3.1.46 \(\int \frac {x^3 (a+b \log (c x^n))}{(d+e x)^3} \, dx\) [46]

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

[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 = 0.17, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2384, 45, 2393, 2332, 2354, 2438} \begin {gather*} -\frac {3 b d n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4}-\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 n x}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*n*x)/e^3 + ((6*a + 5*b*n)*x)/(2*e^3) + (3*b*x*Log[c*x^n])/e^3 - (x^3*(a + b*Log[c*x^n]))/(2*e*(d + e*x)^
2) - (x^2*(3*a + b*n + 3*b*Log[c*x^n]))/(2*e^2*(d + e*x)) - (d*(6*a + 5*b*n + 6*b*Log[c*x^n])*Log[1 + (e*x)/d]
)/(2*e^4) - (3*b*d*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 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 2384

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(f*x
)^m*(d + e*x)^(q + 1)*((a + b*Log[c*x^n])/(e*(q + 1))), x] - Dist[f/(e*(q + 1)), Int[(f*x)^(m - 1)*(d + e*x)^(
q + 1)*(a*m + b*n + b*m*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && ILtQ[q, -1] && GtQ[m, 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*} \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 = 0.10, size = 150, normalized size = 1.01 \begin {gather*} \frac {2 a e x-2 b e n x+2 b e x \log \left (c x^n\right )+\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 b d n (\log (x)-\log (d+e x))-b d n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )-6 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-6 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{2 e^4} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 764, normalized size = 5.13

method	result	size
risch	\(\frac {3 b n d \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{4}}+\frac {3 b n d \dilog \left (-\frac {e x}{d}\right )}{e^{4}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x}{2 e^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x}{2 e^{3}}+\frac {3 i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d \ln \left (e x +d \right )}{2 e^{4}}+\frac {3 i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d^{2}}{2 e^{4} \left (e x +d \right )}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d^{3}}{4 e^{4} \left (e x +d \right )^{2}}+\frac {a x}{e^{3}}-\frac {3 i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (e x +d \right )}{2 e^{4}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) x}{2 e^{3}}-\frac {3 i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (e x +d \right )}{2 e^{4}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} x}{2 e^{3}}-\frac {3 b \ln \left (x^{n}\right ) d^{2}}{e^{4} \left (e x +d \right )}+\frac {b \ln \left (x^{n}\right ) d^{3}}{2 e^{4} \left (e x +d \right )^{2}}-\frac {3 b \ln \left (x^{n}\right ) d \ln \left (e x +d \right )}{e^{4}}-\frac {3 b \ln \left (c \right ) d^{2}}{e^{4} \left (e x +d \right )}+\frac {b \ln \left (c \right ) d^{3}}{2 e^{4} \left (e x +d \right )^{2}}-\frac {3 b \ln \left (c \right ) d \ln \left (e x +d \right )}{e^{4}}-\frac {3 i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d^{2}}{2 e^{4} \left (e x +d \right )}-\frac {3 a d \ln \left (e x +d \right )}{e^{4}}-\frac {3 a \,d^{2}}{e^{4} \left (e x +d \right )}+\frac {a \,d^{3}}{2 e^{4} \left (e x +d \right )^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d^{3}}{4 e^{4} \left (e x +d \right )^{2}}+\frac {3 i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d \ln \left (e x +d \right )}{2 e^{4}}+\frac {3 i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d^{2}}{2 e^{4} \left (e x +d \right )}+\frac {5 b n d \ln \left (e x \right )}{2 e^{4}}-\frac {5 b n d \ln \left (e x +d \right )}{2 e^{4}}-\frac {b n \,d^{2}}{2 e^{4} \left (e x +d \right )}+\frac {b \ln \left (x^{n}\right ) x}{e^{3}}+\frac {b \ln \left (c \right ) x}{e^{3}}-\frac {b n d}{e^{4}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d^{3}}{4 e^{4} \left (e x +d \right )^{2}}-\frac {3 i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d^{2}}{2 e^{4} \left (e x +d \right )}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d^{3}}{4 e^{4} \left (e x +d \right )^{2}}-\frac {b n x}{e^{3}}\)	\(764\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)/(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3), x)

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Sympy [A]
time = 29.62, size = 391, normalized size = 2.62 \begin {gather*} - \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 {\left (x \right )}}{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 {\left (x \right )}}{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} - \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 {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}} \end {gather*}

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((-polylog(2, e*x*exp_polar(I*pi)/d), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*log(x) - polylog(2, e*x*exp_pol
ar(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 - 3*b*d*Piecewise((x/d, Eq(e, 0)), (log(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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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/(x*e + d)^3, x)

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

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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