3.452 \(\int (a+b \log (c (d+e \sqrt [3]{x})^n))^2 \, dx\)

Optimal. Leaf size=267 \[ \frac {2 b d^3 n \log \left (d+e \sqrt [3]{x}\right ) \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )}{e^3}-\frac {6 b d^2 n \left (d+e \sqrt [3]{x}\right ) \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )}{e^3}+\frac {3 b d n \left (d+e \sqrt [3]{x}\right )^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )}{e^3}-\frac {2 b n \left (d+e \sqrt [3]{x}\right )^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )}{3 e^3}+x \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )^2-\frac {b^2 d^3 n^2 \log ^2\left (d+e \sqrt [3]{x}\right )}{e^3}+\frac {6 b^2 d^2 n^2 \sqrt [3]{x}}{e^2}-\frac {3 b^2 d n^2 \left (d+e \sqrt [3]{x}\right )^2}{2 e^3}+\frac {2 b^2 n^2 \left (d+e \sqrt [3]{x}\right )^3}{9 e^3} \]

[Out]

-3/2*b^2*d*n^2*(d+e*x^(1/3))^2/e^3+2/9*b^2*n^2*(d+e*x^(1/3))^3/e^3+6*b^2*d^2*n^2*x^(1/3)/e^2-b^2*d^3*n^2*ln(d+
e*x^(1/3))^2/e^3-6*b*d^2*n*(d+e*x^(1/3))*(a+b*ln(c*(d+e*x^(1/3))^n))/e^3+3*b*d*n*(d+e*x^(1/3))^2*(a+b*ln(c*(d+
e*x^(1/3))^n))/e^3-2/3*b*n*(d+e*x^(1/3))^3*(a+b*ln(c*(d+e*x^(1/3))^n))/e^3+2*b*d^3*n*ln(d+e*x^(1/3))*(a+b*ln(c
*(d+e*x^(1/3))^n))/e^3+x*(a+b*ln(c*(d+e*x^(1/3))^n))^2

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Rubi [A]  time = 0.29, antiderivative size = 210, normalized size of antiderivative = 0.79, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2451, 2398, 2411, 43, 2334, 12, 14, 2301} \[ -\frac {1}{3} b n \left (\frac {18 d^2 \left (d+e \sqrt [3]{x}\right )}{e^3}-\frac {6 d^3 \log \left (d+e \sqrt [3]{x}\right )}{e^3}-\frac {9 d \left (d+e \sqrt [3]{x}\right )^2}{e^3}+\frac {2 \left (d+e \sqrt [3]{x}\right )^3}{e^3}\right ) \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )+x \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )^2+\frac {6 b^2 d^2 n^2 \sqrt [3]{x}}{e^2}-\frac {b^2 d^3 n^2 \log ^2\left (d+e \sqrt [3]{x}\right )}{e^3}-\frac {3 b^2 d n^2 \left (d+e \sqrt [3]{x}\right )^2}{2 e^3}+\frac {2 b^2 n^2 \left (d+e \sqrt [3]{x}\right )^3}{9 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2451

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_), x_Symbol] :> With[{k = Denominator[n]}, Di
st[k, Subst[Int[x^(k - 1)*(a + b*Log[c*(d + e*x^(k*n))^p])^q, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p,
q}, x] && FractionQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 197, normalized size = 0.74 \[ \frac {18 a^2 \left (d^3+e^3 x\right )+6 b \left (6 a \left (d^3+e^3 x\right )-b n \left (11 d^3+6 d^2 e \sqrt [3]{x}-3 d e^2 x^{2/3}+2 e^3 x\right )\right ) \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )+6 a b n \left (7 d^3-6 d^2 e \sqrt [3]{x}+3 d e^2 x^{2/3}-2 e^3 x\right )+18 b^2 \left (d^3+e^3 x\right ) \log ^2\left (c \left (d+e \sqrt [3]{x}\right )^n\right )+b^2 e n^2 \sqrt [3]{x} \left (66 d^2-15 d e \sqrt [3]{x}+4 e^2 x^{2/3}\right )}{18 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*e*n^2*(66*d^2 - 15*d*e*x^(1/3) + 4*e^2*x^(2/3))*x^(1/3) + 6*a*b*n*(7*d^3 - 6*d^2*e*x^(1/3) + 3*d*e^2*x^(2
/3) - 2*e^3*x) + 18*a^2*(d^3 + e^3*x) + 6*b*(6*a*(d^3 + e^3*x) - b*n*(11*d^3 + 6*d^2*e*x^(1/3) - 3*d*e^2*x^(2/
3) + 2*e^3*x))*Log[c*(d + e*x^(1/3))^n] + 18*b^2*(d^3 + e^3*x)*Log[c*(d + e*x^(1/3))^n]^2)/(18*e^3)

