3.1.53 \(\int \frac {x^5 (a+b \log (c x^n))}{(d+e x)^4} \, dx\) [53]

Optimal. Leaf size=229 \[ \frac {10 b d n x}{e^5}-\frac {d (60 a+47 b n) x}{6 e^5}-\frac {5 b n x^2}{2 e^4}-\frac {10 b d x \log \left (c x^n\right )}{e^5}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}-\frac {x^4 \left (5 a+b n+5 b \log \left (c x^n\right )\right )}{6 e^2 (d+e x)^2}-\frac {x^3 \left (20 a+9 b n+20 b \log \left (c x^n\right )\right )}{6 e^3 (d+e x)}+\frac {x^2 \left (60 a+47 b n+60 b \log \left (c x^n\right )\right )}{12 e^4}+\frac {d^2 \left (60 a+47 b n+60 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{6 e^6}+\frac {10 b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^6} \]

[Out]

10*b*d*n*x/e^5-1/6*d*(47*b*n+60*a)*x/e^5-5/2*b*n*x^2/e^4-10*b*d*x*ln(c*x^n)/e^5-1/3*x^5*(a+b*ln(c*x^n))/e/(e*x
+d)^3-1/6*x^4*(5*a+b*n+5*b*ln(c*x^n))/e^2/(e*x+d)^2-1/6*x^3*(20*a+9*b*n+20*b*ln(c*x^n))/e^3/(e*x+d)+1/12*x^2*(
60*a+47*b*n+60*b*ln(c*x^n))/e^4+1/6*d^2*(60*a+47*b*n+60*b*ln(c*x^n))*ln(1+e*x/d)/e^6+10*b*d^2*n*polylog(2,-e*x
/d)/e^6

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Rubi [A]
time = 0.28, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2384, 45, 2393, 2332, 2341, 2354, 2438} \begin {gather*} \frac {10 b d^2 n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^6}+\frac {d^2 \log \left (\frac {e x}{d}+1\right ) \left (60 a+60 b \log \left (c x^n\right )+47 b n\right )}{6 e^6}-\frac {x^3 \left (20 a+20 b \log \left (c x^n\right )+9 b n\right )}{6 e^3 (d+e x)}-\frac {x^4 \left (5 a+5 b \log \left (c x^n\right )+b n\right )}{6 e^2 (d+e x)^2}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e (d+e x)^3}+\frac {x^2 \left (60 a+60 b \log \left (c x^n\right )+47 b n\right )}{12 e^4}-\frac {d x (60 a+47 b n)}{6 e^5}-\frac {10 b d x \log \left (c x^n\right )}{e^5}+\frac {10 b d n x}{e^5}-\frac {5 b n x^2}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*Log[c*x^n]))/(d + e*x)^4,x]

[Out]

(10*b*d*n*x)/e^5 - (d*(60*a + 47*b*n)*x)/(6*e^5) - (5*b*n*x^2)/(2*e^4) - (10*b*d*x*Log[c*x^n])/e^5 - (x^5*(a +
b*Log[c*x^n]))/(3*e*(d + e*x)^3) - (x^4*(5*a + b*n + 5*b*Log[c*x^n]))/(6*e^2*(d + e*x)^2) - (x^3*(20*a + 9*b*
n + 20*b*Log[c*x^n]))/(6*e^3*(d + e*x)) + (x^2*(60*a + 47*b*n + 60*b*Log[c*x^n]))/(12*e^4) + (d^2*(60*a + 47*b
*n + 60*b*Log[c*x^n])*Log[1 + (e*x)/d])/(6*e^6) + (10*b*d^2*n*PolyLog[2, -((e*x)/d)])/e^6

