3.1.59 \(\int \frac {a+b \log (c x^n)}{x (d+e x)^4} \, dx\) [59]

Optimal. Leaf size=174 \[ -\frac {b n}{6 d^2 (d+e x)^2}-\frac {5 b n}{6 d^3 (d+e x)}-\frac {5 b n \log (x)}{6 d^4}+\frac {a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac {a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac {\log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {11 b n \log (d+e x)}{6 d^4}+\frac {b n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^4} \]

[Out]

-1/6*b*n/d^2/(e*x+d)^2-5/6*b*n/d^3/(e*x+d)-5/6*b*n*ln(x)/d^4+1/3*(a+b*ln(c*x^n))/d/(e*x+d)^3+1/2*(a+b*ln(c*x^n
))/d^2/(e*x+d)^2-e*x*(a+b*ln(c*x^n))/d^4/(e*x+d)-ln(1+d/e/x)*(a+b*ln(c*x^n))/d^4+11/6*b*n*ln(e*x+d)/d^4+b*n*po
lylog(2,-d/e/x)/d^4

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Rubi [A]
time = 0.21, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2389, 2379, 2438, 2351, 31, 2356, 46} \begin {gather*} \frac {b n \text {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4}-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac {a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}+\frac {a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac {11 b n \log (d+e x)}{6 d^4}-\frac {5 b n \log (x)}{6 d^4}-\frac {5 b n}{6 d^3 (d+e x)}-\frac {b n}{6 d^2 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*n)/(d^2*(d + e*x)^2) - (5*b*n)/(6*d^3*(d + e*x)) - (5*b*n*Log[x])/(6*d^4) + (a + b*Log[c*x^n])/(3*d*(d
+ e*x)^3) + (a + b*Log[c*x^n])/(2*d^2*(d + e*x)^2) - (e*x*(a + b*Log[c*x^n]))/(d^4*(d + e*x)) - (Log[1 + d/(e
*x)]*(a + b*Log[c*x^n]))/d^4 + (11*b*n*Log[d + e*x])/(6*d^4) + (b*n*PolyLog[2, -(d/(e*x))])/d^4

Rule 31

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2351

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 2356

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 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]

Rule 2389

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[(d
+ e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x), x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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 {a+b \log \left (c x^n\right )}{x (d+e x)^4} \, dx &=\frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx}{d}\\ &=\frac {a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx}{d^2}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{d^2}-\frac {(b n) \int \frac {1}{x (d+e x)^3} \, dx}{3 d}\\ &=\frac {a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac {a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}+\frac {\int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx}{d^3}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{d^3}-\frac {(b n) \int \frac {1}{x (d+e x)^2} \, dx}{2 d^2}-\frac {(b n) \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 d}\\ &=-\frac {b n}{6 d^2 (d+e x)^2}-\frac {b n}{3 d^3 (d+e x)}-\frac {b n \log (x)}{3 d^4}+\frac {a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac {a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac {b n \log (d+e x)}{3 d^4}+\frac {\int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{d^4}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{d^4}-\frac {(b n) \int \left (\frac {1}{d^2 x}-\frac {e}{d (d+e x)^2}-\frac {e}{d^2 (d+e x)}\right ) \, dx}{2 d^2}+\frac {(b e n) \int \frac {1}{d+e x} \, dx}{d^4}\\ &=-\frac {b n}{6 d^2 (d+e x)^2}-\frac {5 b n}{6 d^3 (d+e x)}-\frac {5 b n \log (x)}{6 d^4}+\frac {a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac {a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}+\frac {11 b n \log (d+e x)}{6 d^4}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^4}+\frac {(b n) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{d^4}\\ &=-\frac {b n}{6 d^2 (d+e x)^2}-\frac {5 b n}{6 d^3 (d+e x)}-\frac {5 b n \log (x)}{6 d^4}+\frac {a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac {a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}+\frac {11 b n \log (d+e x)}{6 d^4}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^4}-\frac {b n \text {Li}_2\left (-\frac {e x}{d}\right )}{d^4}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 222, normalized size = 1.28 \begin {gather*} \frac {\frac {3 a^2}{b n}+\frac {2 a d^3}{(d+e x)^3}+\frac {3 a d^2}{(d+e x)^2}-\frac {b d^2 n}{(d+e x)^2}+\frac {6 a d}{d+e x}-\frac {5 b d n}{d+e x}-11 b n \log (x)+\frac {6 a \log \left (c x^n\right )}{n}+\frac {2 b d^3 \log \left (c x^n\right )}{(d+e x)^3}+\frac {3 b d^2 \log \left (c x^n\right )}{(d+e x)^2}+\frac {6 b d \log \left (c x^n\right )}{d+e x}+\frac {3 b \log ^2\left (c x^n\right )}{n}+11 b n \log (d+e x)-6 a \log \left (1+\frac {e x}{d}\right )-6 b \log \left (c x^n\right ) \log \left (1+\frac {e x}{d}\right )-6 b n \text {Li}_2\left (-\frac {e x}{d}\right )}{6 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2)/(b*n) + (2*a*d^3)/(d + e*x)^3 + (3*a*d^2)/(d + e*x)^2 - (b*d^2*n)/(d + e*x)^2 + (6*a*d)/(d + e*x) - (
5*b*d*n)/(d + e*x) - 11*b*n*Log[x] + (6*a*Log[c*x^n])/n + (2*b*d^3*Log[c*x^n])/(d + e*x)^3 + (3*b*d^2*Log[c*x^
n])/(d + e*x)^2 + (6*b*d*Log[c*x^n])/(d + e*x) + (3*b*Log[c*x^n]^2)/n + 11*b*n*Log[d + e*x] - 6*a*Log[1 + (e*x
)/d] - 6*b*Log[c*x^n]*Log[1 + (e*x)/d] - 6*b*n*PolyLog[2, -((e*x)/d)])/(6*d^4)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 884, normalized size = 5.08

