3.2.1 \(\int \frac {x^2 (a+b \log (c x^n))^2}{(d+e x)^2} \, dx\) [101]

Optimal. Leaf size=203 \[ -\frac {2 a b n x}{e^2}+\frac {2 b^2 n^2 x}{e^2}-\frac {2 b^2 n x \log \left (c x^n\right )}{e^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{e^2}+\frac {d x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}-\frac {2 b d n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{e^3}-\frac {2 b^2 d n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}-\frac {4 b d n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}+\frac {4 b^2 d n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2395, 2333, 2332, 2355, 2354, 2438, 2421, 6724} \begin {gather*} -\frac {4 b d n \text {PolyLog}\left (2,-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 b^2 d n^2 \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3}+\frac {4 b^2 d n^2 \text {PolyLog}\left (3,-\frac {e x}{d}\right )}{e^3}-\frac {2 b d n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e^3}+\frac {d x \left (a+b \log \left (c x^n\right )\right )^2}{e^2 (d+e x)}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{e^2}-\frac {2 a b n x}{e^2}-\frac {2 b^2 n x \log \left (c x^n\right )}{e^2}+\frac {2 b^2 n^2 x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

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 2355

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 2395

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 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-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*L
og[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 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]

Rule 6724

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

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Mathematica [A]
time = 0.12, size = 186, normalized size = 0.92 \begin {gather*} \frac {d \left (a+b \log \left (c x^n\right )\right )^2+e x \left (a+b \log \left (c x^n\right )\right )^2-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}-2 b e n x \left (a-b n+b \log \left (c x^n\right )\right )-2 b d n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-2 d \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )-2 b^2 d n^2 \text {Li}_2\left (-\frac {e x}{d}\right )-4 b d n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )+4 b^2 d n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.20, size = 3778, normalized size = 18.61

