3.1.54 \(\int \frac {x^4 (a+b \log (c x^n))}{(d+e x)^4} \, dx\) [54]

Optimal. Leaf size=183 \[ -\frac {4 b n x}{e^4}+\frac {(12 a+13 b n) x}{3 e^4}+\frac {4 b x \log \left (c x^n\right )}{e^4}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^3 \left (4 a+b n+4 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x^2 \left (12 a+7 b n+12 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}-\frac {d \left (12 a+13 b n+12 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^5}-\frac {4 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^5} \]

[Out]

-4*b*n*x/e^4+1/3*(13*b*n+12*a)*x/e^4+4*b*x*ln(c*x^n)/e^4-1/3*x^4*(a+b*ln(c*x^n))/e/(e*x+d)^3-1/6*x^3*(4*a+b*n+
4*b*ln(c*x^n))/e^2/(e*x+d)^2-1/6*x^2*(12*a+7*b*n+12*b*ln(c*x^n))/e^3/(e*x+d)-1/3*d*(12*a+13*b*n+12*b*ln(c*x^n)
)*ln(1+e*x/d)/e^5-4*b*d*n*polylog(2,-e*x/d)/e^5

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Rubi [A]
time = 0.25, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 9, 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 {4 b d n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^5}-\frac {d \log \left (\frac {e x}{d}+1\right ) \left (12 a+12 b \log \left (c x^n\right )+13 b n\right )}{3 e^5}-\frac {x^2 \left (12 a+12 b \log \left (c x^n\right )+7 b n\right )}{6 e^3 (d+e x)}-\frac {x^3 \left (4 a+4 b \log \left (c x^n\right )+b n\right )}{6 e^2 (d+e x)^2}-\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}+\frac {x (12 a+13 b n)}{3 e^4}+\frac {4 b x \log \left (c x^n\right )}{e^4}-\frac {4 b n x}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*Log[c*x^n]))/(d + e*x)^4,x]

[Out]

(-4*b*n*x)/e^4 + ((12*a + 13*b*n)*x)/(3*e^4) + (4*b*x*Log[c*x^n])/e^4 - (x^4*(a + b*Log[c*x^n]))/(3*e*(d + e*x
)^3) - (x^3*(4*a + b*n + 4*b*Log[c*x^n]))/(6*e^2*(d + e*x)^2) - (x^2*(12*a + 7*b*n + 12*b*Log[c*x^n]))/(6*e^3*
(d + e*x)) - (d*(12*a + 13*b*n + 12*b*Log[c*x^n])*Log[1 + (e*x)/d])/(3*e^5) - (4*b*d*n*PolyLog[2, -((e*x)/d)])
/e^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 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^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{e^4}+\frac {d^4 \left (a+b \log \left (c x^n\right )\right )}{e^4 (d+e x)^4}-\frac {4 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 (d+e x)^3}+\frac {6 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 (d+e x)^2}-\frac {4 d \left (a+b \log \left (c x^n\right )\right )}{e^4 (d+e x)}\right ) \, dx\\ &=\frac {\int \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^4}-\frac {(4 d) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^4}+\frac {\left (6 d^2\right ) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^4}-\frac {\left (4 d^3\right ) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{e^4}+\frac {d^4 \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx}{e^4}\\ &=\frac {a x}{e^4}-\frac {d^4 \left (a+b \log \left (c x^n\right )\right )}{3 e^5 (d+e x)^3}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)^2}+\frac {6 d x \left (a+b \log \left (c x^n\right )\right )}{e^4 (d+e x)}-\frac {4 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^5}+\frac {b \int \log \left (c x^n\right ) \, dx}{e^4}+\frac {(4 b d n) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^5}-\frac {\left (2 b d^3 n\right ) \int \frac {1}{x (d+e x)^2} \, dx}{e^5}+\frac {\left (b d^4 n\right ) \int \frac {1}{x (d+e x)^3} \, dx}{3 e^5}-\frac {(6 b d n) \int \frac {1}{d+e x} \, dx}{e^4}\\ &=\frac {a x}{e^4}-\frac {b n x}{e^4}+\frac {b x \log \left (c x^n\right )}{e^4}-\frac {d^4 \left (a+b \log \left (c x^n\right )\right )}{3 e^5 (d+e x)^3}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)^2}+\frac {6 d x \left (a+b \log \left (c x^n\right )\right )}{e^4 (d+e x)}-\frac {6 b d n \log (d+e x)}{e^5}-\frac {4 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^5}-\frac {4 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^5}-\frac {\left (2 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}{e^5}+\frac {\left (b d^4 n\right ) \int \left (\frac {1}{d^3 x}-\frac {e}{d (d+e x)^3}-\frac {e}{d^2 (d+e x)^2}-\frac {e}{d^3 (d+e x)}\right ) \, dx}{3 e^5}\\ &=\frac {a x}{e^4}-\frac {b n x}{e^4}+\frac {b d^3 n}{6 e^5 (d+e x)^2}-\frac {5 b d^2 n}{3 e^5 (d+e x)}-\frac {5 b d n \log (x)}{3 e^5}+\frac {b x \log \left (c x^n\right )}{e^4}-\frac {d^4 \left (a+b \log \left (c x^n\right )\right )}{3 e^5 (d+e x)^3}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)^2}+\frac {6 d x \left (a+b \log \left (c x^n\right )\right )}{e^4 (d+e x)}-\frac {13 b d n \log (d+e x)}{3 e^5}-\frac {4 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^5}-\frac {4 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^5}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 207, normalized size = 1.13 \begin {gather*} \frac {6 a e x-6 b e n x+6 b e x \log \left (c x^n\right )-\frac {2 d^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}+\frac {12 d^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac {36 d^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}+b d n \left (\frac {d (3 d+2 e x)}{(d+e x)^2}+2 \log (x)-2 \log (d+e x)\right )+36 b d n (\log (x)-\log (d+e x))-12 b d n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )-24 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-24 b d n \text {Li}_2\left (-\frac {e x}{d}\right )}{6 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*Log[c*x^n]))/(d + e*x)^4,x]

