\(\int \log (c (a+b x)^p) \, dx\) [179]

Optimal result

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

[Out]

-p*x+(b*x+a)*ln(c*(b*x+a)^p)/b

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} \[ \int \log \left (c (a+b x)^p\right ) \, dx=\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b}-p x \]

[In]

Int[Log[c*(a + b*x)^p],x]

[Out]

-(p*x) + ((a + b*x)*Log[c*(a + b*x)^p])/b

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^p\right ) \, dx,x,a+b x\right )}{b} \\ & = -p x+\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[Log[c*(a + b*x)^p],x]

[Out]

-(p*x) + ((a + b*x)*Log[c*(a + b*x)^p])/b

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33

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

[In]

int(ln(c*(b*x+a)^p),x,method=_RETURNVERBOSE)

[Out]

x*ln(c*exp(p*ln(b*x+a)))+p*a/b*ln(b*x+a)-p*x

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \log \left (c (a+b x)^p\right ) \, dx=-\frac {b p x - b x \log \left (c\right ) - {\left (b p x + a p\right )} \log \left (b x + a\right )}{b} \]

[In]

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

[Out]

-(b*p*x - b*x*log(c) - (b*p*x + a*p)*log(b*x + a))/b

Sympy [A] (verification not implemented)

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

[In]

integrate(ln(c*(b*x+a)**p),x)

[Out]

Piecewise((a*log(c*(a + b*x)**p)/b - p*x + x*log(c*(a + b*x)**p), Ne(b, 0)), (x*log(a**p*c), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \log \left (c (a+b x)^p\right ) \, dx=-b p {\left (\frac {x}{b} - \frac {a \log \left (b x + a\right )}{b^{2}}\right )} + x \log \left ({\left (b x + a\right )}^{p} c\right ) \]

[In]

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

[Out]

-b*p*(x/b - a*log(b*x + a)/b^2) + x*log((b*x + a)^p*c)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \log \left (c (a+b x)^p\right ) \, dx=\frac {{\left (b x + a\right )} p \log \left (b x + a\right )}{b} - \frac {{\left (b x + a\right )} p}{b} + \frac {{\left (b x + a\right )} \log \left (c\right )}{b} \]

[In]

integrate(log(c*(b*x+a)^p),x, algorithm="giac")

[Out]

(b*x + a)*p*log(b*x + a)/b - (b*x + a)*p/b + (b*x + a)*log(c)/b

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \log \left (c (a+b x)^p\right ) \, dx=x\,\ln \left (c\,{\left (a+b\,x\right )}^p\right )-p\,x+\frac {a\,p\,\ln \left (a+b\,x\right )}{b} \]

[In]

int(log(c*(a + b*x)^p),x)

[Out]

x*log(c*(a + b*x)^p) - p*x + (a*p*log(a + b*x))/b