method	result	size
risch	\(\text {Expression too large to display}\)	\(3778\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^2*ln(x^n)^2/e^3*d*ln(e*x+d)-2/e^2*n*x*b^2*ln(c)+1/4*d^2/e^3/(e*x+d)*Pi^2*b^2*csgn(I*x^n)^2*csgn(I*c*x^n)^
4+I/e^3*n*d*b^2*Pi*csgn(I*c*x^n)^3-I/e^3*n*d*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2+2*I/e^3*d*ln(e*x+d)*ln(c)*Pi*b^2
*csgn(I*c*x^n)^3-I/e^2*n*x*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*d^2/e^3/(e*x+d)*Pi^2*b^2*csgn(I*c)^2*csgn(I*
x^n)^2*csgn(I*c*x^n)^2-1/2*d^2/e^3/(e*x+d)*Pi^2*b^2*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3-1/e^3*d*ln(e*x+d)*
Pi^2*b^2*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3-1/e^3*d*ln(e*x+d)*Pi^2*b^2*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x
^n)^3-x/e^2*Pi^2*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4+1/2/e^3*d*ln(e*x+d)*Pi^2*b^2*csgn(I*c*x^n)^6-1/4*x/
e^2*Pi^2*b^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4+1/2*x/e^2*Pi^2*b^2*csgn(I*x^n)*csgn(I*c*x^n)^5+2*b^2/e^3*n^2*dilog(
-e*x/d)*d-b^2/e^3*n^2*d*ln(x)^2+4*b/e^3*n*ln(e*x+d)*ln(-e*x/d)*d*a+2*I/e^3*n*dilog(-e*x/d)*d*b^2*Pi*csgn(I*c)*
csgn(I*c*x^n)^2-I*ln(x^n)*d^2/e^3/(e*x+d)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-4*b^2/e^3*d*ln(x)*ln(e*x+d)*ln(-e
*x/d)*n^2+a^2*x/e^2-1/2*d^2/e^3/(e*x+d)*Pi^2*b^2*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3+1/2/e^3*d*ln(e*x+d)*P
i^2*b^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2+2/e^3*d*ln(e*x+d)*Pi^2*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^
n)^4+I*x/e^2*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I*c*x^n)^2-2*a^2/e^3*d*ln(e*x+d)-a^2*d^2/e^3/(e*x+d)+2*b^2/e^3*n^2*
ln(e*x+d)*ln(-e*x/d)*d-4*b^2/e^3*d*ln(x)*dilog(-e*x/d)*n^2+I/e^2*ln(x^n)*x*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+
I*x/e^2*Pi*a*b*csgn(I*c)*csgn(I*c*x^n)^2-2/e^3*n*d*b^2*ln(c)-2*n/e^3*d*ln(e*x+d)*b^2*ln(c)+2*n/e^3*d*ln(e*x)*b
^2*ln(c)+4/e^3*n*dilog(-e*x/d)*d*b^2*ln(c)+d^2/e^3/(e*x+d)*Pi^2*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4+2*I/
e^3*ln(e*x+d)*ln(x^n)*d*b^2*Pi*csgn(I*c*x^n)^3+2*I/e^3*d*ln(e*x+d)*Pi*a*b*csgn(I*c*x^n)^3-2*I/e^3*n*dilog(-e*x
/d)*d*b^2*Pi*csgn(I*c*x^n)^3-I*n/e^3*d*ln(e*x)*b^2*Pi*csgn(I*c*x^n)^3-I/e^2*n*x*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)
^2+2*b^2*n/e^3*ln(x)*ln(x^n)*d-2*b^2*n/e^3*ln(e*x+d)*ln(x^n)*d-4/e^3*ln(e*x+d)*ln(x^n)*d*b^2*ln(c)-2*ln(x^n)*d
^2/e^3/(e*x+d)*b^2*ln(c)+I*x/e^2*Pi*a*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*x/e^2*Pi*a*b*csgn(I*c*x^n)^3-2*I/e^3*n*l
n(e*x+d)*ln(-e*x/d)*d*b^2*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*d^2/e^3/(e*x+d)*ln(c)*Pi*b^2*csgn(I*c*x^n)^
3+I*x/e^2*ln(c)*Pi*b^2*csgn(I*c)*csgn(I*c*x^n)^2+I*d^2/e^3/(e*x+d)*Pi*a*b*csgn(I*c*x^n)^3-I/e^3*n*d*b^2*Pi*csg
n(I*x^n)*csgn(I*c*x^n)^2+I*ln(x^n)*d^2/e^3/(e*x+d)*b^2*Pi*csgn(I*c*x^n)^3+I*n/e^3*d*ln(e*x+d)*b^2*Pi*csgn(I*c*
x^n)^3-2*a*b*n*x/e^2+4*b^2*d*n^2*polylog(3,-e*x/d)/e^3+2*I/e^3*n*ln(e*x+d)*ln(-e*x/d)*d*b^2*Pi*csgn(I*x^n)*csg