[Out]

(6*a*e*x - 6*b*e*n*x + 6*b*e*x*Log[c*x^n] - (2*d^4*(a + b*Log[c*x^n]))/(d + e*x)^3 + (12*d^3*(a + b*Log[c*x^n]
))/(d + e*x)^2 - (36*d^2*(a + b*Log[c*x^n]))/(d + e*x) + b*d*n*((d*(3*d + 2*e*x))/(d + e*x)^2 + 2*Log[x] - 2*L
og[d + e*x]) + 36*b*d*n*(Log[x] - Log[d + e*x]) - 12*b*d*n*(d/(d + e*x) + Log[x] - Log[d + e*x]) - 24*d*(a + b
*Log[c*x^n])*Log[1 + (e*x)/d] - 24*b*d*n*PolyLog[2, -((e*x)/d)])/(6*e^5)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 969, normalized size = 5.30

method	result	size
risch	\(\text {Expression too large to display}\)	\(969\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*ln(c*x^n))/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

4*b*n/e^5*d*ln(e*x+d)*ln(-e*x/d)+1/6*I*b*Pi*csgn(I*c*x^n)^3*d^4/e^5/(e*x+d)^3+2*I*b*Pi*csgn(I*c*x^n)^3/e^5*d*l
n(e*x+d)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^4*x-I*b*Pi*csgn(I*c*x^n)^3/e^5*d^3/(e*x+d)^2+3*I*b*Pi*csgn(I
*c*x^n)^3/e^5*d^2/(e*x+d)-1/2*I*b*Pi*csgn(I*c*x^n)^3/e^4*x-1/6*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2*d^4/e^5/(e*x+d
)^3-3*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^5*d^2/(e*x+d)+a/e^4*x-2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^5*d*ln(e
*x+d)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^5*d^3/(e*x+d)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^5*d^3/(e*x+d)^2-
1/3*b*ln(x^n)*d^4/e^5/(e*x+d)^3-6*b*ln(x^n)/e^5*d^2/(e*x+d)-1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*d^4/e^5/(e*
x+d)^3-2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^5*d*ln(e*x+d)-4*b*ln(x^n)/e^5*d*ln(e*x+d)-4*a/e^5*d*ln(e*x+d)-6*
a/e^5*d^2/(e*x+d)+2*a/e^5*d^3/(e*x+d)^2-1/3*a*d^4/e^5/(e*x+d)^3-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e^5
*d^3/(e*x+d)^2+3*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e^5*d^2/(e*x+d)+1/6*I*b*Pi*csgn(I*c)*csgn(I*x^n)*c
sgn(I*c*x^n)*d^4/e^5/(e*x+d)^3+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^4*x+2*b*ln(c)/e^5*d^3/(e*x+d)^2-4*b*ln(c
)/e^5*d*ln(e*x+d)-1/3*b*ln(c)*d^4/e^5/(e*x+d)^3-6*b*ln(c)/e^5*d^2/(e*x+d)+13/3*b*n/e^5*d*ln(e*x)+1/6*b*n/e^5*d
^3/(e*x+d)^2-13/3*b*n/e^5*d*ln(e*x+d)-5/3*b*n/e^5*d^2/(e*x+d)+4*b*n/e^5*d*dilog(-e*x/d)+b*ln(c)/e^4*x-b*n/e^5*
d+2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e^5*d*ln(e*x+d)+b*ln(x^n)/e^4*x-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n
)*csgn(I*c*x^n)/e^4*x-3*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^5*d^2/(e*x+d)+2*b*ln(x^n)/e^5*d^3/(e*x+d)^2-b*n*x/e
^4

