3.130 \(\int x^2 \log ^2(c (d+e x^3)^p) \, dx\)

Optimal. Leaf size=66 \[ \frac{\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e}-\frac{2 p \left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 e}+\frac{2 p^2 x^3}{3} \]

[Out]

(2*p^2*x^3)/3 - (2*p*(d + e*x^3)*Log[c*(d + e*x^3)^p])/(3*e) + ((d + e*x^3)*Log[c*(d + e*x^3)^p]^2)/(3*e)

________________________________________________________________________________________

Rubi [A]  time = 0.0556588, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2454, 2389, 2296, 2295} \[ \frac{\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e}-\frac{2 p \left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 e}+\frac{2 p^2 x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Log[c*(d + e*x^3)^p]^2,x]

[Out]

(2*p^2*x^3)/3 - (2*p*(d + e*x^3)*Log[c*(d + e*x^3)^p])/(3*e) + ((d + e*x^3)*Log[c*(d + e*x^3)^p]^2)/(3*e)

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin{align*} \int x^2 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^3\right )\\ &=\frac{\operatorname{Subst}\left (\int \log ^2\left (c x^p\right ) \, dx,x,d+e x^3\right )}{3 e}\\ &=\frac{\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e}-\frac{(2 p) \operatorname{Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^3\right )}{3 e}\\ &=\frac{2 p^2 x^3}{3}-\frac{2 p \left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 e}+\frac{\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e}\\ \end{align*}

Mathematica [A]  time = 0.0095782, size = 63, normalized size = 0.95 \[ \frac{1}{3} \left (\frac{\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{e}-2 p \left (\frac{\left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{e}-p x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[c*(d + e*x^3)^p]^2,x]

[Out]

(((d + e*x^3)*Log[c*(d + e*x^3)^p]^2)/e - 2*p*(-(p*x^3) + ((d + e*x^3)*Log[c*(d + e*x^3)^p])/e))/3

________________________________________________________________________________________

Maple [C]  time = 0.556, size = 1036, normalized size = 15.7 \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*(e*x^3+d)^p)^2,x)

[Out]

-2/3*ln(c)*p*x^3-1/12*Pi^2*x^3*csgn(I*c*(e*x^3+d)^p)^6+1/3*(I*Pi*e*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^
p)^2-I*Pi*e*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)-I*Pi*e*x^3*csgn(I*c*(e*x^3+d)^p)^3+I*Pi*e*
x^3*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)+2*ln(c)*e*x^3-2*x^3*p*e+2*d*p*ln(e*x^3+d))/e*ln((e*x^3+d)^p)+2/3*p^2*x^3
+1/3*ln(c)^2*x^3-1/3/e*d*p^2*ln(e*x^3+d)^2-1/12*Pi^2*x^3*csgn(I*(e*x^3+d)^p)^2*csgn(I*c*(e*x^3+d)^p)^4+1/6*Pi^
2*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^5+1/6*Pi^2*x^3*csgn(I*c*(e*x^3+d)^p)^5*csgn(I*c)-1/12*Pi^2*x^3
*csgn(I*c*(e*x^3+d)^p)^4*csgn(I*c)^2+1/3*x^3*ln((e*x^3+d)^p)^2-2/3*d*p^2/e*ln(e*x^3+d)+2/3/e*ln(c)*ln(e*x^3+d)
*d*p+1/6*Pi^2*x^3*csgn(I*(e*x^3+d)^p)^2*csgn(I*c*(e*x^3+d)^p)^3*csgn(I*c)-1/12*Pi^2*x^3*csgn(I*(e*x^3+d)^p)^2*
csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)^2-1/3*Pi^2*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^4*csgn(I*c)+1/6*Pi^
2*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^3*csgn(I*c)^2-1/3*I*ln(c)*Pi*x^3*csgn(I*c*(e*x^3+d)^p)^3+1/3*I
*Pi*p*x^3*csgn(I*c*(e*x^3+d)^p)^3-1/3*I/e*Pi*ln(e*x^3+d)*d*p*csgn(I*c*(e*x^3+d)^p)^3-1/3*I*ln(c)*Pi*x^3*csgn(I
*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)+1/3*I*Pi*p*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*
c)+1/3*I/e*Pi*ln(e*x^3+d)*d*p*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2+1/3*I/e*Pi*ln(e*x^3+d)*d*p*csgn(I*c*
(e*x^3+d)^p)^2*csgn(I*c)-1/3*I/e*Pi*ln(e*x^3+d)*d*p*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)+1/3*I*
ln(c)*Pi*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2+1/3*I*ln(c)*Pi*x^3*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)-
1/3*I*Pi*p*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2-1/3*I*Pi*p*x^3*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)

