3.11.41 \(\int \frac {24+224 x+4 x^2+\sqrt [4]{e} (-16+4 x)+(-16 x+44 \sqrt [4]{e} x+48 x^2) \log (x)+(-2 x+2 \sqrt {e} x+4 \sqrt [4]{e} x^2+2 x^3) \log ^2(x)}{144 x+(24 \sqrt [4]{e} x+24 x^2) \log (x)+(\sqrt {e} x+2 \sqrt [4]{e} x^2+x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ 2 \left (x+\frac {8-2 x+\log (x)}{12+\left (\sqrt [4]{e}+x\right ) \log (x)}\right ) \]

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Rubi [F]  time = 2.42, 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 {gather*} \int \frac {24+224 x+4 x^2+\sqrt [4]{e} (-16+4 x)+\left (-16 x+44 \sqrt [4]{e} x+48 x^2\right ) \log (x)+\left (-2 x+2 \sqrt {e} x+4 \sqrt [4]{e} x^2+2 x^3\right ) \log ^2(x)}{144 x+\left (24 \sqrt [4]{e} x+24 x^2\right ) \log (x)+\left (\sqrt {e} x+2 \sqrt [4]{e} x^2+x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(24 + 224*x + 4*x^2 + E^(1/4)*(-16 + 4*x) + (-16*x + 44*E^(1/4)*x + 48*x^2)*Log[x] + (-2*x + 2*Sqrt[E]*x +
4*E^(1/4)*x^2 + 2*x^3)*Log[x]^2)/(144*x + (24*E^(1/4)*x + 24*x^2)*Log[x] + (Sqrt[E]*x + 2*E^(1/4)*x^2 + x^3)*
Log[x]^2),x]

[Out]

2*x + 2/(E^(1/4) + x) - 4*(16 - E^(1/4))*Defer[Int][(12 + E^(1/4)*Log[x] + x*Log[x])^(-2), x] + 8*(3 - 2*E^(1/
4))*Defer[Int][1/(x*(12 + E^(1/4)*Log[x] + x*Log[x])^2), x] + 4*Defer[Int][x/(12 + E^(1/4)*Log[x] + x*Log[x])^
2, x] - 288*Defer[Int][1/((E^(1/4) + x)^2*(12 + E^(1/4)*Log[x] + x*Log[x])^2), x] + 48*(4 + E^(1/4))*Defer[Int
][1/((E^(1/4) + x)*(12 + E^(1/4)*Log[x] + x*Log[x])^2), x] + 48*Defer[Int][1/((E^(1/4) + x)^2*(12 + E^(1/4)*Lo
g[x] + x*Log[x])), x] - 4*(4 + E^(1/4))*Defer[Int][1/((E^(1/4) + x)*(12 + E^(1/4)*Log[x] + x*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (6+\sqrt [4]{e} (-4+x)+56 x+x^2\right )+4 x \left (-4+11 \sqrt [4]{e}+12 x\right ) \log (x)+2 x \left (-1+\sqrt {e}+2 \sqrt [4]{e} x+x^2\right ) \log ^2(x)}{x \left (12+\left (\sqrt [4]{e}+x\right ) \log (x)\right )^2} \, dx\\ &=\int \left (\frac {2 \left (-1+\sqrt {e}+2 \sqrt [4]{e} x+x^2\right )}{\left (\sqrt [4]{e}+x\right )^2}+\frac {4 \left (2 \left (3-2 \sqrt [4]{e}\right ) \sqrt {e}-\left (72-60 \sqrt [4]{e}+12 \sqrt {e}-e^{3/4}\right ) x+3 \left (18-8 \sqrt [4]{e}+\sqrt {e}\right ) x^2-\left (16-3 \sqrt [4]{e}\right ) x^3+x^4\right )}{x \left (\sqrt [4]{e}+x\right )^2 \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2}+\frac {4 \left (12-4 \sqrt [4]{e}-\sqrt {e}-\left (4+\sqrt [4]{e}\right ) x\right )}{\left (\sqrt [4]{e}+x\right )^2 \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {-1+\sqrt {e}+2 \sqrt [4]{e} x+x^2}{\left (\sqrt [4]{e}+x\right )^2} \, dx+4 \int \frac {2 \left (3-2 \sqrt [4]{e}\right ) \sqrt {e}-\left (72-60 \sqrt [4]{e}+12 \sqrt {e}-e^{3/4}\right ) x+3 \left (18-8 \sqrt [4]{e}+\sqrt {e}\right ) x^2-\left (16-3 \sqrt [4]{e}\right ) x^3+x^4}{x \left (\sqrt [4]{e}+x\right )^2 \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2} \, dx+4 \int \frac {12-4 \sqrt [4]{e}-\sqrt {e}-\left (4+\sqrt [4]{e}\right ) x}{\left (\sqrt [4]{e}+x\right )^2 \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )} \, dx\\ &=2 \int \left (1-\frac {1}{\left (\sqrt [4]{e}+x\right )^2}\right ) \, dx+4 \int \left (-\frac {16 \left (1-\frac {\sqrt [4]{e}}{16}\right )}{\left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2}-\frac {2 \left (-3+2 \sqrt [4]{e}\right )}{x \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2}+\frac {x}{\left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2}-\frac {72}{\left (\sqrt [4]{e}+x\right )^2 \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2}+\frac {12 \left (4+\sqrt [4]{e}\right )}{\left (\sqrt [4]{e}+x\right ) \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2}\right ) \, dx+4 \int \left (\frac {12}{\left (\sqrt [4]{e}+x\right )^2 \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )}+\frac {-4-\sqrt [4]{e}}{\left (\sqrt [4]{e}+x\right ) \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )}\right ) \, dx\\ &=2 x+\frac {2}{\sqrt [4]{e}+x}+4 \int \frac {x}{\left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2} \, dx+48 \int \frac {1}{\left (\sqrt [4]{e}+x\right )^2 \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )} \, dx-288 \int \frac {1}{\left (\sqrt [4]{e}+x\right )^2 \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2} \, dx+\left (8 \left (3-2 \sqrt [4]{e}\right )\right ) \int \frac {1}{x \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2} \, dx-\left (4 \left (16-\sqrt [4]{e}\right )\right ) \int \frac {1}{\left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2} \, dx-\left (4 \left (4+\sqrt [4]{e}\right )\right ) \int \frac {1}{\left (\sqrt [4]{e}+x\right ) \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )} \, dx+\left (48 \left (4+\sqrt [4]{e}\right )\right ) \int \frac {1}{\left (\sqrt [4]{e}+x\right ) \left (12+\sqrt [4]{e} \log (x)+x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 36, normalized size = 1.38 \begin {gather*} \frac {2 \left (8+10 x+\left (1+\sqrt [4]{e} x+x^2\right ) \log (x)\right )}{12+\left (\sqrt [4]{e}+x\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24 + 224*x + 4*x^2 + E^(1/4)*(-16 + 4*x) + (-16*x + 44*E^(1/4)*x + 48*x^2)*Log[x] + (-2*x + 2*Sqrt[
E]*x + 4*E^(1/4)*x^2 + 2*x^3)*Log[x]^2)/(144*x + (24*E^(1/4)*x + 24*x^2)*Log[x] + (Sqrt[E]*x + 2*E^(1/4)*x^2 +
x^3)*Log[x]^2),x]

