mathrm e}^{2}-16 \,{\mathrm e}^{2} x -8 \,3^{\frac {1}{4}} x +4 x^{2}+32 \,{\mathrm e}^{2}+4 \sqrt {3}+16 \,3^{\frac {1}{4}}-16 x +25\right )\)	\(50\)

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 {gather*} \log \left (4 \, x^{2} - 16 \, {\left (x - 2\right )} e^{2} - 8 \cdot 3^{\frac {1}{4}} {\left (x - 2 \, e^{2} - 2\right )} - 16 \, x + 4 \, \sqrt {3} + 16 \, e^{4} + 25\right ) \end {gather*}

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 {gather*} \int -\frac {16\,{\mathrm {e}}^2-8\,x+8\,3^{1/4}+16}{16\,{\mathrm {e}}^4-16\,x+4\,\sqrt {3}+3^{1/4}\,\left (16\,{\mathrm {e}}^2-8\,x+16\right )+4\,x^2-{\mathrm {e}}^2\,\left (16\,x-32\right )+25} \,d x \end {gather*}

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 {gather*} \log {\left (4 x^{2} + x \left (- 16 e^{2} - 16 - 8 \sqrt [4]{3}\right ) + 4 \sqrt {3} + 16 \sqrt [4]{3} + 25 + 16 \sqrt [4]{3} e^{2} + 32 e^{2} + 16 e^{4} \right )} \end {gather*}

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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