________________________________________________________________________________________

Maxima [A]  time = 1.06436, size = 131, normalized size = 1.98 \begin{align*} \frac{1}{3} \, x^{3} \log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2} - \frac{2}{3} \,{\left (\frac{x^{3}}{e} - \frac{d \log \left (e x^{3} + d\right )}{e^{2}}\right )} e p \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) + \frac{{\left (2 \, e x^{3} - d \log \left (e x^{3} + d\right )^{2} - 2 \, d \log \left (e x^{3} + d\right )\right )} p^{2}}{3 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(c*(e*x^3+d)^p)^2,x, algorithm="maxima")

[Out]

1/3*x^3*log((e*x^3 + d)^p*c)^2 - 2/3*(x^3/e - d*log(e*x^3 + d)/e^2)*e*p*log((e*x^3 + d)^p*c) + 1/3*(2*e*x^3 -
d*log(e*x^3 + d)^2 - 2*d*log(e*x^3 + d))*p^2/e

________________________________________________________________________________________

Fricas [A]  time = 1.88889, size = 216, normalized size = 3.27 \begin{align*} \frac{2 \, e p^{2} x^{3} - 2 \, e p x^{3} \log \left (c\right ) + e x^{3} \log \left (c\right )^{2} +{\left (e p^{2} x^{3} + d p^{2}\right )} \log \left (e x^{3} + d\right )^{2} - 2 \,{\left (e p^{2} x^{3} + d p^{2} -{\left (e p x^{3} + d p\right )} \log \left (c\right )\right )} \log \left (e x^{3} + d\right )}{3 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(c*(e*x^3+d)^p)^2,x, algorithm="fricas")

[Out]

1/3*(2*e*p^2*x^3 - 2*e*p*x^3*log(c) + e*x^3*log(c)^2 + (e*p^2*x^3 + d*p^2)*log(e*x^3 + d)^2 - 2*(e*p^2*x^3 + d
*p^2 - (e*p*x^3 + d*p)*log(c))*log(e*x^3 + d))/e

________________________________________________________________________________________

Sympy [A]  time = 12.3171, size = 160, normalized size = 2.42 \begin{align*} \begin{cases} \frac{d p^{2} \log{\left (d + e x^{3} \right )}^{2}}{3 e} - \frac{2 d p^{2} \log{\left (d + e x^{3} \right )}}{3 e} + \frac{2 d p \log{\left (c \right )} \log{\left (d + e x^{3} \right )}}{3 e} + \frac{p^{2} x^{3} \log{\left (d + e x^{3} \right )}^{2}}{3} - \frac{2 p^{2} x^{3} \log{\left (d + e x^{3} \right )}}{3} + \frac{2 p^{2} x^{3}}{3} + \frac{2 p x^{3} \log{\left (c \right )} \log{\left (d + e x^{3} \right )}}{3} - \frac{2 p x^{3} \log{\left (c \right )}}{3} + \frac{x^{3} \log{\left (c \right )}^{2}}{3} & \text{for}\: e \neq 0 \\\frac{x^{3} \log{\left (c d^{p} \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*(e*x**3+d)**p)**2,x)

[Out]

Piecewise((d*p**2*log(d + e*x**3)**2/(3*e) - 2*d*p**2*log(d + e*x**3)/(3*e) + 2*d*p*log(c)*log(d + e*x**3)/(3*
e) + p**2*x**3*log(d + e*x**3)**2/3 - 2*p**2*x**3*log(d + e*x**3)/3 + 2*p**2*x**3/3 + 2*p*x**3*log(c)*log(d +
e*x**3)/3 - 2*p*x**3*log(c)/3 + x**3*log(c)**2/3, Ne(e, 0)), (x**3*log(c*d**p)**2/3, True))

________________________________________________________________________________________

Giac [A]  time = 1.24961, size = 140, normalized size = 2.12 \begin{align*} \frac{1}{3} \,{\left ({\left (2 \, x^{3} e +{\left (x^{3} e + d\right )} \log \left (x^{3} e + d\right )^{2} - 2 \,{\left (x^{3} e + d\right )} \log \left (x^{3} e + d\right ) + 2 \, d\right )} p^{2} - 2 \,{\left (x^{3} e -{\left (x^{3} e + d\right )} \log \left (x^{3} e + d\right ) + d\right )} p \log \left (c\right ) +{\left (x^{3} e + d\right )} \log \left (c\right )^{2}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(c*(e*x^3+d)^p)^2,x, algorithm="giac")

[Out]

1/3*((2*x^3*e + (x^3*e + d)*log(x^3*e + d)^2 - 2*(x^3*e + d)*log(x^3*e + d) + 2*d)*p^2 - 2*(x^3*e - (x^3*e + d
)*log(x^3*e + d) + d)*p*log(c) + (x^3*e + d)*log(c)^2)*e^(-1)