3.3.70 \(\int \frac {e^{10} (6-3 x)-3 x+7 x^2-3 x^3+e^5 (-3+16 x-7 x^2)+(3 x-x^2) \log (x)}{-3 x^3+x^4+e^{10} (-3 x+x^2)+e^5 (-6 x^2+2 x^3)} \, dx\)

Optimal. Leaf size=28 \[ -2+\log (2)+\frac {-x+\log (x)}{e^5+x}-\log \left ((-3+x) x^2\right ) \]

________________________________________________________________________________________

Rubi [B]  time = 1.02, antiderivative size = 246, normalized size of antiderivative = 8.79, number of steps used = 11, number of rules used = 7, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6688, 6742, 148, 893, 1612, 2314, 31} \begin {gather*} \frac {3+16 e^5+7 e^{10}}{\left (3+e^5\right ) \left (x+e^5\right )}-\frac {3+7 e^5+3 e^{10}}{\left (3+e^5\right ) \left (x+e^5\right )}-\frac {3 e^5 \left (2+e^5\right )}{\left (3+e^5\right ) \left (x+e^5\right )}-\frac {e^{10} \log (3-x)}{\left (3+e^5\right )^2}-\frac {6 e^5 \log (3-x)}{\left (3+e^5\right )^2}-\frac {9 \log (3-x)}{\left (3+e^5\right )^2}-\frac {x \log (x)}{e^5 \left (x+e^5\right )}+\frac {\log (x)}{e^5}-2 \log (x)-\frac {3 \left (6+6 e^5+e^{10}\right ) \log \left (x+e^5\right )}{\left (3+e^5\right )^2}+\frac {3 \left (6+4 e^5+e^{10}\right ) \log \left (x+e^5\right )}{\left (3+e^5\right )^2}-\frac {\left (9+6 e^5-5 e^{10}\right ) \log \left (x+e^5\right )}{e^5 \left (3+e^5\right )^2}+\frac {\log \left (x+e^5\right )}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^10*(6 - 3*x) - 3*x + 7*x^2 - 3*x^3 + E^5*(-3 + 16*x - 7*x^2) + (3*x - x^2)*Log[x])/(-3*x^3 + x^4 + E^10
*(-3*x + x^2) + E^5*(-6*x^2 + 2*x^3)),x]

[Out]

(-3*E^5*(2 + E^5))/((3 + E^5)*(E^5 + x)) - (3 + 7*E^5 + 3*E^10)/((3 + E^5)*(E^5 + x)) + (3 + 16*E^5 + 7*E^10)/
((3 + E^5)*(E^5 + x)) - (9*Log[3 - x])/(3 + E^5)^2 - (6*E^5*Log[3 - x])/(3 + E^5)^2 - (E^10*Log[3 - x])/(3 + E
^5)^2 - 2*Log[x] + Log[x]/E^5 - (x*Log[x])/(E^5*(E^5 + x)) + Log[E^5 + x]/E^5 - ((9 + 6*E^5 - 5*E^10)*Log[E^5
+ x])/(E^5*(3 + E^5)^2) + (3*(6 + 4*E^5 + E^10)*Log[E^5 + x])/(3 + E^5)^2 - (3*(6 + 6*E^5 + E^10)*Log[E^5 + x]
)/(3 + E^5)^2

Rule 31

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

Rule 148

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g
, h, m}, x] && (IntegersQ[m, n, p] || (IGtQ[n, 0] && IGtQ[p, 0]))

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 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