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 2341

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 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^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^4} \, dx &=\int \left (-\frac {4 d \left (a+b \log \left (c x^n\right )\right )}{e^5}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d^5 \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)^4}+\frac {5 d^4 \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)^3}-\frac {10 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)^2}+\frac {10 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)}\right ) \, dx\\ &=-\frac {(4 d) \int \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^5}+\frac {\left (10 d^2\right ) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^5}-\frac {\left (10 d^3\right ) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^5}+\frac {\left (5 d^4\right ) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{e^5}-\frac {d^5 \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx}{e^5}+\frac {\int x \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^4}\\ &=-\frac {4 a d x}{e^5}-\frac {b n x^2}{4 e^4}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^4}+\frac {d^5 \left (a+b \log \left (c x^n\right )\right )}{3 e^6 (d+e x)^3}-\frac {5 d^4 \left (a+b \log \left (c x^n\right )\right )}{2 e^6 (d+e x)^2}-\frac {10 d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)}+\frac {10 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^6}-\frac {(4 b d) \int \log \left (c x^n\right ) \, dx}{e^5}-\frac {\left (10 b d^2 n\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^6}+\frac {\left (5 b d^4 n\right ) \int \frac {1}{x (d+e x)^2} \, dx}{2 e^6}-\frac {\left (b d^5 n\right ) \int \frac {1}{x (d+e x)^3} \, dx}{3 e^6}+\frac {\left (10 b d^2 n\right ) \int \frac {1}{d+e x} \, dx}{e^5}\\ &=-\frac {4 a d x}{e^5}+\frac {4 b d n x}{e^5}-\frac {b n x^2}{4 e^4}-\frac {4 b d x \log \left (c x^n\right )}{e^5}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^4}+\frac {d^5 \left (a+b \log \left (c x^n\right )\right )}{3 e^6 (d+e x)^3}-\frac {5 d^4 \left (a+b \log \left (c x^n\right )\right )}{2 e^6 (d+e x)^2}-\frac {10 d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)}+\frac {10 b d^2 n \log (d+e x)}{e^6}+\frac {10 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^6}+\frac {10 b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^6}+\frac {\left (5 b d^4 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^6}-\frac {\left (b d^5 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^6}\\ &=-\frac {4 a d x}{e^5}+\frac {4 b d n x}{e^5}-\frac {b n x^2}{4 e^4}-\frac {b d^4 n}{6 e^6 (d+e x)^2}+\frac {13 b d^3 n}{6 e^6 (d+e x)}+\frac {13 b d^2 n \log (x)}{6 e^6}-\frac {4 b d x \log \left (c x^n\right )}{e^5}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^4}+\frac {d^5 \left (a+b \log \left (c x^n\right )\right )}{3 e^6 (d+e x)^3}-\frac {5 d^4 \left (a+b \log \left (c x^n\right )\right )}{2 e^6 (d+e x)^2}-\frac {10 d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^5 (d+e x)}+\frac {47 b d^2 n \log (d+e x)}{6 e^6}+\frac {10 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^6}+\frac {10 b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^6}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 249, normalized size = 1.09 \begin {gather*} \frac {-48 a d e x+48 b d e n x-3 b e^2 n x^2-48 b d e x \log \left (c x^n\right )+6 e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {4 d^5 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}-\frac {30 d^4 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}+\frac {120 d^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x}-2 b d^2 n \left (\frac {d (3 d+2 e x)}{(d+e x)^2}+2 \log (x)-2 \log (d+e x)\right )-120 b d^2 n (\log (x)-\log (d+e x))+30 b d^2 n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )+120 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+120 b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{12 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*Log[c*x^n]))/(d + e*x)^4,x]

[Out]

(-48*a*d*e*x + 48*b*d*e*n*x - 3*b*e^2*n*x^2 - 48*b*d*e*x*Log[c*x^n] + 6*e^2*x^2*(a + b*Log[c*x^n]) + (4*d^5*(a
+ b*Log[c*x^n]))/(d + e*x)^3 - (30*d^4*(a + b*Log[c*x^n]))/(d + e*x)^2 + (120*d^3*(a + b*Log[c*x^n]))/(d + e*
x) - 2*b*d^2*n*((d*(3*d + 2*e*x))/(d + e*x)^2 + 2*Log[x] - 2*Log[d + e*x]) - 120*b*d^2*n*(Log[x] - Log[d + e*x
]) + 30*b*d^2*n*(d/(d + e*x) + Log[x] - Log[d + e*x]) + 120*d^2*(a + b*Log[c*x^n])*Log[1 + (e*x)/d] + 120*b*d^
2*n*PolyLog[2, -((e*x)/d)])/(12*e^6)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 1153, normalized size = 5.03

method	result	size
risch	\(\text {Expression too large to display}\)	\(1153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*ln(c*x^n))/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

