\(\int \log (c (d+e x)^n) \, dx\) [70]

Optimal result

Integrand size = 10, antiderivative size = 24 \[ \int \log \left (c (d+e x)^n\right ) \, dx=-n x+\frac {(d+e x) \log \left (c (d+e x)^n\right )}{e} \]

[Out]

-n*x+(e*x+d)*ln(c*(e*x+d)^n)/e

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2436, 2332} \[ \int \log \left (c (d+e x)^n\right ) \, dx=\frac {(d+e x) \log \left (c (d+e x)^n\right )}{e}-n x \]

[In]

Int[Log[c*(d + e*x)^n],x]

[Out]

-(n*x) + ((d + e*x)*Log[c*(d + e*x)^n])/e

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \log \left (c x^n\right ) \, dx,x,d+e x\right )}{e} \\ & = -n x+\frac {(d+e x) \log \left (c (d+e x)^n\right )}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \log \left (c (d+e x)^n\right ) \, dx=-n x+\frac {(d+e x) \log \left (c (d+e x)^n\right )}{e} \]

[In]

Integrate[Log[c*(d + e*x)^n],x]

[Out]

-(n*x) + ((d + e*x)*Log[c*(d + e*x)^n])/e

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33

method	result	size
norman	\(x \ln \left (c \,{\mathrm e}^{n \ln \left (e x +d \right )}\right )+\frac {n d \ln \left (e x +d \right )}{e}-n x\)	\(32\)
default	\(\ln \left (c \left (e x +d \right )^{n}\right ) x -e n \left (\frac {x}{e}-\frac {d \ln \left (e x +d \right )}{e^{2}}\right )\)	\(36\)
parts	\(\ln \left (c \left (e x +d \right )^{n}\right ) x -e n \left (\frac {x}{e}-\frac {d \ln \left (e x +d \right )}{e^{2}}\right )\)	\(36\)
parallelrisch	\(\frac {x \ln \left (c \left (e x +d \right )^{n}\right ) d e n -x d e \,n^{2}+\ln \left (c \left (e x +d \right )^{n}\right ) d^{2} n}{d e n}\)	\(50\)
risch	\(x \ln \left (\left (e x +d \right )^{n}\right )-\frac {i \pi x \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i \pi x \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+x \ln \left (c \right )+\frac {n d \ln \left (e x +d \right )}{e}-n x\)	\(138\)

[In]

int(ln(c*(e*x+d)^n),x,method=_RETURNVERBOSE)

[Out]

x*ln(c*exp(n*ln(e*x+d)))+n*d/e*ln(e*x+d)-n*x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \log \left (c (d+e x)^n\right ) \, dx=-\frac {e n x - e x \log \left (c\right ) - {\left (e n x + d n\right )} \log \left (e x + d\right )}{e} \]

[In]

integrate(log(c*(e*x+d)^n),x, algorithm="fricas")

[Out]

-(e*n*x - e*x*log(c) - (e*n*x + d*n)*log(e*x + d))/e

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \log \left (c (d+e x)^n\right ) \, dx=\begin {cases} \frac {d \log {\left (c \left (d + e x\right )^{n} \right )}}{e} - n x + x \log {\left (c \left (d + e x\right )^{n} \right )} & \text {for}\: e \neq 0 \\x \log {\left (c d^{n} \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(ln(c*(e*x+d)**n),x)

[Out]

Piecewise((d*log(c*(d + e*x)**n)/e - n*x + x*log(c*(d + e*x)**n), Ne(e, 0)), (x*log(c*d**n), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \log \left (c (d+e x)^n\right ) \, dx=-e n {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} + x \log \left ({\left (e x + d\right )}^{n} c\right ) \]

[In]

integrate(log(c*(e*x+d)^n),x, algorithm="maxima")

[Out]

-e*n*(x/e - d*log(e*x + d)/e^2) + x*log((e*x + d)^n*c)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \log \left (c (d+e x)^n\right ) \, dx=\frac {{\left (e x + d\right )} n \log \left (e x + d\right )}{e} - \frac {{\left (e x + d\right )} n}{e} + \frac {{\left (e x + d\right )} \log \left (c\right )}{e} \]

[In]

integrate(log(c*(e*x+d)^n),x, algorithm="giac")

[Out]

(e*x + d)*n*log(e*x + d)/e - (e*x + d)*n/e + (e*x + d)*log(c)/e

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \log \left (c (d+e x)^n\right ) \, dx=x\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )-n\,x+\frac {d\,n\,\ln \left (d+e\,x\right )}{e} \]

[In]

int(log(c*(d + e*x)^n),x)

[Out]

x*log(c*(d + e*x)^n) - n*x + (d*n*log(d + e*x))/e