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 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 {3 e^{10} (-2+x)-e^5 \left (-3+16 x-7 x^2\right )-x \left (-3+7 x-3 x^2\right )+(-3+x) x \log (x)}{(3-x) x \left (e^5+x\right )^2} \, dx\\ &=\int \left (-\frac {3 e^{10} (-2+x)}{(-3+x) x \left (e^5+x\right )^2}+\frac {-3+7 x-3 x^2}{(-3+x) \left (e^5+x\right )^2}-\frac {e^5 \left (3-16 x+7 x^2\right )}{(-3+x) x \left (e^5+x\right )^2}-\frac {\log (x)}{\left (e^5+x\right )^2}\right ) \, dx\\ &=-\left (e^5 \int \frac {3-16 x+7 x^2}{(-3+x) x \left (e^5+x\right )^2} \, dx\right )-\left (3 e^{10}\right ) \int \frac {-2+x}{(-3+x) x \left (e^5+x\right )^2} \, dx+\int \frac {-3+7 x-3 x^2}{(-3+x) \left (e^5+x\right )^2} \, dx-\int \frac {\log (x)}{\left (e^5+x\right )^2} \, dx\\ &=-\frac {x \log (x)}{e^5 \left (e^5+x\right )}+\frac {\int \frac {1}{e^5+x} \, dx}{e^5}-e^5 \int \left (\frac {6}{\left (3+e^5\right )^2 (-3+x)}-\frac {1}{e^{10} x}+\frac {3+16 e^5+7 e^{10}}{e^5 \left (3+e^5\right ) \left (e^5+x\right )^2}+\frac {9+6 e^5-5 e^{10}}{e^{10} \left (3+e^5\right )^2 \left (e^5+x\right )}\right ) \, dx-\left (3 e^{10}\right ) \int \left (\frac {1}{3 \left (3+e^5\right )^2 (-3+x)}+\frac {2}{3 e^{10} x}+\frac {-2-e^5}{e^5 \left (3+e^5\right ) \left (e^5+x\right )^2}+\frac {-6-4 e^5-e^{10}}{e^{10} \left (3+e^5\right )^2 \left (e^5+x\right )}\right ) \, dx+\int \left (-\frac {9}{\left (3+e^5\right )^2 (-3+x)}+\frac {3+7 e^5+3 e^{10}}{\left (3+e^5\right ) \left (e^5+x\right )^2}-\frac {3 \left (6+6 e^5+e^{10}\right )}{\left (3+e^5\right )^2 \left (e^5+x\right )}\right ) \, dx\\ &=-\frac {3 e^5 \left (2+e^5\right )}{\left (3+e^5\right ) \left (e^5+x\right )}-\frac {3+7 e^5+3 e^{10}}{\left (3+e^5\right ) \left (e^5+x\right )}+\frac {3+16 e^5+7 e^{10}}{\left (3+e^5\right ) \left (e^5+x\right )}-\frac {9 \log (3-x)}{\left (3+e^5\right )^2}-\frac {6 e^5 \log (3-x)}{\left (3+e^5\right )^2}-\frac {e^{10} \log (3-x)}{\left (3+e^5\right )^2}-2 \log (x)+\frac {\log (x)}{e^5}-\frac {x \log (x)}{e^5 \left (e^5+x\right )}+\frac {\log \left (e^5+x\right )}{e^5}-\frac {\left (9+6 e^5-5 e^{10}\right ) \log \left (e^5+x\right )}{e^5 \left (3+e^5\right )^2}+\frac {3 \left (6+4 e^5+e^{10}\right ) \log \left (e^5+x\right )}{\left (3+e^5\right )^2}-\frac {3 \left (6+6 e^5+e^{10}\right ) \log \left (e^5+x\right )}{\left (3+e^5\right )^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 34, normalized size = 1.21 \begin {gather*} \frac {e^5}{e^5+x}-\log (3-x)-2 \log (x)+\frac {\log (x)}{e^5+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^10*(6 - 3*x) - 3*x + 7*x^2 - 3*x^3 + E^5*(-3 + 16*x - 7*x^2) + (3*x - x^2)*Log[x])/(-3*x^3 + x^4
+ E^10*(-3*x + x^2) + E^5*(-6*x^2 + 2*x^3)),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.72, size = 34, normalized size = 1.21 \begin {gather*} -\frac {{\left (x + e^{5}\right )} \log \left (x - 3\right ) + {\left (2 \, x + 2 \, e^{5} - 1\right )} \log \relax (x) - e^{5}}{x + e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+3*x)*log(x)+(-3*x+6)*exp(5)^2+(-7*x^2+16*x-3)*exp(5)-3*x^3+7*x^2-3*x)/((x^2-3*x)*exp(5)^2+(2*
x^3-6*x^2)*exp(5)+x^4-3*x^3),x, algorithm="fricas")

[Out]

-((x + e^5)*log(x - 3) + (2*x + 2*e^5 - 1)*log(x) - e^5)/(x + e^5)

