\(\int \frac {20 e^{-25+5 x} \log (7-2 e^{-25+5 x})}{63-18 e^{-25+5 x}+(-7+2 e^{-25+5 x}) \log ^2(7-2 e^{-25+5 x})} \, dx\) [7393]

Optimal result

Integrand size = 60, antiderivative size = 21 \[ \int \frac {20 e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx=\log \left (4 \left (-9+\log ^2\left (1-2 \left (-3+e^{5 (-5+x)}\right )\right )\right )\right ) \]

[Out]

ln(4*ln(-2*exp(5*x-25)+7)^2-36)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2320, 6828, 212, 6816} \[ \int \frac {20 e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx=\log \left (9-\log ^2\left (7-2 e^{5 x-25}\right )\right ) \]

[In]

Int[(20*E^(-25 + 5*x)*Log[7 - 2*E^(-25 + 5*x)])/(63 - 18*E^(-25 + 5*x) + (-7 + 2*E^(-25 + 5*x))*Log[7 - 2*E^(-
25 + 5*x)]^2),x]

[Out]

Log[9 - Log[7 - 2*E^(-25 + 5*x)]^2]

Rule 12

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6828

Int[(u_.)*((a_.) + (b_.)*(y_)^(n_))^(p_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Dist[q, Subst[In
t[(a + b*x^n)^p, x], x, y], x] /;  !FalseQ[q]] /; FreeQ[{a, b, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = 20 \int \frac {e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx \\ & = 4 \text {Subst}\left (\int \frac {\log (7-2 x)}{(7-2 x) \left (9-\log ^2(7-2 x)\right )} \, dx,x,e^{-25+5 x}\right ) \\ & = \log \left (9-\log ^2\left (7-2 e^{-25+5 x}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {20 e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx=\log \left (9-\log ^2\left (7-2 e^{-25+5 x}\right )\right ) \]

[In]

Integrate[(20*E^(-25 + 5*x)*Log[7 - 2*E^(-25 + 5*x)])/(63 - 18*E^(-25 + 5*x) + (-7 + 2*E^(-25 + 5*x))*Log[7 -
2*E^(-25 + 5*x)]^2),x]

[Out]

Log[9 - Log[7 - 2*E^(-25 + 5*x)]^2]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\ln \left (\ln \left (-2 \,{\mathrm e}^{5 x -25}+7\right )^{2}-9\right )\)	\(17\)
default	\(\ln \left (\ln \left (-2 \,{\mathrm e}^{5 x -25}+7\right )^{2}-9\right )\)	\(17\)
risch	\(\ln \left (\ln \left (-2 \,{\mathrm e}^{5 x -25}+7\right )^{2}-9\right )\)	\(17\)
norman	\(\ln \left (\ln \left (-2 \,{\mathrm e}^{5 x -25}+7\right )-3\right )+\ln \left (\ln \left (-2 \,{\mathrm e}^{5 x -25}+7\right )+3\right )\)	\(30\)
parallelrisch	\(\ln \left (\ln \left (-2 \,{\mathrm e}^{5 x -25}+7\right )-3\right )+\ln \left (\ln \left (-2 \,{\mathrm e}^{5 x -25}+7\right )+3\right )\)	\(30\)

[In]

int(20*exp(5*x-25)*ln(-2*exp(5*x-25)+7)/((2*exp(5*x-25)-7)*ln(-2*exp(5*x-25)+7)^2-18*exp(5*x-25)+63),x,method=
_RETURNVERBOSE)

[Out]

ln(ln(-2*exp(5*x-25)+7)^2-9)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {20 e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx=\log \left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) + 3\right ) + \log \left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) - 3\right ) \]

[In]

integrate(20*exp(5*x-25)*log(-2*exp(5*x-25)+7)/((2*exp(5*x-25)-7)*log(-2*exp(5*x-25)+7)^2-18*exp(5*x-25)+63),x
, algorithm="fricas")

[Out]

log(log(-2*e^(5*x - 25) + 7) + 3) + log(log(-2*e^(5*x - 25) + 7) - 3)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {20 e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx=\log {\left (\log {\left (7 - 2 e^{5 x - 25} \right )}^{2} - 9 \right )} \]

[In]

integrate(20*exp(5*x-25)*ln(-2*exp(5*x-25)+7)/((2*exp(5*x-25)-7)*ln(-2*exp(5*x-25)+7)**2-18*exp(5*x-25)+63),x)

[Out]

log(log(7 - 2*exp(5*x - 25))**2 - 9)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (18) = 36\).

Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.95 \[ \int \frac {20 e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx=-\frac {1}{3} \, {\left (\log \left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) + 3\right ) - \log \left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) - 3\right )\right )} \log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) + \frac {1}{3} \, {\left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) + 3\right )} \log \left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) + 3\right ) - \frac {1}{3} \, {\left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) - 3\right )} \log \left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right ) - 3\right ) - 2 \]

[In]

integrate(20*exp(5*x-25)*log(-2*exp(5*x-25)+7)/((2*exp(5*x-25)-7)*log(-2*exp(5*x-25)+7)^2-18*exp(5*x-25)+63),x
, algorithm="maxima")

[Out]

-1/3*(log(log(-2*e^(5*x - 25) + 7) + 3) - log(log(-2*e^(5*x - 25) + 7) - 3))*log(-2*e^(5*x - 25) + 7) + 1/3*(l
og(-2*e^(5*x - 25) + 7) + 3)*log(log(-2*e^(5*x - 25) + 7) + 3) - 1/3*(log(-2*e^(5*x - 25) + 7) - 3)*log(log(-2
*e^(5*x - 25) + 7) - 3) - 2

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {20 e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx=\log \left (\log \left (-2 \, e^{\left (5 \, x - 25\right )} + 7\right )^{2} - 9\right ) \]

[In]

integrate(20*exp(5*x-25)*log(-2*exp(5*x-25)+7)/((2*exp(5*x-25)-7)*log(-2*exp(5*x-25)+7)^2-18*exp(5*x-25)+63),x
, algorithm="giac")

[Out]

log(log(-2*e^(5*x - 25) + 7)^2 - 9)

Mupad [B] (verification not implemented)

Time = 11.44 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {20 e^{-25+5 x} \log \left (7-2 e^{-25+5 x}\right )}{63-18 e^{-25+5 x}+\left (-7+2 e^{-25+5 x}\right ) \log ^2\left (7-2 e^{-25+5 x}\right )} \, dx=\ln \left ({\ln \left (7-2\,{\mathrm {e}}^{5\,x-25}\right )}^2-9\right ) \]

[In]

int((20*exp(5*x - 25)*log(7 - 2*exp(5*x - 25)))/(log(7 - 2*exp(5*x - 25))^2*(2*exp(5*x - 25) - 7) - 18*exp(5*x
- 25) + 63),x)

[Out]

log(log(7 - 2*exp(5*x - 25))^2 - 9)