\(\int \frac {e^2 (2-3 x)+10 x+(-3 e^2 x-15 x^2) \log (\frac {15 x}{e^2+5 x})}{2 e^2 x+10 x^2} \, dx\) [8917]

Optimal result

Integrand size = 54, antiderivative size = 25 \[ \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{2 e^2 x+10 x^2} \, dx=\log (3 x)-\frac {3}{2} x \log \left (\frac {3 x}{\frac {e^2}{5}+x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {1607, 6874, 78, 2536, 31} \[ \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{2 e^2 x+10 x^2} \, dx=\log (x)-\frac {3}{2} x \log \left (\frac {15 x}{5 x+e^2}\right ) \]

[In]

Int[(E^2*(2 - 3*x) + 10*x + (-3*E^2*x - 15*x^2)*Log[(15*x)/(E^2 + 5*x)])/(2*E^2*x + 10*x^2),x]

[Out]

Log[x] - (3*x*Log[(15*x)/(E^2 + 5*x)])/2

Rule 31

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

Rule 78

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 1607

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 2536

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.), x_Symbol] :> Simp[
(a + b*x)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^p/b), x] - Dist[B*n*p*((b*c - a*d)/b), Int[(A + B*Log[e*((
a + b*x)^n/(c + d*x)^n)])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && EqQ[n + mn, 0] &&
NeQ[b*c - a*d, 0] && IGtQ[p, 0]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{x \left (2 e^2+10 x\right )} \, dx \\ & = \int \left (\frac {2 e^2+\left (10-3 e^2\right ) x}{2 x \left (e^2+5 x\right )}-\frac {3}{2} \log \left (\frac {15 x}{e^2+5 x}\right )\right ) \, dx \\ & = \frac {1}{2} \int \frac {2 e^2+\left (10-3 e^2\right ) x}{x \left (e^2+5 x\right )} \, dx-\frac {3}{2} \int \log \left (\frac {15 x}{e^2+5 x}\right ) \, dx \\ & = -\frac {3}{2} x \log \left (\frac {15 x}{e^2+5 x}\right )+\frac {1}{2} \int \left (\frac {2}{x}-\frac {3 e^2}{e^2+5 x}\right ) \, dx+\frac {1}{2} \left (3 e^2\right ) \int \frac {1}{e^2+5 x} \, dx \\ & = \log (x)-\frac {3}{2} x \log \left (\frac {15 x}{e^2+5 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{2 e^2 x+10 x^2} \, dx=\frac {1}{2} \left (2 \log (x)-3 x \log \left (\frac {15 x}{e^2+5 x}\right )\right ) \]

[In]

Integrate[(E^2*(2 - 3*x) + 10*x + (-3*E^2*x - 15*x^2)*Log[(15*x)/(E^2 + 5*x)])/(2*E^2*x + 10*x^2),x]

[Out]

(2*Log[x] - 3*x*Log[(15*x)/(E^2 + 5*x)])/2

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
norman	\(-\frac {3 x \ln \left (\frac {15 x}{{\mathrm e}^{2}+5 x}\right )}{2}+\ln \left (x \right )\)	\(19\)
risch	\(-\frac {3 x \ln \left (\frac {15 x}{{\mathrm e}^{2}+5 x}\right )}{2}+\ln \left (x \right )\)	\(19\)
parallelrisch	\(-\frac {3 \,{\mathrm e}^{2} \ln \left (x \right )}{5}+\frac {3 \,{\mathrm e}^{2} \ln \left (x +\frac {{\mathrm e}^{2}}{5}\right )}{5}+\frac {3 \,{\mathrm e}^{2} \ln \left (\frac {15 x}{{\mathrm e}^{2}+5 x}\right )}{5}-\frac {3 x \ln \left (\frac {15 x}{{\mathrm e}^{2}+5 x}\right )}{2}+\ln \left (x \right )\)	\(52\)
parts	\(\ln \left (x \right )-\frac {3 \,{\mathrm e}^{2} \ln \left ({\mathrm e}^{2}+5 x \right )}{10}-\frac {9 \,{\mathrm e}^{2} \left (\frac {\ln \left (-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )}{3}+\frac {\ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) {\mathrm e}^{-2} \left ({\mathrm e}^{2}+5 x \right )}{9}\right )}{10}\)	\(76\)
derivativedivides	\(\frac {3 \,{\mathrm e}^{2} \left ({\mathrm e}^{-2} \left (\frac {10 \ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )}{3}+\left ({\mathrm e}^{2}-\frac {10}{3}\right ) \ln \left (-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )\right )-\ln \left (-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )-\frac {\ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) {\mathrm e}^{-2} \left ({\mathrm e}^{2}+5 x \right )}{3}\right )}{10}\)	\(103\)
default	\(\frac {3 \,{\mathrm e}^{2} \left ({\mathrm e}^{-2} \left (\frac {10 \ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )}{3}+\left ({\mathrm e}^{2}-\frac {10}{3}\right ) \ln \left (-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )\right )-\ln \left (-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right )-\frac {\ln \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) \left (3-\frac {3 \,{\mathrm e}^{2}}{{\mathrm e}^{2}+5 x}\right ) {\mathrm e}^{-2} \left ({\mathrm e}^{2}+5 x \right )}{3}\right )}{10}\)	\(103\)