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fricas [A]  time = 0.46, size = 287, normalized size = 1.07 \[ \frac {18 \, b^{2} e^{3} x \log \relax (c)^{2} + 18 \, {\left (b^{2} e^{3} n^{2} x + b^{2} d^{3} n^{2}\right )} \log \left (e x^{\frac {1}{3}} + d\right )^{2} - 12 \, {\left (b^{2} e^{3} n - 3 \, a b e^{3}\right )} x \log \relax (c) + 2 \, {\left (2 \, b^{2} e^{3} n^{2} - 6 \, a b e^{3} n + 9 \, a^{2} e^{3}\right )} x + 6 \, {\left (3 \, b^{2} d e^{2} n^{2} x^{\frac {2}{3}} - 6 \, b^{2} d^{2} e n^{2} x^{\frac {1}{3}} - 11 \, b^{2} d^{3} n^{2} + 6 \, a b d^{3} n - 2 \, {\left (b^{2} e^{3} n^{2} - 3 \, a b e^{3} n\right )} x + 6 \, {\left (b^{2} e^{3} n x + b^{2} d^{3} n\right )} \log \relax (c)\right )} \log \left (e x^{\frac {1}{3}} + d\right ) - 3 \, {\left (5 \, b^{2} d e^{2} n^{2} - 6 \, b^{2} d e^{2} n \log \relax (c) - 6 \, a b d e^{2} n\right )} x^{\frac {2}{3}} + 6 \, {\left (11 \, b^{2} d^{2} e n^{2} - 6 \, b^{2} d^{2} e n \log \relax (c) - 6 \, a b d^{2} e n\right )} x^{\frac {1}{3}}}{18 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))^2,x, algorithm="fricas")

[Out]

1/18*(18*b^2*e^3*x*log(c)^2 + 18*(b^2*e^3*n^2*x + b^2*d^3*n^2)*log(e*x^(1/3) + d)^2 - 12*(b^2*e^3*n - 3*a*b*e^
3)*x*log(c) + 2*(2*b^2*e^3*n^2 - 6*a*b*e^3*n + 9*a^2*e^3)*x + 6*(3*b^2*d*e^2*n^2*x^(2/3) - 6*b^2*d^2*e*n^2*x^(
1/3) - 11*b^2*d^3*n^2 + 6*a*b*d^3*n - 2*(b^2*e^3*n^2 - 3*a*b*e^3*n)*x + 6*(b^2*e^3*n*x + b^2*d^3*n)*log(c))*lo
g(e*x^(1/3) + d) - 3*(5*b^2*d*e^2*n^2 - 6*b^2*d*e^2*n*log(c) - 6*a*b*d*e^2*n)*x^(2/3) + 6*(11*b^2*d^2*e*n^2 -
6*b^2*d^2*e*n*log(c) - 6*a*b*d^2*e*n)*x^(1/3))/e^3

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giac [B]  time = 0.21, size = 479, normalized size = 1.79 \[ \frac {1}{18} \, {\left (18 \, b^{2} x e \log \relax (c)^{2} + {\left (18 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right )^{2} - 54 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right )^{2} + 54 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right )^{2} - 12 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) + 54 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) - 108 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) + 4 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} - 27 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} + 108 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )}\right )} b^{2} n^{2} + 6 \, {\left (6 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) - 18 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) + 18 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) - 2 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} + 9 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} - 18 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )}\right )} b^{2} n \log \relax (c) + 36 \, a b x e \log \relax (c) + 6 \, {\left (6 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) - 18 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) + 18 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )} \log \left (x^{\frac {1}{3}} e + d\right ) - 2 \, {\left (x^{\frac {1}{3}} e + d\right )}^{3} e^{\left (-2\right )} + 9 \, {\left (x^{\frac {1}{3}} e + d\right )}^{2} d e^{\left (-2\right )} - 18 \, {\left (x^{\frac {1}{3}} e + d\right )} d^{2} e^{\left (-2\right )}\right )} a b n + 18 \, a^{2} x e\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))^2,x, algorithm="giac")

[Out]

