3.47.45 \(\int \frac {-11 x+4 \log ^2(-5+e^5)+x \log (\frac {x^4}{16})}{x^3-2 x^2 \log ^2(-5+e^5)+x \log ^4(-5+e^5)} \, dx\)

Optimal. Leaf size=25 \[ \frac {-7+\log \left (\frac {x^4}{16}\right )}{-x+\log ^2\left (-5+e^5\right )} \]

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Rubi [B]  time = 0.22, antiderivative size = 62, normalized size of antiderivative = 2.48, number of steps used = 8, number of rules used = 6, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1594, 27, 6742, 77, 2314, 31} \begin {gather*} -\frac {x \log \left (\frac {x^4}{16}\right )}{\log ^2\left (e^5-5\right ) \left (x-\log ^2\left (e^5-5\right )\right )}+\frac {4 \log (x)}{\log ^2\left (e^5-5\right )}+\frac {7}{x-\log ^2\left (e^5-5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-11*x + 4*Log[-5 + E^5]^2 + x*Log[x^4/16])/(x^3 - 2*x^2*Log[-5 + E^5]^2 + x*Log[-5 + E^5]^4),x]

[Out]

7/(x - Log[-5 + E^5]^2) + (4*Log[x])/Log[-5 + E^5]^2 - (x*Log[x^4/16])/(Log[-5 + E^5]^2*(x - Log[-5 + E^5]^2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 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 {-11 x+4 \log ^2\left (-5+e^5\right )+x \log \left (\frac {x^4}{16}\right )}{x \left (x^2-2 x \log ^2\left (-5+e^5\right )+\log ^4\left (-5+e^5\right )\right )} \, dx\\ &=\int \frac {-11 x+4 \log ^2\left (-5+e^5\right )+x \log \left (\frac {x^4}{16}\right )}{x \left (x-\log ^2\left (-5+e^5\right )\right )^2} \, dx\\ &=\int \left (\frac {-11 x+4 \log ^2\left (-5+e^5\right )}{x \left (x-\log ^2\left (-5+e^5\right )\right )^2}+\frac {\log \left (\frac {x^4}{16}\right )}{\left (x-\log ^2\left (-5+e^5\right )\right )^2}\right ) \, dx\\ &=\int \frac {-11 x+4 \log ^2\left (-5+e^5\right )}{x \left (x-\log ^2\left (-5+e^5\right )\right )^2} \, dx+\int \frac {\log \left (\frac {x^4}{16}\right )}{\left (x-\log ^2\left (-5+e^5\right )\right )^2} \, dx\\ &=-\frac {x \log \left (\frac {x^4}{16}\right )}{\log ^2\left (-5+e^5\right ) \left (x-\log ^2\left (-5+e^5\right )\right )}+\frac {4 \int \frac {1}{x-\log ^2\left (-5+e^5\right )} \, dx}{\log ^2\left (-5+e^5\right )}+\int \left (\frac {4}{x \log ^2\left (-5+e^5\right )}-\frac {7}{\left (x-\log ^2\left (-5+e^5\right )\right )^2}-\frac {4}{\log ^2\left (-5+e^5\right ) \left (x-\log ^2\left (-5+e^5\right )\right )}\right ) \, dx\\ &=\frac {7}{x-\log ^2\left (-5+e^5\right )}+\frac {4 \log (x)}{\log ^2\left (-5+e^5\right )}-\frac {x \log \left (\frac {x^4}{16}\right )}{\log ^2\left (-5+e^5\right ) \left (x-\log ^2\left (-5+e^5\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 1.00 \begin {gather*} \frac {7+\log (16)-\log \left (x^4\right )}{x-\log ^2\left (-5+e^5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-11*x + 4*Log[-5 + E^5]^2 + x*Log[x^4/16])/(x^3 - 2*x^2*Log[-5 + E^5]^2 + x*Log[-5 + E^5]^4),x]

[Out]

(7 + Log[16] - Log[x^4])/(x - Log[-5 + E^5]^2)

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fricas [A]  time = 0.64, size = 22, normalized size = 0.88 \begin {gather*} \frac {\log \left (\frac {1}{16} \, x^{4}\right ) - 7}{\log \left (e^{5} - 5\right )^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(exp(5)-5)^2+x*log(1/16*x^4)-11*x)/(x*log(exp(5)-5)^4-2*x^2*log(exp(5)-5)^2+x^3),x, algorithm=
"fricas")

[Out]

(log(1/16*x^4) - 7)/(log(e^5 - 5)^2 - x)

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giac [A]  time = 1.20, size = 27, normalized size = 1.08 \begin {gather*} -\frac {4 \, \log \relax (2) - \log \left (x^{4}\right ) + 7}{\log \left (e^{5} - 5\right )^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(exp(5)-5)^2+x*log(1/16*x^4)-11*x)/(x*log(exp(5)-5)^4-2*x^2*log(exp(5)-5)^2+x^3),x, algorithm=
"giac")

