Integrand size = 21, antiderivative size = 121 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=-\frac {b}{4 e x^2}+\frac {b d}{e^2 x}-\frac {a+b \log (c x)}{2 e x^2}+\frac {d (a+b \log (c x))}{e^2 x}+\frac {d^2 (a+b \log (c x))^2}{2 b e^3}-\frac {d^2 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{e^3}-\frac {b d^2 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^3} \]
-1/4*b/e/x^2+b*d/e^2/x+1/2*(-a-b*ln(c*x))/e/x^2+d*(a+b*ln(c*x))/e^2/x+1/2* d^2*(a+b*ln(c*x))^2/b/e^3-d^2*(a+b*ln(c*x))*ln(1+d*x/e)/e^3-b*d^2*polylog( 2,-d*x/e)/e^3
Time = 0.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=-\frac {\frac {b e^2}{x^2}-\frac {4 b d e}{x}+\frac {2 e^2 (a+b \log (c x))}{x^2}-\frac {4 d e (a+b \log (c x))}{x}-\frac {2 d^2 (a+b \log (c x))^2}{b}+4 d^2 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )+4 b d^2 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{4 e^3} \]
-1/4*((b*e^2)/x^2 - (4*b*d*e)/x + (2*e^2*(a + b*Log[c*x]))/x^2 - (4*d*e*(a + b*Log[c*x]))/x - (2*d^2*(a + b*Log[c*x])^2)/b + 4*d^2*(a + b*Log[c*x])* Log[1 + (d*x)/e] + 4*b*d^2*PolyLog[2, -((d*x)/e)])/e^3
Time = 0.56 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2005, 2780, 2741, 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.
| \(\displaystyle \int \frac {a+b \log (c x)}{x^4 \left (d+\frac {e}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \int \frac {a+b \log (c x)}{x^3 (d x+e)}dx\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {\int \frac {a+b \log (c x)}{x^3}dx}{e}-\frac {d \int \frac {a+b \log (c x)}{x^2 (e+d x)}dx}{e}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {-\frac {a+b \log (c x)}{2 x^2}-\frac {b}{4 x^2}}{e}-\frac {d \int \frac {a+b \log (c x)}{x^2 (e+d x)}dx}{e}\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {-\frac {a+b \log (c x)}{2 x^2}-\frac {b}{4 x^2}}{e}-\frac {d \left (\frac {\int \frac {a+b \log (c x)}{x^2}dx}{e}-\frac {d \int \frac {a+b \log (c x)}{x (e+d x)}dx}{e}\right )}{e}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {-\frac {a+b \log (c x)}{2 x^2}-\frac {b}{4 x^2}}{e}-\frac {d \left (\frac {-\frac {a+b \log (c x)}{x}-\frac {b}{x}}{e}-\frac {d \int \frac {a+b \log (c x)}{x (e+d x)}dx}{e}\right )}{e}\) |
| \(\Big \downarrow \) 2779 |
| \(\displaystyle \frac {-\frac {a+b \log (c x)}{2 x^2}-\frac {b}{4 x^2}}{e}-\frac {d \left (\frac {-\frac {a+b \log (c x)}{x}-\frac {b}{x}}{e}-\frac {d \left (\frac {b \int \frac {\log \left (\frac {e}{d x}+1\right )}{x}dx}{e}-\frac {\log \left (\frac {e}{d x}+1\right ) (a+b \log (c x))}{e}\right )}{e}\right )}{e}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {-\frac {a+b \log (c x)}{2 x^2}-\frac {b}{4 x^2}}{e}-\frac {d \left (\frac {-\frac {a+b \log (c x)}{x}-\frac {b}{x}}{e}-\frac {d \left (\frac {b \operatorname {PolyLog}\left (2,-\frac {e}{d x}\right )}{e}-\frac {\log \left (\frac {e}{d x}+1\right ) (a+b \log (c x))}{e}\right )}{e}\right )}{e}\) |
(-1/4*b/x^2 - (a + b*Log[c*x])/(2*x^2))/e - (d*((-(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))/e
3.4.44.3.1 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 = 0.06 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.30
| method | result | size |
| parts | \(a \left (-\frac {d^{2} \ln \left (d x +e \right )}{e^{3}}-\frac {1}{2 e \,x^{2}}+\frac {d^{2} \ln \left (x \right )}{e^{3}}+\frac {d}{e^{2} x}\right )-\frac {b \,d^{2} \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{e^{3}}-\frac {b \,d^{2} \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{e^{3}}+\frac {b d \ln \left (x c \right )}{e^{2} x}+\frac {b d}{e^{2} x}+\frac {b \,d^{2} \ln \left (x c \right )^{2}}{2 e^{3}}-\frac {b \ln \left (x c \right )}{2 e \,x^{2}}-\frac {b}{4 e \,x^{2}}\) | \(157\) |
| risch | \(-\frac {a \,d^{2} \ln \left (d x +e \right )}{e^{3}}-\frac {a}{2 e \,x^{2}}+\frac {a \,d^{2} \ln \left (x \right )}{e^{3}}+\frac {a d}{e^{2} x}-\frac {b \,d^{2} \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{e^{3}}-\frac {b \,d^{2} \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{e^{3}}+\frac {b d \ln \left (x c \right )}{e^{2} x}+\frac {b d}{e^{2} x}+\frac {b \,d^{2} \ln \left (x c \right )^{2}}{2 e^{3}}-\frac {b \ln \left (x c \right )}{2 e \,x^{2}}-\frac {b}{4 e \,x^{2}}\) | \(158\) |
