3.3.79 \(\int \frac {e^6 (-5+x+3 x^2)-16 e^3 \log (2)+(e^6 (5-2 x)+16 e^3 \log (2)) \log (x)}{e^6 (25 x^2-10 x^3+x^4)+e^3 (160 x^2-32 x^3) \log (2)+256 x^2 \log ^2(2)} \, dx\)

Optimal. Leaf size=24 \[ \frac {3-\frac {\log (x)}{x}}{5-x+\frac {16 \log (2)}{e^3}} \]

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Rubi [B]  time = 0.99, antiderivative size = 103, normalized size of antiderivative = 4.29, number of steps used = 11, number of rules used = 8, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6688, 12, 6742, 893, 2357, 2304, 2314, 31} \begin {gather*} -\frac {e^9 x \log (x)}{\left (5 e^3+16 \log (2)\right )^2 \left (-e^3 x+5 e^3+16 \log (2)\right )}-\frac {e^3 \log (x)}{x \left (5 e^3+16 \log (2)\right )}-\frac {e^6 \log (x)}{\left (5 e^3+16 \log (2)\right )^2}+\frac {3 e^3}{-e^3 x+5 e^3+16 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^6*(-5 + x + 3*x^2) - 16*E^3*Log[2] + (E^6*(5 - 2*x) + 16*E^3*Log[2])*Log[x])/(E^6*(25*x^2 - 10*x^3 + x^
4) + E^3*(160*x^2 - 32*x^3)*Log[2] + 256*x^2*Log[2]^2),x]

[Out]

(3*E^3)/(5*E^3 - E^3*x + 16*Log[2]) - (E^6*Log[x])/(5*E^3 + 16*Log[2])^2 - (E^3*Log[x])/(x*(5*E^3 + 16*Log[2])
) - (E^9*x*Log[x])/((5*E^3 + 16*Log[2])^2*(5*E^3 - E^3*x + 16*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2304

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

Rule 2314

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

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 \left (e^3 \left (-5+x+3 x^2\right )-16 \log (2)+\left (e^3 (5-2 x)+16 \log (2)\right ) \log (x)\right )}{x^2 \left (5 e^3-e^3 x+16 \log (2)\right )^2} \, dx\\ &=e^3 \int \frac {e^3 \left (-5+x+3 x^2\right )-16 \log (2)+\left (e^3 (5-2 x)+16 \log (2)\right ) \log (x)}{x^2 \left (5 e^3-e^3 x+16 \log (2)\right )^2} \, dx\\ &=e^3 \int \left (\frac {-5 e^3+e^3 x+3 e^3 x^2-16 \log (2)}{x^2 \left (-5 e^3+e^3 x-16 \log (2)\right )^2}-\frac {\left (-5 e^3+2 e^3 x-16 \log (2)\right ) \log (x)}{x^2 \left (-5 e^3+e^3 x-16 \log (2)\right )^2}\right ) \, dx\\ &=e^3 \int \frac {-5 e^3+e^3 x+3 e^3 x^2-16 \log (2)}{x^2 \left (-5 e^3+e^3 x-16 \log (2)\right )^2} \, dx-e^3 \int \frac {\left (-5 e^3+2 e^3 x-16 \log (2)\right ) \log (x)}{x^2 \left (-5 e^3+e^3 x-16 \log (2)\right )^2} \, dx\\ &=e^3 \int \left (\frac {3 e^3}{\left (-5 e^3+e^3 x-16 \log (2)\right )^2}-\frac {e^3}{x \left (5 e^3+16 \log (2)\right )^2}+\frac {e^6}{\left (-5 e^3+e^3 x-16 \log (2)\right ) \left (5 e^3+16 \log (2)\right )^2}-\frac {1}{x^2 \left (5 e^3+16 \log (2)\right )}\right ) \, dx-e^3 \int \left (-\frac {\log (x)}{x^2 \left (5 e^3+16 \log (2)\right )}+\frac {e^6 \log (x)}{\left (-5 e^3+e^3 x-16 \log (2)\right )^2 \left (5 e^3+16 \log (2)\right )}\right ) \, dx\\ &=\frac {e^3}{x \left (5 e^3+16 \log (2)\right )}+\frac {3 e^3}{5 e^3-e^3 x+16 \log (2)}-\frac {e^6 \log (x)}{\left (5 e^3+16 \log (2)\right )^2}+\frac {e^6 \log \left (5 e^3-e^3 x+16 \log (2)\right )}{\left (5 e^3+16 \log (2)\right )^2}+\frac {e^3 \int \frac {\log (x)}{x^2} \, dx}{5 e^3+16 \log (2)}-\frac {e^9 \int \frac {\log (x)}{\left (-5 e^3+e^3 x-16 \log (2)\right )^2} \, dx}{5 e^3+16 \log (2)}\\ &=\frac {3 e^3}{5 e^3-e^3 x+16 \log (2)}-\frac {e^6 \log (x)}{\left (5 e^3+16 \log (2)\right )^2}-\frac {e^3 \log (x)}{x \left (5 e^3+16 \log (2)\right )}-\frac {e^9 x \log (x)}{\left (5 e^3+16 \log (2)\right )^2 \left (5 e^3-e^3 x+16 \log (2)\right )}+\frac {e^6 \log \left (5 e^3-e^3 x+16 \log (2)\right )}{\left (5 e^3+16 \log (2)\right )^2}-\frac {e^9 \int \frac {1}{-5 e^3+e^3 x-16 \log (2)} \, dx}{\left (5 e^3+16 \log (2)\right )^2}\\ &=\frac {3 e^3}{5 e^3-e^3 x+16 \log (2)}-\frac {e^6 \log (x)}{\left (5 e^3+16 \log (2)\right )^2}-\frac {e^3 \log (x)}{x \left (5 e^3+16 \log (2)\right )}-\frac {e^9 x \log (x)}{\left (5 e^3+16 \log (2)\right )^2 \left (5 e^3-e^3 x+16 \log (2)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 27, normalized size = 1.12 \begin {gather*} \frac {e^3 (-3 x+\log (x))}{x \left (e^3 (-5+x)-16 \log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^6*(-5 + x + 3*x^2) - 16*E^3*Log[2] + (E^6*(5 - 2*x) + 16*E^3*Log[2])*Log[x])/(E^6*(25*x^2 - 10*x^
3 + x^4) + E^3*(160*x^2 - 32*x^3)*Log[2] + 256*x^2*Log[2]^2),x]

