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

Optimal. Leaf size=168 \[ -\frac {2 n \log \left (c (a+b x)^n\right ) \text {Li}_2\left (-\frac {e (a+b x)}{b d-a e}\right )}{d}-\frac {\log ^2\left (c (a+b x)^n\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {2 n \text {Li}_2\left (\frac {b x}{a}+1\right ) \log \left (c (a+b x)^n\right )}{d}+\frac {\log \left (-\frac {b x}{a}\right ) \log ^2\left (c (a+b x)^n\right )}{d}+\frac {2 n^2 \text {Li}_3\left (-\frac {e (a+b x)}{b d-a e}\right )}{d}-\frac {2 n^2 \text {Li}_3\left (\frac {b x}{a}+1\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1593, 2416, 2396, 2433, 2374, 6589} \[ -\frac {2 n \log \left (c (a+b x)^n\right ) \text {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d}+\frac {2 n \text {PolyLog}\left (2,\frac {b x}{a}+1\right ) \log \left (c (a+b x)^n\right )}{d}+\frac {2 n^2 \text {PolyLog}\left (3,-\frac {e (a+b x)}{b d-a e}\right )}{d}-\frac {2 n^2 \text {PolyLog}\left (3,\frac {b x}{a}+1\right )}{d}-\frac {\log ^2\left (c (a+b x)^n\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d}+\frac {\log \left (-\frac {b x}{a}\right ) \log ^2\left (c (a+b x)^n\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2396

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

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

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Mathematica [A]  time = 0.14, size = 292, normalized size = 1.74 \[ \frac {-2 n \left (n \log (a+b x)-\log \left (c (a+b x)^n\right )\right ) \left (-\text {Li}_2\left (\frac {e (a+b x)}{a e-b d}\right )-\log (a+b x) \log \left (\frac {b (d+e x)}{b d-a e}\right )-\text {Li}_2\left (-\frac {b x}{a}\right )+\log (x) \left (\log (a+b x)-\log \left (\frac {b x}{a}+1\right )\right )\right )-\log (d+e x) \left (\log \left (c (a+b x)^n\right )-n \log (a+b x)\right )^2+\log (x) \left (\log \left (c (a+b x)^n\right )-n \log (a+b x)\right )^2+n^2 \left (2 \text {Li}_3\left (\frac {e (a+b x)}{a e-b d}\right )-2 \log (a+b x) \text {Li}_2\left (\frac {e (a+b x)}{a e-b d}\right )-\log ^2(a+b x) \log \left (\frac {b (d+e x)}{b d-a e}\right )-2 \text {Li}_3\left (\frac {b x}{a}+1\right )+2 \text {Li}_2\left (\frac {b x}{a}+1\right ) \log (a+b x)+\log \left (-\frac {b x}{a}\right ) \log ^2(a+b x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x + a\right )}^{n} c\right )^{2}}{e x^{2} + d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.52, size = 2679, normalized size = 15.95 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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