Integrand size = 21, antiderivative size = 84 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=-\frac {b}{e x}-\frac {a+b \log (c x)}{e x}-\frac {d (a+b \log (c x))^2}{2 b e^2}+\frac {d (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{e^2}+\frac {b d \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^2} \]
-b/e/x+(-a-b*ln(c*x))/e/x-1/2*d*(a+b*ln(c*x))^2/b/e^2+d*(a+b*ln(c*x))*ln(1 +d*x/e)/e^2+b*d*polylog(2,-d*x/e)/e^2
Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=-\frac {\frac {2 b e}{x}+\frac {2 e (a+b \log (c x))}{x}+\frac {d (a+b \log (c x))^2}{b}-2 d (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )-2 b d \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{2 e^2} \]
-1/2*((2*b*e)/x + (2*e*(a + b*Log[c*x]))/x + (d*(a + b*Log[c*x])^2)/b - 2* d*(a + b*Log[c*x])*Log[1 + (d*x)/e] - 2*b*d*PolyLog[2, -((d*x)/e)])/e^2
Time = 0.40 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2005, 2780, 2741, 2779, 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.
\(\displaystyle \int \frac {a+b \log (c x)}{x^3 \left (d+\frac {e}{x}\right )} \, dx\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \int \frac {a+b \log (c x)}{x^2 (d x+e)}dx\) |
\(\Big \downarrow \) 2780 |
\(\displaystyle \frac {\int \frac {a+b \log (c x)}{x^2}dx}{e}-\frac {d \int \frac {a+b \log (c x)}{x (e+d x)}dx}{e}\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {-\frac {a+b \log (c x)}{x}-\frac {b}{x}}{e}-\frac {d \int \frac {a+b \log (c x)}{x (e+d x)}dx}{e}\) |
\(\Big \downarrow \) 2779 |
\(\displaystyle \frac {-\frac {a+b \log (c x)}{x}-\frac {b}{x}}{e}-\frac {d \left (\frac {b \int \frac {\log \left (\frac {e}{d x}+1\right )}{x}dx}{e}-\frac {\log \left (\frac {e}{d x}+1\right ) (a+b \log (c x))}{e}\right )}{e}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {-\frac {a+b \log (c x)}{x}-\frac {b}{x}}{e}-\frac {d \left (\frac {b \operatorname {PolyLog}\left (2,-\frac {e}{d x}\right )}{e}-\frac {\log \left (\frac {e}{d x}+1\right ) (a+b \log (c x))}{e}\right )}{e}\) |
(-(b/x) - (a + b*Log[c*x])/x)/e - (d*(-((Log[1 + e/(d*x)]*(a + b*Log[c*x]) )/e) + (b*PolyLog[2, -(e/(d*x))])/e))/e
3.4.43.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)* (x_)^(r_.)), x_Symbol] :> Simp[1/d Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Simp[e/d Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /; Fre eQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]
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]
Time = 0.05 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.37
method | result | size |
risch | \(\frac {a d \ln \left (d x +e \right )}{e^{2}}-\frac {a}{e x}-\frac {a d \ln \left (x \right )}{e^{2}}+\frac {b d \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{e^{2}}+\frac {b d \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{e^{2}}-\frac {b \ln \left (x c \right )}{e x}-\frac {b}{e x}-\frac {b d \ln \left (x c \right )^{2}}{2 e^{2}}\) | \(115\) |
parts | \(a \left (\frac {d \ln \left (d x +e \right )}{e^{2}}-\frac {1}{e x}-\frac {d \ln \left (x \right )}{e^{2}}\right )+\frac {b d \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{e^{2}}+\frac {b d \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{e^{2}}-\frac {b \ln \left (x c \right )}{e x}-\frac {b}{e x}-\frac {b d \ln \left (x c \right )^{2}}{2 e^{2}}\) | \(115\) |
derivativedivides | \(c^{2} \left (a \left (-\frac {1}{e \,c^{2} x}-\frac {d \ln \left (x c \right )}{e^{2} c^{2}}+\frac {d \ln \left (c d x +c e \right )}{e^{2} c^{2}}\right )+b \left (-\frac {d \ln \left (x c \right )^{2}}{2 e^{2} c^{2}}+\frac {-\frac {\ln \left (x c \right )}{x c}-\frac {1}{x c}}{e c}+\frac {d^{2} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{e^{2} c^{2}}\right )\right )\) | \(151\) |
default | \(c^{2} \left (a \left (-\frac {1}{e \,c^{2} x}-\frac {d \ln \left (x c \right )}{e^{2} c^{2}}+\frac {d \ln \left (c d x +c e \right )}{e^{2} c^{2}}\right )+b \left (-\frac {d \ln \left (x c \right )^{2}}{2 e^{2} c^{2}}+\frac {-\frac {\ln \left (x c \right )}{x c}-\frac {1}{x c}}{e c}+\frac {d^{2} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{e^{2} c^{2}}\right )\right )\) | \(151\) |
a*d/e^2*ln(d*x+e)-a/e/x-a*d/e^2*ln(x)+b/e^2*d*dilog((c*d*x+c*e)/e/c)+b/e^2 *d*ln(x*c)*ln((c*d*x+c*e)/e/c)-b/e*ln(x*c)/x-b/e/x-1/2*b/e^2*d*ln(x*c)^2
\[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\int { \frac {b \log \left (c x\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{3}} \,d x } \]
Time = 37.73 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.45 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\frac {a d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {a d \log {\left (x \right )}}{e^{2}} - \frac {a}{e x} - \frac {b d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{e^{2}} + \frac {b d \log {\left (x \right )}^{2}}{2 e^{2}} - \frac {b d \log {\left (x \right )} \log {\left (c x \right )}}{e^{2}} - \frac {b \log {\left (c x \right )}}{e x} - \frac {b}{e x} \]
a*d**2*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/e**2 - a*d*log(x )/e**2 - a/(e*x) - b*d**2*Piecewise((x/e, Eq(d, 0)), (Piecewise((-polylog( 2, d*x*exp_polar(I*pi)/e), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(e)*log(x) - polylog(2, d*x*exp_polar(I*pi)/e), Abs(x) < 1), (-log(e)*log(1/x) - poly log(2, d*x*exp_polar(I*pi)/e), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(e) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(e) - poly log(2, d*x*exp_polar(I*pi)/e), True))/d, True))/e**2 + b*d**2*Piecewise((x /e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x)/e**2 + b*d*log(x)**2/(2*e* *2) - b*d*log(x)*log(c*x)/e**2 - b*log(c*x)/(e*x) - b/(e*x)
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b d}{e^{2}} + \frac {{\left (b d \log \left (c\right ) + a d\right )} \log \left (d x + e\right )}{e^{2}} - \frac {b d x \log \left (x\right )^{2} + 2 \, {\left (e \log \left (c\right ) + e\right )} b + 2 \, a e + 2 \, {\left (b e + {\left (b d \log \left (c\right ) + a d\right )} x\right )} \log \left (x\right )}{2 \, e^{2} x} \]
(log(d*x/e + 1)*log(x) + dilog(-d*x/e))*b*d/e^2 + (b*d*log(c) + a*d)*log(d *x + e)/e^2 - 1/2*(b*d*x*log(x)^2 + 2*(e*log(c) + e)*b + 2*a*e + 2*(b*e + (b*d*log(c) + a*d)*x)*log(x))/(e^2*x)
\[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\int { \frac {b \log \left (c x\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\int \frac {a+b\,\ln \left (c\,x\right )}{x^3\,\left (d+\frac {e}{x}\right )} \,d x \]