3.85 \(\int x^2 \log ^2(c (a+b x)^n) \, dx\)

Optimal. Leaf size=187 \[ \frac{2 a^3 n \log (a+b x) \log \left (c (a+b x)^n\right )}{3 b^3}-\frac{2 a^2 n (a+b x) \log \left (c (a+b x)^n\right )}{b^3}+\frac{2 a^2 n^2 x}{b^2}-\frac{a^3 n^2 \log ^2(a+b x)}{3 b^3}+\frac{a n (a+b x)^2 \log \left (c (a+b x)^n\right )}{b^3}-\frac{2 n (a+b x)^3 \log \left (c (a+b x)^n\right )}{9 b^3}-\frac{a n^2 (a+b x)^2}{2 b^3}+\frac{2 n^2 (a+b x)^3}{27 b^3}+\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right ) \]

[Out]

(2*a^2*n^2*x)/b^2 - (a*n^2*(a + b*x)^2)/(2*b^3) + (2*n^2*(a + b*x)^3)/(27*b^3) - (a^3*n^2*Log[a + b*x]^2)/(3*b
^3) - (2*a^2*n*(a + b*x)*Log[c*(a + b*x)^n])/b^3 + (a*n*(a + b*x)^2*Log[c*(a + b*x)^n])/b^3 - (2*n*(a + b*x)^3
*Log[c*(a + b*x)^n])/(9*b^3) + (2*a^3*n*Log[a + b*x]*Log[c*(a + b*x)^n])/(3*b^3) + (x^3*Log[c*(a + b*x)^n]^2)/
3

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Rubi [A]  time = 0.191644, antiderivative size = 156, normalized size of antiderivative = 0.83, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {2398, 2411, 43, 2334, 12, 14, 2301} \[ -\frac{1}{9} n \left (\frac{18 a^2 (a+b x)}{b^3}-\frac{6 a^3 \log (a+b x)}{b^3}-\frac{9 a (a+b x)^2}{b^3}+\frac{2 (a+b x)^3}{b^3}\right ) \log \left (c (a+b x)^n\right )+\frac{2 a^2 n^2 x}{b^2}-\frac{a^3 n^2 \log ^2(a+b x)}{3 b^3}-\frac{a n^2 (a+b x)^2}{2 b^3}+\frac{2 n^2 (a+b x)^3}{27 b^3}+\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Log[c*(a + b*x)^n]^2,x]

[Out]

(2*a^2*n^2*x)/b^2 - (a*n^2*(a + b*x)^2)/(2*b^3) + (2*n^2*(a + b*x)^3)/(27*b^3) - (a^3*n^2*Log[a + b*x]^2)/(3*b
^3) - (n*((18*a^2*(a + b*x))/b^3 - (9*a*(a + b*x)^2)/b^3 + (2*(a + b*x)^3)/b^3 - (6*a^3*Log[a + b*x])/b^3)*Log
[c*(a + b*x)^n])/9 + (x^3*Log[c*(a + b*x)^n]^2)/3

Rule 2398

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
f + g*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(g*(q + 1)), x] - Dist[(b*e*n*p)/(g*(q + 1)), Int[((f + g*x)^(q
+ 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rubi steps

