\(\int \frac {-4 x^2-10 x^3-3 x^4+(4 x+10 x^2+3 x^3) \log (5)+\log (5) \log (e^{6+4 x+5 x^2+x^3})}{x^2-2 x \log (5)+\log ^2(5)} \, dx\) [9435]

Optimal result

Integrand size = 68, antiderivative size = 25 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=\frac {x \log \left (e^{6+x+x (3+x (5+x))}\right )}{-x+\log (5)} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(234\) vs. \(2(25)=50\).

Time = 0.21 (sec) , antiderivative size = 234, normalized size of antiderivative = 9.36, number of steps used = 14, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {27, 6874, 45, 785, 2631, 712} \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-x^3-5 x^2-\frac {3}{2} x^2 \log (25)+3 x^2 \log (5)-\frac {\log (5) \log \left (e^{x^3+5 x^2+4 x+6}\right )}{x-\log (5)}-4 x+\frac {3 \log ^4(5)}{x-\log (5)}-12 \log ^3(5) \log (x-\log (5))+\frac {10 \log ^3(5)}{x-\log (5)}-9 x \log ^2(5)+\log (5) \left (4+3 \log ^2(5)+10 \log (5)\right ) \log (x-\log (5))-30 \log ^2(5) \log (x-\log (5))-\frac {\log ^2(5) \left (4+3 \log ^2(5)+10 \log (5)\right )}{x-\log (5)}+\frac {4 \log ^2(5)}{x-\log (5)}+x \log (5) (10+\log (125))+2 x \log (5) (5+\log (125))-10 x \log (25)-4 \log (25) \log (x-\log (5))+\log (5) (2+\log (5)) (2+9 \log (5)) \log (x-\log (5)) \]

[In]

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

[Out]

-4*x - 5*x^2 - x^3 + 3*x^2*Log[5] - 9*x*Log[5]^2 + (4*Log[5]^2)/(x - Log[5]) + (10*Log[5]^3)/(x - Log[5]) + (3
*Log[5]^4)/(x - Log[5]) - (Log[5]^2*(4 + 10*Log[5] + 3*Log[5]^2))/(x - Log[5]) - 10*x*Log[25] - (3*x^2*Log[25]
)/2 + 2*x*Log[5]*(5 + Log[125]) + x*Log[5]*(10 + Log[125]) - (Log[5]*Log[E^(6 + 4*x + 5*x^2 + x^3)])/(x - Log[
5]) - 30*Log[5]^2*Log[x - Log[5]] - 12*Log[5]^3*Log[x - Log[5]] + Log[5]*(2 + Log[5])*(2 + 9*Log[5])*Log[x - L
og[5]] + Log[5]*(4 + 10*Log[5] + 3*Log[5]^2)*Log[x - Log[5]] - 4*Log[25]*Log[x - Log[5]]

