3.108 \(\int \frac {x^2 (a+b \log (c x^n))^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=232 \[ -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^3 (d+e x)^2}+\frac {2 b n \text {Li}_2\left (-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {3 b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e^3}-\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}-\frac {2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 e^3}+\frac {3 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}-\frac {2 b^2 n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{e^3}+\frac {b^2 n^2 \log (d+e x)}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.45, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {2353, 2319, 2347, 2344, 2301, 2317, 2391, 2314, 31, 2318, 2374, 6589} \[ \frac {2 b n \text {PolyLog}\left (2,-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {3 b^2 n^2 \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3}-\frac {2 b^2 n^2 \text {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^3 (d+e x)^2}-\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}+\frac {3 b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}+\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e^3}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 e^3}+\frac {b^2 n^2 \log (d+e x)}{e^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*Log[c*x^n])^2)/(d + e*x)^3,x]

[Out]

-((b*n*x*(a + b*Log[c*x^n]))/(e^2*(d + e*x))) + (a + b*Log[c*x^n])^2/(2*e^3) - (d^2*(a + b*Log[c*x^n])^2)/(2*e
^3*(d + e*x)^2) - (2*x*(a + b*Log[c*x^n])^2)/(e^2*(d + e*x)) + (b^2*n^2*Log[d + e*x])/e^3 + (3*b*n*(a + b*Log[
c*x^n])*Log[1 + (e*x)/d])/e^3 + ((a + b*Log[c*x^n])^2*Log[1 + (e*x)/d])/e^3 + (3*b^2*n^2*PolyLog[2, -((e*x)/d)
])/e^3 + (2*b*n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)])/e^3 - (2*b^2*n^2*PolyLog[3, -((e*x)/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 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 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

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 6589

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx &=\int \left (\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)^3}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)^2}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx}{e^2}-\frac {(2 d) \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx}{e^2}+\frac {d^2 \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^3} \, dx}{e^2}\\ &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^3 (d+e x)^2}-\frac {2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^3}-\frac {(2 b n) \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^3}+\frac {\left (b d^2 n\right ) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx}{e^3}+\frac {(4 b n) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^2}\\ &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^3 (d+e x)^2}-\frac {2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}+\frac {4 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}+\frac {(b d n) \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx}{e^3}-\frac {(b d n) \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{e^2}-\frac {\left (2 b^2 n^2\right ) \int \frac {\text {Li}_2\left (-\frac {e x}{d}\right )}{x} \, dx}{e^3}-\frac {\left (4 b^2 n^2\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^3}\\ &=-\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^3 (d+e x)^2}-\frac {2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}+\frac {4 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {4 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}-\frac {2 b^2 n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{e^3}+\frac {(b n) \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{e^3}-\frac {(b n) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^2}+\frac {\left (b^2 n^2\right ) \int \frac {1}{d+e x} \, dx}{e^2}\\ &=-\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^3 (d+e x)^2}-\frac {2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}+\frac {b^2 n^2 \log (d+e x)}{e^3}+\frac {3 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {4 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}-\frac {2 b^2 n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{e^3}+\frac {\left (b^2 n^2\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{e^3}\\ &=-\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{e^2 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e^3 (d+e x)^2}-\frac {2 x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}+\frac {b^2 n^2 \log (d+e x)}{e^3}+\frac {3 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {3 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}-\frac {2 b^2 n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{e^3}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 212, normalized size = 0.91 \[ \frac {-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2}+4 b n \text {Li}_2\left (-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {2 b d n \left (a+b \log \left (c x^n\right )\right )}{d+e x}+6 b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {4 d \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}+2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2-3 \left (a+b \log \left (c x^n\right )\right )^2+6 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )-4 b^2 n^2 \text {Li}_3\left (-\frac {e x}{d}\right )-2 b^2 n^2 (\log (x)-\log (d+e x))}{2 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*Log[c*x^n])^2)/(d + e*x)^3,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2}}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))^2/(e*x+d)^3,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.77, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right )^{2} x^{2}}{\left (e x +d \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))^2/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*ln(c*x**n))**2/(e*x+d)**3,x)

[Out]

Integral(x**2*(a + b*log(c*x**n))**2/(d + e*x)**3, x)

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