1/18*(18*b^2*x*e*log(c)^2 + (18*(x^(1/3)*e + d)^3*e^(-2)*log(x^(1/3)*e + d)^2 - 54*(x^(1/3)*e + d)^2*d*e^(-2)*
log(x^(1/3)*e + d)^2 + 54*(x^(1/3)*e + d)*d^2*e^(-2)*log(x^(1/3)*e + d)^2 - 12*(x^(1/3)*e + d)^3*e^(-2)*log(x^
(1/3)*e + d) + 54*(x^(1/3)*e + d)^2*d*e^(-2)*log(x^(1/3)*e + d) - 108*(x^(1/3)*e + d)*d^2*e^(-2)*log(x^(1/3)*e
+ d) + 4*(x^(1/3)*e + d)^3*e^(-2) - 27*(x^(1/3)*e + d)^2*d*e^(-2) + 108*(x^(1/3)*e + d)*d^2*e^(-2))*b^2*n^2 +
6*(6*(x^(1/3)*e + d)^3*e^(-2)*log(x^(1/3)*e + d) - 18*(x^(1/3)*e + d)^2*d*e^(-2)*log(x^(1/3)*e + d) + 18*(x^(
1/3)*e + d)*d^2*e^(-2)*log(x^(1/3)*e + d) - 2*(x^(1/3)*e + d)^3*e^(-2) + 9*(x^(1/3)*e + d)^2*d*e^(-2) - 18*(x^
(1/3)*e + d)*d^2*e^(-2))*b^2*n*log(c) + 36*a*b*x*e*log(c) + 6*(6*(x^(1/3)*e + d)^3*e^(-2)*log(x^(1/3)*e + d) -
18*(x^(1/3)*e + d)^2*d*e^(-2)*log(x^(1/3)*e + d) + 18*(x^(1/3)*e + d)*d^2*e^(-2)*log(x^(1/3)*e + d) - 2*(x^(1
/3)*e + d)^3*e^(-2) + 9*(x^(1/3)*e + d)^2*d*e^(-2) - 18*(x^(1/3)*e + d)*d^2*e^(-2))*a*b*n + 18*a^2*x*e)*e^(-1)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \left (e \,x^{\frac {1}{3}}+d \right )^{n}\right )+a \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 217, normalized size = 0.81 \[ \frac {1}{3} \, {\left (e n {\left (\frac {6 \, d^{3} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{4}} - \frac {2 \, e^{2} x - 3 \, d e x^{\frac {2}{3}} + 6 \, d^{2} x^{\frac {1}{3}}}{e^{3}}\right )} + 6 \, x \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right )\right )} a b + \frac {1}{18} \, {\left (6 \, e n {\left (\frac {6 \, d^{3} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{4}} - \frac {2 \, e^{2} x - 3 \, d e x^{\frac {2}{3}} + 6 \, d^{2} x^{\frac {1}{3}}}{e^{3}}\right )} \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right ) + 18 \, x \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right )^{2} - \frac {{\left (18 \, d^{3} \log \left (e x^{\frac {1}{3}} + d\right )^{2} - 4 \, e^{3} x + 66 \, d^{3} \log \left (e x^{\frac {1}{3}} + d\right ) + 15 \, d e^{2} x^{\frac {2}{3}} - 66 \, d^{2} e x^{\frac {1}{3}}\right )} n^{2}}{e^{3}}\right )} b^{2} + a^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.51, size = 290, normalized size = 1.09 \[ \ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )\,\left (\frac {2\,b\,x\,\left (3\,a-b\,n\right )}{3}-x^{2/3}\,\left (\frac {b\,d\,\left (3\,a-b\,n\right )}{e}-\frac {3\,a\,b\,d}{e}\right )+\frac {d\,x^{1/3}\,\left (\frac {2\,b\,d\,\left (3\,a-b\,n\right )}{e}-\frac {6\,a\,b\,d}{e}\right )}{e}\right )-x^{2/3}\,\left (\frac {d\,\left (3\,a^2-2\,a\,b\,n+\frac {2\,b^2\,n^2}{3}\right )}{2\,e}-\frac {d\,\left (3\,a^2-b^2\,n^2\right )}{2\,e}\right )+x^{1/3}\,\left (\frac {d\,\left (\frac {d\,\left (3\,a^2-2\,a\,b\,n+\frac {2\,b^2\,n^2}{3}\right )}{e}-\frac {d\,\left (3\,a^2-b^2\,n^2\right )}{e}\right )}{e}+\frac {2\,b^2\,d^2\,n^2}{e^2}\right )+x\,\left (a^2-\frac {2\,a\,b\,n}{3}+\frac {2\,b^2\,n^2}{9}\right )+{\ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )}^2\,\left (b^2\,x+\frac {b^2\,d^3}{e^3}\right )-\frac {\ln \left (d+e\,x^{1/3}\right )\,\left (11\,b^2\,d^3\,n^2-6\,a\,b\,d^3\,n\right )}{3\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d + e*x^(1/3))^n))^2,x)

[Out]

log(c*(d + e*x^(1/3))^n)*((2*b*x*(3*a - b*n))/3 - x^(2/3)*((b*d*(3*a - b*n))/e - (3*a*b*d)/e) + (d*x^(1/3)*((2
*b*d*(3*a - b*n))/e - (6*a*b*d)/e))/e) - x^(2/3)*((d*(3*a^2 + (2*b^2*n^2)/3 - 2*a*b*n))/(2*e) - (d*(3*a^2 - b^
2*n^2))/(2*e)) + x^(1/3)*((d*((d*(3*a^2 + (2*b^2*n^2)/3 - 2*a*b*n))/e - (d*(3*a^2 - b^2*n^2))/e))/e + (2*b^2*d
^2*n^2)/e^2) + x*(a^2 + (2*b^2*n^2)/9 - (2*a*b*n)/3) + log(c*(d + e*x^(1/3))^n)^2*(b^2*x + (b^2*d^3)/e^3) - (l
og(d + e*x^(1/3))*(11*b^2*d^3*n^2 - 6*a*b*d^3*n))/(3*e^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \log {\left (c \left (d + e \sqrt [3]{x}\right )^{n} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d+e*x**(1/3))**n))**2,x)

[Out]

Integral((a + b*log(c*(d + e*x**(1/3))**n))**2, x)

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