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

Optimal. Leaf size=23 \[ \frac{\log (a+b \log (c (e+f x)))}{b d f} \]

[Out]

Log[a + b*Log[c*(e + f*x)]]/(b*d*f)

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Rubi [A]  time = 0.0663648, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2390, 12, 2302, 29} \[ \frac{\log (a+b \log (c (e+f x)))}{b d f} \]

Antiderivative was successfully verified.

[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 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 12

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

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 29

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

Rubi steps

\begin{align*} \int \frac{1}{(d e+d f x) (a+b \log (c (e+f x)))} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{d x (a+b \log (c x))} \, dx,x,e+f x\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b \log (c x))} \, dx,x,e+f x\right )}{d f}\\ &=\frac{\operatorname{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]  time = 0.0174128, size = 23, normalized size = 1. \[ \frac{\log (a+b \log (c (e+f x)))}{b d f} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.066, size = 25, normalized size = 1.1 \begin{align*}{\frac{\ln \left ( a+b\ln \left ( cfx+ce \right ) \right ) }{bdf}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.18346, size = 39, normalized size = 1.7 \begin{align*} \frac{\log \left (\frac{b \log \left (f x + e\right ) + b \log \left (c\right ) + a}{b}\right )}{b d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.65417, size = 50, normalized size = 2.17 \begin{align*} \frac{\log \left (b \log \left (c f x + c e\right ) + a\right )}{b d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.214731, size = 17, normalized size = 0.74 \begin{align*} \frac{\log{\left (\frac{a}{b} + \log{\left (c \left (e + f x\right ) \right )} \right )}}{b d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B]  time = 1.23041, size = 113, normalized size = 4.91 \begin{align*} \frac{\log \left (\frac{1}{4} \,{\left (\pi{\left (\mathrm{sgn}\left (f x + e\right ) - 1\right )} + \pi{\left (\mathrm{sgn}\left (c\right ) - 1\right )} + 4 \, \pi \left \lfloor -\frac{\pi{\left (\mathrm{sgn}\left (f x + e\right ) - 1\right )} + \pi{\left (\mathrm{sgn}\left (c\right ) - 1\right )}}{4 \, \pi } + \frac{1}{2} \right \rfloor \right )}^{2} b^{2} +{\left (b \log \left ({\left | f x + e \right |}{\left | c \right |}\right ) + a\right )}^{2}\right )}{2 \, b d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(1/4*(pi*(sgn(f*x + e) - 1) + pi*(sgn(c) - 1) + 4*pi*floor(-1/4*(pi*(sgn(f*x + e) - 1) + pi*(sgn(c) - 1
))/pi + 1/2))^2*b^2 + (b*log(abs(f*x + e)*abs(c)) + a)^2)/(b*d*f)