3.61 \(\int \frac {a+b \log (c x^n)}{x^3 (d+e x)^4} \, dx\)

Optimal. Leaf size=263 \[ -\frac {6 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^6 (d+e x)}-\frac {10 e^2 \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^6}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^4 (d+e x)^2}-\frac {a+b \log \left (c x^n\right )}{2 d^4 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^3}+\frac {10 b e^2 n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^6}-\frac {11 b e^2 n \log (x)}{6 d^6}+\frac {47 b e^2 n \log (d+e x)}{6 d^6}-\frac {11 b e^2 n}{6 d^5 (d+e x)}+\frac {4 b e n}{d^5 x}-\frac {b e^2 n}{6 d^4 (d+e x)^2}-\frac {b n}{4 d^4 x^2} \]

[Out]

-1/4*b*n/d^4/x^2+4*b*e*n/d^5/x-1/6*b*e^2*n/d^4/(e*x+d)^2-11/6*b*e^2*n/d^5/(e*x+d)-11/6*b*e^2*n*ln(x)/d^6+1/2*(
-a-b*ln(c*x^n))/d^4/x^2+4*e*(a+b*ln(c*x^n))/d^5/x+1/3*e^2*(a+b*ln(c*x^n))/d^3/(e*x+d)^3+3/2*e^2*(a+b*ln(c*x^n)
)/d^4/(e*x+d)^2-6*e^3*x*(a+b*ln(c*x^n))/d^6/(e*x+d)-10*e^2*ln(1+d/e/x)*(a+b*ln(c*x^n))/d^6+47/6*b*e^2*n*ln(e*x
+d)/d^6+10*b*e^2*n*polylog(2,-d/e/x)/d^6

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 285, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {44, 2351, 2304, 2301, 2319, 2314, 31, 2317, 2391} \[ -\frac {10 b e^2 n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^6}-\frac {6 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^6 (d+e x)}+\frac {5 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b d^6 n}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^4 (d+e x)^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^3}-\frac {10 e^2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^6}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x}-\frac {a+b \log \left (c x^n\right )}{2 d^4 x^2}-\frac {11 b e^2 n}{6 d^5 (d+e x)}-\frac {b e^2 n}{6 d^4 (d+e x)^2}-\frac {11 b e^2 n \log (x)}{6 d^6}+\frac {47 b e^2 n \log (d+e x)}{6 d^6}+\frac {4 b e n}{d^5 x}-\frac {b n}{4 d^4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x^3*(d + e*x)^4),x]

[Out]

-(b*n)/(4*d^4*x^2) + (4*b*e*n)/(d^5*x) - (b*e^2*n)/(6*d^4*(d + e*x)^2) - (11*b*e^2*n)/(6*d^5*(d + e*x)) - (11*
b*e^2*n*Log[x])/(6*d^6) - (a + b*Log[c*x^n])/(2*d^4*x^2) + (4*e*(a + b*Log[c*x^n]))/(d^5*x) + (e^2*(a + b*Log[
c*x^n]))/(3*d^3*(d + e*x)^3) + (3*e^2*(a + b*Log[c*x^n]))/(2*d^4*(d + e*x)^2) - (6*e^3*x*(a + b*Log[c*x^n]))/(
d^6*(d + e*x)) + (5*e^2*(a + b*Log[c*x^n])^2)/(b*d^6*n) + (47*b*e^2*n*Log[d + e*x])/(6*d^6) - (10*e^2*(a + b*L
og[c*x^n])*Log[1 + (e*x)/d])/d^6 - (10*b*e^2*n*PolyLog[2, -((e*x)/d)])/d^6