n(I*c*x^n)^2-I*n/e^3*d*ln(e*x)*b^2*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/e^3*d*ln(e*x+d)*Pi^2*b^2*csgn(I*x^
n)*csgn(I*c*x^n)^5+b^2*ln(x^n)^2*x/e^2+x/e^2*ln(c)^2*b^2-2*b/e^3*n*d*a-4/e^3*d*ln(e*x+d)*ln(c)*a*b+1/4*d^2/e^3
/(e*x+d)*Pi^2*b^2*csgn(I*c*x^n)^6-1/4*x/e^2*Pi^2*b^2*csgn(I*c)^2*csgn(I*c*x^n)^4+1/2*x/e^2*Pi^2*b^2*csgn(I*c)*
csgn(I*c*x^n)^5+4/e^3*n*ln(e*x+d)*ln(-e*x/d)*d*b^2*ln(c)+2*b/e^2*ln(x^n)*x*a-2*b*n/e^3*d*ln(e*x+d)*a+2*b*n/e^3
*d*ln(e*x)*a+4*b/e^3*n*dilog(-e*x/d)*d*a-2*d^2/e^3/(e*x+d)*ln(c)*a*b-2*b^2/e^3*d*n^2*ln(x)^2*ln(1+e*x/d)-4*b^2
/e^3*d*n^2*ln(x)*polylog(2,-e*x/d)+2*b^2/e^3*d*n^2*ln(x)^2*ln(e*x+d)-4*b/e^3*ln(e*x+d)*ln(x^n)*d*a-2*b*ln(x^n)
*d^2/e^3/(e*x+d)*a+4*b^2*n/e^3*d*ln(x^n)*dilog(-e*x/d)-d^2/e^3/(e*x+d)*ln(c)^2*b^2-2/e^3*d*ln(e*x+d)*ln(c)^2*b
^2+I/e^2*ln(x^n)*x*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2-2*I/e^3*d*ln(e*x+d)*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I*c*x^n)
^2-2*b^2*n/e^2*ln(x^n)*x+2*x/e^2*ln(c)*a*b-I*x/e^2*ln(c)*Pi*b^2*csgn(I*c*x^n)^3-1/4*x/e^2*Pi^2*b^2*csgn(I*c*x^
n)^6+I/e^2*n*x*b^2*Pi*csgn(I*c*x^n)^3-I*d^2/e^3/(e*x+d)*Pi*a*b*csgn(I*c)*csgn(I*c*x^n)^2+I/e^2*n*x*b^2*Pi*csgn
(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2/e^3*d*ln(e*x+d)*Pi^2*b^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4-1/2*d^2/e^3/(e*x+d)
*Pi^2*b^2*csgn(I*c)*csgn(I*c*x^n)^5+1/4*d^2/e^3/(e*x+d)*Pi^2*b^2*csgn(I*c)^2*csgn(I*c*x^n)^4-I/e^2*ln(x^n)*x*b
^2*Pi*csgn(I*c*x^n)^3+1/2/e^3*d*ln(e*x+d)*Pi^2*b^2*csgn(I*c)^2*csgn(I*c*x^n)^4+1/2*x/e^2*Pi^2*b^2*csgn(I*c)^2*
csgn(I*x^n)*csgn(I*c*x^n)^3+2*I/e^3*n*dilog(-e*x/d)*d*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I/e^2*ln(x^n)*x*b^2*P
i*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-2*I/e^3*d*ln(e*x+d)*Pi*a*b*csgn(I*c)*csgn(I*c*x^n)^2-2*I/e^3*d*ln(e*x+d)
*Pi*a*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*x/e^2*ln(c)*Pi*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*n/e^3*d*ln(e*x)
*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+I*n/e^3*d*ln(e*x)*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2-2*I/e^3*d*ln(e*x+d)*ln(
c)*Pi*b^2*csgn(I*c)*csgn(I*c*x^n)^2-I*d^2/e^3/(e*x+d)*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I*c*x^n)^2-2*I/e^3*ln(e*x+
d)*ln(x^n)*d*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+2/e^2*ln(x^n)*x*b^2*ln(c)-2*I/e^3*n*ln(e*x+d)*ln(-e*x/d)*d*b^2
*Pi*csgn(I*c*x^n)^3-2*I/e^3*ln(e*x+d)*ln(x^n)*d*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2+2*b^2*n^2*x/e^2-b^2*ln(x^n)^2
*d^2/e^3/(e*x+d)-I*x/e^2*Pi*a*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I/e^3*n*d*b^2*Pi*csgn(I*c)*csgn(I*x^n)*csg
n(I*c*x^n)-I*d^2/e^3/(e*x+d)*Pi*a*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*n/e^3*d*ln(e*x+d)*b^2*Pi*csgn(I*c)*csgn(I*c*
x^n)^2-I*n/e^3*d*ln(e*x+d)*b^2*Pi*csgn(I*x^n)*c...

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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