3.22.54 \(\int \frac {-4 x-15 x^3+(4+x+16 x^2) \log (2)+(-5 x+5 \log (2)) \log (5)+(-x^3+x^2 \log (2)) \log (-\frac {4 x}{-x+\log (2)})}{-x^3+x^2 \log (2)} \, dx\)

Optimal. Leaf size=33 \[ (1+x) \left (5-\frac {4+5 (-2 x+\log (5))}{x}+\log \left (\frac {4 x}{x-\log (2)}\right )\right ) \]

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Rubi [A]  time = 0.43, antiderivative size = 54, normalized size of antiderivative = 1.64, number of steps used = 7, number of rules used = 5, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {1593, 6742, 1620, 2486, 31} \begin {gather*} 15 x+x \log \left (\frac {4 x}{x-\log (2)}\right )+\log (x)-(1+\log (2)) \log (x-\log (2))+\log (2) \log (x-\log (2))-\frac {4+5 \log (5)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*x - 15*x^3 + (4 + x + 16*x^2)*Log[2] + (-5*x + 5*Log[2])*Log[5] + (-x^3 + x^2*Log[2])*Log[(-4*x)/(-x +
Log[2])])/(-x^3 + x^2*Log[2]),x]

[Out]

15*x - (4 + 5*Log[5])/x + Log[x] + x*Log[(4*x)/(x - Log[2])] + Log[2]*Log[x - Log[2]] - (1 + Log[2])*Log[x - L
og[2]]

Rule 31

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

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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2486

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p*
(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] &&
EqQ[p + q, 0] && IGtQ[s, 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 {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{x^2 (-x+\log (2))} \, dx\\ &=\int \left (\frac {15 x^3-16 x^2 \log (2)-\log (2) (4+5 \log (5))+x \left (4+\log \left (\frac {3125}{2}\right )\right )}{x^2 (x-\log (2))}+\log \left (\frac {4 x}{x-\log (2)}\right )\right ) \, dx\\ &=\int \frac {15 x^3-16 x^2 \log (2)-\log (2) (4+5 \log (5))+x \left (4+\log \left (\frac {3125}{2}\right )\right )}{x^2 (x-\log (2))} \, dx+\int \log \left (\frac {4 x}{x-\log (2)}\right ) \, dx\\ &=x \log \left (\frac {4 x}{x-\log (2)}\right )+\log (2) \int \frac {1}{x-\log (2)} \, dx+\int \left (15+\frac {1}{x}+\frac {-1-\log (2)}{x-\log (2)}+\frac {4+5 \log (5)}{x^2}\right ) \, dx\\ &=15 x-\frac {4+5 \log (5)}{x}+\log (x)+x \log \left (\frac {4 x}{x-\log (2)}\right )+\log (2) \log (x-\log (2))-(1+\log (2)) \log (x-\log (2))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 41, normalized size = 1.24 \begin {gather*} -\frac {4}{x}+15 x-\frac {5 \log (5)}{x}+\log (x)+x \log \left (\frac {4 x}{x-\log (2)}\right )-\log (x-\log (2)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x - 15*x^3 + (4 + x + 16*x^2)*Log[2] + (-5*x + 5*Log[2])*Log[5] + (-x^3 + x^2*Log[2])*Log[(-4*x)
/(-x + Log[2])])/(-x^3 + x^2*Log[2]),x]

[Out]

-4/x + 15*x - (5*Log[5])/x + Log[x] + x*Log[(4*x)/(x - Log[2])] - Log[x - Log[2]]

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fricas [A]  time = 0.54, size = 33, normalized size = 1.00 \begin {gather*} \frac {15 \, x^{2} + {\left (x^{2} + x\right )} \log \left (\frac {4 \, x}{x - \log \relax (2)}\right ) - 5 \, \log \relax (5) - 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*log(2)-x^3)*log(-4*x/(log(2)-x))+(5*log(2)-5*x)*log(5)+(16*x^2+x+4)*log(2)-15*x^3-4*x)/(x^2*lo
g(2)-x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 45, normalized size = 1.36 \begin {gather*} x {\left (2 \, \log \relax (2) + 15\right )} - x \log \left (x - \log \relax (2)\right ) + x \log \relax (x) - \frac {5 \, \log \relax (5) + 4}{x} - \log \left (x - \log \relax (2)\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*log(2)-x^3)*log(-4*x/(log(2)-x))+(5*log(2)-5*x)*log(5)+(16*x^2+x+4)*log(2)-15*x^3-4*x)/(x^2*lo
g(2)-x^3),x, algorithm="giac")

[Out]

x*(2*log(2) + 15) - x*log(x - log(2)) + x*log(x) - (5*log(5) + 4)/x - log(x - log(2)) + log(x)