Rule 31

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

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2304

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 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 {a+b \log \left (c x^n\right )}{x^3 (d+e x)^4} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{d^4 x^3}-\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x^2}+\frac {10 e^2 \left (a+b \log \left (c x^n\right )\right )}{d^6 x}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)^4}-\frac {3 e^3 \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)^3}-\frac {6 e^3 \left (a+b \log \left (c x^n\right )\right )}{d^5 (d+e x)^2}-\frac {10 e^3 \left (a+b \log \left (c x^n\right )\right )}{d^6 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \log \left (c x^n\right )}{x^3} \, dx}{d^4}-\frac {(4 e) \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{d^5}+\frac {\left (10 e^2\right ) \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{d^6}-\frac {\left (10 e^3\right ) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{d^6}-\frac {\left (6 e^3\right ) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{d^5}-\frac {\left (3 e^3\right ) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{d^4}-\frac {e^3 \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx}{d^3}\\ &=-\frac {b n}{4 d^4 x^2}+\frac {4 b e n}{d^5 x}-\frac {a+b \log \left (c x^n\right )}{2 d^4 x^2}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^3}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^4 (d+e x)^2}-\frac {6 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^6 (d+e x)}+\frac {5 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b d^6 n}-\frac {10 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^6}+\frac {\left (10 b e^2 n\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{d^6}-\frac {\left (3 b e^2 n\right ) \int \frac {1}{x (d+e x)^2} \, dx}{2 d^4}-\frac {\left (b e^2 n\right ) \int \frac {1}{x (d+e x)^3} \, dx}{3 d^3}+\frac {\left (6 b e^3 n\right ) \int \frac {1}{d+e x} \, dx}{d^6}\\ &=-\frac {b n}{4 d^4 x^2}+\frac {4 b e n}{d^5 x}-\frac {a+b \log \left (c x^n\right )}{2 d^4 x^2}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^3}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^4 (d+e x)^2}-\frac {6 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^6 (d+e x)}+\frac {5 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b d^6 n}+\frac {6 b e^2 n \log (d+e x)}{d^6}-\frac {10 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^6}-\frac {10 b e^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{d^6}-\frac {\left (3 b e^2 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 d^4}-\frac {\left (b e^2 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 d^3}\\ &=-\frac {b n}{4 d^4 x^2}+\frac {4 b e n}{d^5 x}-\frac {b e^2 n}{6 d^4 (d+e x)^2}-\frac {11 b e^2 n}{6 d^5 (d+e x)}-\frac {11 b e^2 n \log (x)}{6 d^6}-\frac {a+b \log \left (c x^n\right )}{2 d^4 x^2}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{d^5 x}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 d^3 (d+e x)^3}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^4 (d+e x)^2}-\frac {6 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^6 (d+e x)}+\frac {5 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b d^6 n}+\frac {47 b e^2 n \log (d+e x)}{6 d^6}-\frac {10 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^6}-\frac {10 b e^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{d^6}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.38, size = 276, normalized size = 1.05 \[ \frac {\frac {4 d^3 e^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}+\frac {18 d^2 e^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac {6 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}+\frac {72 d e^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}-120 e^2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {48 d e \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {60 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}-\frac {3 b d^2 n}{x^2}-120 b e^2 n \text {Li}_2\left (-\frac {e x}{d}\right )-\frac {2 b d e^2 n (3 d+2 e x)}{(d+e x)^2}-\frac {18 b d e^2 n}{d+e x}-72 b e^2 n (\log (x)-\log (d+e x))+22 b e^2 n \log (d+e x)+\frac {48 b d e n}{x}-22 b e^2 n \log (x)}{12 d^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x^3*(d + e*x)^4),x]

[Out]

((-3*b*d^2*n)/x^2 + (48*b*d*e*n)/x - (18*b*d*e^2*n)/(d + e*x) - (2*b*d*e^2*n*(3*d + 2*e*x))/(d + e*x)^2 - 22*b
*e^2*n*Log[x] - (6*d^2*(a + b*Log[c*x^n]))/x^2 + (48*d*e*(a + b*Log[c*x^n]))/x + (4*d^3*e^2*(a + b*Log[c*x^n])
)/(d + e*x)^3 + (18*d^2*e^2*(a + b*Log[c*x^n]))/(d + e*x)^2 + (72*d*e^2*(a + b*Log[c*x^n]))/(d + e*x) + (60*e^
2*(a + b*Log[c*x^n])^2)/(b*n) - 72*b*e^2*n*(Log[x] - Log[d + e*x]) + 22*b*e^2*n*Log[d + e*x] - 120*e^2*(a + b*
Log[c*x^n])*Log[1 + (e*x)/d] - 120*b*e^2*n*PolyLog[2, -((e*x)/d)])/(12*d^6)