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 45

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 712

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

Rule 785

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 2631

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1)*(Log[u]/(b*(m + 1))), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[(a + b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{(x-\log (5))^2} \, dx \\ & = \int \left (-\frac {4 x^2}{(x-\log (5))^2}-\frac {10 x^3}{(x-\log (5))^2}-\frac {3 x^4}{(x-\log (5))^2}+\frac {x \left (4+10 x+3 x^2\right ) \log (5)}{(x-\log (5))^2}+\frac {\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{(x-\log (5))^2}\right ) \, dx \\ & = -\left (3 \int \frac {x^4}{(x-\log (5))^2} \, dx\right )-4 \int \frac {x^2}{(x-\log (5))^2} \, dx-10 \int \frac {x^3}{(x-\log (5))^2} \, dx+\log (5) \int \frac {x \left (4+10 x+3 x^2\right )}{(x-\log (5))^2} \, dx+\log (5) \int \frac {\log \left (e^{6+4 x+5 x^2+x^3}\right )}{(x-\log (5))^2} \, dx \\ & = -\frac {\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x-\log (5)}-3 \int \left (x^2+3 \log ^2(5)+\frac {4 \log ^3(5)}{x-\log (5)}+\frac {\log ^4(5)}{(x-\log (5))^2}+x \log (25)\right ) \, dx-4 \int \left (1+\frac {\log ^2(5)}{(x-\log (5))^2}+\frac {\log (25)}{x-\log (5)}\right ) \, dx-10 \int \left (x+\frac {3 \log ^2(5)}{x-\log (5)}+\frac {\log ^3(5)}{(x-\log (5))^2}+\log (25)\right ) \, dx+\log (5) \int \frac {4+10 x+3 x^2}{x-\log (5)} \, dx+\log (5) \int \left (3 x+\frac {(2+\log (5)) (2+9 \log (5))}{x-\log (5)}+\frac {\log (5) \left (4+10 \log (5)+3 \log ^2(5)\right )}{(x-\log (5))^2}+2 (5+\log (125))\right ) \, dx \\ & = -4 x-5 x^2-x^3+\frac {3}{2} x^2 \log (5)-9 x \log ^2(5)+\frac {4 \log ^2(5)}{x-\log (5)}+\frac {10 \log ^3(5)}{x-\log (5)}+\frac {3 \log ^4(5)}{x-\log (5)}-\frac {\log ^2(5) \left (4+10 \log (5)+3 \log ^2(5)\right )}{x-\log (5)}-10 x \log (25)-\frac {3}{2} x^2 \log (25)+2 x \log (5) (5+\log (125))-\frac {\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x-\log (5)}-30 \log ^2(5) \log (x-\log (5))-12 \log ^3(5) \log (x-\log (5))+\log (5) (2+\log (5)) (2+9 \log (5)) \log (x-\log (5))-4 \log (25) \log (x-\log (5))+\log (5) \int \left (3 x+10 \left (1+\frac {3 \log (5)}{10}\right )+\frac {4+10 \log (5)+3 \log ^2(5)}{x-\log (5)}\right ) \, dx \\ & = -4 x-5 x^2-x^3+3 x^2 \log (5)-9 x \log ^2(5)+\frac {4 \log ^2(5)}{x-\log (5)}+\frac {10 \log ^3(5)}{x-\log (5)}+\frac {3 \log ^4(5)}{x-\log (5)}-\frac {\log ^2(5) \left (4+10 \log (5)+3 \log ^2(5)\right )}{x-\log (5)}-10 x \log (25)-\frac {3}{2} x^2 \log (25)+2 x \log (5) (5+\log (125))+x \log (5) (10+\log (125))-\frac {\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x-\log (5)}-30 \log ^2(5) \log (x-\log (5))-12 \log ^3(5) \log (x-\log (5))+\log (5) (2+\log (5)) (2+9 \log (5)) \log (x-\log (5))+\log (5) \left (4+10 \log (5)+3 \log ^2(5)\right ) \log (x-\log (5))-4 \log (25) \log (x-\log (5)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-x \left (4+5 x+x^2\right )-\frac {\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x-\log (5)} \]

[In]

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

[Out]

-(x*(4 + 5*x + x^2)) - (Log[5]*Log[E^(6 + 4*x + 5*x^2 + x^3)])/(x - Log[5])

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
norman	\(\frac {x \ln \left ({\mathrm e}^{6} {\mathrm e}^{x^{3}+5 x^{2}+4 x}\right )}{\ln \left (5\right )-x}\)	\(30\)
parallelrisch	\(\frac {-x^{3} \ln \left (5\right )+x^{4}-5 x^{2} \ln \left (5\right )+5 x^{3}-4 \ln \left (5\right )^{2}+\ln \left (5\right ) \ln \left ({\mathrm e}^{6} {\mathrm e}^{x^{3}+5 x^{2}+4 x}\right )+4 x^{2}}{\ln \left (5\right )-x}\)	\(66\)
risch	\(\frac {\ln \left (5\right ) \ln \left ({\mathrm e}^{x \left (1+x \right ) \left (4+x \right )}\right )}{\ln \left (5\right )-x}-\frac {-2 x^{3} \ln \left (5\right )+2 x^{4}-10 x^{2} \ln \left (5\right )+10 x^{3}-8 x \ln \left (5\right )+8 x^{2}+12 \ln \left (5\right )}{2 \left (-\ln \left (5\right )+x \right )}\)	\(72\)
default	\(-3 x \ln \left (5\right )^{2}-\frac {3 x^{2} \ln \left (5\right )}{2}-x^{3}-10 x \ln \left (5\right )-5 x^{2}-4 x -\left (3 \ln \left (5\right )^{2}+10 \ln \left (5\right )+4\right ) \ln \left (5\right ) \ln \left (-\ln \left (5\right )+x \right )+\ln \left (5\right ) \left (-\frac {\ln \left ({\mathrm e}^{6} {\mathrm e}^{x^{3}+5 x^{2}+4 x}\right )}{-\ln \left (5\right )+x}+\frac {3 x^{2}}{2}+3 x \ln \left (5\right )+10 x +\left (3 \ln \left (5\right )^{2}+10 \ln \left (5\right )+4\right ) \ln \left (-\ln \left (5\right )+x \right )\right )\)	\(123\)
parts	\(-3 x \ln \left (5\right )^{2}-\frac {3 x^{2} \ln \left (5\right )}{2}-x^{3}-10 x \ln \left (5\right )-5 x^{2}-4 x -\left (3 \ln \left (5\right )^{2}+10 \ln \left (5\right )+4\right ) \ln \left (5\right ) \ln \left (-\ln \left (5\right )+x \right )+\ln \left (5\right ) \left (-\frac {\ln \left ({\mathrm e}^{6} {\mathrm e}^{x^{3}+5 x^{2}+4 x}\right )}{-\ln \left (5\right )+x}+\frac {3 x^{2}}{2}+3 x \ln \left (5\right )+10 x +\left (3 \ln \left (5\right )^{2}+10 \ln \left (5\right )+4\right ) \ln \left (-\ln \left (5\right )+x \right )\right )\)	\(123\)

