\(\int \frac {a+b \log (c x^n)}{x (d+e x)^3} \, dx\) [50]

Optimal result

Integrand size = 21, antiderivative size = 134 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=-\frac {b n}{2 d^2 (d+e x)}-\frac {b n \log (x)}{2 d^3}+\frac {a+b \log \left (c x^n\right )}{2 d (d+e x)^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}-\frac {\log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {3 b n \log (d+e x)}{2 d^3}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2389, 2379, 2438, 2351, 31, 2356, 46} \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)}+\frac {a+b \log \left (c x^n\right )}{2 d (d+e x)^2}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^3}+\frac {3 b n \log (d+e x)}{2 d^3}-\frac {b n \log (x)}{2 d^3}-\frac {b n}{2 d^2 (d+e x)} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 2351

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

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)
*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Dist[b*n*(p/(e*(q + 1))), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2379

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] + Dist[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]

Rule 2389

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[(d
+ e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x), x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=\frac {\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{b n}-2 b n (\log (x)-\log (d+e x))+b n \left (-\frac {d}{d+e x}-\log (x)+\log (d+e x)\right )-2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-2 b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 d^3} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.90 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.04

method	result	size
risch	\(-\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{d^{3}}+\frac {b \ln \left (x^{n}\right )}{d^{2} \left (e x +d \right )}+\frac {b \ln \left (x^{n}\right )}{2 d \left (e x +d \right )^{2}}+\frac {b \ln \left (x^{n}\right ) \ln \left (x \right )}{d^{3}}-\frac {b n}{2 d^{2} \left (e x +d \right )}+\frac {3 b n \ln \left (e x +d \right )}{2 d^{3}}-\frac {3 b n \ln \left (x \right )}{2 d^{3}}-\frac {b n \ln \left (x \right )^{2}}{2 d^{3}}+\frac {b n \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{3}}+\frac {b n \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {\ln \left (e x +d \right )}{d^{3}}+\frac {1}{d^{2} \left (e x +d \right )}+\frac {1}{2 d \left (e x +d \right )^{2}}+\frac {\ln \left (x \right )}{d^{3}}\right )\)	\(273\)

[In]

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

[Out]

-b*ln(x^n)/d^3*ln(e*x+d)+b*ln(x^n)/d^2/(e*x+d)+1/2*b*ln(x^n)/d/(e*x+d)^2+b*ln(x^n)/d^3*ln(x)-1/2*b*n/d^2/(e*x+
d)+3/2*b*n*ln(e*x+d)/d^3-3/2*b*n*ln(x)/d^3-1/2*b*n/d^3*ln(x)^2+b*n/d^3*ln(e*x+d)*ln(-e*x/d)+b*n/d^3*dilog(-e*x
/d)+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/2*I*b*Pi*csgn(I*x^
n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3+b*ln(c)+a)*(-1/d^3*ln(e*x+d)+1/d^2/(e*x+d)+1/2/d/(e*x+d)^2+1/d^3
*ln(x))

Fricas [F]

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

[In]

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

[Out]

integral((b*log(c*x^n) + a)/(e^3*x^4 + 3*d*e^2*x^3 + 3*d^2*e*x^2 + d^3*x), x)

Sympy [A] (verification not implemented)

Time = 37.00 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.63 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx=- \frac {a e \left (\begin {cases} \frac {x}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x\right )^{2}} & \text {otherwise} \end {cases}\right )}{d} - \frac {a e \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a e \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {a \log {\left (x \right )}}{d^{3}} + \frac {b e^{2} n \left (\begin {cases} - \frac {1}{e^{3} x} & \text {for}\: d = 0 \\- \frac {1}{2 d e^{2} + 2 e^{3} x} - \frac {\log {\left (d + e x \right )}}{2 d e^{2}} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {b e^{2} \left (\begin {cases} \frac {1}{e^{3} x} & \text {for}\: d = 0 \\- \frac {1}{2 d \left (\frac {d}{x} + e\right )^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} - \frac {2 b e n \left (\begin {cases} - \frac {1}{e^{2} x} & \text {for}\: d = 0 \\- \frac {\log {\left (d^{2} + d e x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{d^{2}} + \frac {2 b e \left (\begin {cases} \frac {1}{e^{2} x} & \text {for}\: d = 0 \\- \frac {1}{\frac {d^{2}}{x} + d e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} + \frac {b n \left (\begin {cases} - \frac {1}{e x} & \text {for}\: d = 0 \\\frac {\begin {cases} \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\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 e^{i \pi }}{e x}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\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 e^{i \pi }}{e x}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {b \left (\begin {cases} \frac {1}{e x} & \text {for}\: d = 0 \\\frac {\log {\left (\frac {d}{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} \]

[In]

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

[Out]

-a*e*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/d - a*e*Piecewise((x/d**2, Eq(e, 0)), (-1/(d
*e + e**2*x), True))/d**2 - a*e*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/d**3 + a*log(x)/d**3 + b*e*
*2*n*Piecewise((-1/(e**3*x), Eq(d, 0)), (-1/(2*d*e**2 + 2*e**3*x) - log(d + e*x)/(2*d*e**2), True))/d**2 - b*e
**2*Piecewise((1/(e**3*x), Eq(d, 0)), (-1/(2*d*(d/x + e)**2), True))*log(c*x**n)/d**2 - 2*b*e*n*Piecewise((-1/
(e**2*x), Eq(d, 0)), (-log(d**2 + d*e*x)/(d*e), True))/d**2 + 2*b*e*Piecewise((1/(e**2*x), Eq(d, 0)), (-1/(d**
2/x + d*e), True))*log(c*x**n)/d**2 + b*n*Piecewise((-1/(e*x), Eq(d, 0)), (Piecewise((polylog(2, d*exp_polar(I
*pi)/(e*x)), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(e)*log(x) + polylog(2, d*exp_polar(I*pi)/(e*x)), Abs(x) < 1)
, (-log(e)*log(1/x) + polylog(2, d*exp_polar(I*pi)/(e*x)), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ())
, x)*log(e) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(e) + polylog(2, d*exp_polar(I*pi)/(e*x)), True))/d, T
rue))/d**2 - b*Piecewise((1/(e*x), Eq(d, 0)), (log(d/x + e)/d, True))*log(c*x**n)/d**2

Maxima [F]

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

[In]

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

[Out]

1/2*a*((2*e*x + 3*d)/(d^2*e^2*x^2 + 2*d^3*e*x + d^4) - 2*log(e*x + d)/d^3 + 2*log(x)/d^3) + b*integrate((log(c
) + log(x^n))/(e^3*x^4 + 3*d*e^2*x^3 + 3*d^2*e*x^2 + d^3*x), x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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