\(\int \frac {x^3 (a+b \log (c x^n))}{d+e x} \, dx\) [31]

Optimal result

Integrand size = 21, antiderivative size = 148 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\frac {a d^2 x}{e^3}-\frac {b d^2 n x}{e^3}+\frac {b d n x^2}{4 e^2}-\frac {b n x^3}{9 e}+\frac {b d^2 x \log \left (c x^n\right )}{e^3}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}-\frac {b d^3 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {45, 2393, 2332, 2341, 2354, 2438} \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=-\frac {d^3 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}+\frac {a d^2 x}{e^3}+\frac {b d^2 x \log \left (c x^n\right )}{e^3}-\frac {b d^3 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}-\frac {b d^2 n x}{e^3}+\frac {b d n x^2}{4 e^2}-\frac {b n x^3}{9 e} \]

[In]

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

[Out]

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

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 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 {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {d x \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}\right ) \, dx \\ & = \frac {d^2 \int \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^3}-\frac {d^3 \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^3}-\frac {d \int x \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^2}+\frac {\int x^2 \left (a+b \log \left (c x^n\right )\right ) \, dx}{e} \\ & = \frac {a d^2 x}{e^3}+\frac {b d n x^2}{4 e^2}-\frac {b n x^3}{9 e}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}+\frac {\left (b d^2\right ) \int \log \left (c x^n\right ) \, dx}{e^3}+\frac {\left (b d^3 n\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^4} \\ & = \frac {a d^2 x}{e^3}-\frac {b d^2 n x}{e^3}+\frac {b d n x^2}{4 e^2}-\frac {b n x^3}{9 e}+\frac {b d^2 x \log \left (c x^n\right )}{e^3}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}-\frac {b d^3 n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\frac {36 a d^2 e x-36 b d^2 e n x-18 a d e^2 x^2+9 b d e^2 n x^2+12 a e^3 x^3-4 b e^3 n x^3-36 a d^3 \log \left (1+\frac {e x}{d}\right )+6 b \log \left (c x^n\right ) \left (e x \left (6 d^2-3 d e x+2 e^2 x^2\right )-6 d^3 \log \left (1+\frac {e x}{d}\right )\right )-36 b d^3 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{36 e^4} \]

[In]

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

[Out]

(36*a*d^2*e*x - 36*b*d^2*e*n*x - 18*a*d*e^2*x^2 + 9*b*d*e^2*n*x^2 + 12*a*e^3*x^3 - 4*b*e^3*n*x^3 - 36*a*d^3*Lo
g[1 + (e*x)/d] + 6*b*Log[c*x^n]*(e*x*(6*d^2 - 3*d*e*x + 2*e^2*x^2) - 6*d^3*Log[1 + (e*x)/d]) - 36*b*d^3*n*Poly
Log[2, -((e*x)/d)])/(36*e^4)

Maple [C] (warning: unable to verify)

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

Time = 0.67 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.84

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) x^{3}}{3 e}-\frac {b \ln \left (x^{n}\right ) d \,x^{2}}{2 e^{2}}+\frac {b \ln \left (x^{n}\right ) x \,d^{2}}{e^{3}}-\frac {b \ln \left (x^{n}\right ) d^{3} \ln \left (e x +d \right )}{e^{4}}-\frac {b n \,x^{3}}{9 e}+\frac {b d n \,x^{2}}{4 e^{2}}-\frac {b \,d^{2} n x}{e^{3}}-\frac {49 b n \,d^{3}}{36 e^{4}}+\frac {b n \,d^{3} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{4}}+\frac {b n \,d^{3} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{3} e^{2} x^{3}-\frac {1}{2} d e \,x^{2}+d^{2} x}{e^{3}}-\frac {d^{3} \ln \left (e x +d \right )}{e^{4}}\right )\)	\(272\)

[In]

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

[Out]

1/3*b*ln(x^n)/e*x^3-1/2*b*ln(x^n)/e^2*d*x^2+b*ln(x^n)/e^3*x*d^2-b*ln(x^n)*d^3/e^4*ln(e*x+d)-1/9*b*n*x^3/e+1/4*
b*d*n*x^2/e^2-b*d^2*n*x/e^3-49/36*b*n*d^3/e^4+b*n*d^3/e^4*ln(e*x+d)*ln(-e*x/d)+b*n*d^3/e^4*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/e^3*(1/3*e^2*x^3-1/2*d*e*x^2+d^2*x)-d^3/e^4*ln(e*x+d))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 16.24 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.80 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=- \frac {a d^{3} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {a d^{2} x}{e^{3}} - \frac {a d x^{2}}{2 e^{2}} + \frac {a x^{3}}{3 e} + \frac {b d^{3} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} - \frac {b d^{2} n x}{e^{3}} + \frac {b d^{2} x \log {\left (c x^{n} \right )}}{e^{3}} + \frac {b d n x^{2}}{4 e^{2}} - \frac {b d x^{2} \log {\left (c x^{n} \right )}}{2 e^{2}} - \frac {b n x^{3}}{9 e} + \frac {b x^{3} \log {\left (c x^{n} \right )}}{3 e} \]

[In]

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

[Out]

-a*d**3*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**3 + a*d**2*x/e**3 - a*d*x**2/(2*e**2) + a*x**3/(
3*e) + b*d**3*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((-polylog(2, e*x*exp_polar(I*pi)/d), (Abs(x) < 1) & (1/A
bs(x) < 1)), (log(d)*log(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))/e**3 - b*d**3*Piecewise((x/d, Eq
(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**3 - b*d**2*n*x/e**3 + b*d**2*x*log(c*x**n)/e**3 + b*d*n*x**2/(
4*e**2) - b*d*x**2*log(c*x**n)/(2*e**2) - b*n*x**3/(9*e) + b*x**3*log(c*x**n)/(3*e)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)