3.83.35 \(\int \frac {3+3 e^3+8 x^2+(1+e^3+6 x^2) \log (2)}{x^2+x^2 \log (2)} \, dx\)

Optimal. Leaf size=30 \[ 5 x-\left (4+\frac {1+e^3}{x}-x\right ) \left (1+\frac {2}{1+\log (2)}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6, 12, 14} \begin {gather*} \frac {2 x (4+\log (8))}{1+\log (2)}-\frac {\left (1+e^3\right ) (3+\log (2))}{x (1+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 3*E^3 + 8*x^2 + (1 + E^3 + 6*x^2)*Log[2])/(x^2 + x^2*Log[2]),x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2 (1+\log (2))} \, dx\\ &=\frac {\int \frac {3+3 e^3+8 x^2+\left (1+e^3+6 x^2\right ) \log (2)}{x^2} \, dx}{1+\log (2)}\\ &=\frac {\int \left (8 \left (1+\frac {3 \log (2)}{4}\right )+\frac {\left (1+e^3\right ) (3+\log (2))}{x^2}\right ) \, dx}{1+\log (2)}\\ &=-\frac {\left (1+e^3\right ) (3+\log (2))}{x (1+\log (2))}+\frac {2 x (4+\log (8))}{1+\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 32, normalized size = 1.07 \begin {gather*} -\frac {3+\log (2)+e^3 (3+\log (2))-x^2 (8+\log (64))}{x (1+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 3*E^3 + 8*x^2 + (1 + E^3 + 6*x^2)*Log[2])/(x^2 + x^2*Log[2]),x]

[Out]

-((3 + Log[2] + E^3*(3 + Log[2]) - x^2*(8 + Log[64]))/(x*(1 + Log[2])))

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fricas [A]  time = 0.68, size = 34, normalized size = 1.13 \begin {gather*} \frac {8 \, x^{2} + {\left (6 \, x^{2} - e^{3} - 1\right )} \log \relax (2) - 3 \, e^{3} - 3}{x \log \relax (2) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(3)+6*x^2+1)*log(2)+3*exp(3)+8*x^2+3)/(x^2*log(2)+x^2),x, algorithm="fricas")

[Out]

(8*x^2 + (6*x^2 - e^3 - 1)*log(2) - 3*e^3 - 3)/(x*log(2) + x)

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giac [A]  time = 0.16, size = 42, normalized size = 1.40 \begin {gather*} \frac {2 \, {\left (3 \, x \log \relax (2) + 4 \, x\right )}}{\log \relax (2) + 1} - \frac {e^{3} \log \relax (2) + 3 \, e^{3} + \log \relax (2) + 3}{x {\left (\log \relax (2) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(3)+6*x^2+1)*log(2)+3*exp(3)+8*x^2+3)/(x^2*log(2)+x^2),x, algorithm="giac")

[Out]

2*(3*x*log(2) + 4*x)/(log(2) + 1) - (e^3*log(2) + 3*e^3 + log(2) + 3)/(x*(log(2) + 1))

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maple [A]  time = 0.34, size = 35, normalized size = 1.17

method	result	size
default	\(\frac {6 x \ln \relax (2)+8 x -\frac {{\mathrm e}^{3} \ln \relax (2)+\ln \relax (2)+3 \,{\mathrm e}^{3}+3}{x}}{1+\ln \relax (2)}\)	\(35\)
gosper	\(-\frac {-6 x^{2} \ln \relax (2)+{\mathrm e}^{3} \ln \relax (2)-8 x^{2}+\ln \relax (2)+3 \,{\mathrm e}^{3}+3}{x \left (1+\ln \relax (2)\right )}\)	\(37\)
norman	\(\frac {-\frac {{\mathrm e}^{3} \ln \relax (2)+\ln \relax (2)+3 \,{\mathrm e}^{3}+3}{1+\ln \relax (2)}+\frac {2 \left (3 \ln \relax (2)+4\right ) x^{2}}{1+\ln \relax (2)}}{x}\)	\(44\)
risch	\(\frac {6 x \ln \relax (2)}{1+\ln \relax (2)}+\frac {8 x}{1+\ln \relax (2)}-\frac {\ln \relax (2)^{2} {\mathrm e}^{3}}{\left (1+\ln \relax (2)\right )^{2} x}-\frac {\ln \relax (2)^{2}}{\left (1+\ln \relax (2)\right )^{2} x}-\frac {4 \ln \relax (2) {\mathrm e}^{3}}{\left (1+\ln \relax (2)\right )^{2} x}-\frac {4 \ln \relax (2)}{\left (1+\ln \relax (2)\right )^{2} x}-\frac {3 \,{\mathrm e}^{3}}{\left (1+\ln \relax (2)\right )^{2} x}-\frac {3}{\left (1+\ln \relax (2)\right )^{2} x}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(3)+6*x^2+1)*ln(2)+3*exp(3)+8*x^2+3)/(x^2*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

1/(1+ln(2))*(6*x*ln(2)+8*x-(exp(3)*ln(2)+ln(2)+3*exp(3)+3)/x)

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maxima [A]  time = 0.38, size = 40, normalized size = 1.33 \begin {gather*} \frac {2 \, x {\left (3 \, \log \relax (2) + 4\right )}}{\log \relax (2) + 1} - \frac {{\left (e^{3} + 1\right )} \log \relax (2) + 3 \, e^{3} + 3}{x {\left (\log \relax (2) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(3)+6*x^2+1)*log(2)+3*exp(3)+8*x^2+3)/(x^2*log(2)+x^2),x, algorithm="maxima")

[Out]

2*x*(3*log(2) + 4)/(log(2) + 1) - ((e^3 + 1)*log(2) + 3*e^3 + 3)/(x*(log(2) + 1))

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mupad [B]  time = 0.10, size = 37, normalized size = 1.23 \begin {gather*} \frac {x\,\left (\ln \left (64\right )+8\right )}{\ln \relax (2)+1}-\frac {3\,{\mathrm {e}}^3+\ln \relax (2)+{\mathrm {e}}^3\,\ln \relax (2)+3}{x\,\left (\ln \relax (2)+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(3) + log(2)*(exp(3) + 6*x^2 + 1) + 8*x^2 + 3)/(x^2*log(2) + x^2),x)

[Out]

(x*(log(64) + 8))/(log(2) + 1) - (3*exp(3) + log(2) + exp(3)*log(2) + 3)/(x*(log(2) + 1))

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sympy [A]  time = 0.12, size = 32, normalized size = 1.07 \begin {gather*} \frac {x \left (6 \log {\relax (2 )} + 8\right ) + \frac {- 3 e^{3} - e^{3} \log {\relax (2 )} - 3 - \log {\relax (2 )}}{x}}{\log {\relax (2 )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(3)+6*x**2+1)*ln(2)+3*exp(3)+8*x**2+3)/(x**2*ln(2)+x**2),x)

[Out]

(x*(6*log(2) + 8) + (-3*exp(3) - exp(3)*log(2) - 3 - log(2))/x)/(log(2) + 1)

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