\begin{align*} \int x^2 \log ^2\left (c (a+b x)^n\right ) \, dx &=\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right )-\frac{1}{3} (2 b n) \int \frac{x^3 \log \left (c (a+b x)^n\right )}{a+b x} \, dx\\ &=\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right )-\frac{1}{3} (2 n) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^3 \log \left (c x^n\right )}{x} \, dx,x,a+b x\right )\\ &=-\frac{1}{9} n \left (\frac{18 a^2 (a+b x)}{b^3}-\frac{9 a (a+b x)^2}{b^3}+\frac{2 (a+b x)^3}{b^3}-\frac{6 a^3 \log (a+b x)}{b^3}\right ) \log \left (c (a+b x)^n\right )+\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right )+\frac{1}{3} \left (2 n^2\right ) \operatorname{Subst}\left (\int \frac{18 a^2 x-9 a x^2+2 x^3-6 a^3 \log (x)}{6 b^3 x} \, dx,x,a+b x\right )\\ &=-\frac{1}{9} n \left (\frac{18 a^2 (a+b x)}{b^3}-\frac{9 a (a+b x)^2}{b^3}+\frac{2 (a+b x)^3}{b^3}-\frac{6 a^3 \log (a+b x)}{b^3}\right ) \log \left (c (a+b x)^n\right )+\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right )+\frac{n^2 \operatorname{Subst}\left (\int \frac{18 a^2 x-9 a x^2+2 x^3-6 a^3 \log (x)}{x} \, dx,x,a+b x\right )}{9 b^3}\\ &=-\frac{1}{9} n \left (\frac{18 a^2 (a+b x)}{b^3}-\frac{9 a (a+b x)^2}{b^3}+\frac{2 (a+b x)^3}{b^3}-\frac{6 a^3 \log (a+b x)}{b^3}\right ) \log \left (c (a+b x)^n\right )+\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right )+\frac{n^2 \operatorname{Subst}\left (\int \left (18 a^2-9 a x+2 x^2-\frac{6 a^3 \log (x)}{x}\right ) \, dx,x,a+b x\right )}{9 b^3}\\ &=\frac{2 a^2 n^2 x}{b^2}-\frac{a n^2 (a+b x)^2}{2 b^3}+\frac{2 n^2 (a+b x)^3}{27 b^3}-\frac{1}{9} n \left (\frac{18 a^2 (a+b x)}{b^3}-\frac{9 a (a+b x)^2}{b^3}+\frac{2 (a+b x)^3}{b^3}-\frac{6 a^3 \log (a+b x)}{b^3}\right ) \log \left (c (a+b x)^n\right )+\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right )-\frac{\left (2 a^3 n^2\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,a+b x\right )}{3 b^3}\\ &=\frac{2 a^2 n^2 x}{b^2}-\frac{a n^2 (a+b x)^2}{2 b^3}+\frac{2 n^2 (a+b x)^3}{27 b^3}-\frac{a^3 n^2 \log ^2(a+b x)}{3 b^3}-\frac{1}{9} n \left (\frac{18 a^2 (a+b x)}{b^3}-\frac{9 a (a+b x)^2}{b^3}+\frac{2 (a+b x)^3}{b^3}-\frac{6 a^3 \log (a+b x)}{b^3}\right ) \log \left (c (a+b x)^n\right )+\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right )\\ \end{align*}

Mathematica [A]  time = 0.0483155, size = 163, normalized size = 0.87 \[ \frac{a^3 \log ^2\left (c (a+b x)^n\right )}{3 b^3}-\frac{11 a^3 n \log \left (c (a+b x)^n\right )}{9 b^3}-\frac{2 a^2 n x \log \left (c (a+b x)^n\right )}{3 b^2}+\frac{11 a^2 n^2 x}{9 b^2}+\frac{1}{3} x^3 \log ^2\left (c (a+b x)^n\right )+\frac{a n x^2 \log \left (c (a+b x)^n\right )}{3 b}-\frac{2}{9} n x^3 \log \left (c (a+b x)^n\right )-\frac{5 a n^2 x^2}{18 b}+\frac{2 n^2 x^3}{27} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[c*(a + b*x)^n]^2,x]

[Out]

(11*a^2*n^2*x)/(9*b^2) - (5*a*n^2*x^2)/(18*b) + (2*n^2*x^3)/27 - (11*a^3*n*Log[c*(a + b*x)^n])/(9*b^3) - (2*a^
2*n*x*Log[c*(a + b*x)^n])/(3*b^2) + (a*n*x^2*Log[c*(a + b*x)^n])/(3*b) - (2*n*x^3*Log[c*(a + b*x)^n])/9 + (a^3
*Log[c*(a + b*x)^n]^2)/(3*b^3) + (x^3*Log[c*(a + b*x)^n]^2)/3

