\(\int \frac {x^3 (a+b \log (c x^n))}{d+\frac {e}{x}} \, dx\) [329]

Optimal result

Integrand size = 23, antiderivative size = 185 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=-\frac {a e^3 x}{d^4}+\frac {b e^3 n x}{d^4}-\frac {b e^2 n x^2}{4 d^3}+\frac {b e n x^3}{9 d^2}-\frac {b n x^4}{16 d}-\frac {b e^3 x \log \left (c x^n\right )}{d^4}+\frac {e^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^3}-\frac {e x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d}+\frac {e^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^5}+\frac {b e^4 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^5} \]

[Out]

-a*e^3*x/d^4+b*e^3*n*x/d^4-1/4*b*e^2*n*x^2/d^3+1/9*b*e*n*x^3/d^2-1/16*b*n*x^4/d-b*e^3*x*ln(c*x^n)/d^4+1/2*e^2*
x^2*(a+b*ln(c*x^n))/d^3-1/3*e*x^3*(a+b*ln(c*x^n))/d^2+1/4*x^4*(a+b*ln(c*x^n))/d+e^4*(a+b*ln(c*x^n))*ln(1+d*x/e
)/d^5+b*e^4*n*polylog(2,-d*x/e)/d^5

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {269, 45, 2393, 2332, 2341, 2354, 2438} \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {e^4 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^5}+\frac {e^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^3}-\frac {e x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d}-\frac {a e^3 x}{d^4}-\frac {b e^3 x \log \left (c x^n\right )}{d^4}+\frac {b e^4 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^5}+\frac {b e^3 n x}{d^4}-\frac {b e^2 n x^2}{4 d^3}+\frac {b e n x^3}{9 d^2}-\frac {b n x^4}{16 d} \]

[In]

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

[Out]

-((a*e^3*x)/d^4) + (b*e^3*n*x)/d^4 - (b*e^2*n*x^2)/(4*d^3) + (b*e*n*x^3)/(9*d^2) - (b*n*x^4)/(16*d) - (b*e^3*x
*Log[c*x^n])/d^4 + (e^2*x^2*(a + b*Log[c*x^n]))/(2*d^3) - (e*x^3*(a + b*Log[c*x^n]))/(3*d^2) + (x^4*(a + b*Log
[c*x^n]))/(4*d) + (e^4*(a + b*Log[c*x^n])*Log[1 + (d*x)/e])/d^5 + (b*e^4*n*PolyLog[2, -((d*x)/e)])/d^5

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{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 {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {e x^2 \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d}+\frac {e^4 \left (a+b \log \left (c x^n\right )\right )}{d^4 (e+d x)}\right ) \, dx \\ & = \frac {\int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx}{d}-\frac {e \int x^2 \left (a+b \log \left (c x^n\right )\right ) \, dx}{d^2}+\frac {e^2 \int x \left (a+b \log \left (c x^n\right )\right ) \, dx}{d^3}-\frac {e^3 \int \left (a+b \log \left (c x^n\right )\right ) \, dx}{d^4}+\frac {e^4 \int \frac {a+b \log \left (c x^n\right )}{e+d x} \, dx}{d^4} \\ & = -\frac {a e^3 x}{d^4}-\frac {b e^2 n x^2}{4 d^3}+\frac {b e n x^3}{9 d^2}-\frac {b n x^4}{16 d}+\frac {e^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^3}-\frac {e x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d}+\frac {e^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^5}-\frac {\left (b e^3\right ) \int \log \left (c x^n\right ) \, dx}{d^4}-\frac {\left (b e^4 n\right ) \int \frac {\log \left (1+\frac {d x}{e}\right )}{x} \, dx}{d^5} \\ & = -\frac {a e^3 x}{d^4}+\frac {b e^3 n x}{d^4}-\frac {b e^2 n x^2}{4 d^3}+\frac {b e n x^3}{9 d^2}-\frac {b n x^4}{16 d}-\frac {b e^3 x \log \left (c x^n\right )}{d^4}+\frac {e^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^3}-\frac {e x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d}+\frac {e^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^5}+\frac {b e^4 n \text {Li}_2\left (-\frac {d x}{e}\right )}{d^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {-144 a d e^3 x+144 b d e^3 n x-36 b d^2 e^2 n x^2+16 b d^3 e n x^3-9 b d^4 n x^4-144 b d e^3 x \log \left (c x^n\right )+72 d^2 e^2 x^2 \left (a+b \log \left (c x^n\right )\right )-48 d^3 e x^3 \left (a+b \log \left (c x^n\right )\right )+36 d^4 x^4 \left (a+b \log \left (c x^n\right )\right )+144 e^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )+144 b e^4 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{144 d^5} \]

[In]

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

[Out]

