Integrand size = 18, antiderivative size = 58 \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e} \] Output:
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
Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )}{e} \] Input:
Integrate[Log[c*(a + b*x)^p]/(d + e*x),x]
Output:
(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
Time = 0.47 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2841, 2840, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx\) |
\(\Big \downarrow \) 2841 |
\(\displaystyle \frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}-\frac {b p \int \frac {\log \left (\frac {b (d+e x)}{b d-a e}\right )}{a+b x}dx}{e}\) |
\(\Big \downarrow \) 2840 |
\(\displaystyle \frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}-\frac {p \int \frac {\log \left (\frac {e (a+b x)}{b d-a e}+1\right )}{a+b x}d(a+b x)}{e}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {\log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e}\) |
Input:
Int[Log[c*(a + b*x)^p]/(d + e*x),x]
Output:
(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
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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_ Symbol] :> Simp[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]
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] - Simp[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]
Time = 2.90 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.69
method | result | size |
parts | \(\frac {\ln \left (c \left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{e}-\frac {p b \left (\frac {\operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{b}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{b}\right )}{e}\) | \(98\) |
risch | \(\frac {\ln \left (\left (b x +a \right )^{p}\right ) \ln \left (e x +d \right )}{e}-\frac {p \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{e}-\frac {p \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{e}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \ln \left (e x +d \right )}{e}\) | \(207\) |
Input:
int(ln(c*(b*x+a)^p)/(e*x+d),x,method=_RETURNVERBOSE)
Output:
ln(c*(b*x+a)^p)/e*ln(e*x+d)-1/e*p*b*(dilog(((e*x+d)*b+e*a-b*d)/(a*e-b*d))/ b+ln(e*x+d)*ln(((e*x+d)*b+e*a-b*d)/(a*e-b*d))/b)
\[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d} \,d x } \] Input:
integrate(log(c*(b*x+a)^p)/(e*x+d),x, algorithm="fricas")
Output:
integral(log((b*x + a)^p*c)/(e*x + d), x)
\[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\int \frac {\log {\left (c \left (a + b x\right )^{p} \right )}}{d + e x}\, dx \] Input:
integrate(ln(c*(b*x+a)**p)/(e*x+d),x)
Output:
Integral(log(c*(a + b*x)**p)/(d + e*x), x)
Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (57) = 114\).
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.03 \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {b p {\left (\frac {\log \left (b x + a\right ) \log \left (e x + d\right )}{b} - \frac {\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )}{b}\right )}}{e} - \frac {p \log \left (b x + a\right ) \log \left (e x + d\right )}{e} + \frac {\log \left ({\left (b x + a\right )}^{p} c\right ) \log \left (e x + d\right )}{e} \] Input:
integrate(log(c*(b*x+a)^p)/(e*x+d),x, algorithm="maxima")
Output:
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
\[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (b x + a\right )}^{p} c\right )}{e x + d} \,d x } \] Input:
integrate(log(c*(b*x+a)^p)/(e*x+d),x, algorithm="giac")
Output:
integrate(log((b*x + a)^p*c)/(e*x + d), x)
Timed out. \[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\int \frac {\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{d+e\,x} \,d x \] Input:
int(log(c*(a + b*x)^p)/(d + e*x),x)
Output:
int(log(c*(a + b*x)^p)/(d + e*x), x)
\[ \int \frac {\log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (\left (b x +a \right )^{p} c \right )}{b e \,x^{2}+a e x +b d x +a d}d x \right ) a e p -2 \left (\int \frac {\mathrm {log}\left (\left (b x +a \right )^{p} c \right )}{b e \,x^{2}+a e x +b d x +a d}d x \right ) b d p +\mathrm {log}\left (\left (b x +a \right )^{p} c \right )^{2}}{2 e p} \] Input:
int(log(c*(b*x+a)^p)/(e*x+d),x)
Output:
(2*int(log((a + b*x)**p*c)/(a*d + a*e*x + b*d*x + b*e*x**2),x)*a*e*p - 2*i nt(log((a + b*x)**p*c)/(a*d + a*e*x + b*d*x + b*e*x**2),x)*b*d*p + log((a + b*x)**p*c)**2)/(2*e*p)