3.54.6 \(\int \frac {-4-16 e^{16}+(-4-16 e^{16}+16 x) \log (\frac {1}{4} (-1-4 e^{16}+4 x))}{x^2+4 e^{16} x^2-4 x^3} \, dx\)

Optimal. Leaf size=19 \[ \frac {4 \left (1+x+\log \left (-\frac {1}{4}-e^{16}+x\right )\right )}{x} \]

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Rubi [A]  time = 0.41, antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 11, number of rules used = 9, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6, 1593, 6741, 6742, 44, 2395, 36, 29, 31} \begin {gather*} \frac {4}{x}+\frac {4 \log \left (x+\frac {1}{4} \left (-1-4 e^{16}\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 16*E^16 + (-4 - 16*E^16 + 16*x)*Log[(-1 - 4*E^16 + 4*x)/4])/(x^2 + 4*E^16*x^2 - 4*x^3),x]

[Out]

4/x + (4*Log[(-1 - 4*E^16)/4 + x])/x

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1593

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

Rule 2395

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

Rule 6741

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

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 {-4-16 e^{16}+\left (-4-16 e^{16}+16 x\right ) \log \left (\frac {1}{4} \left (-1-4 e^{16}+4 x\right )\right )}{\left (1+4 e^{16}\right ) x^2-4 x^3} \, dx\\ &=\int \frac {-4-16 e^{16}+\left (-4-16 e^{16}+16 x\right ) \log \left (\frac {1}{4} \left (-1-4 e^{16}+4 x\right )\right )}{\left (1+4 e^{16}-4 x\right ) x^2} \, dx\\ &=\int \frac {-4 \left (1+4 e^{16}\right )+\left (-4-16 e^{16}+16 x\right ) \log \left (\frac {1}{4} \left (-1-4 e^{16}+4 x\right )\right )}{\left (1+4 e^{16}-4 x\right ) x^2} \, dx\\ &=\int \left (-\frac {4 \left (1+4 e^{16}\right )}{\left (1+4 e^{16}-4 x\right ) x^2}-\frac {4 \log \left (-\frac {1}{4}-e^{16}+x\right )}{x^2}\right ) \, dx\\ &=-\left (4 \int \frac {\log \left (-\frac {1}{4}-e^{16}+x\right )}{x^2} \, dx\right )-\left (4 \left (1+4 e^{16}\right )\right ) \int \frac {1}{\left (1+4 e^{16}-4 x\right ) x^2} \, dx\\ &=\frac {4 \log \left (\frac {1}{4} \left (-1-4 e^{16}\right )+x\right )}{x}-4 \int \frac {1}{x \left (-\frac {1}{4}-e^{16}+x\right )} \, dx-\left (4 \left (1+4 e^{16}\right )\right ) \int \left (\frac {16}{\left (1+4 e^{16}\right )^2 \left (1+4 e^{16}-4 x\right )}+\frac {1}{\left (1+4 e^{16}\right ) x^2}+\frac {4}{\left (1+4 e^{16}\right )^2 x}\right ) \, dx\\ &=\frac {4}{x}+\frac {16 \log \left (1+4 e^{16}-4 x\right )}{1+4 e^{16}}-\frac {16 \log (x)}{1+4 e^{16}}+\frac {4 \log \left (\frac {1}{4} \left (-1-4 e^{16}\right )+x\right )}{x}+\frac {16 \int \frac {1}{x} \, dx}{1+4 e^{16}}-\frac {16 \int \frac {1}{-\frac {1}{4}-e^{16}+x} \, dx}{1+4 e^{16}}\\ &=\frac {4}{x}+\frac {4 \log \left (\frac {1}{4} \left (-1-4 e^{16}\right )+x\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 18, normalized size = 0.95 \begin {gather*} \frac {4 \left (1+\log \left (-\frac {1}{4}-e^{16}+x\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 16*E^16 + (-4 - 16*E^16 + 16*x)*Log[(-1 - 4*E^16 + 4*x)/4])/(x^2 + 4*E^16*x^2 - 4*x^3),x]

[Out]

(4*(1 + Log[-1/4 - E^16 + x]))/x

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fricas [A]  time = 0.94, size = 15, normalized size = 0.79 \begin {gather*} \frac {4 \, {\left (\log \left (x - e^{16} - \frac {1}{4}\right ) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*exp(8)^2+16*x-4)*log(-exp(8)^2+x-1/4)-16*exp(8)^2-4)/(4*x^2*exp(8)^2-4*x^3+x^2),x, algorithm="
fricas")

[Out]

4*(log(x - e^16 - 1/4) + 1)/x

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giac [A]  time = 0.98, size = 15, normalized size = 0.79 \begin {gather*} \frac {4 \, {\left (\log \left (x - e^{16} - \frac {1}{4}\right ) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*exp(8)^2+16*x-4)*log(-exp(8)^2+x-1/4)-16*exp(8)^2-4)/(4*x^2*exp(8)^2-4*x^3+x^2),x, algorithm="
giac")

[Out]