[Out]

(2*(8 + 10*x + (1 + E^(1/4)*x + x^2)*Log[x]))/(12 + (E^(1/4) + x)*Log[x])

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fricas [A]  time = 0.60, size = 30, normalized size = 1.15 \begin {gather*} \frac {2 \, {\left ({\left (x^{2} + x e^{\frac {1}{4}} + 1\right )} \log \relax (x) + 10 \, x + 8\right )}}{{\left (x + e^{\frac {1}{4}}\right )} \log \relax (x) + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(1/4)^2+4*x^2*exp(1/4)+2*x^3-2*x)*log(x)^2+(44*x*exp(1/4)+48*x^2-16*x)*log(x)+(4*x-16)*exp(
1/4)+4*x^2+224*x+24)/((x*exp(1/4)^2+2*x^2*exp(1/4)+x^3)*log(x)^2+(24*x*exp(1/4)+24*x^2)*log(x)+144*x),x, algor
ithm="fricas")

[Out]

2*((x^2 + x*e^(1/4) + 1)*log(x) + 10*x + 8)/((x + e^(1/4))*log(x) + 12)

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giac [A]  time = 0.83, size = 34, normalized size = 1.31 \begin {gather*} \frac {2 \, {\left (x^{2} \log \relax (x) + x e^{\frac {1}{4}} \log \relax (x) + 10 \, x + \log \relax (x) + 8\right )}}{x \log \relax (x) + e^{\frac {1}{4}} \log \relax (x) + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(1/4)^2+4*x^2*exp(1/4)+2*x^3-2*x)*log(x)^2+(44*x*exp(1/4)+48*x^2-16*x)*log(x)+(4*x-16)*exp(
1/4)+4*x^2+224*x+24)/((x*exp(1/4)^2+2*x^2*exp(1/4)+x^3)*log(x)^2+(24*x*exp(1/4)+24*x^2)*log(x)+144*x),x, algor
ithm="giac")