[Out]

(E^3*(-3*x + Log[x]))/(x*(E^3*(-5 + x) - 16*Log[2]))

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fricas [A]  time = 1.11, size = 32, normalized size = 1.33 \begin {gather*} -\frac {3 \, x e^{3} - e^{3} \log \relax (x)}{{\left (x^{2} - 5 \, x\right )} e^{3} - 16 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(3)*log(2)+(-2*x+5)*exp(3)^2)*log(x)-16*exp(3)*log(2)+(3*x^2+x-5)*exp(3)^2)/(256*x^2*log(2)^
2+(-32*x^3+160*x^2)*exp(3)*log(2)+(x^4-10*x^3+25*x^2)*exp(3)^2),x, algorithm="fricas")

[Out]

-(3*x*e^3 - e^3*log(x))/((x^2 - 5*x)*e^3 - 16*x*log(2))

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giac [A]  time = 0.47, size = 33, normalized size = 1.38 \begin {gather*} -\frac {3 \, x e^{3} - e^{3} \log \relax (x)}{x^{2} e^{3} - 5 \, x e^{3} - 16 \, x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(3)*log(2)+(-2*x+5)*exp(3)^2)*log(x)-16*exp(3)*log(2)+(3*x^2+x-5)*exp(3)^2)/(256*x^2*log(2)^
2+(-32*x^3+160*x^2)*exp(3)*log(2)+(x^4-10*x^3+25*x^2)*exp(3)^2),x, algorithm="giac")

[Out]

-(3*x*e^3 - e^3*log(x))/(x^2*e^3 - 5*x*e^3 - 16*x*log(2))

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maple [A]  time = 0.33, size = 31, normalized size = 1.29

method	result	size
norman	\(\frac {\ln \relax (x ) {\mathrm e}^{3}-3 x \,{\mathrm e}^{3}}{x \left (x \,{\mathrm e}^{3}-5 \,{\mathrm e}^{3}-16 \ln \relax (2)\right )}\)	\(31\)
risch	\(\frac {{\mathrm e}^{3} \ln \relax (x )}{\left (x \,{\mathrm e}^{3}-5 \,{\mathrm e}^{3}-16 \ln \relax (2)\right ) x}-\frac {3 \,{\mathrm e}^{3}}{x \,{\mathrm e}^{3}-5 \,{\mathrm e}^{3}-16 \ln \relax (2)}\)	\(44\)
default	\(\text {Expression too large to display}\)	\(2282\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*exp(3)*ln(2)+(-2*x+5)*exp(3)^2)*ln(x)-16*exp(3)*ln(2)+(3*x^2+x-5)*exp(3)^2)/(256*x^2*ln(2)^2+(-32*x^3
+160*x^2)*exp(3)*ln(2)+(x^4-10*x^3+25*x^2)*exp(3)^2),x,method=_RETURNVERBOSE)

[Out]

