3.379 \(\int x^2 (a+b \log (c (d+e x)^n)) (f+g \log (c (d+e x)^n)) \, dx\)

Optimal. Leaf size=258 \[ \frac {d^3 n \log (d+e x) \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{3 e^3}-\frac {d^2 n (d+e x) \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{e^3}+\frac {d n (d+e x)^2 \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{2 e^3}-\frac {n (d+e x)^3 \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{9 e^3}+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (g \log \left (c (d+e x)^n\right )+f\right )-\frac {b d^3 g n^2 \log ^2(d+e x)}{3 e^3}+\frac {2 b d^2 g n^2 x}{e^2}-\frac {b d g n^2 (d+e x)^2}{2 e^3}+\frac {2 b g n^2 (d+e x)^3}{27 e^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.44, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2439, 2411, 43, 2334, 12, 14, 2301} \[ -\frac {1}{18} g n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (g \log \left (c (d+e x)^n\right )+f\right )-\frac {1}{18} b n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}\right ) \left (g \log \left (c (d+e x)^n\right )+f\right )+\frac {2 b d^2 g n^2 x}{e^2}-\frac {b d^3 g n^2 \log ^2(d+e x)}{3 e^3}-\frac {b d g n^2 (d+e x)^2}{2 e^3}+\frac {2 b g n^2 (d+e x)^3}{27 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d^2*g*n^2*x)/e^2 - (b*d*g*n^2*(d + e*x)^2)/(2*e^3) + (2*b*g*n^2*(d + e*x)^3)/(27*e^3) - (b*d^3*g*n^2*Log[
d + e*x]^2)/(3*e^3) - (g*n*((18*d^2*(d + e*x))/e^3 - (9*d*(d + e*x)^2)/e^3 + (2*(d + e*x)^3)/e^3 - (6*d^3*Log[
d + e*x])/e^3)*(a + b*Log[c*(d + e*x)^n]))/18 - (b*n*((18*d^2*(d + e*x))/e^3 - (9*d*(d + e*x)^2)/e^3 + (2*(d +
e*x)^3)/e^3 - (6*d^3*Log[d + e*x])/e^3)*(f + g*Log[c*(d + e*x)^n]))/18 + (x^3*(a + b*Log[c*(d + e*x)^n])*(f +
g*Log[c*(d + e*x)^n]))/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 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 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 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 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 2439