________________________________________________________________________________________

giac [A]  time = 0.54, size = 41, normalized size = 1.46 \begin {gather*} -\frac {x \log \left (x - 3\right ) + e^{5} \log \left (x - 3\right ) + 2 \, x \log \relax (x) + 2 \, e^{5} \log \relax (x) - e^{5} - \log \relax (x)}{x + e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+3*x)*log(x)+(-3*x+6)*exp(5)^2+(-7*x^2+16*x-3)*exp(5)-3*x^3+7*x^2-3*x)/((x^2-3*x)*exp(5)^2+(2*
x^3-6*x^2)*exp(5)+x^4-3*x^3),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.33, size = 32, normalized size = 1.14

method	result	size
norman	\(\frac {\left (1-2 \,{\mathrm e}^{5}\right ) \ln \relax (x )-2 x \ln \relax (x )+{\mathrm e}^{5}}{{\mathrm e}^{5}+x}-\ln \left (x -3\right )\)	\(32\)
risch	\(\frac {\ln \relax (x )}{{\mathrm e}^{5}+x}-\frac {2 \ln \left (-x \right ) {\mathrm e}^{5}+\ln \left (x -3\right ) {\mathrm e}^{5}+2 \ln \left (-x \right ) x +\ln \left (x -3\right ) x -{\mathrm e}^{5}}{{\mathrm e}^{5}+x}\)	\(52\)
default	\(-\frac {\ln \relax (x ) \left (\ln \left (\frac {{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )-\ln \left (\frac {{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )\right )}{2 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}-\frac {\dilog \left (\frac {{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{2 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}+\frac {\dilog \left (\frac {{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{2 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}+\frac {{\mathrm e}^{5}}{{\mathrm e}^{5}+x}-\frac {\ln \left ({\mathrm e}^{5}+x \right )}{{\mathrm e}^{5}}-2 \ln \relax (x )+\frac {\ln \relax (x )}{{\mathrm e}^{5}}-\ln \left (x -3\right )\)	\(223\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+3*x)*ln(x)+(-3*x+6)*exp(5)^2+(-7*x^2+16*x-3)*exp(5)-3*x^3+7*x^2-3*x)/((x^2-3*x)*exp(5)^2+(2*x^3-6*x
^2)*exp(5)+x^4-3*x^3),x,method=_RETURNVERBOSE)

[Out]

((1-2*exp(5))*ln(x)-2*x*ln(x)+exp(5))/(exp(5)+x)-ln(x-3)

________________________________________________________________________________________

maxima [B]  time = 0.65, size = 443, normalized size = 15.82 \begin {gather*} -2 \, {\left (e^{\left (-10\right )} \log \relax (x) - \frac {3 \, {\left (2 \, e^{5} + 3\right )} \log \left (x + e^{5}\right )}{e^{20} + 6 \, e^{15} + 9 \, e^{10}} - \frac {\log \left (x - 3\right )}{e^{10} + 6 \, e^{5} + 9} + \frac {3}{x {\left (e^{10} + 3 \, e^{5}\right )} + e^{15} + 3 \, e^{10}}\right )} e^{10} + 3 \, {\left (\frac {\log \left (x + e^{5}\right )}{e^{10} + 6 \, e^{5} + 9} - \frac {\log \left (x - 3\right )}{e^{10} + 6 \, e^{5} + 9} - \frac {1}{x {\left (e^{5} + 3\right )} + e^{10} + 3 \, e^{5}}\right )} e^{10} + {\left (e^{\left (-10\right )} \log \relax (x) - \frac {3 \, {\left (2 \, e^{5} + 3\right )} \log \left (x + e^{5}\right )}{e^{20} + 6 \, e^{15} + 9 \, e^{10}} - \frac {\log \left (x - 3\right )}{e^{10} + 6 \, e^{5} + 9} + \frac {3}{x {\left (e^{10} + 3 \, e^{5}\right )} + e^{15} + 3 \, e^{10}}\right )} e^{5} + 7 \, {\left (\frac {e^{5}}{x {\left (e^{5} + 3\right )} + e^{10} + 3 \, e^{5}} + \frac {3 \, \log \left (x + e^{5}\right )}{e^{10} + 6 \, e^{5} + 9} - \frac {3 \, \log \left (x - 3\right )}{e^{10} + 6 \, e^{5} + 9}\right )} e^{5} - 16 \, {\left (\frac {\log \left (x + e^{5}\right )}{e^{10} + 6 \, e^{5} + 9} - \frac {\log \left (x - 3\right )}{e^{10} + 6 \, e^{5} + 9} - \frac {1}{x {\left (e^{5} + 3\right )} + e^{10} + 3 \, e^{5}}\right )} e^{5} + e^{\left (-5\right )} \log \left (x + e^{5}\right ) - e^{\left (-5\right )} \log \relax (x) - \frac {3 \, {\left (e^{10} + 6 \, e^{5}\right )} \log \left (x + e^{5}\right )}{e^{10} + 6 \, e^{5} + 9} - \frac {3 \, e^{10}}{x {\left (e^{5} + 3\right )} + e^{10} + 3 \, e^{5}} - \frac {7 \, e^{5}}{x {\left (e^{5} + 3\right )} + e^{10} + 3 \, e^{5}} - \frac {18 \, \log \left (x + e^{5}\right )}{e^{10} + 6 \, e^{5} + 9} - \frac {9 \, \log \left (x - 3\right )}{e^{10} + 6 \, e^{5} + 9} + \frac {\log \relax (x)}{x + e^{5}} - \frac {3}{x {\left (e^{5} + 3\right )} + e^{10} + 3 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+3*x)*log(x)+(-3*x+6)*exp(5)^2+(-7*x^2+16*x-3)*exp(5)-3*x^3+7*x^2-3*x)/((x^2-3*x)*exp(5)^2+(2*
x^3-6*x^2)*exp(5)+x^4-3*x^3),x, algorithm="maxima")

