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

Optimal. Leaf size=149 \[ -\frac {d \log \left (\frac {e x}{d}+1\right ) \left (6 a+6 b \log \left (c x^n\right )+5 b n\right )}{2 e^4}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{2 e^2 (d+e x)}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}+\frac {x (6 a+5 b n)}{2 e^3}+\frac {3 b x \log \left (c x^n\right )}{e^3}-\frac {3 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4}-\frac {3 b n x}{e^3} \]

[Out]

-3*b*n*x/e^3+1/2*(5*b*n+6*a)*x/e^3+3*b*x*ln(c*x^n)/e^3-1/2*x^3*(a+b*ln(c*x^n))/e/(e*x+d)^2-1/2*x^2*(3*a+b*n+3*
b*ln(c*x^n))/e^2/(e*x+d)-1/2*d*(6*a+5*b*n+6*b*ln(c*x^n))*ln(1+e*x/d)/e^4-3*b*d*n*polylog(2,-e*x/d)/e^4

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Rubi [A]  time = 0.22, antiderivative size = 167, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {43, 2351, 2295, 2319, 44, 2314, 31, 2317, 2391} \[ -\frac {3 b d n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 e^4 (d+e x)^2}+\frac {3 d x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}-\frac {3 d \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {a x}{e^3}+\frac {b x \log \left (c x^n\right )}{e^3}-\frac {b d^2 n}{2 e^4 (d+e x)}-\frac {b d n \log (x)}{2 e^4}-\frac {5 b d n \log (d+e x)}{2 e^4}-\frac {b n x}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 43

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 44

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] && L
tQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2314

