\(\int \frac {a+b \log (c x^n)}{x^3 (d+e x)} \, dx\) [37]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)} \, dx=-\frac {b n}{4 d x^2}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{2 d x^2}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {e^2 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2380, 2341, 2379, 2438} \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)} \, dx=-\frac {e^2 \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{2 d x^2}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^3}+\frac {b e n}{d^2 x}-\frac {b n}{4 d x^2} \]

[In]

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

[Out]

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

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 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 2380

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

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^3} \, dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)} \, dx}{d} \\ & = -\frac {b n}{4 d x^2}-\frac {a+b \log \left (c x^n\right )}{2 d x^2}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{d^2}+\frac {e^2 \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx}{d^2} \\ & = -\frac {b n}{4 d x^2}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{2 d x^2}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {e^2 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {\left (b e^2 n\right ) \int \frac {\log \left (1+\frac {d}{e x}\right )}{x} \, dx}{d^3} \\ & = -\frac {b n}{4 d x^2}+\frac {b e n}{d^2 x}-\frac {a+b \log \left (c x^n\right )}{2 d x^2}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2 x}-\frac {e^2 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {b e^2 n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.13 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)} \, dx=-\frac {\frac {b d^2 n}{x^2}-\frac {4 b d e n}{x}+\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {4 d e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+4 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+4 b e^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 d^3} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.45 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.40

method	result	size
risch	\(-\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (e x +d \right )}{d^{3}}-\frac {b \ln \left (x^{n}\right )}{2 d \,x^{2}}+\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {b \ln \left (x^{n}\right ) e}{d^{2} x}+\frac {b e n}{d^{2} x}-\frac {b n}{4 d \,x^{2}}-\frac {b n \,e^{2} \ln \left (x \right )^{2}}{2 d^{3}}+\frac {b n \,e^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{3}}+\frac {b n \,e^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {e^{2} \ln \left (e x +d \right )}{d^{3}}-\frac {1}{2 d \,x^{2}}+\frac {e^{2} \ln \left (x \right )}{d^{3}}+\frac {e}{d^{2} x}\right )\)	\(264\)

[In]

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

[Out]

-b*ln(x^n)*e^2/d^3*ln(e*x+d)-1/2*b*ln(x^n)/d/x^2+b*ln(x^n)*e^2/d^3*ln(x)+b*ln(x^n)*e/d^2/x+b*e*n/d^2/x-1/4*b*n
/d/x^2-1/2*b*n*e^2/d^3*ln(x)^2+b*n*e^2/d^3*ln(e*x+d)*ln(-e*x/d)+b*n*e^2/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)*(-e^2/d^3*ln(e*x+d)-1/2/d/x^2+e^2/d^3*ln(x)+e/d^2/x)

Fricas [F]

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 35.16 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.41 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)} \, dx=- \frac {a}{2 d x^{2}} + \frac {a e}{d^{2} x} - \frac {a e^{3} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {a e^{2} \log {\left (x \right )}}{d^{3}} - \frac {b n}{4 d x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 d x^{2}} + \frac {b e n}{d^{2} x} + \frac {b e \log {\left (c x^{n} \right )}}{d^{2} x} + \frac {b e^{3} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {b e^{3} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} - \frac {b e^{2} n \log {\left (x \right )}^{2}}{2 d^{3}} + \frac {b e^{2} \log {\left (x \right )} \log {\left (c x^{n} \right )}}{d^{3}} \]

[In]

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

[Out]

-a/(2*d*x**2) + a*e/(d**2*x) - a*e**3*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/d**3 + a*e**2*log(x)/
d**3 - b*n/(4*d*x**2) - b*log(c*x**n)/(2*d*x**2) + b*e*n/(d**2*x) + b*e*log(c*x**n)/(d**2*x) + b*e**3*n*Piecew
ise((x/d, Eq(e, 0)), (Piecewise((-polylog(2, e*x*exp_polar(I*pi)/d), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*l
og(x) - polylog(2, e*x*exp_polar(I*pi)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi)/d),
1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d
) - polylog(2, e*x*exp_polar(I*pi)/d), True))/e, True))/d**3 - b*e**3*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)
/e, True))*log(c*x**n)/d**3 - b*e**2*n*log(x)**2/(2*d**3) + b*e**2*log(x)*log(c*x**n)/d**3

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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