[Out]

-2*(e^(-10)*log(x) - 3*(2*e^5 + 3)*log(x + e^5)/(e^20 + 6*e^15 + 9*e^10) - log(x - 3)/(e^10 + 6*e^5 + 9) + 3/(
x*(e^10 + 3*e^5) + e^15 + 3*e^10))*e^10 + 3*(log(x + e^5)/(e^10 + 6*e^5 + 9) - log(x - 3)/(e^10 + 6*e^5 + 9) -
1/(x*(e^5 + 3) + e^10 + 3*e^5))*e^10 + (e^(-10)*log(x) - 3*(2*e^5 + 3)*log(x + e^5)/(e^20 + 6*e^15 + 9*e^10)
- log(x - 3)/(e^10 + 6*e^5 + 9) + 3/(x*(e^10 + 3*e^5) + e^15 + 3*e^10))*e^5 + 7*(e^5/(x*(e^5 + 3) + e^10 + 3*e
^5) + 3*log(x + e^5)/(e^10 + 6*e^5 + 9) - 3*log(x - 3)/(e^10 + 6*e^5 + 9))*e^5 - 16*(log(x + e^5)/(e^10 + 6*e^
5 + 9) - log(x - 3)/(e^10 + 6*e^5 + 9) - 1/(x*(e^5 + 3) + e^10 + 3*e^5))*e^5 + e^(-5)*log(x + e^5) - e^(-5)*lo
g(x) - 3*(e^10 + 6*e^5)*log(x + e^5)/(e^10 + 6*e^5 + 9) - 3*e^10/(x*(e^5 + 3) + e^10 + 3*e^5) - 7*e^5/(x*(e^5
+ 3) + e^10 + 3*e^5) - 18*log(x + e^5)/(e^10 + 6*e^5 + 9) - 9*log(x - 3)/(e^10 + 6*e^5 + 9) + log(x)/(x + e^5)
- 3/(x*(e^5 + 3) + e^10 + 3*e^5)

________________________________________________________________________________________

mupad [B]  time = 0.67, size = 29, normalized size = 1.04 \begin {gather*} \frac {{\mathrm {e}}^5}{x+{\mathrm {e}}^5}-2\,\ln \relax (x)-\ln \left (x-3\right )+\frac {\ln \relax (x)}{x+{\mathrm {e}}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + exp(5)*(7*x^2 - 16*x + 3) - log(x)*(3*x - x^2) - 7*x^2 + 3*x^3 + exp(10)*(3*x - 6))/(exp(10)*(3*x -
x^2) + exp(5)*(6*x^2 - 2*x^3) + 3*x^3 - x^4),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.75, size = 26, normalized size = 0.93 \begin {gather*} - 2 \log {\relax (x )} - \log {\left (x - 3 \right )} + \frac {\log {\relax (x )}}{x + e^{5}} + \frac {e^{5}}{x + e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+3*x)*ln(x)+(-3*x+6)*exp(5)**2+(-7*x**2+16*x-3)*exp(5)-3*x**3+7*x**2-3*x)/((x**2-3*x)*exp(5)*
*2+(2*x**3-6*x**2)*exp(5)+x**4-3*x**3),x)

[Out]

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

________________________________________________________________________________________