3.57.7 \(\int \frac {e^2 (-3+x)-4 x+(-2 e^2-4 x) \log (x)}{4 x^3 \log (2)} \, dx\)

Optimal. Leaf size=24 \[ \frac {\left (\frac {e^2}{4}+x\right ) (2-x+\log (x))}{x^2 \log (2)} \]

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Rubi [B]  time = 0.07, antiderivative size = 90, normalized size of antiderivative = 3.75, number of steps used = 8, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {12, 14, 37, 2334, 43} \begin {gather*} \frac {\left (2 x+e^2\right )^2 \log (x)}{4 e^2 x^2 \log (2)}+\frac {\left (\left (4-e^2\right ) x+3 e^2\right )^2}{24 e^2 x^2 \log (2)}+\frac {e^2}{8 x^2 \log (2)}-\frac {\log (x)}{e^2 \log (2)}+\frac {1}{x \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-3 + x) - 4*x + (-2*E^2 - 4*x)*Log[x])/(4*x^3*Log[2]),x]

[Out]

E^2/(8*x^2*Log[2]) + 1/(x*Log[2]) + (3*E^2 + (4 - E^2)*x)^2/(24*E^2*x^2*Log[2]) - Log[x]/(E^2*Log[2]) + ((E^2
+ 2*x)^2*Log[x])/(4*E^2*x^2*Log[2])

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]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^2 (-3+x)-4 x+\left (-2 e^2-4 x\right ) \log (x)}{x^3} \, dx}{4 \log (2)}\\ &=\frac {\int \left (\frac {-3 e^2-\left (4-e^2\right ) x}{x^3}-\frac {2 \left (e^2+2 x\right ) \log (x)}{x^3}\right ) \, dx}{4 \log (2)}\\ &=\frac {\int \frac {-3 e^2-\left (4-e^2\right ) x}{x^3} \, dx}{4 \log (2)}-\frac {\int \frac {\left (e^2+2 x\right ) \log (x)}{x^3} \, dx}{2 \log (2)}\\ &=\frac {\left (3 e^2+\left (4-e^2\right ) x\right )^2}{24 e^2 x^2 \log (2)}+\frac {\left (e^2+2 x\right )^2 \log (x)}{4 e^2 x^2 \log (2)}+\frac {\int -\frac {\left (e^2+2 x\right )^2}{2 e^2 x^3} \, dx}{2 \log (2)}\\ &=\frac {\left (3 e^2+\left (4-e^2\right ) x\right )^2}{24 e^2 x^2 \log (2)}+\frac {\left (e^2+2 x\right )^2 \log (x)}{4 e^2 x^2 \log (2)}-\frac {\int \frac {\left (e^2+2 x\right )^2}{x^3} \, dx}{4 e^2 \log (2)}\\ &=\frac {\left (3 e^2+\left (4-e^2\right ) x\right )^2}{24 e^2 x^2 \log (2)}+\frac {\left (e^2+2 x\right )^2 \log (x)}{4 e^2 x^2 \log (2)}-\frac {\int \left (\frac {e^4}{x^3}+\frac {4 e^2}{x^2}+\frac {4}{x}\right ) \, dx}{4 e^2 \log (2)}\\ &=\frac {e^2}{8 x^2 \log (2)}+\frac {1}{x \log (2)}+\frac {\left (3 e^2+\left (4-e^2\right ) x\right )^2}{24 e^2 x^2 \log (2)}-\frac {\log (x)}{e^2 \log (2)}+\frac {\left (e^2+2 x\right )^2 \log (x)}{4 e^2 x^2 \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 1.79 \begin {gather*} \frac {\frac {2 e^2}{x^2}+\frac {8}{x}-\frac {e^2}{x}+\frac {e^2 \log (x)}{x^2}+\frac {4 \log (x)}{x}}{\log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-3 + x) - 4*x + (-2*E^2 - 4*x)*Log[x])/(4*x^3*Log[2]),x]

[Out]

