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

Optimal. Leaf size=156 \[ -\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b e^2 n \log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{2 d^2}-\frac {b e^2 m n \text {Li}_2\left (-\frac {d}{e x}\right )}{2 d^2}-\frac {b e^2 m n \log (x)}{4 d^2}+\frac {b e^2 m n \log (d+e x)}{4 d^2}-\frac {b e n \log \left (f x^m\right )}{2 d x}-\frac {3 b e m n}{4 d x} \]

[Out]

-3/4*b*e*m*n/d/x-1/4*b*e^2*m*n*ln(x)/d^2-1/2*b*e*n*ln(f*x^m)/d/x+1/2*b*e^2*n*ln(1+d/e/x)*ln(f*x^m)/d^2+1/4*b*e
^2*m*n*ln(e*x+d)/d^2-1/4*(m/x^2+2*ln(f*x^m)/x^2)*(a+b*ln(c*(e*x+d)^n))-1/2*b*e^2*m*n*polylog(2,-d/e/x)/d^2

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Rubi [A]  time = 0.15, antiderivative size = 175, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2426, 44, 2351, 2304, 2301, 2317, 2391} \[ \frac {b e^2 m n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 d^2}-\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b e^2 n \log ^2\left (f x^m\right )}{4 d^2 m}+\frac {b e^2 n \log \left (\frac {e x}{d}+1\right ) \log \left (f x^m\right )}{2 d^2}-\frac {b e^2 m n \log (x)}{4 d^2}+\frac {b e^2 m n \log (d+e x)}{4 d^2}-\frac {b e n \log \left (f x^m\right )}{2 d x}-\frac {3 b e m n}{4 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rule 2426