(ln(x)*exp(3)-3*x*exp(3))/x/(x*exp(3)-5*exp(3)-16*ln(2))

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maxima [B]  time = 0.66, size = 589, normalized size = 24.54 \begin {gather*} 16 \, {\left (\frac {2 \, e^{3} \log \left (x e^{3} - 5 \, e^{3} - 16 \, \log \relax (2)\right )}{3840 \, e^{3} \log \relax (2)^{2} + 4096 \, \log \relax (2)^{3} + 1200 \, e^{6} \log \relax (2) + 125 \, e^{9}} - \frac {2 \, e^{3} \log \relax (x)}{3840 \, e^{3} \log \relax (2)^{2} + 4096 \, \log \relax (2)^{3} + 1200 \, e^{6} \log \relax (2) + 125 \, e^{9}} + \frac {2 \, x e^{3} - 5 \, e^{3} - 16 \, \log \relax (2)}{{\left (256 \, e^{3} \log \relax (2)^{2} + 160 \, e^{6} \log \relax (2) + 25 \, e^{9}\right )} x^{2} - {\left (3840 \, e^{3} \log \relax (2)^{2} + 4096 \, \log \relax (2)^{3} + 1200 \, e^{6} \log \relax (2) + 125 \, e^{9}\right )} x}\right )} e^{3} \log \relax (2) + 5 \, {\left (\frac {2 \, e^{3} \log \left (x e^{3} - 5 \, e^{3} - 16 \, \log \relax (2)\right )}{3840 \, e^{3} \log \relax (2)^{2} + 4096 \, \log \relax (2)^{3} + 1200 \, e^{6} \log \relax (2) + 125 \, e^{9}} - \frac {2 \, e^{3} \log \relax (x)}{3840 \, e^{3} \log \relax (2)^{2} + 4096 \, \log \relax (2)^{3} + 1200 \, e^{6} \log \relax (2) + 125 \, e^{9}} + \frac {2 \, x e^{3} - 5 \, e^{3} - 16 \, \log \relax (2)}{{\left (256 \, e^{3} \log \relax (2)^{2} + 160 \, e^{6} \log \relax (2) + 25 \, e^{9}\right )} x^{2} - {\left (3840 \, e^{3} \log \relax (2)^{2} + 4096 \, \log \relax (2)^{3} + 1200 \, e^{6} \log \relax (2) + 125 \, e^{9}\right )} x}\right )} e^{6} - {\left (\frac {\log \left (x e^{3} - 5 \, e^{3} - 16 \, \log \relax (2)\right )}{160 \, e^{3} \log \relax (2) + 256 \, \log \relax (2)^{2} + 25 \, e^{6}} - \frac {\log \relax (x)}{160 \, e^{3} \log \relax (2) + 256 \, \log \relax (2)^{2} + 25 \, e^{6}} + \frac {1}{{\left (16 \, e^{3} \log \relax (2) + 5 \, e^{6}\right )} x - 160 \, e^{3} \log \relax (2) - 256 \, \log \relax (2)^{2} - 25 \, e^{6}}\right )} e^{6} - \frac {e^{6} \log \left (x e^{3} - 5 \, e^{3} - 16 \, \log \relax (2)\right )}{160 \, e^{3} \log \relax (2) + 256 \, \log \relax (2)^{2} + 25 \, e^{6}} + \frac {256 \, e^{3} \log \relax (2)^{2} - {\left (16 \, e^{6} \log \relax (2) + 5 \, e^{9}\right )} x + 160 \, e^{6} \log \relax (2) + {\left (x^{2} e^{9} + 256 \, e^{3} \log \relax (2)^{2} - {\left (16 \, e^{6} \log \relax (2) + 5 \, e^{9}\right )} x + 160 \, e^{6} \log \relax (2) + 25 \, e^{9}\right )} \log \relax (x) + 25 \, e^{9}}{{\left (256 \, e^{3} \log \relax (2)^{2} + 160 \, e^{6} \log \relax (2) + 25 \, e^{9}\right )} x^{2} - {\left (3840 \, e^{3} \log \relax (2)^{2} + 4096 \, \log \relax (2)^{3} + 1200 \, e^{6} \log \relax (2) + 125 \, e^{9}\right )} x} - \frac {3 \, e^{6}}{x e^{6} - 16 \, e^{3} \log \relax (2) - 5 \, e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(3)*log(2)+(-2*x+5)*exp(3)^2)*log(x)-16*exp(3)*log(2)+(3*x^2+x-5)*exp(3)^2)/(256*x^2*log(2)^
2+(-32*x^3+160*x^2)*exp(3)*log(2)+(x^4-10*x^3+25*x^2)*exp(3)^2),x, algorithm="maxima")