| derivativedivides | \(c^{3} \left (a \left (-\frac {1}{2 e \,c^{3} x^{2}}+\frac {d^{2} \ln \left (x c \right )}{e^{3} c^{3}}+\frac {d}{e^{2} c^{3} x}-\frac {d^{2} \ln \left (c d x +c e \right )}{e^{3} c^{3}}\right )+b \left (\frac {-\frac {\ln \left (x c \right )}{2 x^{2} c^{2}}-\frac {1}{4 x^{2} c^{2}}}{e c}+\frac {d^{2} \ln \left (x c \right )^{2}}{2 e^{3} c^{3}}-\frac {d \left (-\frac {\ln \left (x c \right )}{x c}-\frac {1}{x c}\right )}{e^{2} c^{2}}-\frac {d^{3} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{e^{3} c^{3}}\right )\right )\) | \(199\) |
| default | \(c^{3} \left (a \left (-\frac {1}{2 e \,c^{3} x^{2}}+\frac {d^{2} \ln \left (x c \right )}{e^{3} c^{3}}+\frac {d}{e^{2} c^{3} x}-\frac {d^{2} \ln \left (c d x +c e \right )}{e^{3} c^{3}}\right )+b \left (\frac {-\frac {\ln \left (x c \right )}{2 x^{2} c^{2}}-\frac {1}{4 x^{2} c^{2}}}{e c}+\frac {d^{2} \ln \left (x c \right )^{2}}{2 e^{3} c^{3}}-\frac {d \left (-\frac {\ln \left (x c \right )}{x c}-\frac {1}{x c}\right )}{e^{2} c^{2}}-\frac {d^{3} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{e^{3} c^{3}}\right )\right )\) | \(199\) |
a*(-d^2/e^3*ln(d*x+e)-1/2/e/x^2+d^2/e^3*ln(x)+d/e^2/x)-b/e^3*d^2*dilog((c* d*x+c*e)/e/c)-b/e^3*d^2*ln(x*c)*ln((c*d*x+c*e)/e/c)+b/e^2*d*ln(x*c)/x+b*d/ e^2/x+1/2*b/e^3*d^2*ln(x*c)^2-1/2*b/e*ln(x*c)/x^2-1/4*b/e/x^2
\[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int { \frac {b \log \left (c x\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{4}} \,d x } \]
Time = 42.65 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.08 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=- \frac {a d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {a d^{2} \log {\left (x \right )}}{e^{3}} + \frac {a d}{e^{2} x} - \frac {a}{2 e x^{2}} + \frac {b d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{e^{3}} - \frac {b d^{2} \log {\left (x \right )}^{2}}{2 e^{3}} + \frac {b d^{2} \log {\left (x \right )} \log {\left (c x \right )}}{e^{3}} + \frac {b d \log {\left (c x \right )}}{e^{2} x} + \frac {b d}{e^{2} x} - \frac {b \log {\left (c x \right )}}{2 e x^{2}} - \frac {b}{4 e x^{2}} \]
-a*d**3*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/e**3 + a*d**2*l og(x)/e**3 + a*d/(e**2*x) - a/(2*e*x**2) + b*d**3*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) - polylog(2, d*x*exp_polar(I*pi)/e), 1/Abs(x) < 1), (-me ijerg(((), (1, 1)), ((0, 0), ()), x)*log(e) + meijerg(((1, 1), ()), ((), ( 0, 0)), x)*log(e) - polylog(2, d*x*exp_polar(I*pi)/e), True))/d, True))/e* *3 - b*d**3*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x)/e* *3 - b*d**2*log(x)**2/(2*e**3) + b*d**2*log(x)*log(c*x)/e**3 + b*d*log(c*x )/(e**2*x) + b*d/(e**2*x) - b*log(c*x)/(2*e*x**2) - b/(4*e*x**2)
Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.25 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=-\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b d^{2}}{e^{3}} - \frac {{\left (b d^{2} \log \left (c\right ) + a d^{2}\right )} \log \left (d x + e\right )}{e^{3}} + \frac {2 \, b d^{2} x^{2} \log \left (x\right )^{2} - 2 \, a e^{2} - {\left (2 \, e^{2} \log \left (c\right ) + e^{2}\right )} b + 4 \, {\left (a d e + {\left (d e \log \left (c\right ) + d e\right )} b\right )} x + 2 \, {\left (2 \, b d e x - b e^{2} + 2 \, {\left (b d^{2} \log \left (c\right ) + a d^{2}\right )} x^{2}\right )} \log \left (x\right )}{4 \, e^{3} x^{2}} \]
-(log(d*x/e + 1)*log(x) + dilog(-d*x/e))*b*d^2/e^3 - (b*d^2*log(c) + a*d^2 )*log(d*x + e)/e^3 + 1/4*(2*b*d^2*x^2*log(x)^2 - 2*a*e^2 - (2*e^2*log(c) + e^2)*b + 4*(a*d*e + (d*e*log(c) + d*e)*b)*x + 2*(2*b*d*e*x - b*e^2 + 2*(b *d^2*log(c) + a*d^2)*x^2)*log(x))/(e^3*x^2)
\[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int { \frac {b \log \left (c x\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int \frac {a+b\,\ln \left (c\,x\right )}{x^4\,\left (d+\frac {e}{x}\right )} \,d x \]