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

Rubi steps

\begin {align*} \int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right ) \, dx &=\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )-\frac {1}{3} (b e n) \int \frac {x^3 \left (f+g \log \left (c (d+e x)^n\right )\right )}{d+e x} \, dx-\frac {1}{3} (e g n) \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{d+e x} \, dx\\ &=\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )-\frac {1}{3} (b n) \operatorname {Subst}\left (\int \frac {\left (-\frac {d}{e}+\frac {x}{e}\right )^3 \left (f+g \log \left (c x^n\right )\right )}{x} \, dx,x,d+e x\right )-\frac {1}{3} (g n) \operatorname {Subst}\left (\int \frac {\left (-\frac {d}{e}+\frac {x}{e}\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x} \, dx,x,d+e x\right )\\ &=-\frac {1}{18} g n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{18} b n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+2 \left (\frac {1}{3} \left (b g n^2\right ) \operatorname {Subst}\left (\int \frac {18 d^2 x-9 d x^2+2 x^3-6 d^3 \log (x)}{6 e^3 x} \, dx,x,d+e x\right )\right )\\ &=-\frac {1}{18} g n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{18} b n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+2 \frac {\left (b g n^2\right ) \operatorname {Subst}\left (\int \frac {18 d^2 x-9 d x^2+2 x^3-6 d^3 \log (x)}{x} \, dx,x,d+e x\right )}{18 e^3}\\ &=-\frac {1}{18} g n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{18} b n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+2 \frac {\left (b g n^2\right ) \operatorname {Subst}\left (\int \left (18 d^2-9 d x+2 x^2-\frac {6 d^3 \log (x)}{x}\right ) \, dx,x,d+e x\right )}{18 e^3}\\ &=-\frac {1}{18} g n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{18} b n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+2 \left (\frac {b d^2 g n^2 x}{e^2}-\frac {b d g n^2 (d+e x)^2}{4 e^3}+\frac {b g n^2 (d+e x)^3}{27 e^3}-\frac {\left (b d^3 g n^2\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,d+e x\right )}{3 e^3}\right )\\ &=2 \left (\frac {b d^2 g n^2 x}{e^2}-\frac {b d g n^2 (d+e x)^2}{4 e^3}+\frac {b g n^2 (d+e x)^3}{27 e^3}-\frac {b d^3 g n^2 \log ^2(d+e x)}{6 e^3}\right )-\frac {1}{18} g n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{18} b n \left (\frac {18 d^2 (d+e x)}{e^3}-\frac {9 d (d+e x)^2}{e^3}+\frac {2 (d+e x)^3}{e^3}-\frac {6 d^3 \log (d+e x)}{e^3}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{3} x^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 342, normalized size = 1.33 \[ \frac {1}{3} a g x^3 \log \left (c (d+e x)^n\right )+\frac {a d^3 g n \log (d+e x)}{3 e^3}-\frac {a d^2 g n x}{3 e^2}+\frac {a d g n x^2}{6 e}+\frac {1}{3} a f x^3-\frac {1}{9} a g n x^3+\frac {b d^3 g \log ^2\left (c (d+e x)^n\right )}{3 e^3}-\frac {11 b d^3 g n \log \left (c (d+e x)^n\right )}{9 e^3}-\frac {2 b d^2 g n x \log \left (c (d+e x)^n\right )}{3 e^2}+\frac {1}{3} b f x^3 \log \left (c (d+e x)^n\right )+\frac {1}{3} b g x^3 \log ^2\left (c (d+e x)^n\right )-\frac {2}{9} b g n x^3 \log \left (c (d+e x)^n\right )+\frac {b d g n x^2 \log \left (c (d+e x)^n\right )}{3 e}+\frac {b d^3 f n \log (d+e x)}{3 e^3}-\frac {b d^2 f n x}{3 e^2}+\frac {11 b d^2 g n^2 x}{9 e^2}+\frac {b d f n x^2}{6 e}-\frac {5 b d g n^2 x^2}{18 e}-\frac {1}{9} b f n x^3+\frac {2}{27} b g n^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b*d^2*f*n*x)/e^2 - (a*d^2*g*n*x)/(3*e^2) + (11*b*d^2*g*n^2*x)/(9*e^2) + (b*d*f*n*x^2)/(6*e) + (a*d*g*n*x
^2)/(6*e) - (5*b*d*g*n^2*x^2)/(18*e) + (a*f*x^3)/3 - (b*f*n*x^3)/9 - (a*g*n*x^3)/9 + (2*b*g*n^2*x^3)/27 + (b*d
^3*f*n*Log[d + e*x])/(3*e^3) + (a*d^3*g*n*Log[d + e*x])/(3*e^3) - (11*b*d^3*g*n*Log[c*(d + e*x)^n])/(9*e^3) -
(2*b*d^2*g*n*x*Log[c*(d + e*x)^n])/(3*e^2) + (b*d*g*n*x^2*Log[c*(d + e*x)^n])/(3*e) + (b*f*x^3*Log[c*(d + e*x)
^n])/3 + (a*g*x^3*Log[c*(d + e*x)^n])/3 - (2*b*g*n*x^3*Log[c*(d + e*x)^n])/9 + (b*d^3*g*Log[c*(d + e*x)^n]^2)/
(3*e^3) + (b*g*x^3*Log[c*(d + e*x)^n]^2)/3

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 329, normalized size = 1.28 \[ \frac {18 \, b e^{3} g x^{3} \log \relax (c)^{2} + 2 \, {\left (2 \, b e^{3} g n^{2} + 9 \, a e^{3} f - 3 \, {\left (b e^{3} f + a e^{3} g\right )} n\right )} x^{3} - 3 \, {\left (5 \, b d e^{2} g n^{2} - 3 \, {\left (b d e^{2} f + a d e^{2} g\right )} n\right )} x^{2} + 18 \, {\left (b e^{3} g n^{2} x^{3} + b d^{3} g n^{2}\right )} \log \left (e x + d\right )^{2} + 6 \, {\left (11 \, b d^{2} e g n^{2} - 3 \, {\left (b d^{2} e f + a d^{2} e g\right )} n\right )} x + 6 \, {\left (3 \, b d e^{2} g n^{2} x^{2} - 6 \, b d^{2} e g n^{2} x - 11 \, b d^{3} g n^{2} - {\left (2 \, b e^{3} g n^{2} - 3 \, {\left (b e^{3} f + a e^{3} g\right )} n\right )} x^{3} + 3 \, {\left (b d^{3} f + a d^{3} g\right )} n + 6 \, {\left (b e^{3} g n x^{3} + b d^{3} g n\right )} \log \relax (c)\right )} \log \left (e x + d\right ) + 6 \, {\left (3 \, b d e^{2} g n x^{2} - 6 \, b d^{2} e g n x - {\left (2 \, b e^{3} g n - 3 \, b e^{3} f - 3 \, a e^{3} g\right )} x^{3}\right )} \log \relax (c)}{54 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(18*b*e^3*g*x^3*log(c)^2 + 2*(2*b*e^3*g*n^2 + 9*a*e^3*f - 3*(b*e^3*f + a*e^3*g)*n)*x^3 - 3*(5*b*d*e^2*g*n
^2 - 3*(b*d*e^2*f + a*d*e^2*g)*n)*x^2 + 18*(b*e^3*g*n^2*x^3 + b*d^3*g*n^2)*log(e*x + d)^2 + 6*(11*b*d^2*e*g*n^
2 - 3*(b*d^2*e*f + a*d^2*e*g)*n)*x + 6*(3*b*d*e^2*g*n^2*x^2 - 6*b*d^2*e*g*n^2*x - 11*b*d^3*g*n^2 - (2*b*e^3*g*
n^2 - 3*(b*e^3*f + a*e^3*g)*n)*x^3 + 3*(b*d^3*f + a*d^3*g)*n + 6*(b*e^3*g*n*x^3 + b*d^3*g*n)*log(c))*log(e*x +
d) + 6*(3*b*d*e^2*g*n*x^2 - 6*b*d^2*e*g*n*x - (2*b*e^3*g*n - 3*b*e^3*f - 3*a*e^3*g)*x^3)*log(c))/e^3

