3.32.67 \(\int \frac {-x+(2+e^{5/4}+x) \log (10+5 e^{5/4}+5 x)}{(2 x+e^{5/4} x+x^2) \log (10+5 e^{5/4}+5 x)} \, dx\)

Optimal. Leaf size=17 \[ \log \left (\frac {33 x}{\log \left (5 \left (2+e^{5/4}+x\right )\right )}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6, 1593, 6688, 2390, 12, 2302, 29} \begin {gather*} \log (x)-\log \left (\log \left (5 \left (x+e^{5/4}+2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - Log[Log[5*(2 + E^(5/4) + 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

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

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x+\left (2+e^{5/4}+x\right ) \log \left (10+5 e^{5/4}+5 x\right )}{\left (\left (2+e^{5/4}\right ) x+x^2\right ) \log \left (10+5 e^{5/4}+5 x\right )} \, dx\\ &=\int \frac {-x+\left (2+e^{5/4}+x\right ) \log \left (10+5 e^{5/4}+5 x\right )}{x \left (2+e^{5/4}+x\right ) \log \left (10+5 e^{5/4}+5 x\right )} \, dx\\ &=\int \left (\frac {1}{x}-\frac {1}{\left (2+e^{5/4}+x\right ) \log \left (5 \left (2+e^{5/4}\right )+5 x\right )}\right ) \, dx\\ &=\log (x)-\int \frac {1}{\left (2+e^{5/4}+x\right ) \log \left (5 \left (2+e^{5/4}\right )+5 x\right )} \, dx\\ &=\log (x)-\frac {1}{5} \operatorname {Subst}\left (\int \frac {5}{x \log (x)} \, dx,x,5 \left (2+e^{5/4}\right )+5 x\right )\\ &=\log (x)-\operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,5 \left (2+e^{5/4}\right )+5 x\right )\\ &=\log (x)-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (5 \left (2+e^{5/4}+x\right )\right )\right )\\ &=\log (x)-\log \left (\log \left (5 \left (2+e^{5/4}+x\right )\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 17, normalized size = 1.00 \begin {gather*} \log (x)-\log \left (\log \left (5 \left (2+e^{5/4}+x\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 16, normalized size = 0.94 \begin {gather*} \log \relax (x) - \log \left (\log \left (5 \, x + 5 \, e^{\frac {5}{4}} + 10\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - log(log(5*x + 5*e^(5/4) + 10))

________________________________________________________________________________________

giac [A]  time = 0.20, size = 16, normalized size = 0.94 \begin {gather*} \log \relax (x) - \log \left (\log \left (5 \, x + 5 \, e^{\frac {5}{4}} + 10\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - log(log(5*x + 5*e^(5/4) + 10))

________________________________________________________________________________________

maple [A]  time = 0.50, size = 17, normalized size = 1.00

method	result	size
norman	\(-\ln \left (\ln \left (5 \,{\mathrm e}^{\frac {5}{4}}+5 x +10\right )\right )+\ln \relax (x )\)	\(17\)
risch	\(-\ln \left (\ln \left (5 \,{\mathrm e}^{\frac {5}{4}}+5 x +10\right )\right )+\ln \relax (x )\)	\(17\)
derivativedivides	\(\ln \left (-5 x \right )-\ln \left (\ln \left (5 \,{\mathrm e}^{\frac {5}{4}}+5 x +10\right )\right )\)	\(19\)
default	\(\ln \left (-5 x \right )-\ln \left (\ln \left (5 \,{\mathrm e}^{\frac {5}{4}}+5 x +10\right )\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(ln(5*exp(5/4)+5*x+10))+ln(x)

________________________________________________________________________________________

maxima [B]  time = 0.89, size = 113, normalized size = 6.65 \begin {gather*} -{\left (\frac {\log \left (x + e^{\frac {5}{4}} + 2\right )}{e^{\frac {5}{4}} + 2} - \frac {\log \relax (x)}{e^{\frac {5}{4}} + 2}\right )} e^{\frac {5}{4}} - {\left (\log \relax (5) + \log \left (x + e^{\frac {5}{4}} + 2\right )\right )} \log \left (\log \relax (5) + \log \left (x + e^{\frac {5}{4}} + 2\right )\right ) + \log \left (5 \, x + 5 \, e^{\frac {5}{4}} + 10\right ) \log \left (\log \relax (5) + \log \left (x + e^{\frac {5}{4}} + 2\right )\right ) - \frac {2 \, \log \left (x + e^{\frac {5}{4}} + 2\right )}{e^{\frac {5}{4}} + 2} + \frac {2 \, \log \relax (x)}{e^{\frac {5}{4}} + 2} + \log \left (x + e^{\frac {5}{4}} + 2\right ) - \log \left (\log \relax (5) + \log \left (x + e^{\frac {5}{4}} + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(x + e^(5/4) + 2)/(e^(5/4) + 2) - log(x)/(e^(5/4) + 2))*e^(5/4) - (log(5) + log(x + e^(5/4) + 2))*log(log
(5) + log(x + e^(5/4) + 2)) + log(5*x + 5*e^(5/4) + 10)*log(log(5) + log(x + e^(5/4) + 2)) - 2*log(x + e^(5/4)
+ 2)/(e^(5/4) + 2) + 2*log(x)/(e^(5/4) + 2) + log(x + e^(5/4) + 2) - log(log(5) + log(x + e^(5/4) + 2))

________________________________________________________________________________________

mupad [B]  time = 0.49, size = 16, normalized size = 0.94 \begin {gather*} \ln \relax (x)-\ln \left (\ln \left (5\,x+5\,{\mathrm {e}}^{5/4}+10\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - log(log(5*x + 5*exp(5/4) + 10))

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 17, normalized size = 1.00 \begin {gather*} \log {\relax (x )} - \log {\left (\log {\left (5 x + 10 + 5 e^{\frac {5}{4}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - log(log(5*x + 10 + 5*exp(5/4)))

________________________________________________________________________________________