Int[Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :
> -Simp[(((m*(g*x)^(q + 1))/(q + 1) - (g*x)^(q + 1)*Log[f*x^m])*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] +
(-Dist[(b*e*n)/(g*(q + 1)), Int[((g*x)^(q + 1)*Log[f*x^m])/(d + e*x), x], x] + Dist[(b*e*m*n)/(g*(q + 1)^2), I
nt[(g*x)^(q + 1)/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, m, n, q}, x] && NeQ[q, -1]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 204, normalized size = 1.31 \[ -\frac {2 a d^2 \log \left (f x^m\right )+a d^2 m+2 b d^2 \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )+b d^2 m \log \left (c (d+e x)^n\right )-2 b e^2 n x^2 \log (d+e x) \log \left (f x^m\right )+b e^2 n x^2 \log (x) \left (2 m \log (d+e x)-2 m \log \left (\frac {e x}{d}+1\right )+2 \log \left (f x^m\right )+m\right )-2 b e^2 m n x^2 \text {Li}_2\left (-\frac {e x}{d}\right )-b e^2 m n x^2 \log (d+e x)+2 b d e n x \log \left (f x^m\right )+3 b d e m n x-b e^2 m n x^2 \log ^2(x)}{4 d^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) \log \left (f x^{m}\right ) + a \log \left (f x^{m}\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log((e*x + d)^n*c)*log(f*x^m) + a*log(f*x^m))/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.77, size = 2051, normalized size = 13.15 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/x^2*a*m+1/8*b*Pi^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*x^m)*csgn(I*f*x^m)^2-1/4*I/d^2*b*e^
2*n*ln(e*x+d)*Pi*csgn(I*f*x^m)^3-1/4*I/d^2*b*e^2*n*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/2/x^2*ln
(f)*a-1/2*m*b*e^2*n/d^2*ln(e*x+d)*ln(-1/d*e*x)+1/4/d^2*b*e^2*m*n*ln(x)^2-1/2*a/x^2*ln(x^m)+(-1/2*b/x^2*ln(x^m)
-1/4*(-I*Pi*b*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+I*Pi*b*csgn(I*f)*csgn(I*f*x^m)^2+I*Pi*b*csgn(I*x^m)*csgn(I*f
*x^m)^2-I*Pi*b*csgn(I*f*x^m)^3+2*b*ln(f)+b*m)/x^2)*ln((e*x+d)^n)-1/2/x^2*ln(f)*ln(c)*b-1/4/x^2*ln(c)*b*m+1/8*b
*Pi^2*csgn(I*c*(e*x+d)^n)^3/x^2*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/8*b*Pi^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csg
n(I*c*(e*x+d)^n)/x^2*csgn(I*f*x^m)^3-1/2/d*e*b*n/x*ln(f)+1/2/d^2*b*e^2*n*ln(e*x+d)*ln(f)-1/4*I*b*Pi*csgn(I*c)*
csgn(I*c*(e*x+d)^n)^2/x^2*ln(x^m)-1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/x^2*ln(x^m)-1/4*I/x^2*ln(
f)*Pi*b*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/4*I/x^2*ln(f)*Pi*b*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/4*I/d^2
*b*e^2*n*ln(x)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/4*b*e^2*m*n*ln(x)/d^2+1/4*b*e^2*m*n*ln(e*x+d)/d^2-1/2*
m*b*e^2*n/d^2*dilog(-1/d*e*x)+1/4*I/d*e*b*n/x*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/4*I/x^2*ln(f)*Pi*b*csgn
(I*c*(e*x+d)^n)^3+1/8*I/x^2*Pi*b*m*csgn(I*c*(e*x+d)^n)^3+1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/x^2*ln(x^m)-1/4*I/x^
2*Pi*a*csgn(I*f)*csgn(I*f*x^m)^2-1/4*I/x^2*Pi*a*csgn(I*x^m)*csgn(I*f*x^m)^2+1/4*I/x^2*Pi*ln(c)*b*csgn(I*f*x^m)
^3-1/8*b*Pi^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f*x^m)^3-1/8*b*Pi^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d
)^n)^2/x^2*csgn(I*f*x^m)^3-1/8*b*Pi^2*csgn(I*c*(e*x+d)^n)^3/x^2*csgn(I*f)*csgn(I*f*x^m)^2-1/8*b*Pi^2*csgn(I*c*
(e*x+d)^n)^3/x^2*csgn(I*x^m)*csgn(I*f*x^m)^2-1/4*I/x^2*Pi*ln(c)*b*csgn(I*f)*csgn(I*f*x^m)^2-1/2/d^2*b*e^2*n*ln
(x)*ln(f)-1/2*e^2*n*b*ln(x^m)/d^2*ln(x)+1/2*e^2*n*b*ln(x^m)/d^2*ln(e*x+d)-1/2*e*n*b*ln(x^m)/d/x+1/8*b*Pi^2*csg
n(I*c)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f)*csgn(I*f*x^m)^2+1/8*b*Pi^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn
(I*x^m)*csgn(I*f*x^m)^2+1/8*b*Pi^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f)*csgn(I*f*x^m)^2-3/4*b
*e*m*n/d/x+1/4*I/d*e*b*n/x*Pi*csgn(I*f*x^m)^3+1/4*I/x^2*ln(f)*Pi*b*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d
)^n)+1/8*b*Pi^2*csgn(I*c*(e*x+d)^n)^3/x^2*csgn(I*f*x^m)^3+1/4*I/x^2*Pi*a*csgn(I*f*x^m)^3-1/4*I/d^2*b*e^2*n*ln(
x)*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/8*b*Pi^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/x^2*csgn(I*f)*csgn(
I*x^m)*csgn(I*f*x^m)-1/4*I/d*e*b*n/x*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/4*I/d*e*b*n/x*Pi*csgn(I*x^m)*csgn(I*f*x^m)
^2+1/4*I/d^2*b*e^2*n*ln(e*x+d)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/4*I/d^2*b*e^2*n*ln(x)*Pi*csgn(I*x^m)*csgn(I*f*
x^m)^2+1/4*I/d^2*b*e^2*n*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/8*I/x^2*Pi*b*m*csgn(I*(e*x+d)^n)*csgn(I*c*(e
*x+d)^n)^2+1/4*I/x^2*Pi*a*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/8*b*Pi^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)
^2/x^2*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/x^2*ln(x
^m)-1/8*b*Pi^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/x^2*csgn(I*f)*csgn(I*f*x^m)^2+1/4*I/x^2*Pi*ln(c
)*b*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/8*I/x^2*Pi*b*m*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/4*I
/d^2*b*e^2*n*ln(x)*Pi*csgn(I*f*x^m)^3-1/8*b*Pi^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/x^2*csgn(I*x^
m)*csgn(I*f*x^m)^2-1/8*b*Pi^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/4*I/x^
2*Pi*ln(c)*b*csgn(I*x^m)*csgn(I*f*x^m)^2-1/8*I/x^2*Pi*b*m*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/2*b*ln(c)/x^2*ln(x
^m)

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maxima [A]  time = 0.68, size = 198, normalized size = 1.27 \[ \frac {1}{4} \, {\left (\frac {2 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} b e^{2} n}{d^{2}} + \frac {b e^{2} n \log \left (e x + d\right )}{d^{2}} - \frac {2 \, b e^{2} n x^{2} \log \left (e x + d\right ) \log \relax (x) - b e^{2} n x^{2} \log \relax (x)^{2} + b e^{2} n x^{2} \log \relax (x) + 3 \, b d e n x + b d^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + b d^{2} \log \relax (c) + a d^{2}}{d^{2} x^{2}}\right )} m + \frac {1}{2} \, {\left (b e n {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \relax (x)}{d^{2}} - \frac {1}{d x}\right )} - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{x^{2}} - \frac {a}{x^{2}}\right )} \log \left (f x^{m}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(log(e*x/d + 1)*log(x) + dilog(-e*x/d))*b*e^2*n/d^2 + b*e^2*n*log(e*x + d)/d^2 - (2*b*e^2*n*x^2*log(e*x
+ d)*log(x) - b*e^2*n*x^2*log(x)^2 + b*e^2*n*x^2*log(x) + 3*b*d*e*n*x + b*d^2*log((e*x + d)^n) + b*d^2*log(c)
+ a*d^2)/(d^2*x^2))*m + 1/2*(b*e*n*(e*log(e*x + d)/d^2 - e*log(x)/d^2 - 1/(d*x)) - b*log((e*x + d)^n*c)/x^2 -
a/x^2)*log(f*x^m)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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