3.1.85 \(\int x^2 \log ^2(c (a+b x)^n) \, dx\) [85]

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

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 187, normalized size of antiderivative = 1.00, 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 = {2445, 2458, 45, 2372, 12, 14, 2338} \begin {gather*} \frac {2 a^3 n \log (a+b x) \log \left (c (a+b x)^n\right )}{3 b^3}-\frac {a^3 n^2 \log ^2(a+b x)}{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 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 ) \end {gather*}

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) - (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

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 45

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 2338

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 2372

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]}, Dist[a + b*Log[c*x^n], u, 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 2445

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 2458

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]

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) \text {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 ) \text {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 \text {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 \text {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 ) \text {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*}

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Mathematica [A]
time = 0.07, size = 131, normalized size = 0.70 \begin {gather*} \frac {-18 a^3 n^2 \log ^2(a+b x)+6 a^3 n \log (a+b x) \left (-11 n+6 \log \left (c (a+b x)^n\right )\right )+b x \left (n^2 \left (66 a^2-15 a b x+4 b^2 x^2\right )-6 n \left (6 a^2-3 a b x+2 b^2 x^2\right ) \log \left (c (a+b x)^n\right )+18 b^2 x^2 \log ^2\left (c (a+b x)^n\right )\right )}{54 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[c*(a + b*x)^n]^2,x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.41, size = 1300, normalized size = 6.95

method	result	size
risch	\(\text {Expression too large to display}\)	\(1300\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 131, normalized size = 0.70 \begin {gather*} \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 {gather*}

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 = 0.35, size = 179, normalized size = 0.96 \begin {gather*} \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 {gather*}

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 = 0.92, size = 173, normalized size = 0.93 \begin {gather*} \begin {cases} - \frac {11 a^{3} n \log {\left (c \left (a + b x\right )^{n} \right )}}{9 b^{3}} + \frac {a^{3} \log {\left (c \left (a + b x\right )^{n} \right )}^{2}}{3 b^{3}} + \frac {11 a^{2} n^{2} x}{9 b^{2}} - \frac {2 a^{2} n x \log {\left (c \left (a + b x\right )^{n} \right )}}{3 b^{2}} - \frac {5 a n^{2} x^{2}}{18 b} + \frac {a n x^{2} \log {\left (c \left (a + b x\right )^{n} \right )}}{3 b} + \frac {2 n^{2} x^{3}}{27} - \frac {2 n x^{3} \log {\left (c \left (a + b x\right )^{n} \right )}}{9} + \frac {x^{3} \log {\left (c \left (a + b x\right )^{n} \right )}^{2}}{3} & \text {for}\: b \neq 0 \\\frac {x^{3} \log {\left (a^{n} c \right )}^{2}}{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 6.19, size = 342, normalized size = 1.83 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.24, size = 116, normalized size = 0.62 \begin {gather*} \frac {2\,n^2\,x^3}{27}+{\ln \left (c\,{\left (a+b\,x\right )}^n\right )}^2\,\left (\frac {x^3}{3}+\frac {a^3}{3\,b^3}\right )-\ln \left (c\,{\left (a+b\,x\right )}^n\right )\,\left (\frac {2\,n\,x^3}{9}-\frac {a\,n\,x^2}{3\,b}+\frac {2\,a^2\,n\,x}{3\,b^2}\right )-\frac {11\,a^3\,n^2\,\ln \left (a+b\,x\right )}{9\,b^3}-\frac {5\,a\,n^2\,x^2}{18\,b}+\frac {11\,a^2\,n^2\,x}{9\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*log(c*(a + b*x)^n)^2,x)

[Out]

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

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