(-144*a*d*e^3*x + 144*b*d*e^3*n*x - 36*b*d^2*e^2*n*x^2 + 16*b*d^3*e*n*x^3 - 9*b*d^4*n*x^4 - 144*b*d*e^3*x*Log[
c*x^n] + 72*d^2*e^2*x^2*(a + b*Log[c*x^n]) - 48*d^3*e*x^3*(a + b*Log[c*x^n]) + 36*d^4*x^4*(a + b*Log[c*x^n]) +
144*e^4*(a + b*Log[c*x^n])*Log[1 + (d*x)/e] + 144*b*e^4*n*PolyLog[2, -((d*x)/e)])/(144*d^5)

Maple [C] (warning: unable to verify)

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

Time = 0.34 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.69

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) x^{4}}{4 d}-\frac {b \ln \left (x^{n}\right ) e \,x^{3}}{3 d^{2}}+\frac {b \ln \left (x^{n}\right ) x^{2} e^{2}}{2 d^{3}}-\frac {b \ln \left (x^{n}\right ) x \,e^{3}}{d^{4}}+\frac {b \ln \left (x^{n}\right ) e^{4} \ln \left (d x +e \right )}{d^{5}}-\frac {b n \,e^{4} \ln \left (d x +e \right ) \ln \left (-\frac {d x}{e}\right )}{d^{5}}-\frac {b n \,e^{4} \operatorname {dilog}\left (-\frac {d x}{e}\right )}{d^{5}}-\frac {b n \,x^{4}}{16 d}+\frac {b e n \,x^{3}}{9 d^{2}}-\frac {b \,e^{2} n \,x^{2}}{4 d^{3}}+\frac {b \,e^{3} n x}{d^{4}}+\frac {205 b n \,e^{4}}{144 d^{5}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{4} d^{3} x^{4}-\frac {1}{3} e \,d^{2} x^{3}+\frac {1}{2} d \,e^{2} x^{2}-x \,e^{3}}{d^{4}}+\frac {e^{4} \ln \left (d x +e \right )}{d^{5}}\right )\)	\(313\)

[In]

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

[Out]

1/4*b*ln(x^n)/d*x^4-1/3*b*ln(x^n)/d^2*e*x^3+1/2*b*ln(x^n)/d^3*x^2*e^2-b*ln(x^n)/d^4*x*e^3+b*ln(x^n)*e^4/d^5*ln
(d*x+e)-b*n*e^4/d^5*ln(d*x+e)*ln(-d*x/e)-b*n*e^4/d^5*dilog(-d*x/e)-1/16*b*n*x^4/d+1/9*b*e*n*x^3/d^2-1/4*b*e^2*
n*x^2/d^3+b*e^3*n*x/d^4+205/144*b*n*e^4/d^5+(-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^4*(1/4*d
^3*x^4-1/3*e*d^2*x^3+1/2*d*e^2*x^2-x*e^3)+e^4/d^5*ln(d*x+e))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 75.86 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.71 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {a x^{4}}{4 d} - \frac {a e x^{3}}{3 d^{2}} + \frac {a e^{2} x^{2}}{2 d^{3}} + \frac {a e^{4} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{4}} - \frac {a e^{3} x}{d^{4}} - \frac {b n x^{4}}{16 d} + \frac {b x^{4} \log {\left (c x^{n} \right )}}{4 d} + \frac {b e n x^{3}}{9 d^{2}} - \frac {b e x^{3} \log {\left (c x^{n} \right )}}{3 d^{2}} - \frac {b e^{2} n x^{2}}{4 d^{3}} + \frac {b e^{2} x^{2} \log {\left (c x^{n} \right )}}{2 d^{3}} - \frac {b e^{4} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{4}} + \frac {b e^{4} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{4}} + \frac {b e^{3} n x}{d^{4}} - \frac {b e^{3} x \log {\left (c x^{n} \right )}}{d^{4}} \]

[In]

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

[Out]

a*x**4/(4*d) - a*e*x**3/(3*d**2) + a*e**2*x**2/(2*d**3) + a*e**4*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, T
rue))/d**4 - a*e**3*x/d**4 - b*n*x**4/(16*d) + b*x**4*log(c*x**n)/(4*d) + b*e*n*x**3/(9*d**2) - b*e*x**3*log(c
*x**n)/(3*d**2) - b*e**2*n*x**2/(4*d**3) + b*e**2*x**2*log(c*x**n)/(2*d**3) - b*e**4*n*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))/d**4 + b*e**4*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*
x**n)/d**4 + b*e**3*n*x/d**4 - b*e**3*x*log(c*x**n)/d**4

Maxima [F]

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

[In]

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

[Out]

1/12*a*(12*e^4*log(d*x + e)/d^5 + (3*d^3*x^4 - 4*d^2*e*x^3 + 6*d*e^2*x^2 - 12*e^3*x)/d^4) + b*integrate((x^4*l
og(c) + x^4*log(x^n))/(d*x + e), x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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