[Out]

2*(x^2*log(x) + x*e^(1/4)*log(x) + 10*x + log(x) + 8)/(x*log(x) + e^(1/4)*log(x) + 12)

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maple [A]  time = 0.12, size = 42, normalized size = 1.62

method	result	size
norman	\(\frac {\left (-2 \,{\mathrm e}^{\frac {1}{2}}+2\right ) \ln \relax (x )+20 x +2 x^{2} \ln \relax (x )+16-24 \,{\mathrm e}^{\frac {1}{4}}}{\ln \relax (x ) {\mathrm e}^{\frac {1}{4}}+x \ln \relax (x )+12}\)	\(42\)
risch	\(\frac {2 x \,{\mathrm e}^{\frac {1}{4}}+2 x^{2}+2}{x +{\mathrm e}^{\frac {1}{4}}}-\frac {4 \left (x \,{\mathrm e}^{\frac {1}{4}}+x^{2}-4 \,{\mathrm e}^{\frac {1}{4}}-4 x +6\right )}{\left (x +{\mathrm e}^{\frac {1}{4}}\right ) \left (\ln \relax (x ) {\mathrm e}^{\frac {1}{4}}+x \ln \relax (x )+12\right )}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(1/4)^2+4*x^2*exp(1/4)+2*x^3-2*x)*ln(x)^2+(44*x*exp(1/4)+48*x^2-16*x)*ln(x)+(4*x-16)*exp(1/4)+4*x
^2+224*x+24)/((x*exp(1/4)^2+2*x^2*exp(1/4)+x^3)*ln(x)^2+(24*x*exp(1/4)+24*x^2)*ln(x)+144*x),x,method=_RETURNVE
RBOSE)

[Out]

((-2*exp(1/4)^2+2)*ln(x)+20*x+2*x^2*ln(x)+16-24*exp(1/4))/(ln(x)*exp(1/4)+x*ln(x)+12)

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maxima [A]  time = 1.99, size = 30, normalized size = 1.15 \begin {gather*} \frac {2 \, {\left ({\left (x^{2} + x e^{\frac {1}{4}} + 1\right )} \log \relax (x) + 10 \, x + 8\right )}}{{\left (x + e^{\frac {1}{4}}\right )} \log \relax (x) + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(1/4)^2+4*x^2*exp(1/4)+2*x^3-2*x)*log(x)^2+(44*x*exp(1/4)+48*x^2-16*x)*log(x)+(4*x-16)*exp(
1/4)+4*x^2+224*x+24)/((x*exp(1/4)^2+2*x^2*exp(1/4)+x^3)*log(x)^2+(24*x*exp(1/4)+24*x^2)*log(x)+144*x),x, algor
ithm="maxima")

[Out]

2*((x^2 + x*e^(1/4) + 1)*log(x) + 10*x + 8)/((x + e^(1/4))*log(x) + 12)

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mupad [B]  time = 1.74, size = 34, normalized size = 1.31 \begin {gather*} \frac {2\,\left (10\,x+\ln \relax (x)+x^2\,\ln \relax (x)+x\,{\mathrm {e}}^{1/4}\,\ln \relax (x)+8\right )}{{\mathrm {e}}^{1/4}\,\ln \relax (x)+x\,\ln \relax (x)+12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((224*x + 4*x^2 + log(x)^2*(2*x*exp(1/2) - 2*x + 4*x^2*exp(1/4) + 2*x^3) + log(x)*(44*x*exp(1/4) - 16*x + 4
8*x^2) + exp(1/4)*(4*x - 16) + 24)/(144*x + log(x)*(24*x*exp(1/4) + 24*x^2) + log(x)^2*(x*exp(1/2) + 2*x^2*exp
(1/4) + x^3)),x)

[Out]

(2*(10*x + log(x) + x^2*log(x) + x*exp(1/4)*log(x) + 8))/(exp(1/4)*log(x) + x*log(x) + 12)

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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(((2*x*exp(1/4)**2+4*x**2*exp(1/4)+2*x**3-2*x)*ln(x)**2+(44*x*exp(1/4)+48*x**2-16*x)*ln(x)+(4*x-16)*e
xp(1/4)+4*x**2+224*x+24)/((x*exp(1/4)**2+2*x**2*exp(1/4)+x**3)*ln(x)**2+(24*x*exp(1/4)+24*x**2)*ln(x)+144*x),x
)

[Out]

Timed out

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