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 2317

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 2319

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 2351

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 2391

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*} \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{e^3}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)^3}+\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)^2}-\frac {3 d \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}\right ) \, dx\\ &=\frac {\int \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^3}-\frac {(3 d) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^3}+\frac {\left (3 d^2\right ) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^3}-\frac {d^3 \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{e^3}\\ &=\frac {a x}{e^3}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 e^4 (d+e x)^2}+\frac {3 d x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}-\frac {3 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}+\frac {b \int \log \left (c x^n\right ) \, dx}{e^3}+\frac {(3 b d n) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^4}-\frac {\left (b d^3 n\right ) \int \frac {1}{x (d+e x)^2} \, dx}{2 e^4}-\frac {(3 b d n) \int \frac {1}{d+e x} \, dx}{e^3}\\ &=\frac {a x}{e^3}-\frac {b n x}{e^3}+\frac {b x \log \left (c x^n\right )}{e^3}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 e^4 (d+e x)^2}+\frac {3 d x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}-\frac {3 b d n \log (d+e x)}{e^4}-\frac {3 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}-\frac {3 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4}-\frac {\left (b d^3 n\right ) \int \left (\frac {1}{d^2 x}-\frac {e}{d (d+e x)^2}-\frac {e}{d^2 (d+e x)}\right ) \, dx}{2 e^4}\\ &=\frac {a x}{e^3}-\frac {b n x}{e^3}-\frac {b d^2 n}{2 e^4 (d+e x)}-\frac {b d n \log (x)}{2 e^4}+\frac {b x \log \left (c x^n\right )}{e^3}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 e^4 (d+e x)^2}+\frac {3 d x \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}-\frac {5 b d n \log (d+e x)}{2 e^4}-\frac {3 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^4}-\frac {3 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^4}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 150, normalized size = 1.01 \[ \frac {\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac {6 d^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}-6 d \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+2 a e x+2 b e x \log \left (c x^n\right )-6 b d n \text {Li}_2\left (-\frac {e x}{d}\right )+6 b d n (\log (x)-\log (d+e x))-b d n \left (\frac {d}{d+e x}-\log (d+e x)+\log (x)\right )-2 b e n x}{2 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*e*x - 2*b*e*n*x + 2*b*e*x*Log[c*x^n] + (d^3*(a + b*Log[c*x^n]))/(d + e*x)^2 - (6*d^2*(a + b*Log[c*x^n]))/
(d + e*x) + 6*b*d*n*(Log[x] - Log[d + e*x]) - b*d*n*(d/(d + e*x) + Log[x] - Log[d + e*x]) - 6*d*(a + b*Log[c*x
^n])*Log[1 + (e*x)/d] - 6*b*d*n*PolyLog[2, -((e*x)/d)])/(2*e^4)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \log \left (c x^{n}\right ) + a x^{3}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.21, size = 764, normalized size = 5.13 \[ \frac {3 b d n \dilog \left (-\frac {e x}{d}\right )}{e^{4}}-\frac {3 b \,d^{2} \ln \left (x^{n}\right )}{\left (e x +d \right ) e^{4}}-\frac {3 b d \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{e^{4}}+\frac {b \,d^{3} \ln \left (x^{n}\right )}{2 \left (e x +d \right )^{2} e^{4}}-\frac {3 b \,d^{2} \ln \relax (c )}{\left (e x +d \right ) e^{4}}-\frac {3 b d \ln \relax (c ) \ln \left (e x +d \right )}{e^{4}}+\frac {b \,d^{3} \ln \relax (c )}{2 \left (e x +d \right )^{2} e^{4}}+\frac {5 b d n \ln \left (e x \right )}{2 e^{4}}-\frac {5 b d n \ln \left (e x +d \right )}{2 e^{4}}-\frac {b \,d^{2} n}{2 \left (e x +d \right ) e^{4}}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 \left (e x +d \right )^{2} e^{4}}+\frac {a \,d^{3}}{2 \left (e x +d \right )^{2} e^{4}}-\frac {3 a \,d^{2}}{\left (e x +d \right ) e^{4}}-\frac {3 a d \ln \left (e x +d \right )}{e^{4}}+\frac {b x \ln \relax (c )}{e^{3}}-\frac {b d n}{e^{4}}+\frac {b x \ln \left (x^{n}\right )}{e^{3}}+\frac {3 b d n \ln \left (-\frac {e x}{d}\right ) \ln \left (e x +d \right )}{e^{4}}+\frac {a x}{e^{3}}-\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 \left (e x +d \right )^{2} e^{4}}+\frac {3 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 \left (e x +d \right ) e^{4}}-\frac {3 i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 e^{4}}-\frac {3 i \pi b d \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 e^{4}}-\frac {i \pi b x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 e^{3}}-\frac {3 i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \left (e x +d \right ) e^{4}}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 \left (e x +d \right )^{2} e^{4}}-\frac {3 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \left (e x +d \right ) e^{4}}+\frac {3 i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x +d \right )}{2 e^{4}}-\frac {b n x}{e^{3}}-\frac {i \pi b x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 e^{3}}-\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 \left (e x +d \right )^{2} e^{4}}+\frac {3 i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 \left (e x +d \right ) e^{4}}+\frac {3 i \pi b d \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +d \right )}{2 e^{4}}+\frac {i \pi b x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 e^{3}}+\frac {i \pi b x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*ln(x^n)/e^4*d^2/(e*x+d)-3*b*ln(x^n)/e^4*d*ln(e*x+d)+1/2*b*ln(x^n)*d^3/e^4/(e*x+d)^2-3*b*ln(c)/e^4*d^2/(e*
x+d)-3*b*ln(c)/e^4*d*ln(e*x+d)+1/2*b*ln(c)*d^3/e^4/(e*x+d)^2+5/2*b*n/e^4*d*ln(e*x)-5/2*b*n/e^4*d*ln(e*x+d)-1/2
*b*n/e^4*d^2/(e*x+d)+3*b*n/e^4*d*dilog(-1/d*e*x)-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*d^3/e^4/(e*x+d
)^2+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*d^3/e^4/(e*x+d)^2+1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*d^3/e^4/(e*x
+d)^2-3/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^4*d^2/(e*x+d)-3/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e^4*d*ln(e*x
+d)-3/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e^4*d^2/(e*x+d)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e^3*x+
1/2*a*d^3/e^4/(e*x+d)^2-3*a/e^4*d^2/(e*x+d)-3*a/e^4*d*ln(e*x+d)+b*ln(c)/e^3*x-b*n/e^4*d+b*ln(x^n)/e^3*x-1/2*I*
b*Pi*csgn(I*c*x^n)^3/e^3*x+3*b*n/e^4*d*ln(e*x+d)*ln(-1/d*e*x)+a/e^3*x-3/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e
^4*d*ln(e*x+d)+3/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e^4*d*ln(e*x+d)+3/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*
x^n)*csgn(I*c)/e^4*d^2/(e*x+d)-b*n*x/e^3+1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e^3*x+1/2*I*b*Pi*csgn(I*x^n)*csg
n(I*c*x^n)^2/e^3*x-1/4*I*b*Pi*csgn(I*c*x^n)^3*d^3/e^4/(e*x+d)^2+3/2*I*b*Pi*csgn(I*c*x^n)^3/e^4*d^2/(e*x+d)+3/2
*I*b*Pi*csgn(I*c*x^n)^3/e^4*d*ln(e*x+d)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {6 \, d^{2} e x + 5 \, d^{3}}{e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}} - \frac {2 \, x}{e^{3}} + \frac {6 \, d \log \left (e x + d\right )}{e^{4}}\right )} + b \int \frac {x^{3} \log \relax (c) + x^{3} \log \left (x^{n}\right )}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 48.37, size = 372, normalized size = 2.50 \[ - \frac {a d^{3} \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 )}{e^{3}} + \frac {3 a d^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {3 a d \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 x}{e^{3}} + \frac {b d^{3} n \left (\begin {cases} \frac {x}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 d^{2} e + 2 d e^{2} x} - \frac {\log {\relax (x )}}{2 d^{2} e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \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 ) \log {\left (c x^{n} \right )}}{e^{3}} - \frac {3 b d^{2} n \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\relax (x )}}{d e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {3 b d^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} + \frac {3 b d n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} \log {\relax (d )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (d )} \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 {\relax (d )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (d )} - \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 {3 b d \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 n x}{e^{3}} + \frac {b x \log {\left (c x^{n} \right )}}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d**3*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/e**3 + 3*a*d**2*Piecewise((x/d**2, Eq(e,
0)), (-1/(d*e + e**2*x), True))/e**3 - 3*a*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**3 + a*x/e**
3 + b*d**3*n*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*d**2*e + 2*d*e**2*x) - log(x)/(2*d**2*e) + log(d/e + x)/(2*d
**2*e), True))/e**3 - b*d**3*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))*log(c*x**n)/e**3 - 3
*b*d**2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e), True))/e**3 + 3*b*d**2*Piecewise(
(x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/e**3 + 3*b*d*n*Piecewise((x/d, Eq(e, 0)), (Piecewis
e((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_pola
r(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 - 3*b*d*Piecewise((x/d, Eq(e, 0)), (lo
g(d + e*x)/e, True))*log(c*x**n)/e**3 - b*n*x/e**3 + b*x*log(c*x**n)/e**3

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