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

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fricas [A]  time = 1.01, size = 29, normalized size = 1.21 \begin {gather*} -\frac {{\left (x - 2\right )} e^{2} - {\left (4 \, x + e^{2}\right )} \log \relax (x) - 8 \, x}{4 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*exp(2)-4*x)*log(x)+(x-3)*exp(2)-4*x)/x^3/log(2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 32, normalized size = 1.33 \begin {gather*} -\frac {x e^{2} - 4 \, x \log \relax (x) - e^{2} \log \relax (x) - 8 \, x - 2 \, e^{2}}{4 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*exp(2)-4*x)*log(x)+(x-3)*exp(2)-4*x)/x^3/log(2),x, algorithm="giac")

[Out]

-1/4*(x*e^2 - 4*x*log(x) - e^2*log(x) - 8*x - 2*e^2)/(x^2*log(2))

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maple [A]  time = 0.05, size = 40, normalized size = 1.67

method	result	size
risch	\(\frac {\left ({\mathrm e}^{2}+4 x \right ) \ln \relax (x )}{4 \ln \relax (2) x^{2}}-\frac {{\mathrm e}^{2} x -2 \,{\mathrm e}^{2}-8 x}{4 \ln \relax (2) x^{2}}\)	\(40\)
norman	\(\frac {\frac {x \ln \relax (x )}{\ln \relax (2)}+\frac {{\mathrm e}^{2}}{2 \ln \relax (2)}-\frac {\left (-8+{\mathrm e}^{2}\right ) x}{4 \ln \relax (2)}+\frac {{\mathrm e}^{2} \ln \relax (x )}{4 \ln \relax (2)}}{x^{2}}\)	\(43\)
default	\(\frac {-2 \,{\mathrm e}^{2} \left (-\frac {\ln \relax (x )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )-\frac {{\mathrm e}^{2}}{x}+\frac {4 \ln \relax (x )}{x}+\frac {8}{x}+\frac {3 \,{\mathrm e}^{2}}{2 x^{2}}}{4 \ln \relax (2)}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-2*exp(2)-4*x)*ln(x)+(x-3)*exp(2)-4*x)/x^3/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/4/ln(2)*(exp(2)+4*x)/x^2*ln(x)-1/4/ln(2)*(exp(2)*x-2*exp(2)-8*x)/x^2

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maxima [B]  time = 0.43, size = 47, normalized size = 1.96 \begin {gather*} \frac {{\left (\frac {2 \, \log \relax (x)}{x^{2}} + \frac {1}{x^{2}}\right )} e^{2} - \frac {2 \, e^{2}}{x} + \frac {8 \, \log \relax (x)}{x} + \frac {16}{x} + \frac {3 \, e^{2}}{x^{2}}}{8 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*exp(2)-4*x)*log(x)+(x-3)*exp(2)-4*x)/x^3/log(2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.57, size = 34, normalized size = 1.42 \begin {gather*} \frac {x^2\,\left (\ln \relax (x)-\frac {{\mathrm {e}}^2}{4}+2\right )+x\,\left (\frac {{\mathrm {e}}^2}{2}+\frac {{\mathrm {e}}^2\,\ln \relax (x)}{4}\right )}{x^3\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + (log(x)*(4*x + 2*exp(2)))/4 - (exp(2)*(x - 3))/4)/(x^3*log(2)),x)

[Out]

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

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sympy [A]  time = 0.20, size = 37, normalized size = 1.54 \begin {gather*} \frac {\left (4 x + e^{2}\right ) \log {\relax (x )}}{4 x^{2} \log {\relax (2 )}} - \frac {x \left (-8 + e^{2}\right ) - 2 e^{2}}{4 x^{2} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*exp(2)-4*x)*ln(x)+(x-3)*exp(2)-4*x)/x**3/ln(2),x)

[Out]

(4*x + exp(2))*log(x)/(4*x**2*log(2)) - (x*(-8 + exp(2)) - 2*exp(2))/(4*x**2*log(2))

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