[Out]

-(4*log(2) - log(x^4) + 7)/(log(e^5 - 5)^2 - x)

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maple [A]  time = 0.11, size = 23, normalized size = 0.92

method	result	size
norman	\(\frac {-7+\ln \left (\frac {x^{4}}{16}\right )}{\ln \left ({\mathrm e}^{5}-5\right )^{2}-x}\)	\(23\)
risch	\(\frac {\ln \left (\frac {x^{4}}{16}\right )}{\ln \left ({\mathrm e}^{5}-5\right )^{2}-x}-\frac {7}{\ln \left ({\mathrm e}^{5}-5\right )^{2}-x}\)	\(37\)
default	\(\frac {x \ln \left (x^{4}\right )}{\ln \left ({\mathrm e}^{5}-5\right )^{2} \left (\ln \left ({\mathrm e}^{5}-5\right )^{2}-x \right )}+\frac {4 \ln \left (\ln \left ({\mathrm e}^{5}-5\right )^{2}-x \right )}{\ln \left ({\mathrm e}^{5}-5\right )^{2}}+\frac {7}{-\ln \left ({\mathrm e}^{5}-5\right )^{2}+x}-\frac {4 \ln \left (-\ln \left ({\mathrm e}^{5}-5\right )^{2}+x \right )}{\ln \left ({\mathrm e}^{5}-5\right )^{2}}+\frac {4 \ln \relax (x )}{\ln \left ({\mathrm e}^{5}-5\right )^{2}}+\frac {4 \ln \relax (2)}{-\ln \left ({\mathrm e}^{5}-5\right )^{2}+x}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(exp(5)-5)^2+x*ln(1/16*x^4)-11*x)/(x*ln(exp(5)-5)^4-2*x^2*ln(exp(5)-5)^2+x^3),x,method=_RETURNVERBOSE
)

[Out]

(-7+ln(1/16*x^4))/(ln(exp(5)-5)^2-x)

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maxima [B]  time = 0.37, size = 129, normalized size = 5.16 \begin {gather*} 4 \, {\left (\frac {1}{\log \left (e^{5} - 5\right )^{4} - x \log \left (e^{5} - 5\right )^{2}} - \frac {\log \left (-\log \left (e^{5} - 5\right )^{2} + x\right )}{\log \left (e^{5} - 5\right )^{4}} + \frac {\log \relax (x)}{\log \left (e^{5} - 5\right )^{4}}\right )} \log \left (e^{5} - 5\right )^{2} + \frac {\log \left (\frac {1}{16} \, x^{4}\right )}{\log \left (e^{5} - 5\right )^{2} - x} - \frac {11}{\log \left (e^{5} - 5\right )^{2} - x} + \frac {4 \, \log \left (-\log \left (e^{5} - 5\right )^{2} + x\right )}{\log \left (e^{5} - 5\right )^{2}} - \frac {4 \, \log \relax (x)}{\log \left (e^{5} - 5\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(exp(5)-5)^2+x*log(1/16*x^4)-11*x)/(x*log(exp(5)-5)^4-2*x^2*log(exp(5)-5)^2+x^3),x, algorithm=
"maxima")

[Out]

4*(1/(log(e^5 - 5)^4 - x*log(e^5 - 5)^2) - log(-log(e^5 - 5)^2 + x)/log(e^5 - 5)^4 + log(x)/log(e^5 - 5)^4)*lo
g(e^5 - 5)^2 + log(1/16*x^4)/(log(e^5 - 5)^2 - x) - 11/(log(e^5 - 5)^2 - x) + 4*log(-log(e^5 - 5)^2 + x)/log(e
^5 - 5)^2 - 4*log(x)/log(e^5 - 5)^2

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mupad [B]  time = 3.44, size = 23, normalized size = 0.92 \begin {gather*} -\frac {\ln \left (\frac {x^4}{16}\right )-7}{x-{\ln \left ({\mathrm {e}}^5-5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(x^4/16) - 11*x + 4*log(exp(5) - 5)^2)/(x*log(exp(5) - 5)^4 - 2*x^2*log(exp(5) - 5)^2 + x^3),x)

[Out]

-(log(x^4/16) - 7)/(x - log(exp(5) - 5)^2)

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sympy [A]  time = 0.22, size = 27, normalized size = 1.08 \begin {gather*} - \frac {\log {\left (\frac {x^{4}}{16} \right )}}{x - \log {\left (-5 + e^{5} \right )}^{2}} + \frac {7}{x - \log {\left (-5 + e^{5} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(exp(5)-5)**2+x*ln(1/16*x**4)-11*x)/(x*ln(exp(5)-5)**4-2*x**2*ln(exp(5)-5)**2+x**3),x)

[Out]

-log(x**4/16)/(x - log(-5 + exp(5))**2) + 7/(x - log(-5 + exp(5))**2)

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