\(\int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2(-\frac {9}{-3+3 x+10 \log (10+e)})} \, dx\) [5035]

Optimal result

Integrand size = 33, antiderivative size = 22 \[ \int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx=\frac {5}{\log \left (\frac {3}{1-x-\frac {10}{3} \log (10+e)}\right )} \]

[Out]

5/ln(3/(1-10/3*ln(10+exp(1))-x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 2437, 2339, 30} \[ \int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx=\frac {5}{\log \left (\frac {9}{-3 x+3-10 \log (10+e)}\right )} \]

[In]

Int[15/((-3 + 3*x + 10*Log[10 + E])*Log[-9/(-3 + 3*x + 10*Log[10 + E])]^2),x]

[Out]

5/Log[9/(3 - 3*x - 10*Log[10 + E])]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2339

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 2437

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]

Rubi steps \begin{align*} \text {integral}& = 15 \int \frac {1}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx \\ & = 5 \text {Subst}\left (\int \frac {1}{x \log ^2\left (-\frac {9}{x}\right )} \, dx,x,-3+3 x+10 \log (10+e)\right ) \\ & = -\left (5 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )\right )\right ) \\ & = \frac {5}{\log \left (\frac {9}{3-3 x-10 \log (10+e)}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx=\frac {5}{\log \left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \]

[In]

Integrate[15/((-3 + 3*x + 10*Log[10 + E])*Log[-9/(-3 + 3*x + 10*Log[10 + E])]^2),x]

[Out]

5/Log[-9/(-3 + 3*x + 10*Log[10 + E])]

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {5}{\ln \left (-\frac {9}{10 \ln \left (10+{\mathrm e}\right )+3 x -3}\right )}\)	\(22\)
default	\(\frac {5}{\ln \left (-\frac {9}{10 \ln \left (10+{\mathrm e}\right )+3 x -3}\right )}\)	\(22\)
norman	\(\frac {5}{\ln \left (-\frac {9}{10 \ln \left (10+{\mathrm e}\right )+3 x -3}\right )}\)	\(22\)
risch	\(\frac {5}{\ln \left (-\frac {9}{10 \ln \left (10+{\mathrm e}\right )+3 x -3}\right )}\)	\(22\)
parallelrisch	\(\frac {5}{\ln \left (-\frac {9}{10 \ln \left (10+{\mathrm e}\right )+3 x -3}\right )}\)	\(22\)

[In]

int(15/(10*ln(10+exp(1))+3*x-3)/ln(-9/(10*ln(10+exp(1))+3*x-3))^2,x,method=_RETURNVERBOSE)

[Out]

5/ln(-9/(10*ln(10+exp(1))+3*x-3))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx=\frac {5}{\log \left (-\frac {9}{3 \, x + 10 \, \log \left (e + 10\right ) - 3}\right )} \]

[In]

integrate(15/(10*log(10+exp(1))+3*x-3)/log(-9/(10*log(10+exp(1))+3*x-3))^2,x, algorithm="fricas")

[Out]

5/log(-9/(3*x + 10*log(e + 10) - 3))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx=\frac {5}{\log {\left (- \frac {9}{3 x - 3 + 10 \log {\left (e + 10 \right )}} \right )}} \]

[In]

integrate(15/(10*ln(10+exp(1))+3*x-3)/ln(-9/(10*ln(10+exp(1))+3*x-3))**2,x)

[Out]

5/log(-9/(3*x - 3 + 10*log(E + 10)))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx=\frac {5}{\log \left (-\frac {9}{3 \, x + 10 \, \log \left (e + 10\right ) - 3}\right )} \]

[In]

integrate(15/(10*log(10+exp(1))+3*x-3)/log(-9/(10*log(10+exp(1))+3*x-3))^2,x, algorithm="maxima")

[Out]

5/log(-9/(3*x + 10*log(e + 10) - 3))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx=\frac {5}{\log \left (-\frac {9}{3 \, x + 10 \, \log \left (e + 10\right ) - 3}\right )} \]

[In]

integrate(15/(10*log(10+exp(1))+3*x-3)/log(-9/(10*log(10+exp(1))+3*x-3))^2,x, algorithm="giac")

[Out]

5/log(-9/(3*x + 10*log(e + 10) - 3))

Mupad [B] (verification not implemented)

Time = 11.69 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {15}{(-3+3 x+10 \log (10+e)) \log ^2\left (-\frac {9}{-3+3 x+10 \log (10+e)}\right )} \, dx=\frac {5}{\ln \left (-\frac {9}{3\,x+10\,\ln \left (\mathrm {e}+10\right )-3}\right )} \]

[In]

int(15/(log(-9/(3*x + 10*log(exp(1) + 10) - 3))^2*(3*x + 10*log(exp(1) + 10) - 3)),x)

[Out]

5/log(-9/(3*x + 10*log(exp(1) + 10) - 3))