\(\int \frac {\log (c (a+b x)^p)}{d+e x} \, dx\) [180]

Optimal result

Integrand size = 18, antiderivative size = 58 \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e} \]

[Out]

ln(c*(b*x+a)^p)*ln(b*(e*x+d)/(-a*e+b*d))/e+p*polylog(2,-e*(b*x+a)/(-a*e+b*d))/e

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2441, 2440, 2438} \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e} \]

[In]

Int[Log[c*(a + b*x)^p]/(d + e*x),x]

[Out]

(Log[c*(a + b*x)^p]*Log[(b*(d + e*x))/(b*d - a*e)])/e + (p*PolyLog[2, -((e*(a + b*x))/(b*d - a*e))])/e

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}-\frac {(b p) \int \frac {\log \left (\frac {b (d+e x)}{b d-a e}\right )}{a+b x} \, dx}{e} \\ & = \frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}-\frac {p \text {Subst}\left (\int \frac {\log \left (1+\frac {e x}{b d-a e}\right )}{x} \, dx,x,a+b x\right )}{e} \\ & = \frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )}{e} \]

[In]

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

[Out]

(Log[c*(a + b*x)^p]*Log[(b*(d + e*x))/(b*d - a*e)])/e + (p*PolyLog[2, (e*(a + b*x))/(-(b*d) + a*e)])/e

Maple [A] (verified)

Time = 1.68 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.69

method	result	size
parts	\(\frac {\ln \left (c \left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{e}-\frac {p b \left (\frac {\operatorname {dilog}\left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{b}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{b}\right )}{e}\)	\(98\)
risch	\(\frac {\ln \left (\left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{e}-\frac {p \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{e}-\frac {p \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +a e -b d}{a e -b d}\right )}{e}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \ln \left (e x +d \right )}{e}\)	\(207\)

[In]

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

[Out]

ln(c*(b*x+a)^p)/e*ln(e*x+d)-1/e*p*b*(dilog(((e*x+d)*b+a*e-b*d)/(a*e-b*d))/b+ln(e*x+d)*ln(((e*x+d)*b+a*e-b*d)/(
a*e-b*d))/b)

Fricas [F]

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

[In]

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

[Out]

integral(log((b*x + a)^p*c)/(e*x + d), x)

Sympy [F]

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

[In]

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

[Out]

Integral(log(c*(a + b*x)**p)/(d + e*x), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (57) = 114\).

Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.03 \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {b p {\left (\frac {\log \left (b x + a\right ) \log \left (e x + d\right )}{b} - \frac {\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )}{b}\right )}}{e} - \frac {p \log \left (b x + a\right ) \log \left (e x + d\right )}{e} + \frac {\log \left ({\left (b x + a\right )}^{p} c\right ) \log \left (e x + d\right )}{e} \]

[In]

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

[Out]

b*p*(log(b*x + a)*log(e*x + d)/b - (log(e*x + d)*log(-(b*e*x + b*d)/(b*d - a*e) + 1) + dilog((b*e*x + b*d)/(b*
d - a*e)))/b)/e - p*log(b*x + a)*log(e*x + d)/e + log((b*x + a)^p*c)*log(e*x + d)/e

Giac [F]

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

[In]

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

[Out]

integrate(log((b*x + a)^p*c)/(e*x + d), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\int \frac {\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{d+e\,x} \,d x \]

[In]

int(log(c*(a + b*x)^p)/(d + e*x),x)

[Out]

int(log(c*(a + b*x)^p)/(d + e*x), x)