\(\int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx\) [195]

Optimal result

Integrand size = 25, antiderivative size = 23 \[ \int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\log (a+b \log (c (e+f x)))}{b d f} \]

[Out]

ln(a+b*ln(c*(f*x+e)))/b/d/f

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2437, 12, 2339, 29} \[ \int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\log (a+b \log (c (e+f x)))}{b d f} \]

[In]

Int[1/((d*e + d*f*x)*(a + b*Log[c*(e + f*x)])),x]

[Out]

Log[a + b*Log[c*(e + f*x)]]/(b*d*f)

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 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}& = \frac {\text {Subst}\left (\int \frac {1}{d x (a+b \log (c x))} \, dx,x,e+f x\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x (a+b \log (c x))} \, dx,x,e+f x\right )}{d f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,a+b \log (c (e+f x))\right )}{b d f} \\ & = \frac {\log (a+b \log (c (e+f x)))}{b d f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\log (a+b \log (c (e+f x)))}{b d f} \]

[In]

Integrate[1/((d*e + d*f*x)*(a + b*Log[c*(e + f*x)])),x]

[Out]

Log[a + b*Log[c*(e + f*x)]]/(b*d*f)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
norman	\(\frac {\ln \left (a +b \ln \left (c \left (f x +e \right )\right )\right )}{b d f}\)	\(24\)
parallelrisch	\(\frac {\ln \left (a +b \ln \left (c \left (f x +e \right )\right )\right )}{b d f}\)	\(24\)
derivativedivides	\(\frac {\ln \left (a +b \ln \left (c f x +c e \right )\right )}{f d b}\)	\(25\)
default	\(\frac {\ln \left (a +b \ln \left (c f x +c e \right )\right )}{f d b}\)	\(25\)
risch	\(\frac {\ln \left (\ln \left (c \left (f x +e \right )\right )+\frac {a}{b}\right )}{b d f}\)	\(26\)

[In]

int(1/(d*f*x+d*e)/(a+b*ln(c*(f*x+e))),x,method=_RETURNVERBOSE)

[Out]

ln(a+b*ln(c*(f*x+e)))/b/d/f

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\log \left (b \log \left (c f x + c e\right ) + a\right )}{b d f} \]

[In]

integrate(1/(d*f*x+d*e)/(a+b*log(c*(f*x+e))),x, algorithm="fricas")

[Out]

log(b*log(c*f*x + c*e) + a)/(b*d*f)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\log {\left (\frac {a}{b} + \log {\left (c \left (e + f x\right ) \right )} \right )}}{b d f} \]

[In]

integrate(1/(d*f*x+d*e)/(a+b*ln(c*(f*x+e))),x)

[Out]

log(a/b + log(c*(e + f*x)))/(b*d*f)

Maxima [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\log \left (\frac {b \log \left (f x + e\right ) + b \log \left (c\right ) + a}{b}\right )}{b d f} \]

[In]

integrate(1/(d*f*x+d*e)/(a+b*log(c*(f*x+e))),x, algorithm="maxima")

[Out]

log((b*log(f*x + e) + b*log(c) + a)/b)/(b*d*f)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\log \left (b \log \left (c f x + c e\right ) + a\right )}{b d f} \]

[In]

integrate(1/(d*f*x+d*e)/(a+b*log(c*(f*x+e))),x, algorithm="giac")

[Out]

log(b*log(c*f*x + c*e) + a)/(b*d*f)

Mupad [B] (verification not implemented)

Time = 2.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx=\frac {\ln \left (a+b\,\ln \left (c\,\left (e+f\,x\right )\right )\right )}{b\,d\,f} \]

[In]

int(1/((d*e + d*f*x)*(a + b*log(c*(e + f*x)))),x)

[Out]

log(a + b*log(c*(e + f*x)))/(b*d*f)