________________________________________________________________________________________

giac [B]  time = 0.21, size = 756, normalized size = 2.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.53, size = 1785, normalized size = 6.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 274, normalized size = 1.06 \[ \frac {1}{3} \, b g x^{3} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + \frac {1}{3} \, b f x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac {1}{3} \, a g x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac {1}{3} \, a f x^{3} + \frac {1}{18} \, b e f n {\left (\frac {6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} + \frac {1}{18} \, a e g n {\left (\frac {6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} + \frac {1}{54} \, {\left (6 \, e n {\left (\frac {6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac {{\left (4 \, e^{3} x^{3} - 15 \, d e^{2} x^{2} - 18 \, d^{3} \log \left (e x + d\right )^{2} + 66 \, d^{2} e x - 66 \, d^{3} \log \left (e x + d\right )\right )} n^{2}}{e^{3}}\right )} b g \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.40, size = 323, normalized size = 1.25 \[ \ln \left (c\,{\left (d+e\,x\right )}^n\right )\,\left (\frac {x^3\,\left (a\,g+b\,f-\frac {2\,b\,g\,n}{3}\right )}{3}+\frac {x^2\,\left (\frac {3\,d\,\left (a\,g+b\,f\right )}{2\,e}-\frac {d\,\left (9\,a\,g+9\,b\,f-6\,b\,g\,n\right )}{6\,e}\right )}{3}-\frac {d\,x\,\left (\frac {9\,d\,\left (a\,g+b\,f\right )}{e}-\frac {d\,\left (9\,a\,g+9\,b\,f-6\,b\,g\,n\right )}{e}\right )}{9\,e}\right )+x^2\,\left (\frac {d\,\left (3\,a\,f-b\,g\,n^2\right )}{6\,e}-\frac {d\,\left (a\,f-\frac {a\,g\,n}{3}-\frac {b\,f\,n}{3}+\frac {2\,b\,g\,n^2}{9}\right )}{2\,e}\right )+{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^2\,\left (\frac {b\,g\,x^3}{3}+\frac {b\,d^3\,g}{3\,e^3}\right )-x\,\left (\frac {d\,\left (\frac {d\,\left (3\,a\,f-b\,g\,n^2\right )}{3\,e}-\frac {d\,\left (a\,f-\frac {a\,g\,n}{3}-\frac {b\,f\,n}{3}+\frac {2\,b\,g\,n^2}{9}\right )}{e}\right )}{e}-\frac {2\,b\,d^2\,g\,n^2}{3\,e^2}\right )+x^3\,\left (\frac {a\,f}{3}-\frac {a\,g\,n}{9}-\frac {b\,f\,n}{9}+\frac {2\,b\,g\,n^2}{27}\right )+\frac {\ln \left (d+e\,x\right )\,\left (3\,a\,d^3\,g\,n+3\,b\,d^3\,f\,n-11\,b\,d^3\,g\,n^2\right )}{9\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*(d + e*x)^n)*((x^3*(a*g + b*f - (2*b*g*n)/3))/3 + (x^2*((3*d*(a*g + b*f))/(2*e) - (d*(9*a*g + 9*b*f - 6*
b*g*n))/(6*e)))/3 - (d*x*((9*d*(a*g + b*f))/e - (d*(9*a*g + 9*b*f - 6*b*g*n))/e))/(9*e)) + x^2*((d*(3*a*f - b*
g*n^2))/(6*e) - (d*(a*f - (a*g*n)/3 - (b*f*n)/3 + (2*b*g*n^2)/9))/(2*e)) + log(c*(d + e*x)^n)^2*((b*g*x^3)/3 +
(b*d^3*g)/(3*e^3)) - x*((d*((d*(3*a*f - b*g*n^2))/(3*e) - (d*(a*f - (a*g*n)/3 - (b*f*n)/3 + (2*b*g*n^2)/9))/e
))/e - (2*b*d^2*g*n^2)/(3*e^2)) + x^3*((a*f)/3 - (a*g*n)/9 - (b*f*n)/9 + (2*b*g*n^2)/27) + (log(d + e*x)*(3*a*
d^3*g*n + 3*b*d^3*f*n - 11*b*d^3*g*n^2))/(9*e^3)

