3.115 \(\int \frac {x^2 (a+b \log (c x^n))^2}{(d+e x)^4} \, dx\)

Optimal. Leaf size=161 \[ -\frac {b n \log \left (\frac {e x}{d}+1\right ) \left (2 a+2 b \log \left (c x^n\right )+3 b n\right )}{3 d e^3}+\frac {b n x \left (2 a+2 b \log \left (c x^n\right )+b n\right )}{3 d e^2 (d+e x)}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 d (d+e x)^3}+\frac {b n x^2 \left (a+b \log \left (c x^n\right )\right )}{3 d e (d+e x)^2}-\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{3 d e^3} \]

[Out]

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

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Rubi [A]  time = 0.72, antiderivative size = 274, normalized size of antiderivative = 1.70, number of steps used = 25, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {2353, 2319, 2347, 2344, 2301, 2317, 2391, 2314, 31, 44, 2318} \[ -\frac {2 b^2 n^2 \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 d e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3 (d+e x)^3}+\frac {4 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d e^2 (d+e x)}+\frac {b d n \left (a+b \log \left (c x^n\right )\right )}{3 e^3 (d+e x)^2}-\frac {2 b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d e^3}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{d e^2 (d+e x)}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}-\frac {2 \left (a+b \log \left (c x^n\right )\right )^2}{3 d e^3}-\frac {b^2 n^2}{3 e^3 (d+e x)}-\frac {b^2 n^2 \log (x)}{3 d e^3}-\frac {b^2 n^2 \log (d+e x)}{d e^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*Log[c*x^n])^2)/(d + e*x)^4,x]

[Out]

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

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 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 2318

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_))^2, x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])
^p)/(d*(d + e*x)), x] - Dist[(b*n*p)/d, Int[(a + b*Log[c*x^n])^(p - 1)/(d + e*x), x], x] /; FreeQ[{a, b, c, d,
e, n, p}, x] && GtQ[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 2344

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

Rule 2347

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 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 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 {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx &=\int \left (\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)^4}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)^3}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)^2}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx}{e^2}-\frac {(2 d) \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx}{e^2}+\frac {d^2 \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx}{e^2}\\ &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3 (d+e x)^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{d e^2 (d+e x)}-\frac {(2 b d n) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx}{e^3}+\frac {\left (2 b d^2 n\right ) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx}{3 e^3}-\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{d e^2}\\ &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3 (d+e x)^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{d e^2 (d+e x)}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d e^3}-\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx}{e^3}+\frac {(2 b d n) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx}{3 e^3}+\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^2}-\frac {(2 b d n) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{3 e^2}+\frac {\left (2 b^2 n^2\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{d e^3}\\ &=\frac {b d n \left (a+b \log \left (c x^n\right )\right )}{3 e^3 (d+e x)^2}+\frac {2 b n x \left (a+b \log \left (c x^n\right )\right )}{d e^2 (d+e x)}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3 (d+e x)^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{d e^2 (d+e x)}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d e^3}-\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{d e^3}+\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx}{3 e^3}-\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{d e^3}-\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{3 e^2}+\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{d e^2}-\frac {\left (b^2 d n^2\right ) \int \frac {1}{x (d+e x)^2} \, dx}{3 e^3}-\frac {\left (2 b^2 n^2\right ) \int \frac {1}{d+e x} \, dx}{d e^2}\\ &=\frac {b d n \left (a+b \log \left (c x^n\right )\right )}{3 e^3 (d+e x)^2}+\frac {4 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d e^2 (d+e x)}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{d e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3 (d+e x)^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{d e^2 (d+e x)}-\frac {2 b^2 n^2 \log (d+e x)}{d e^3}-\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{d e^3}+\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{3 d e^3}-\frac {(2 b n) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{3 d e^2}-\frac {\left (2 b^2 n^2\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{d e^3}-\frac {\left (b^2 d n^2\right ) \int \left (\frac {1}{d^2 x}-\frac {e}{d (d+e x)^2}-\frac {e}{d^2 (d+e x)}\right ) \, dx}{3 e^3}+\frac {\left (2 b^2 n^2\right ) \int \frac {1}{d+e x} \, dx}{3 d e^2}\\ &=-\frac {b^2 n^2}{3 e^3 (d+e x)}-\frac {b^2 n^2 \log (x)}{3 d e^3}+\frac {b d n \left (a+b \log \left (c x^n\right )\right )}{3 e^3 (d+e x)^2}+\frac {4 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d e^2 (d+e x)}-\frac {2 \left (a+b \log \left (c x^n\right )\right )^2}{3 d e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3 (d+e x)^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{d e^2 (d+e x)}-\frac {b^2 n^2 \log (d+e x)}{d e^3}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 d e^3}+\frac {\left (2 b^2 n^2\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{3 d e^3}\\ &=-\frac {b^2 n^2}{3 e^3 (d+e x)}-\frac {b^2 n^2 \log (x)}{3 d e^3}+\frac {b d n \left (a+b \log \left (c x^n\right )\right )}{3 e^3 (d+e x)^2}+\frac {4 b n x \left (a+b \log \left (c x^n\right )\right )}{3 d e^2 (d+e x)}-\frac {2 \left (a+b \log \left (c x^n\right )\right )^2}{3 d e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{3 e^3 (d+e x)^3}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{e^3 (d+e x)^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{d e^2 (d+e x)}-\frac {b^2 n^2 \log (d+e x)}{d e^3}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 d e^3}-\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{3 d e^3}\\ \end {align*}

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Mathematica [B]  time = 0.51, size = 371, normalized size = 2.30 \[ -\frac {\frac {a^2 d^2}{(d+e x)^3}+\frac {3 a^2}{d+e x}-\frac {3 a^2 d}{(d+e x)^2}-\frac {a^2}{d}+\frac {2 a b d^2 \log \left (c x^n\right )}{(d+e x)^3}+\frac {6 a b \log \left (c x^n\right )}{d+e x}-\frac {6 a b d \log \left (c x^n\right )}{(d+e x)^2}-\frac {2 a b \log \left (c x^n\right )}{d}+\frac {4 a b n}{d+e x}-\frac {a b d n}{(d+e x)^2}+\frac {2 a b n \log \left (\frac {e x}{d}+1\right )}{d}+\frac {b^2 d^2 \log ^2\left (c x^n\right )}{(d+e x)^3}+\frac {3 b^2 \log ^2\left (c x^n\right )}{d+e x}-\frac {3 b^2 d \log ^2\left (c x^n\right )}{(d+e x)^2}+\frac {4 b^2 n \log \left (c x^n\right )}{d+e x}-\frac {b^2 d n \log \left (c x^n\right )}{(d+e x)^2}+\frac {2 b^2 n \log \left (c x^n\right ) \log \left (\frac {e x}{d}+1\right )}{d}-\frac {b^2 \log ^2\left (c x^n\right )}{d}+\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{d}+\frac {b^2 n^2}{d+e x}+\frac {3 b^2 n^2 \log (d+e x)}{d}-\frac {3 b^2 n^2 \log (x)}{d}}{3 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*Log[c*x^n])^2)/(d + e*x)^4,x]

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{2} \log \left (c x^{n}\right )^{2} + 2 \, a b x^{2} \log \left (c x^{n}\right ) + a^{2} x^{2}}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2}}{{\left (e x + d\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))^2/(e*x+d)^4,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.31, size = 1658, normalized size = 10.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*ln(c*x^n)+a)^2/(e*x+d)^4,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*ln(c*x**n))**2/(e*x+d)**4,x)

[Out]

Integral(x**2*(a + b*log(c*x**n))**2/(d + e*x)**4, x)

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