[Out]

16*(2*e^3*log(x*e^3 - 5*e^3 - 16*log(2))/(3840*e^3*log(2)^2 + 4096*log(2)^3 + 1200*e^6*log(2) + 125*e^9) - 2*e
^3*log(x)/(3840*e^3*log(2)^2 + 4096*log(2)^3 + 1200*e^6*log(2) + 125*e^9) + (2*x*e^3 - 5*e^3 - 16*log(2))/((25
6*e^3*log(2)^2 + 160*e^6*log(2) + 25*e^9)*x^2 - (3840*e^3*log(2)^2 + 4096*log(2)^3 + 1200*e^6*log(2) + 125*e^9
)*x))*e^3*log(2) + 5*(2*e^3*log(x*e^3 - 5*e^3 - 16*log(2))/(3840*e^3*log(2)^2 + 4096*log(2)^3 + 1200*e^6*log(2
) + 125*e^9) - 2*e^3*log(x)/(3840*e^3*log(2)^2 + 4096*log(2)^3 + 1200*e^6*log(2) + 125*e^9) + (2*x*e^3 - 5*e^3
- 16*log(2))/((256*e^3*log(2)^2 + 160*e^6*log(2) + 25*e^9)*x^2 - (3840*e^3*log(2)^2 + 4096*log(2)^3 + 1200*e^
6*log(2) + 125*e^9)*x))*e^6 - (log(x*e^3 - 5*e^3 - 16*log(2))/(160*e^3*log(2) + 256*log(2)^2 + 25*e^6) - log(x
)/(160*e^3*log(2) + 256*log(2)^2 + 25*e^6) + 1/((16*e^3*log(2) + 5*e^6)*x - 160*e^3*log(2) - 256*log(2)^2 - 25
*e^6))*e^6 - e^6*log(x*e^3 - 5*e^3 - 16*log(2))/(160*e^3*log(2) + 256*log(2)^2 + 25*e^6) + (256*e^3*log(2)^2 -
(16*e^6*log(2) + 5*e^9)*x + 160*e^6*log(2) + (x^2*e^9 + 256*e^3*log(2)^2 - (16*e^6*log(2) + 5*e^9)*x + 160*e^
6*log(2) + 25*e^9)*log(x) + 25*e^9)/((256*e^3*log(2)^2 + 160*e^6*log(2) + 25*e^9)*x^2 - (3840*e^3*log(2)^2 + 4
096*log(2)^3 + 1200*e^6*log(2) + 125*e^9)*x) - 3*e^6/(x*e^6 - 16*e^3*log(2) - 5*e^6)

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mupad [B]  time = 0.80, size = 54, normalized size = 2.25 \begin {gather*} -\frac {{\mathrm {e}}^3\,\left (5\,{\mathrm {e}}^3\,\ln \relax (x)-3\,x^2\,{\mathrm {e}}^3+16\,\ln \relax (2)\,\ln \relax (x)\right )}{x\,\left (5\,{\mathrm {e}}^3+16\,\ln \relax (2)\right )\,\left (5\,{\mathrm {e}}^3+16\,\ln \relax (2)-x\,{\mathrm {e}}^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(16*exp(3)*log(2) - exp(6)*(2*x - 5)) - 16*exp(3)*log(2) + exp(6)*(x + 3*x^2 - 5))/(256*x^2*log(2)
^2 + exp(6)*(25*x^2 - 10*x^3 + x^4) + exp(3)*log(2)*(160*x^2 - 32*x^3)),x)

[Out]

-(exp(3)*(5*exp(3)*log(x) - 3*x^2*exp(3) + 16*log(2)*log(x)))/(x*(5*exp(3) + 16*log(2))*(5*exp(3) + 16*log(2)
- x*exp(3)))

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sympy [B]  time = 0.40, size = 49, normalized size = 2.04 \begin {gather*} \frac {e^{3} \log {\relax (x )}}{x^{2} e^{3} - 5 x e^{3} - 16 x \log {\relax (2 )}} - \frac {3 e^{6}}{x e^{6} - 5 e^{6} - 16 e^{3} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(3)*ln(2)+(-2*x+5)*exp(3)**2)*ln(x)-16*exp(3)*ln(2)+(3*x**2+x-5)*exp(3)**2)/(256*x**2*ln(2)*
*2+(-32*x**3+160*x**2)*exp(3)*ln(2)+(x**4-10*x**3+25*x**2)*exp(3)**2),x)

[Out]

exp(3)*log(x)/(x**2*exp(3) - 5*x*exp(3) - 16*x*log(2)) - 3*exp(6)/(x*exp(6) - 5*exp(6) - 16*exp(3)*log(2))

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