[In]

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

[Out]

x*ln(exp(3)^2*exp(x^3+5*x^2+4*x))/(ln(5)-x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-\frac {x^{4} - {\left (x - 5\right )} \log \left (5\right )^{3} + \log \left (5\right )^{4} + 5 \, x^{3} - {\left (5 \, x - 4\right )} \log \left (5\right )^{2} + 4 \, x^{2} - 2 \, {\left (2 \, x - 3\right )} \log \left (5\right )}{x - \log \left (5\right )} \]

[In]

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

[Out]

-(x^4 - (x - 5)*log(5)^3 + log(5)^4 + 5*x^3 - (5*x - 4)*log(5)^2 + 4*x^2 - 2*(2*x - 3)*log(5))/(x - log(5))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).

Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=- x^{3} - x^{2} \left (\log {\left (5 \right )} + 5\right ) - x \left (\log {\left (5 \right )}^{2} + 4 + 5 \log {\left (5 \right )}\right ) - \frac {\log {\left (5 \right )}^{4} + 6 \log {\left (5 \right )} + 4 \log {\left (5 \right )}^{2} + 5 \log {\left (5 \right )}^{3}}{x - \log {\left (5 \right )}} \]

[In]

integrate((ln(5)*ln(exp(3)**2*exp(x**3+5*x**2+4*x))+(3*x**3+10*x**2+4*x)*ln(5)-3*x**4-10*x**3-4*x**2)/(ln(5)**
2-2*x*ln(5)+x**2),x)

[Out]

-x**3 - x**2*(log(5) + 5) - x*(log(5)**2 + 4 + 5*log(5)) - (log(5)**4 + 6*log(5) + 4*log(5)**2 + 5*log(5)**3)/
(x - log(5))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.36 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-x^{3} - x^{2} {\left (\log \left (5\right ) + 5\right )} - {\left (\log \left (5\right )^{2} + 5 \, \log \left (5\right ) + 4\right )} x - \frac {\log \left (5\right )^{4} + 5 \, \log \left (5\right )^{3} + 4 \, \log \left (5\right )^{2} + 6 \, \log \left (5\right )}{x - \log \left (5\right )} \]

[In]

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

[Out]

-x^3 - x^2*(log(5) + 5) - (log(5)^2 + 5*log(5) + 4)*x - (log(5)^4 + 5*log(5)^3 + 4*log(5)^2 + 6*log(5))/(x - l
og(5))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (25) = 50\).

Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.56 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-x^{3} - x^{2} \log \left (5\right ) - x \log \left (5\right )^{2} - 5 \, x^{2} - 5 \, x \log \left (5\right ) - 4 \, x - \frac {\log \left (5\right )^{4} + 5 \, \log \left (5\right )^{3} + 4 \, \log \left (5\right )^{2} + 6 \, \log \left (5\right )}{x - \log \left (5\right )} \]

[In]

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

[Out]

-x^3 - x^2*log(5) - x*log(5)^2 - 5*x^2 - 5*x*log(5) - 4*x - (log(5)^4 + 5*log(5)^3 + 4*log(5)^2 + 6*log(5))/(x
- log(5))

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.16 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=x\,\left (15\,\ln \left (5\right )-2\,\ln \left (5\right )\,\left (6\,\ln \left (5\right )-\ln \left (625\right )+10\right )+3\,{\ln \left (5\right )}^2-4\right )-\frac {6\,\ln \left (5\right )+\ln \left (5\right )\,\ln \left (625\right )+5\,{\ln \left (5\right )}^3+{\ln \left (5\right )}^4}{x-\ln \left (5\right )}-x^3-x^2\,\left (3\,\ln \left (5\right )-\frac {\ln \left (625\right )}{2}+5\right ) \]

[In]

int(-(4*x^2 - log(5)*log(exp(6)*exp(4*x + 5*x^2 + x^3)) - log(5)*(4*x + 10*x^2 + 3*x^3) + 10*x^3 + 3*x^4)/(log
(5)^2 - 2*x*log(5) + x^2),x)

[Out]

x*(15*log(5) - 2*log(5)*(6*log(5) - log(625) + 10) + 3*log(5)^2 - 4) - (6*log(5) + log(5)*log(625) + 5*log(5)^
3 + log(5)^4)/(x - log(5)) - x^3 - x^2*(3*log(5) - log(625)/2 + 5)