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maple [A]  time = 0.43, size = 45, normalized size = 1.36

method	result	size
risch	\(x \ln \left (-\frac {4 x}{\ln \relax (2)-x}\right )+\frac {x \ln \relax (x )-\ln \left (x -\ln \relax (2)\right ) x +15 x^{2}-5 \ln \relax (5)-4}{x}\)	\(45\)
norman	\(\frac {x^{2} \ln \left (-\frac {4 x}{\ln \relax (2)-x}\right )+x \ln \left (-\frac {4 x}{\ln \relax (2)-x}\right )+15 x^{2}-5 \ln \relax (5)-4}{x}\)	\(46\)
derivativedivides	\(-4 \ln \relax (2) \left (-\frac {\ln \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right ) \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right ) \left (x -\ln \relax (2)\right )}{16 \ln \relax (2)}-\frac {4}{\ln \relax (2)^{2} \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}-\frac {5 \ln \relax (5)}{\ln \relax (2)^{2} \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}-\frac {\ln \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}{4 \ln \relax (2)}-\frac {15 \left (x -\ln \relax (2)\right )}{4 \ln \relax (2)}\right )\)	\(126\)
default	\(-4 \ln \relax (2) \left (-\frac {\ln \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right ) \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right ) \left (x -\ln \relax (2)\right )}{16 \ln \relax (2)}-\frac {4}{\ln \relax (2)^{2} \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}-\frac {5 \ln \relax (5)}{\ln \relax (2)^{2} \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}-\frac {\ln \left (4+\frac {4 \ln \relax (2)}{x -\ln \relax (2)}\right )}{4 \ln \relax (2)}-\frac {15 \left (x -\ln \relax (2)\right )}{4 \ln \relax (2)}\right )\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*ln(2)-x^3)*ln(-4*x/(ln(2)-x))+(5*ln(2)-5*x)*ln(5)+(16*x^2+x+4)*ln(2)-15*x^3-4*x)/(x^2*ln(2)-x^3),x,m
ethod=_RETURNVERBOSE)

[Out]

x*ln(-4*x/(ln(2)-x))+(x*ln(x)-ln(x-ln(2))*x+15*x^2-5*ln(5)-4)/x

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maxima [B]  time = 1.39, size = 178, normalized size = 5.39 \begin {gather*} -5 \, {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)^{2}} - \frac {\log \relax (x)}{\log \relax (2)^{2}} + \frac {1}{x \log \relax (2)}\right )} \log \relax (5) \log \relax (2) + 5 \, {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)} - \frac {\log \relax (x)}{\log \relax (2)}\right )} \log \relax (5) + 2 \, x \log \relax (2) - {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)} - \frac {\log \relax (x)}{\log \relax (2)}\right )} \log \relax (2) - 4 \, {\left (\frac {\log \left (x - \log \relax (2)\right )}{\log \relax (2)^{2}} - \frac {\log \relax (x)}{\log \relax (2)^{2}} + \frac {1}{x \log \relax (2)}\right )} \log \relax (2) - {\left (x - \log \relax (2)\right )} \log \left (x - \log \relax (2)\right ) - \log \relax (2) \log \left (x - \log \relax (2)\right ) + x \log \relax (x) + 15 \, x + \frac {4 \, \log \left (x - \log \relax (2)\right )}{\log \relax (2)} - \frac {4 \, \log \relax (x)}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*log(2)-x^3)*log(-4*x/(log(2)-x))+(5*log(2)-5*x)*log(5)+(16*x^2+x+4)*log(2)-15*x^3-4*x)/(x^2*lo
g(2)-x^3),x, algorithm="maxima")

[Out]

-5*(log(x - log(2))/log(2)^2 - log(x)/log(2)^2 + 1/(x*log(2)))*log(5)*log(2) + 5*(log(x - log(2))/log(2) - log
(x)/log(2))*log(5) + 2*x*log(2) - (log(x - log(2))/log(2) - log(x)/log(2))*log(2) - 4*(log(x - log(2))/log(2)^
2 - log(x)/log(2)^2 + 1/(x*log(2)))*log(2) - (x - log(2))*log(x - log(2)) - log(2)*log(x - log(2)) + x*log(x)
+ 15*x + 4*log(x - log(2))/log(2) - 4*log(x)/log(2)

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mupad [B]  time = 1.78, size = 49, normalized size = 1.48 \begin {gather*} 15\,x+x\,\ln \left (\frac {x}{x-\ln \relax (2)}\right )+2\,x\,\ln \relax (2)-\frac {5\,\ln \relax (5)}{x}-\frac {4}{x}-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{\ln \relax (2)}-\mathrm {i}\right )\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x + log(5)*(5*x - 5*log(2)) + 15*x^3 - log(2)*(x + 16*x^2 + 4) - log((4*x)/(x - log(2)))*(x^2*log(2) -
x^3))/(x^2*log(2) - x^3),x)

[Out]

15*x - atan((x*2i)/log(2) - 1i)*2i + x*log(x/(x - log(2))) + 2*x*log(2) - (5*log(5))/x - 4/x

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sympy [A]  time = 0.38, size = 36, normalized size = 1.09 \begin {gather*} x \log {\left (- \frac {4 x}{- x + \log {\relax (2 )}} \right )} + 15 x + \log {\relax (x )} - \log {\left (x - \log {\relax (2 )} \right )} + \frac {- 5 \log {\relax (5 )} - 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2*ln(2)-x**3)*ln(-4*x/(ln(2)-x))+(5*ln(2)-5*x)*ln(5)+(16*x**2+x+4)*ln(2)-15*x**3-4*x)/(x**2*ln(
2)-x**3),x)

[Out]

x*log(-4*x/(-x + log(2))) + 15*x + log(x) - log(x - log(2)) + (-5*log(5) - 4)/x

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