10*b*ln(x^n)/e^6*d^2*ln(e*x+d)-10*b*n/e^6*d^2*ln(e*x+d)*ln(-e*x/d)-4*a/e^5*d*x+1/3*a*d^5/e^6/(e*x+d)^3+10*a/e^
6*d^2*ln(e*x+d)+10*a/e^6*d^3/(e*x+d)-5/2*a/e^6*d^4/(e*x+d)^2+2*I*b*Pi*csgn(I*c*x^n)^3/e^5*d*x-1/6*I*b*Pi*csgn(
I*c*x^n)^3*d^5/e^6/(e*x+d)^3-5*I*b*Pi*csgn(I*c*x^n)^3/e^6*d^2*ln(e*x+d)+10*b*ln(x^n)/e^6*d^3/(e*x+d)-5/2*b*ln(
x^n)/e^6*d^4/(e*x+d)^2+1/3*b*ln(x^n)*d^5/e^6/(e*x+d)^3-5*I*b*Pi*csgn(I*c*x^n)^3/e^6*d^3/(e*x+d)+1/4*I*b*Pi*csg
n(I*x^n)*csgn(I*c*x^n)^2/e^4*x^2-1/4*I*b*Pi*csgn(I*c*x^n)^3/e^4*x^2+1/6*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2*d^5/e
^6/(e*x+d)^3+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^4*x^2-5/4*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^6*d^4/(e*x+d)
^2-1/4*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e^4*x^2-4*b*ln(x^n)/e^5*d*x+1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*
x^n)^2*d^5/e^6/(e*x+d)^3+5/4*I*b*Pi*csgn(I*c*x^n)^3/e^6*d^4/(e*x+d)^2-2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^5*d
*x+5*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^6*d^3/(e*x+d)+5*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^6*d^2*ln(e*x+d)+5
*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^6*d^2*ln(e*x+d)+5*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e^6*d^3/(e*x+d)-2*I*b
*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^5*d*x+1/2*a/e^4*x^2-5*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e^6*d^2*ln(
e*x+d)-5*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e^6*d^3/(e*x+d)-5/2*b*ln(c)/e^6*d^4/(e*x+d)^2+10*b*ln(c)/e
^6*d^2*ln(e*x+d)-4*b*ln(c)/e^5*d*x+1/3*b*ln(c)*d^5/e^6/(e*x+d)^3+10*b*ln(c)/e^6*d^3/(e*x+d)-47/6*b*n/e^6*d^2*l
n(e*x)+47/6*b*n/e^6*d^2*ln(e*x+d)+13/6*b*n/e^6*d^3/(e*x+d)-1/6*b*n/e^6*d^4/(e*x+d)^2-10*b*n/e^6*d^2*dilog(-e*x
/d)-1/6*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*d^5/e^6/(e*x+d)^3+1/2*b*ln(x^n)/e^4*x^2+17/4*b*n/e^6*d^2+5/
4*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e^6*d^4/(e*x+d)^2+2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e^
5*d*x+1/2*b*ln(c)/e^4*x^2-5/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^6*d^4/(e*x+d)^2+4*b*d*n*x/e^5-1/4*b*n*x^2/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^5*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*(60*d^2*e^(-6)*log(x*e + d) + 3*(x^2*e - 8*d*x)*e^(-5) + (60*d^3*x^2*e^2 + 105*d^4*x*e + 47*d^5)/(x^3*e^9
+ 3*d*x^2*e^8 + 3*d^2*x*e^7 + d^3*e^6))*a + b*integrate((x^5*log(c) + x^5*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^5*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="fricas")

[Out]

integral((b*x^5*log(c*x^n) + a*x^5)/(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 = 65.64, size = 617, normalized size = 2.69 \begin {gather*} - \frac {a d^{5} \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^{5}} + \frac {5 a d^{4} \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^{5}} - \frac {10 a d^{3} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{e^{5}} + \frac {10 a d^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{5}} - \frac {4 a d x}{e^{5}} + \frac {a x^{2}}{2 e^{4}} + \frac {b d^{5} 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^{5}} - \frac {b d^{5} \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^{5}} - \frac {5 b d^{4} 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^{5}} + \frac {5 b d^{4} \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^{5}} + \frac {10 b d^{3} 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^{5}} - \frac {10 b d^{3} \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^{5}} - \frac {10 b d^{2} 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^{5}} + \frac {10 b d^{2} \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^{5}} + \frac {4 b d n x}{e^{5}} - \frac {4 b d x \log {\left (c x^{n} \right )}}{e^{5}} - \frac {b n x^{2}}{4 e^{4}} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*ln(c*x**n))/(e*x+d)**4,x)

[Out]

-a*d**5*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/e**5 + 5*a*d**4*Piecewise((x/d**3, Eq(e,
0)), (-1/(2*e*(d + e*x)**2), True))/e**5 - 10*a*d**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/
e**5 + 10*a*d**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**5 - 4*a*d*x/e**5 + a*x**2/(2*e**4) + b*
d**5*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**5 - b*d**5*Piecewi
se((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))*log(c*x**n)/e**5 - 5*b*d**4*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**5 + 5*b*d**4*Piecewis
e((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))*log(c*x**n)/e**5 + 10*b*d**3*n*Piecewise((x/d**2, Eq(e, 0
)), (-log(x)/(d*e) + log(d/e + x)/(d*e), True))/e**5 - 10*b*d**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2
*x), True))*log(c*x**n)/e**5 - 10*b*d**2*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**5
+ 10*b*d**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**5 + 4*b*d*n*x/e**5 - 4*b*d*x*log
(c*x**n)/e**5 - b*n*x**2/(4*e**4) + b*x**2*log(c*x**n)/(2*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^5*(a+b*log(c*x^n))/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^5/(x*e + d)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^5\,\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^5*(a + b*log(c*x^n)))/(d + e*x)^4,x)

[Out]

int((x^5*(a + b*log(c*x^n)))/(d + e*x)^4, x)

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