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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^4*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="maxima")

[Out]

-1/3*(12*d*e^(-5)*log(x*e + d) - 3*x*e^(-4) + (18*d^2*x^2*e^2 + 30*d^3*x*e + 13*d^4)/(x^3*e^8 + 3*d*x^2*e^7 +
3*d^2*x*e^6 + d^3*e^5))*a + b*integrate((x^4*log(c) + x^4*log(x^n))/(x^4*e^4 + 4*d*x^3*e^3 + 6*d^2*x^2*e^2 + 4
*d^3*x*e + d^4), 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^4*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 32.03, size = 563, normalized size = 3.08 \begin {gather*} \frac {a d^{4} \left (\begin {cases} \frac {x}{d^{4}} & \text {for}\: e = 0 \\- \frac {1}{3 e \left (d + e x\right )^{3}} & \text {otherwise} \end {cases}\right )}{e^{4}} - \frac {4 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^{4}} + \frac {6 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^{4}} - \frac {4 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^{4}} + \frac {a x}{e^{4}} - \frac {b d^{4} n \left (\begin {cases} \frac {x}{d^{4}} & \text {for}\: e = 0 \\- \frac {3 d}{6 d^{4} e + 12 d^{3} e^{2} x + 6 d^{2} e^{3} x^{2}} - \frac {2 e x}{6 d^{4} e + 12 d^{3} e^{2} x + 6 d^{2} e^{3} x^{2}} - \frac {\log {\left (x \right )}}{3 d^{3} e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{3 d^{3} e} & \text {otherwise} \end {cases}\right )}{e^{4}} + \frac {b d^{4} \left (\begin {cases} \frac {x}{d^{4}} & \text {for}\: e = 0 \\- \frac {1}{3 e \left (d + e x\right )^{3}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{4}} + \frac {4 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^{4}} - \frac {4 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^{4}} - \frac {6 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^{4}} + \frac {6 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^{4}} + \frac {4 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^{4}} - \frac {4 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^{4}} - \frac {b n x}{e^{4}} + \frac {b x \log {\left (c x^{n} \right )}}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*ln(c*x**n))/(e*x+d)**4,x)

[Out]

a*d**4*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/e**4 - 4*a*d**3*Piecewise((x/d**3, Eq(e, 0
)), (-1/(2*e*(d + e*x)**2), True))/e**4 + 6*a*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e*
*4 - 4*a*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**4 + a*x/e**4 - b*d**4*n*Piecewise((x/d**4, Eq
(e, 0)), (-3*d/(6*d**4*e + 12*d**3*e**2*x + 6*d**2*e**3*x**2) - 2*e*x/(6*d**4*e + 12*d**3*e**2*x + 6*d**2*e**3
*x**2) - log(x)/(3*d**3*e) + log(d/e + x)/(3*d**3*e), True))/e**4 + b*d**4*Piecewise((x/d**4, Eq(e, 0)), (-1/(
3*e*(d + e*x)**3), True))*log(c*x**n)/e**4 + 4*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**4 - 4*b*d**3*Piecewise((x/d**3, Eq(e, 0)), (-1/(2
*e*(d + e*x)**2), True))*log(c*x**n)/e**4 - 6*b*d**2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e
+ x)/(d*e), True))/e**4 + 6*b*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/e**4 +
4*b*d*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((-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_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**4 - 4*b*d*Piecewise((x/d, Eq(e, 0)),
(log(d + e*x)/e, True))*log(c*x**n)/e**4 - b*n*x/e**4 + b*x*log(c*x**n)/e**4

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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^4*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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