4*(log(x - e^16 - 1/4) + 1)/x

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maple [A]  time = 0.22, size = 19, normalized size = 1.00

method	result	size
norman	\(\frac {4+4 \ln \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right )}{x}\)	\(19\)
risch	\(\frac {4 \ln \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right )}{x}+\frac {4}{x}\)	\(20\)
derivativedivides	\(\frac {16 \ln \left (4 x \right )}{4 \,{\mathrm e}^{16}+1}-\frac {16 \ln \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right ) \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right )}{\left (4 \,{\mathrm e}^{16}+1\right ) x}+\frac {64 \ln \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right ) {\mathrm e}^{16}}{\left (4 \,{\mathrm e}^{16}+1\right )^{2}}+\frac {16 \ln \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right )}{\left (4 \,{\mathrm e}^{16}+1\right )^{2}}-\frac {64 \ln \left (4 x \right ) {\mathrm e}^{16}}{\left (4 \,{\mathrm e}^{16}+1\right )^{2}}-\frac {16 \ln \left (4 x \right )}{\left (4 \,{\mathrm e}^{16}+1\right )^{2}}+\frac {16 \,{\mathrm e}^{16}}{\left (4 \,{\mathrm e}^{16}+1\right ) x}+\frac {4}{\left (4 \,{\mathrm e}^{16}+1\right ) x}\)	\(226\)
default	\(\frac {16 \ln \left (4 x \right )}{4 \,{\mathrm e}^{16}+1}-\frac {16 \ln \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right ) \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right )}{\left (4 \,{\mathrm e}^{16}+1\right ) x}+\frac {64 \ln \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right ) {\mathrm e}^{16}}{\left (4 \,{\mathrm e}^{16}+1\right )^{2}}+\frac {16 \ln \left (-{\mathrm e}^{16}+x -\frac {1}{4}\right )}{\left (4 \,{\mathrm e}^{16}+1\right )^{2}}-\frac {64 \ln \left (4 x \right ) {\mathrm e}^{16}}{\left (4 \,{\mathrm e}^{16}+1\right )^{2}}-\frac {16 \ln \left (4 x \right )}{\left (4 \,{\mathrm e}^{16}+1\right )^{2}}+\frac {16 \,{\mathrm e}^{16}}{\left (4 \,{\mathrm e}^{16}+1\right ) x}+\frac {4}{\left (4 \,{\mathrm e}^{16}+1\right ) x}\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*exp(8)^2+16*x-4)*ln(-exp(8)^2+x-1/4)-16*exp(8)^2-4)/(4*x^2*exp(8)^2-4*x^3+x^2),x,method=_RETURNVERBO
SE)

[Out]

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

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maxima [B]  time = 0.52, size = 167, normalized size = 8.79 \begin {gather*} 16 \, {\left (\frac {4 \, \log \left (4 \, x - 4 \, e^{16} - 1\right )}{16 \, e^{32} + 8 \, e^{16} + 1} - \frac {4 \, \log \relax (x)}{16 \, e^{32} + 8 \, e^{16} + 1} + \frac {1}{x {\left (4 \, e^{16} + 1\right )}}\right )} e^{16} + \frac {16 \, \log \left (4 \, x - 4 \, e^{16} - 1\right )}{16 \, e^{32} + 8 \, e^{16} + 1} - \frac {16 \, \log \relax (x)}{16 \, e^{32} + 8 \, e^{16} + 1} + \frac {16 \, \log \relax (x)}{4 \, e^{16} + 1} - \frac {4 \, {\left (8 \, e^{16} \log \relax (2) + {\left (4 \, x - 4 \, e^{16} - 1\right )} \log \left (4 \, x - 4 \, e^{16} - 1\right ) + 2 \, \log \relax (2)\right )}}{x {\left (4 \, e^{16} + 1\right )}} + \frac {4}{x {\left (4 \, e^{16} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*exp(8)^2+16*x-4)*log(-exp(8)^2+x-1/4)-16*exp(8)^2-4)/(4*x^2*exp(8)^2-4*x^3+x^2),x, algorithm="
maxima")

[Out]

16*(4*log(4*x - 4*e^16 - 1)/(16*e^32 + 8*e^16 + 1) - 4*log(x)/(16*e^32 + 8*e^16 + 1) + 1/(x*(4*e^16 + 1)))*e^1
6 + 16*log(4*x - 4*e^16 - 1)/(16*e^32 + 8*e^16 + 1) - 16*log(x)/(16*e^32 + 8*e^16 + 1) + 16*log(x)/(4*e^16 + 1
) - 4*(8*e^16*log(2) + (4*x - 4*e^16 - 1)*log(4*x - 4*e^16 - 1) + 2*log(2))/(x*(4*e^16 + 1)) + 4/(x*(4*e^16 +
1))

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mupad [B]  time = 3.95, size = 16, normalized size = 0.84 \begin {gather*} \frac {4\,\ln \left (x-{\mathrm {e}}^{16}-\frac {1}{4}\right )+4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(16) + log(x - exp(16) - 1/4)*(16*exp(16) - 16*x + 4) + 4)/(4*x^2*exp(16) + x^2 - 4*x^3),x)

[Out]

(4*log(x - exp(16) - 1/4) + 4)/x

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sympy [A]  time = 0.16, size = 15, normalized size = 0.79 \begin {gather*} \frac {4 \log {\left (x - e^{16} - \frac {1}{4} \right )}}{x} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*exp(8)**2+16*x-4)*ln(-exp(8)**2+x-1/4)-16*exp(8)**2-4)/(4*x**2*exp(8)**2-4*x**3+x**2),x)

[Out]

4*log(x - exp(16) - 1/4)/x + 4/x

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