________________________________________________________________________________________

sympy [A]  time = 6.63, size = 508, normalized size = 1.97 \[ \begin {cases} \frac {a d^{3} g n \log {\left (d + e x \right )}}{3 e^{3}} - \frac {a d^{2} g n x}{3 e^{2}} + \frac {a d g n x^{2}}{6 e} + \frac {a f x^{3}}{3} + \frac {a g n x^{3} \log {\left (d + e x \right )}}{3} - \frac {a g n x^{3}}{9} + \frac {a g x^{3} \log {\relax (c )}}{3} + \frac {b d^{3} f n \log {\left (d + e x \right )}}{3 e^{3}} + \frac {b d^{3} g n^{2} \log {\left (d + e x \right )}^{2}}{3 e^{3}} - \frac {11 b d^{3} g n^{2} \log {\left (d + e x \right )}}{9 e^{3}} + \frac {2 b d^{3} g n \log {\relax (c )} \log {\left (d + e x \right )}}{3 e^{3}} - \frac {b d^{2} f n x}{3 e^{2}} - \frac {2 b d^{2} g n^{2} x \log {\left (d + e x \right )}}{3 e^{2}} + \frac {11 b d^{2} g n^{2} x}{9 e^{2}} - \frac {2 b d^{2} g n x \log {\relax (c )}}{3 e^{2}} + \frac {b d f n x^{2}}{6 e} + \frac {b d g n^{2} x^{2} \log {\left (d + e x \right )}}{3 e} - \frac {5 b d g n^{2} x^{2}}{18 e} + \frac {b d g n x^{2} \log {\relax (c )}}{3 e} + \frac {b f n x^{3} \log {\left (d + e x \right )}}{3} - \frac {b f n x^{3}}{9} + \frac {b f x^{3} \log {\relax (c )}}{3} + \frac {b g n^{2} x^{3} \log {\left (d + e x \right )}^{2}}{3} - \frac {2 b g n^{2} x^{3} \log {\left (d + e x \right )}}{9} + \frac {2 b g n^{2} x^{3}}{27} + \frac {2 b g n x^{3} \log {\relax (c )} \log {\left (d + e x \right )}}{3} - \frac {2 b g n x^{3} \log {\relax (c )}}{9} + \frac {b g x^{3} \log {\relax (c )}^{2}}{3} & \text {for}\: e \neq 0 \\\frac {x^{3} \left (a + b \log {\left (c d^{n} \right )}\right ) \left (f + g \log {\left (c d^{n} \right )}\right )}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**3*g*n*log(d + e*x)/(3*e**3) - a*d**2*g*n*x/(3*e**2) + a*d*g*n*x**2/(6*e) + a*f*x**3/3 + a*g*n*
x**3*log(d + e*x)/3 - a*g*n*x**3/9 + a*g*x**3*log(c)/3 + b*d**3*f*n*log(d + e*x)/(3*e**3) + b*d**3*g*n**2*log(
d + e*x)**2/(3*e**3) - 11*b*d**3*g*n**2*log(d + e*x)/(9*e**3) + 2*b*d**3*g*n*log(c)*log(d + e*x)/(3*e**3) - b*
d**2*f*n*x/(3*e**2) - 2*b*d**2*g*n**2*x*log(d + e*x)/(3*e**2) + 11*b*d**2*g*n**2*x/(9*e**2) - 2*b*d**2*g*n*x*l
og(c)/(3*e**2) + b*d*f*n*x**2/(6*e) + b*d*g*n**2*x**2*log(d + e*x)/(3*e) - 5*b*d*g*n**2*x**2/(18*e) + b*d*g*n*
x**2*log(c)/(3*e) + b*f*n*x**3*log(d + e*x)/3 - b*f*n*x**3/9 + b*f*x**3*log(c)/3 + b*g*n**2*x**3*log(d + e*x)*
*2/3 - 2*b*g*n**2*x**3*log(d + e*x)/9 + 2*b*g*n**2*x**3/27 + 2*b*g*n*x**3*log(c)*log(d + e*x)/3 - 2*b*g*n*x**3
*log(c)/9 + b*g*x**3*log(c)**2/3, Ne(e, 0)), (x**3*(a + b*log(c*d**n))*(f + g*log(c*d**n))/3, True))

________________________________________________________________________________________