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Maple [C]  time = 0.534, size = 1300, normalized size = 7. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/27*n^2*x^3-11/9*a^3*n^2/b^3*ln(b*x+a)-1/3*I/b^2*Pi*a^2*n*x*csgn(I*c*(b*x+a)^n)^2*csgn(I*c)+1/3*ln(c)^2*x^3-2
/3/b^2*ln(c)*a^2*n*x+1/6*Pi^2*x^3*csgn(I*(b*x+a)^n)^2*csgn(I*c*(b*x+a)^n)^3*csgn(I*c)-1/12*Pi^2*x^3*csgn(I*(b*
x+a)^n)^2*csgn(I*c*(b*x+a)^n)^2*csgn(I*c)^2-1/3*Pi^2*x^3*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^4*csgn(I*c)+1/6
*Pi^2*x^3*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^3*csgn(I*c)^2-1/3*I*ln(c)*Pi*x^3*csgn(I*c*(b*x+a)^n)^3+1/9*I*n
*Pi*x^3*csgn(I*c*(b*x+a)^n)^3+1/3/b*ln(c)*a*n*x^2+2/3/b^3*ln(c)*ln(b*x+a)*a^3*n-1/12*Pi^2*x^3*csgn(I*(b*x+a)^n
)^2*csgn(I*c*(b*x+a)^n)^4+1/6*Pi^2*x^3*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^5+1/6*Pi^2*x^3*csgn(I*c*(b*x+a)^n
)^5*csgn(I*c)-1/12*Pi^2*x^3*csgn(I*c*(b*x+a)^n)^4*csgn(I*c)^2-5/18/b*a*n^2*x^2-2/9*n*ln(c)*x^3-1/12*Pi^2*x^3*c
sgn(I*c*(b*x+a)^n)^6+1/3*I*ln(c)*Pi*x^3*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^2+1/3*I*ln(c)*Pi*x^3*csgn(I*c*(b
*x+a)^n)^2*csgn(I*c)-1/9*I*n*Pi*x^3*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^2-1/9*I*n*Pi*x^3*csgn(I*c*(b*x+a)^n)
^2*csgn(I*c)+1/9*(3*I*Pi*b^3*x^3*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^2-3*I*Pi*b^3*x^3*csgn(I*(b*x+a)^n)*csgn
(I*c*(b*x+a)^n)*csgn(I*c)-3*I*Pi*b^3*x^3*csgn(I*c*(b*x+a)^n)^3+3*I*Pi*b^3*x^3*csgn(I*c*(b*x+a)^n)^2*csgn(I*c)+
6*ln(c)*b^3*x^3-2*b^3*n*x^3+3*a*b^2*n*x^2+6*a^3*n*ln(b*x+a)-6*a^2*b*n*x)/b^3*ln((b*x+a)^n)+1/3*x^3*ln((b*x+a)^
n)^2-1/6*I/b*Pi*a*n*x^2*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)*csgn(I*c)-1/3*I/b^3*Pi*ln(b*x+a)*a^3*n*csgn(I*(b
*x+a)^n)*csgn(I*c*(b*x+a)^n)*csgn(I*c)+1/3*I/b^2*Pi*a^2*n*x*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)*csgn(I*c)+1/
6*I/b*Pi*a*n*x^2*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^2+1/6*I/b*Pi*a*n*x^2*csgn(I*c*(b*x+a)^n)^2*csgn(I*c)+1/
3*I/b^3*Pi*ln(b*x+a)*a^3*n*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^2+1/3*I/b^3*Pi*ln(b*x+a)*a^3*n*csgn(I*c*(b*x+
a)^n)^2*csgn(I*c)-1/3*I/b^2*Pi*a^2*n*x*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^2+11/9*a^2*n^2*x/b^2-1/3*a^3*n^2*
ln(b*x+a)^2/b^3+1/9*I*n*Pi*x^3*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)*csgn(I*c)-1/3*I*ln(c)*Pi*x^3*csgn(I*(b*x+
a)^n)*csgn(I*c*(b*x+a)^n)*csgn(I*c)-1/6*I/b*Pi*a*n*x^2*csgn(I*c*(b*x+a)^n)^3-1/3*I/b^3*Pi*ln(b*x+a)*a^3*n*csgn
(I*c*(b*x+a)^n)^3+1/3*I/b^2*Pi*a^2*n*x*csgn(I*c*(b*x+a)^n)^3