[In]

int(((-3*exp(2)*x-15*x^2)*ln(15*x/(exp(2)+5*x))+(2-3*x)*exp(2)+10*x)/(2*exp(2)*x+10*x^2),x,method=_RETURNVERBO
SE)

[Out]

-3/2*x*ln(15*x/(exp(2)+5*x))+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{2 e^2 x+10 x^2} \, dx=-\frac {3}{2} \, x \log \left (\frac {15 \, x}{5 \, x + e^{2}}\right ) + \log \left (x\right ) \]

[In]

integrate(((-3*exp(2)*x-15*x^2)*log(15*x/(exp(2)+5*x))+(2-3*x)*exp(2)+10*x)/(2*exp(2)*x+10*x^2),x, algorithm="
fricas")

[Out]

-3/2*x*log(15*x/(5*x + e^2)) + log(x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{2 e^2 x+10 x^2} \, dx=- \frac {3 x \log {\left (\frac {15 x}{5 x + e^{2}} \right )}}{2} + \log {\left (x \right )} \]

[In]

integrate(((-3*exp(2)*x-15*x**2)*ln(15*x/(exp(2)+5*x))+(2-3*x)*exp(2)+10*x)/(2*exp(2)*x+10*x**2),x)

[Out]

-3*x*log(15*x/(5*x + exp(2)))/2 + log(x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (20) = 40\).

Time = 0.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.72 \[ \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{2 e^2 x+10 x^2} \, dx=-\frac {3}{2} \, x {\left (\log \left (5\right ) + \log \left (3\right )\right )} - {\left (e^{\left (-2\right )} \log \left (5 \, x + e^{2}\right ) - e^{\left (-2\right )} \log \left (x\right )\right )} e^{2} + \frac {3}{10} \, {\left (5 \, x + e^{2}\right )} \log \left (5 \, x + e^{2}\right ) - \frac {3}{10} \, e^{2} \log \left (5 \, x + e^{2}\right ) - \frac {3}{2} \, x \log \left (x\right ) + \log \left (5 \, x + e^{2}\right ) \]

[In]

integrate(((-3*exp(2)*x-15*x^2)*log(15*x/(exp(2)+5*x))+(2-3*x)*exp(2)+10*x)/(2*exp(2)*x+10*x^2),x, algorithm="
maxima")

[Out]

-3/2*x*(log(5) + log(3)) - (e^(-2)*log(5*x + e^2) - e^(-2)*log(x))*e^2 + 3/10*(5*x + e^2)*log(5*x + e^2) - 3/1
0*e^2*log(5*x + e^2) - 3/2*x*log(x) + log(5*x + e^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{2 e^2 x+10 x^2} \, dx=-\frac {3}{2} \, x \log \left (\frac {15 \, x}{5 \, x + e^{2}}\right ) + \log \left (x\right ) \]

[In]

integrate(((-3*exp(2)*x-15*x^2)*log(15*x/(exp(2)+5*x))+(2-3*x)*exp(2)+10*x)/(2*exp(2)*x+10*x^2),x, algorithm="
giac")

[Out]

-3/2*x*log(15*x/(5*x + e^2)) + log(x)

Mupad [B] (verification not implemented)

Time = 13.68 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {e^2 (2-3 x)+10 x+\left (-3 e^2 x-15 x^2\right ) \log \left (\frac {15 x}{e^2+5 x}\right )}{2 e^2 x+10 x^2} \, dx=\ln \left (x\right )-\frac {3\,x\,\ln \left (\frac {15\,x}{5\,x+{\mathrm {e}}^2}\right )}{2} \]

[In]

int(-(log((15*x)/(5*x + exp(2)))*(3*x*exp(2) + 15*x^2) - 10*x + exp(2)*(3*x - 2))/(2*x*exp(2) + 10*x^2),x)

[Out]

log(x) - (3*x*log((15*x)/(5*x + exp(2))))/2