________________________________________________________________________________________

fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e^{4} x^{7} + 4 \, d e^{3} x^{6} + 6 \, d^{2} e^{2} x^{5} + 4 \, d^{3} e x^{4} + d^{4} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^3/(e*x+d)^4,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)/(e^4*x^7 + 4*d*e^3*x^6 + 6*d^2*e^2*x^5 + 4*d^3*e*x^4 + d^4*x^3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{4} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^3/(e*x+d)^4,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.21, size = 1324, normalized size = 5.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*x^n)+a)/x^3/(e*x+d)^4,x)

[Out]

1/4*I*b*Pi*csgn(I*c*x^n)^3/d^4/x^2+10*b*n/d^6*e^2*ln(e*x+d)*ln(-1/d*e*x)-1/2*b*ln(x^n)/d^4/x^2-3/4*I*b*Pi*csgn
(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^4*e^2/(e*x+d)^2+2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^5*e/x-1/2*a/d^4/x^2+3/2
*a/d^4*e^2/(e*x+d)^2+1/3*a*e^2/d^3/(e*x+d)^3+10*a/d^6*e^2*ln(x)-10*a/d^6*e^2*ln(e*x+d)-1/2*b*ln(c)/d^4/x^2+4*a
/d^5*e/x+6*a*e^2/d^5/(e*x+d)-5*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^6*e^2*ln(e*x+d)+1/4*I*b*Pi*csgn(I*x^n)*csgn(
I*c*x^n)*csgn(I*c)/d^4/x^2-1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*e^2/d^3/(e*x+d)^3-5*I*b*Pi*csgn(I*x^
n)*csgn(I*c*x^n)*csgn(I*c)/d^6*e^2*ln(x)-2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^5*e/x+5*I*b*Pi*csgn(I*
x^n)*csgn(I*c*x^n)*csgn(I*c)/d^6*e^2*ln(e*x+d)-3*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*e^2/d^5/(e*x+d)+3/
4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^4*e^2/(e*x+d)^2+3*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e^2/d^5/(e*x+d)+3/
4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^4*e^2/(e*x+d)^2+5*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^6*e^2*ln(x)+1/6*I*
b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*e^2/d^3/(e*x+d)^3+2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^5*e/x+5*I*b*Pi*csgn(I*
c*x^n)^2*csgn(I*c)/d^6*e^2*ln(x)-5*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^6*e^2*ln(e*x+d)+3*I*b*Pi*csgn(I*c*x^n)
^2*csgn(I*c)*e^2/d^5/(e*x+d)+1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e^2/d^3/(e*x+d)^3+1/3*b*ln(c)*e^2/d^3/(e*x
+d)^3+10*b*ln(c)/d^6*e^2*ln(x)-10*b*ln(c)/d^6*e^2*ln(e*x+d)+4*b*ln(c)/d^5*e/x+6*b*ln(c)*e^2/d^5/(e*x+d)+3/2*b*
ln(c)/d^4*e^2/(e*x+d)^2-5*b*n/d^6*e^2*ln(x)^2+10*b*n/d^6*e^2*dilog(-1/d*e*x)-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x
^n)^2/d^4/x^2-5*I*b*Pi*csgn(I*c*x^n)^3/d^6*e^2*ln(x)+4*b*e*n/d^5/x-1/6*b*e^2*n/d^4/(e*x+d)^2-11/6*b*e^2*n/d^5/
(e*x+d)-1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^4/x^2-3*I*b*Pi*csgn(I*c*x^n)^3*e^2/d^5/(e*x+d)-3/4*I*b*Pi*csgn(
I*c*x^n)^3/d^4*e^2/(e*x+d)^2-1/6*I*b*Pi*csgn(I*c*x^n)^3*e^2/d^3/(e*x+d)^3+4*b*ln(x^n)/d^5*e/x-10*b*ln(x^n)/d^6
*e^2*ln(e*x+d)+6*b*ln(x^n)*e^2/d^5/(e*x+d)+3/2*b*ln(x^n)/d^4*e^2/(e*x+d)^2+1/3*b*ln(x^n)*e^2/d^3/(e*x+d)^3+10*
b*ln(x^n)/d^6*e^2*ln(x)-2*I*b*Pi*csgn(I*c*x^n)^3/d^5*e/x+5*I*b*Pi*csgn(I*c*x^n)^3/d^6*e^2*ln(e*x+d)-1/4*b*n/d^
4/x^2-47/6*b*e^2*n*ln(x)/d^6+47/6*b*e^2*n*ln(e*x+d)/d^6

