\(\int \frac {a+b \log (c x)}{(d+\frac {e}{x}) x^3} \, dx\) [343]

Optimal result

Integrand size = 21, antiderivative size = 84 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=-\frac {b}{e x}-\frac {a+b \log (c x)}{e x}-\frac {d (a+b \log (c x))^2}{2 b e^2}+\frac {d (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{e^2}+\frac {b d \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^2} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {269, 46, 2393, 2341, 2338, 2354, 2438} \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=-\frac {d (a+b \log (c x))^2}{2 b e^2}+\frac {d \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{e^2}-\frac {a+b \log (c x)}{e x}+\frac {b d \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^2}-\frac {b}{e x} \]

[In]

Int[(a + b*Log[c*x])/((d + e/x)*x^3),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=-\frac {\frac {2 b e}{x}+\frac {2 e (a+b \log (c x))}{x}+\frac {d (a+b \log (c x))^2}{b}-2 d (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )-2 b d \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{2 e^2} \]

[In]

Integrate[(a + b*Log[c*x])/((d + e/x)*x^3),x]

[Out]

-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

Maple [A] (verified)

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	\(c^{2} \left (a \left (-\frac {1}{e \,c^{2} x}-\frac {d \ln \left (x c \right )}{e^{2} c^{2}}+\frac {d \ln \left (c d x +c e \right )}{e^{2} c^{2}}\right )+b \left (-\frac {d \ln \left (x c \right )^{2}}{2 e^{2} c^{2}}+\frac {-\frac {\ln \left (x c \right )}{x c}-\frac {1}{x c}}{e c}+\frac {d^{2} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{e^{2} c^{2}}\right )\right )\)	\(151\)

[In]

int((a+b*ln(x*c))/(d+e/x)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\int { \frac {b \log \left (c x\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{3}} \,d x } \]

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^3,x, algorithm="fricas")

[Out]

integral((b*log(c*x) + a)/(d*x^3 + e*x^2), x)

Sympy [A] (verification not implemented)

Time = 37.73 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.45 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\frac {a d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {a d \log {\left (x \right )}}{e^{2}} - \frac {a}{e x} - \frac {b d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{e^{2}} + \frac {b d \log {\left (x \right )}^{2}}{2 e^{2}} - \frac {b d \log {\left (x \right )} \log {\left (c x \right )}}{e^{2}} - \frac {b \log {\left (c x \right )}}{e x} - \frac {b}{e x} \]

[In]

integrate((a+b*ln(c*x))/(d+e/x)/x**3,x)

[Out]

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) - polylog(2, d*x*exp_polar(I*pi)/e), 1/A
bs(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**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)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b d}{e^{2}} + \frac {{\left (b d \log \left (c\right ) + a d\right )} \log \left (d x + e\right )}{e^{2}} - \frac {b d x \log \left (x\right )^{2} + 2 \, {\left (e \log \left (c\right ) + e\right )} b + 2 \, a e + 2 \, {\left (b e + {\left (b d \log \left (c\right ) + a d\right )} x\right )} \log \left (x\right )}{2 \, e^{2} x} \]

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^3,x, algorithm="maxima")

[Out]

(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)

Giac [F]

\[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\int { \frac {b \log \left (c x\right ) + a}{{\left (d + \frac {e}{x}\right )} x^{3}} \,d x } \]

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^3,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x^3} \, dx=\int \frac {a+b\,\ln \left (c\,x\right )}{x^3\,\left (d+\frac {e}{x}\right )} \,d x \]

[In]

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

[Out]

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