method	result	size
risch	\(\text {Expression too large to display}\)	\(884\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/d^3/(e*x+d)+1/2*I*b*Pi*csgn(I*c*x^n)^3/d^4*ln(e*x+d)-1/2*I*b*P
i*csgn(I*c*x^n)^3/d^3/(e*x+d)-1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d^4*ln(e*x+d)-1/4*I*b*Pi*csgn(I*c*x^n)^3/d^
2/(e*x+d)^2-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^4*ln(e*x+d)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^4*ln
(x)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^3/(e*x+d)-a/d^4*ln(e*x+d)+a/d^3/(e*x+d)+1/2*a/d^2/(e*x+d)^2+1/3*a
/d/(e*x+d)^3+a/d^4*ln(x)-1/4*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/d^2/(e*x+d)^2-1/2*I*b*Pi*csgn(I*c)*csg
n(I*x^n)*csgn(I*c*x^n)/d^4*ln(x)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d^3/(e*x+d)+1/4*I*b*Pi*csgn(I*x^n)*csgn(
I*c*x^n)^2/d^2/(e*x+d)^2+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d^4*ln(x)+1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(
I*c*x^n)/d^4*ln(e*x+d)-1/6*I*b*Pi*csgn(I*c*x^n)^3/d/(e*x+d)^3-1/2*I*b*Pi*csgn(I*c*x^n)^3/d^4*ln(x)+1/6*I*b*Pi*
csgn(I*c)*csgn(I*c*x^n)^2/d/(e*x+d)^3+1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d/(e*x+d)^3-1/6*b*n/d^2/(e*x+d)^2
-5/6*b*n/d^3/(e*x+d)-1/6*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/d/(e*x+d)^3+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*
x^n)^2/d^2/(e*x+d)^2+b*n/d^4*ln(e*x+d)*ln(-e*x/d)+b*ln(x^n)/d^4*ln(x)-b*ln(x^n)/d^4*ln(e*x+d)+b*ln(x^n)/d^3/(e
*x+d)+1/2*b*ln(x^n)/d^2/(e*x+d)^2+1/3*b*ln(x^n)/d/(e*x+d)^3+b*ln(c)/d^4*ln(x)-1/2*b*n/d^4*ln(x)^2+b*n/d^4*dilo
g(-e*x/d)+b*ln(c)/d^3/(e*x+d)+1/2*b*ln(c)/d^2/(e*x+d)^2+1/3*b*ln(c)/d/(e*x+d)^3-b*ln(c)/d^4*ln(e*x+d)+11/6*b*n
*ln(e*x+d)/d^4-11/6*b*n*ln(x)/d^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((a+b*log(c*x^n))/x/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*a*((6*x^2*e^2 + 15*d*x*e + 11*d^2)/(d^3*x^3*e^3 + 3*d^4*x^2*e^2 + 3*d^5*x*e + d^6) - 6*log(x*e + d)/d^4 +
6*log(x)/d^4) + b*integrate((log(c) + log(x^n))/(x^5*e^4 + 4*d*x^4*e^3 + 6*d^2*x^3*e^2 + 4*d^3*x^2*e + d^4*x),
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((a+b*log(c*x^n))/x/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 63.80, size = 510, normalized size = 2.93 \begin {gather*} - \frac {a e \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 )}{d} - \frac {a e \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 )}{d^{2}} - \frac {a e \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {a e \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{4}} + \frac {a \log {\left (x \right )}}{d^{4}} - \frac {b e^{3} n \left (\begin {cases} - \frac {1}{e^{4} x} & \text {for}\: d = 0 \\- \frac {3 d}{6 d^{2} e^{3} + 12 d e^{4} x + 6 e^{5} x^{2}} - \frac {4 e x}{6 d^{2} e^{3} + 12 d e^{4} x + 6 e^{5} x^{2}} - \frac {\log {\left (d + e x \right )}}{3 d e^{3}} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {b e^{3} \left (\begin {cases} \frac {1}{e^{4} x} & \text {for}\: d = 0 \\- \frac {1}{3 d \left (\frac {d}{x} + e\right )^{3}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} + \frac {3 b e^{2} n \left (\begin {cases} - \frac {1}{e^{3} x} & \text {for}\: d = 0 \\- \frac {1}{2 d e^{2} + 2 e^{3} x} - \frac {\log {\left (d + e x \right )}}{2 d e^{2}} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {3 b e^{2} \left (\begin {cases} \frac {1}{e^{3} x} & \text {for}\: d = 0 \\- \frac {1}{2 d \left (\frac {d}{x} + e\right )^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} - \frac {3 b e n \left (\begin {cases} - \frac {1}{e^{2} x} & \text {for}\: d = 0 \\- \frac {\log {\left (d^{2} + d e x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {3 b e \left (\begin {cases} \frac {1}{e^{2} x} & \text {for}\: d = 0 \\- \frac {1}{\frac {d^{2}}{x} + d e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} + \frac {b n \left (\begin {cases} - \frac {1}{e x} & \text {for}\: d = 0 \\\frac {\begin {cases} \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\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 e^{i \pi }}{e x}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\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 e^{i \pi }}{e x}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {b \left (\begin {cases} \frac {1}{e x} & \text {for}\: d = 0 \\\frac {\log {\left (\frac {d}{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*e*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/d - a*e*Piecewise((x/d**3, Eq(e, 0)), (-1/(2
*e*(d + e*x)**2), True))/d**2 - a*e*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/d**3 - a*e*Piecew
ise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/d**4 + a*log(x)/d**4 - b*e**3*n*Piecewise((-1/(e**4*x), Eq(d, 0))
, (-3*d/(6*d**2*e**3 + 12*d*e**4*x + 6*e**5*x**2) - 4*e*x/(6*d**2*e**3 + 12*d*e**4*x + 6*e**5*x**2) - log(d +
e*x)/(3*d*e**3), True))/d**3 + b*e**3*Piecewise((1/(e**4*x), Eq(d, 0)), (-1/(3*d*(d/x + e)**3), True))*log(c*x
**n)/d**3 + 3*b*e**2*n*Piecewise((-1/(e**3*x), Eq(d, 0)), (-1/(2*d*e**2 + 2*e**3*x) - log(d + e*x)/(2*d*e**2),
True))/d**3 - 3*b*e**2*Piecewise((1/(e**3*x), Eq(d, 0)), (-1/(2*d*(d/x + e)**2), True))*log(c*x**n)/d**3 - 3*
b*e*n*Piecewise((-1/(e**2*x), Eq(d, 0)), (-log(d**2 + d*e*x)/(d*e), True))/d**3 + 3*b*e*Piecewise((1/(e**2*x),
Eq(d, 0)), (-1/(d**2/x + d*e), True))*log(c*x**n)/d**3 + b*n*Piecewise((-1/(e*x), Eq(d, 0)), (Piecewise((poly
log(2, d*exp_polar(I*pi)/(e*x)), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(e)*log(x) + polylog(2, d*exp_polar(I*pi)
/(e*x)), Abs(x) < 1), (-log(e)*log(1/x) + polylog(2, d*exp_polar(I*pi)/(e*x)), 1/Abs(x) < 1), (-meijerg(((), (
1, 1)), ((0, 0), ()), x)*log(e) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(e) + polylog(2, d*exp_polar(I*pi)
/(e*x)), True))/d, True))/d**3 - b*Piecewise((1/(e*x), Eq(d, 0)), (log(d/x + e)/d, True))*log(c*x**n)/d**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((a+b*log(c*x^n))/x/(e*x+d)^4,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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