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Maxima [A]  time = 1.22553, size = 177, normalized size = 0.95 \begin{align*} \frac{1}{3} \, x^{3} \log \left ({\left (b x + a\right )}^{n} c\right )^{2} + \frac{1}{9} \, b n{\left (\frac{6 \, a^{3} \log \left (b x + a\right )}{b^{4}} - \frac{2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{b^{3}}\right )} \log \left ({\left (b x + a\right )}^{n} c\right ) + \frac{{\left (4 \, b^{3} x^{3} - 15 \, a b^{2} x^{2} - 18 \, a^{3} \log \left (b x + a\right )^{2} + 66 \, a^{2} b x - 66 \, a^{3} \log \left (b x + a\right )\right )} n^{2}}{54 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log((b*x + a)^n*c)^2 + 1/9*b*n*(6*a^3*log(b*x + a)/b^4 - (2*b^2*x^3 - 3*a*b*x^2 + 6*a^2*x)/b^3)*log((b
*x + a)^n*c) + 1/54*(4*b^3*x^3 - 15*a*b^2*x^2 - 18*a^3*log(b*x + a)^2 + 66*a^2*b*x - 66*a^3*log(b*x + a))*n^2/
b^3

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Fricas [A]  time = 2.01129, size = 397, normalized size = 2.12 \begin{align*} \frac{4 \, b^{3} n^{2} x^{3} + 18 \, b^{3} x^{3} \log \left (c\right )^{2} - 15 \, a b^{2} n^{2} x^{2} + 66 \, a^{2} b n^{2} x + 18 \,{\left (b^{3} n^{2} x^{3} + a^{3} n^{2}\right )} \log \left (b x + a\right )^{2} - 6 \,{\left (2 \, b^{3} n^{2} x^{3} - 3 \, a b^{2} n^{2} x^{2} + 6 \, a^{2} b n^{2} x + 11 \, a^{3} n^{2} - 6 \,{\left (b^{3} n x^{3} + a^{3} n\right )} \log \left (c\right )\right )} \log \left (b x + a\right ) - 6 \,{\left (2 \, b^{3} n x^{3} - 3 \, a b^{2} n x^{2} + 6 \, a^{2} b n x\right )} \log \left (c\right )}{54 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(4*b^3*n^2*x^3 + 18*b^3*x^3*log(c)^2 - 15*a*b^2*n^2*x^2 + 66*a^2*b*n^2*x + 18*(b^3*n^2*x^3 + a^3*n^2)*log
(b*x + a)^2 - 6*(2*b^3*n^2*x^3 - 3*a*b^2*n^2*x^2 + 6*a^2*b*n^2*x + 11*a^3*n^2 - 6*(b^3*n*x^3 + a^3*n)*log(c))*
log(b*x + a) - 6*(2*b^3*n*x^3 - 3*a*b^2*n*x^2 + 6*a^2*b*n*x)*log(c))/b^3

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Sympy [A]  time = 3.73593, size = 260, normalized size = 1.39 \begin{align*} \begin{cases} \frac{a^{3} n^{2} \log{\left (a + b x \right )}^{2}}{3 b^{3}} - \frac{11 a^{3} n^{2} \log{\left (a + b x \right )}}{9 b^{3}} + \frac{2 a^{3} n \log{\left (c \right )} \log{\left (a + b x \right )}}{3 b^{3}} - \frac{2 a^{2} n^{2} x \log{\left (a + b x \right )}}{3 b^{2}} + \frac{11 a^{2} n^{2} x}{9 b^{2}} - \frac{2 a^{2} n x \log{\left (c \right )}}{3 b^{2}} + \frac{a n^{2} x^{2} \log{\left (a + b x \right )}}{3 b} - \frac{5 a n^{2} x^{2}}{18 b} + \frac{a n x^{2} \log{\left (c \right )}}{3 b} + \frac{n^{2} x^{3} \log{\left (a + b x \right )}^{2}}{3} - \frac{2 n^{2} x^{3} \log{\left (a + b x \right )}}{9} + \frac{2 n^{2} x^{3}}{27} + \frac{2 n x^{3} \log{\left (c \right )} \log{\left (a + b x \right )}}{3} - \frac{2 n x^{3} \log{\left (c \right )}}{9} + \frac{x^{3} \log{\left (c \right )}^{2}}{3} & \text{for}\: b \neq 0 \\\frac{x^{3} \log{\left (a^{n} c \right )}^{2}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*n**2*log(a + b*x)**2/(3*b**3) - 11*a**3*n**2*log(a + b*x)/(9*b**3) + 2*a**3*n*log(c)*log(a + b
*x)/(3*b**3) - 2*a**2*n**2*x*log(a + b*x)/(3*b**2) + 11*a**2*n**2*x/(9*b**2) - 2*a**2*n*x*log(c)/(3*b**2) + a*
n**2*x**2*log(a + b*x)/(3*b) - 5*a*n**2*x**2/(18*b) + a*n*x**2*log(c)/(3*b) + n**2*x**3*log(a + b*x)**2/3 - 2*
n**2*x**3*log(a + b*x)/9 + 2*n**2*x**3/27 + 2*n*x**3*log(c)*log(a + b*x)/3 - 2*n*x**3*log(c)/9 + x**3*log(c)**
2/3, Ne(b, 0)), (x**3*log(a**n*c)**2/3, True))

