\(\int \frac {a+b \log (c x)}{(d+\frac {e}{x}) x^4} \, dx\) [344]

Optimal result

Integrand size = 21, antiderivative size = 121 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=-\frac {b}{4 e x^2}+\frac {b d}{e^2 x}-\frac {a+b \log (c x)}{2 e x^2}+\frac {d (a+b \log (c x))}{e^2 x}+\frac {d^2 (a+b \log (c x))^2}{2 b e^3}-\frac {d^2 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{e^3}-\frac {b d^2 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^3} \]

[Out]

-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*l
n(c*x))*ln(1+d*x/e)/e^3-b*d^2*polylog(2,-d*x/e)/e^3

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {269, 46, 2393, 2341, 2338, 2354, 2438} \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\frac {d^2 (a+b \log (c x))^2}{2 b e^3}-\frac {d^2 \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{e^3}+\frac {d (a+b \log (c x))}{e^2 x}-\frac {a+b \log (c x)}{2 e x^2}-\frac {b d^2 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^3}+\frac {b d}{e^2 x}-\frac {b}{4 e x^2} \]

[In]

Int[(a + b*Log[c*x])/((d + e/x)*x^4),x]

[Out]

-1/4*b/(e*x^2) + (b*d)/(e^2*x) - (a + b*Log[c*x])/(2*e*x^2) + (d*(a + b*Log[c*x]))/(e^2*x) + (d^2*(a + b*Log[c
*x])^2)/(2*b*e^3) - (d^2*(a + b*Log[c*x])*Log[1 + (d*x)/e])/e^3 - (b*d^2*PolyLog[2, -((d*x)/e)])/e^3

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2341

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
]

Rule 2354

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[Log[1 + e*(x/d)]*((a +
b*Log[c*x^n])^p/e), x] - Dist[b*n*(p/e), Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a+b \log (c x)}{e x^3}-\frac {d (a+b \log (c x))}{e^2 x^2}+\frac {d^2 (a+b \log (c x))}{e^3 x}-\frac {d^3 (a+b \log (c x))}{e^3 (e+d x)}\right ) \, dx \\ & = \frac {d^2 \int \frac {a+b \log (c x)}{x} \, dx}{e^3}-\frac {d^3 \int \frac {a+b \log (c x)}{e+d x} \, dx}{e^3}-\frac {d \int \frac {a+b \log (c x)}{x^2} \, dx}{e^2}+\frac {\int \frac {a+b \log (c x)}{x^3} \, dx}{e} \\ & = -\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 {\left (b d^2\right ) \int \frac {\log \left (1+\frac {d x}{e}\right )}{x} \, dx}{e^3} \\ & = -\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 \text {Li}_2\left (-\frac {d x}{e}\right )}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=-\frac {\frac {b e^2}{x^2}-\frac {4 b d e}{x}+\frac {2 e^2 (a+b \log (c x))}{x^2}-\frac {4 d e (a+b \log (c x))}{x}-\frac {2 d^2 (a+b \log (c x))^2}{b}+4 d^2 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )+4 b d^2 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{4 e^3} \]

[In]

Integrate[(a + b*Log[c*x])/((d + e/x)*x^4),x]

[Out]

-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*Lo
g[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

Maple [A] (verified)

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	\(c^{3} \left (a \left (-\frac {1}{2 e \,c^{3} x^{2}}+\frac {d^{2} \ln \left (x c \right )}{e^{3} c^{3}}+\frac {d}{e^{2} c^{3} x}-\frac {d^{2} \ln \left (c d x +c e \right )}{e^{3} c^{3}}\right )+b \left (\frac {-\frac {\ln \left (x c \right )}{2 x^{2} c^{2}}-\frac {1}{4 x^{2} c^{2}}}{e c}+\frac {d^{2} \ln \left (x c \right )^{2}}{2 e^{3} c^{3}}-\frac {d \left (-\frac {\ln \left (x c \right )}{x c}-\frac {1}{x c}\right )}{e^{2} c^{2}}-\frac {d^{3} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{e^{3} c^{3}}\right )\right )\)	\(199\)

[In]

int((a+b*ln(x*c))/(d+e/x)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int { \frac {b \log \left (c x\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^4,x, algorithm="fricas")

[Out]

integral((b*log(c*x) + a)/(d*x^4 + e*x^3), x)

Sympy [A] (verification not implemented)

Time = 42.65 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.08 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=- \frac {a d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {a d^{2} \log {\left (x \right )}}{e^{3}} + \frac {a d}{e^{2} x} - \frac {a}{2 e x^{2}} + \frac {b d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{e^{3}} - \frac {b d^{2} \log {\left (x \right )}^{2}}{2 e^{3}} + \frac {b d^{2} \log {\left (x \right )} \log {\left (c x \right )}}{e^{3}} + \frac {b d \log {\left (c x \right )}}{e^{2} x} + \frac {b d}{e^{2} x} - \frac {b \log {\left (c x \right )}}{2 e x^{2}} - \frac {b}{4 e x^{2}} \]

[In]

integrate((a+b*ln(c*x))/(d+e/x)/x**4,x)

[Out]

-a*d**3*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/e**3 + a*d**2*log(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), (-meijerg(((), (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*l
og(c*x)/(e**2*x) + b*d/(e**2*x) - b*log(c*x)/(2*e*x**2) - b/(4*e*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.25 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=-\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b d^{2}}{e^{3}} - \frac {{\left (b d^{2} \log \left (c\right ) + a d^{2}\right )} \log \left (d x + e\right )}{e^{3}} + \frac {2 \, b d^{2} x^{2} \log \left (x\right )^{2} - 2 \, a e^{2} - {\left (2 \, e^{2} \log \left (c\right ) + e^{2}\right )} b + 4 \, {\left (a d e + {\left (d e \log \left (c\right ) + d e\right )} b\right )} x + 2 \, {\left (2 \, b d e x - b e^{2} + 2 \, {\left (b d^{2} \log \left (c\right ) + a d^{2}\right )} x^{2}\right )} \log \left (x\right )}{4 \, e^{3} x^{2}} \]

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^4,x, algorithm="maxima")

[Out]

-(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)

Giac [F]

\[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int { \frac {b \log \left (c x\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^4,x, algorithm="giac")

[Out]

integrate((b*log(c*x) + a)/((d + e/x)*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^4} \, dx=\int \frac {a+b\,\ln \left (c\,x\right )}{x^4\,\left (d+\frac {e}{x}\right )} \,d x \]

[In]

int((a + b*log(c*x))/(x^4*(d + e/x)),x)

[Out]

int((a + b*log(c*x))/(x^4*(d + e/x)), x)