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, a {\left (\frac {60 \, e^{4} x^{4} + 150 \, d e^{3} x^{3} + 110 \, d^{2} e^{2} x^{2} + 15 \, d^{3} e x - 3 \, d^{4}}{d^{5} e^{3} x^{5} + 3 \, d^{6} e^{2} x^{4} + 3 \, d^{7} e x^{3} + d^{8} x^{2}} - \frac {60 \, e^{2} \log \left (e x + d\right )}{d^{6}} + \frac {60 \, e^{2} \log \relax (x)}{d^{6}}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{e^{4} x^{7} + 4 \, d e^{3} x^{6} + 6 \, d^{2} e^{2} x^{5} + 4 \, d^{3} e x^{4} + d^{4} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^3/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*a*((60*e^4*x^4 + 150*d*e^3*x^3 + 110*d^2*e^2*x^2 + 15*d^3*e*x - 3*d^4)/(d^5*e^3*x^5 + 3*d^6*e^2*x^4 + 3*d^
7*e*x^3 + d^8*x^2) - 60*e^2*log(e*x + d)/d^6 + 60*e^2*log(x)/d^6) + b*integrate((log(c) + log(x^n))/(e^4*x^7 +
4*d*e^3*x^6 + 6*d^2*e^2*x^5 + 4*d^3*e*x^4 + d^4*x^3), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 148.35, size = 649, normalized size = 2.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x**3/(e*x+d)**4,x)

[Out]

-a*e**3*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/d**3 - 3*a*e**3*Piecewise((x/d**3, Eq(e,
0)), (-1/(2*e*(d + e*x)**2), True))/d**4 - a/(2*d**4*x**2) - 6*a*e**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e +
e**2*x), True))/d**5 + 4*a*e/(d**5*x) - 10*a*e**3*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/d**6 + 1
0*a*e**2*log(x)/d**6 + b*e**3*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), Tru
e))/d**3 - b*e**3*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))*log(c*x**n)/d**3 + 3*b*e**3*n*P
iecewise((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))
/d**4 - 3*b*e**3*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))*log(c*x**n)/d**4 - b*n/(4*d**4*x
**2) - b*log(c*x**n)/(2*d**4*x**2) + 6*b*e**3*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d
*e), True))/d**5 - 6*b*e**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/d**5 + 4*b*e*
n/(d**5*x) + 4*b*e*log(c*x**n)/(d**5*x) + 10*b*e**3*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((log(d)*log(x) - p
olylog(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) - polyl
og(2, e*x*exp_polar(I*pi)/d), True))/e, True))/d**6 - 10*b*e**3*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, Tr
ue))*log(c*x**n)/d**6 - 5*b*e**2*n*log(x)**2/d**6 + 10*b*e**2*log(x)*log(c*x**n)/d**6

________________________________________________________________________________________