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Giac [A]  time = 1.18996, size = 462, normalized size = 2.47 \begin{align*} \frac{{\left (b x + a\right )}^{3} n^{2} \log \left (b x + a\right )^{2}}{3 \, b^{3}} - \frac{{\left (b x + a\right )}^{2} a n^{2} \log \left (b x + a\right )^{2}}{b^{3}} + \frac{{\left (b x + a\right )} a^{2} n^{2} \log \left (b x + a\right )^{2}}{b^{3}} - \frac{2 \,{\left (b x + a\right )}^{3} n^{2} \log \left (b x + a\right )}{9 \, b^{3}} + \frac{{\left (b x + a\right )}^{2} a n^{2} \log \left (b x + a\right )}{b^{3}} - \frac{2 \,{\left (b x + a\right )} a^{2} n^{2} \log \left (b x + a\right )}{b^{3}} + \frac{2 \,{\left (b x + a\right )}^{3} n \log \left (b x + a\right ) \log \left (c\right )}{3 \, b^{3}} - \frac{2 \,{\left (b x + a\right )}^{2} a n \log \left (b x + a\right ) \log \left (c\right )}{b^{3}} + \frac{2 \,{\left (b x + a\right )} a^{2} n \log \left (b x + a\right ) \log \left (c\right )}{b^{3}} + \frac{2 \,{\left (b x + a\right )}^{3} n^{2}}{27 \, b^{3}} - \frac{{\left (b x + a\right )}^{2} a n^{2}}{2 \, b^{3}} + \frac{2 \,{\left (b x + a\right )} a^{2} n^{2}}{b^{3}} - \frac{2 \,{\left (b x + a\right )}^{3} n \log \left (c\right )}{9 \, b^{3}} + \frac{{\left (b x + a\right )}^{2} a n \log \left (c\right )}{b^{3}} - \frac{2 \,{\left (b x + a\right )} a^{2} n \log \left (c\right )}{b^{3}} + \frac{{\left (b x + a\right )}^{3} \log \left (c\right )^{2}}{3 \, b^{3}} - \frac{{\left (b x + a\right )}^{2} a \log \left (c\right )^{2}}{b^{3}} + \frac{{\left (b x + a\right )} a^{2} \log \left (c\right )^{2}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x + a)^3*n^2*log(b*x + a)^2/b^3 - (b*x + a)^2*a*n^2*log(b*x + a)^2/b^3 + (b*x + a)*a^2*n^2*log(b*x + a)
^2/b^3 - 2/9*(b*x + a)^3*n^2*log(b*x + a)/b^3 + (b*x + a)^2*a*n^2*log(b*x + a)/b^3 - 2*(b*x + a)*a^2*n^2*log(b
*x + a)/b^3 + 2/3*(b*x + a)^3*n*log(b*x + a)*log(c)/b^3 - 2*(b*x + a)^2*a*n*log(b*x + a)*log(c)/b^3 + 2*(b*x +
a)*a^2*n*log(b*x + a)*log(c)/b^3 + 2/27*(b*x + a)^3*n^2/b^3 - 1/2*(b*x + a)^2*a*n^2/b^3 + 2*(b*x + a)*a^2*n^2
/b^3 - 2/9*(b*x + a)^3*n*log(c)/b^3 + (b*x + a)^2*a*n*log(c)/b^3 - 2*(b*x + a)*a^2*n*log(c)/b^3 + 1/3*(b*x + a
)^3*log(c)^2/b^3 - (b*x + a)^2*a*log(c)^2/b^3 + (b*x + a)*a^2*log(c)^2/b^3