\(\int (a+b \log (c (d+e x)^n)) \, dx\) [20]

Optimal result

Integrand size = 14, antiderivative size = 29 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=a x-b n x+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2436, 2332} \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=a x+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e}-b n x \]

[In]

Int[a + b*Log[c*(d + e*x)^n],x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=a x-b n x+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e} \]

[In]

Integrate[a + b*Log[c*(d + e*x)^n],x]

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24

method	result	size
default	\(a x +b \ln \left (c \left (e x +d \right )^{n}\right ) x -b n x +\frac {b n d \ln \left (e x +d \right )}{e}\)	\(36\)
parts	\(a x +b \ln \left (c \left (e x +d \right )^{n}\right ) x -b n x +\frac {b n d \ln \left (e x +d \right )}{e}\)	\(36\)
norman	\(\left (-b n +a \right ) x +b x \ln \left (c \,{\mathrm e}^{n \ln \left (e x +d \right )}\right )+\frac {b n d \ln \left (e x +d \right )}{e}\)	\(38\)
parallelrisch	\(\frac {b \left (x \ln \left (c \left (e x +d \right )^{n}\right ) d e n -x d e \,n^{2}+\ln \left (c \left (e x +d \right )^{n}\right ) d^{2} n \right )}{d e n}+a x\)	\(55\)
risch	\(a x +b x \ln \left (\left (e x +d \right )^{n}\right )-\frac {i b \pi x \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )}{2}+\frac {i b \pi x \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}+\frac {i b \pi x \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi x \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {b n d \ln \left (e x +d \right )}{e}+\ln \left (c \right ) b x -b n x\)	\(149\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {b e x \log \left (c\right ) - {\left (b e n - a e\right )} x + {\left (b e n x + b d n\right )} \log \left (e x + d\right )}{e} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=a x + b \left (\begin {cases} \frac {d \log {\left (c \left (d + e x\right )^{n} \right )}}{e} - n x + x \log {\left (c \left (d + e x\right )^{n} \right )} & \text {for}\: e \neq 0 \\x \log {\left (c d^{n} \right )} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a*x + b*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 = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-b e n {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} + b x \log \left ({\left (e x + d\right )}^{n} c\right ) + a x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx={\left (\frac {{\left (e x + d\right )} n \log \left (e x + d\right )}{e} - \frac {{\left (e x + d\right )} n}{e} + \frac {{\left (e x + d\right )} \log \left (c\right )}{e}\right )} b + a x \]

[In]

integrate(a+b*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)*b + a*x

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=x\,\left (a-b\,n\right )+b\,x\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )+\frac {b\,d\,n\,\ln \left (d+e\,x\right )}{e} \]

[In]

int(a + b*log(c*(d + e*x)^n),x)

[Out]

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