3.129 \(\int x^5 \log ^2(c (d+e x^3)^p) \, dx\)

Optimal. Leaf size=150 \[ \frac{\left (d+e x^3\right )^2 \log ^2\left (c \left (d+e x^3\right )^p\right )}{6 e^2}-\frac{d \left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e^2}-\frac{p \left (d+e x^3\right )^2 \log \left (c \left (d+e x^3\right )^p\right )}{6 e^2}+\frac{2 d p \left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 e^2}+\frac{p^2 \left (d+e x^3\right )^2}{12 e^2}-\frac{2 d p^2 x^3}{3 e} \]

[Out]

(-2*d*p^2*x^3)/(3*e) + (p^2*(d + e*x^3)^2)/(12*e^2) + (2*d*p*(d + e*x^3)*Log[c*(d + e*x^3)^p])/(3*e^2) - (p*(d
+ e*x^3)^2*Log[c*(d + e*x^3)^p])/(6*e^2) - (d*(d + e*x^3)*Log[c*(d + e*x^3)^p]^2)/(3*e^2) + ((d + e*x^3)^2*Lo
g[c*(d + e*x^3)^p]^2)/(6*e^2)

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Rubi [A]  time = 0.15612, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {2454, 2401, 2389, 2296, 2295, 2390, 2305, 2304} \[ \frac{\left (d+e x^3\right )^2 \log ^2\left (c \left (d+e x^3\right )^p\right )}{6 e^2}-\frac{d \left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e^2}-\frac{p \left (d+e x^3\right )^2 \log \left (c \left (d+e x^3\right )^p\right )}{6 e^2}+\frac{2 d p \left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 e^2}+\frac{p^2 \left (d+e x^3\right )^2}{12 e^2}-\frac{2 d p^2 x^3}{3 e} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Log[c*(d + e*x^3)^p]^2,x]

[Out]

(-2*d*p^2*x^3)/(3*e) + (p^2*(d + e*x^3)^2)/(12*e^2) + (2*d*p*(d + e*x^3)*Log[c*(d + e*x^3)^p])/(3*e^2) - (p*(d
+ e*x^3)^2*Log[c*(d + e*x^3)^p])/(6*e^2) - (d*(d + e*x^3)*Log[c*(d + e*x^3)^p]^2)/(3*e^2) + ((d + e*x^3)^2*Lo
g[c*(d + e*x^3)^p]^2)/(6*e^2)

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 2401

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

Rule 2390

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

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin{align*} \int x^5 \log ^2\left (c \left (d+e x^3\right )^p\right ) \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int x \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{d \log ^2\left (c (d+e x)^p\right )}{e}+\frac{(d+e x) \log ^2\left (c (d+e x)^p\right )}{e}\right ) \, dx,x,x^3\right )\\ &=\frac{\operatorname{Subst}\left (\int (d+e x) \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^3\right )}{3 e}-\frac{d \operatorname{Subst}\left (\int \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^3\right )}{3 e}\\ &=\frac{\operatorname{Subst}\left (\int x \log ^2\left (c x^p\right ) \, dx,x,d+e x^3\right )}{3 e^2}-\frac{d \operatorname{Subst}\left (\int \log ^2\left (c x^p\right ) \, dx,x,d+e x^3\right )}{3 e^2}\\ &=-\frac{d \left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e^2}+\frac{\left (d+e x^3\right )^2 \log ^2\left (c \left (d+e x^3\right )^p\right )}{6 e^2}-\frac{p \operatorname{Subst}\left (\int x \log \left (c x^p\right ) \, dx,x,d+e x^3\right )}{3 e^2}+\frac{(2 d p) \operatorname{Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^3\right )}{3 e^2}\\ &=-\frac{2 d p^2 x^3}{3 e}+\frac{p^2 \left (d+e x^3\right )^2}{12 e^2}+\frac{2 d p \left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 e^2}-\frac{p \left (d+e x^3\right )^2 \log \left (c \left (d+e x^3\right )^p\right )}{6 e^2}-\frac{d \left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 e^2}+\frac{\left (d+e x^3\right )^2 \log ^2\left (c \left (d+e x^3\right )^p\right )}{6 e^2}\\ \end{align*}

Mathematica [A]  time = 0.0637101, size = 105, normalized size = 0.7 \[ \frac{-2 \left (d^2-e^2 x^6\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )+2 p \left (2 d^2+2 d e x^3-e^2 x^6\right ) \log \left (c \left (d+e x^3\right )^p\right )+2 d^2 p^2 \log \left (d+e x^3\right )+e p^2 x^3 \left (e x^3-6 d\right )}{12 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Log[c*(d + e*x^3)^p]^2,x]

[Out]

(e*p^2*x^3*(-6*d + e*x^3) + 2*d^2*p^2*Log[d + e*x^3] + 2*p*(2*d^2 + 2*d*e*x^3 - e^2*x^6)*Log[c*(d + e*x^3)^p]
- 2*(d^2 - e^2*x^6)*Log[c*(d + e*x^3)^p]^2)/(12*e^2)

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Maple [C]  time = 0.681, size = 1242, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*ln(c*(e*x^3+d)^p)^2,x)

[Out]

1/6*(I*Pi*e^2*x^6*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2-I*Pi*e^2*x^6*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3
+d)^p)*csgn(I*c)-I*Pi*e^2*x^6*csgn(I*c*(e*x^3+d)^p)^3+I*Pi*e^2*x^6*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)+2*ln(c)*e
^2*x^6-e^2*p*x^6+2*d*e*p*x^3-2*d^2*p*ln(e*x^3+d))/e^2*ln((e*x^3+d)^p)-1/6*ln(c)*p*x^6+1/6*x^6*ln((e*x^3+d)^p)^
2+1/6*ln(c)^2*x^6+1/12*p^2*x^6+1/6/e^2*d^2*p^2*ln(e*x^3+d)^2-1/24*Pi^2*x^6*csgn(I*c*(e*x^3+d)^p)^4*csgn(I*c)^2
+1/12*Pi^2*x^6*csgn(I*c*(e*x^3+d)^p)^5*csgn(I*c)-1/24*Pi^2*x^6*csgn(I*(e*x^3+d)^p)^2*csgn(I*c*(e*x^3+d)^p)^4+1
/12*Pi^2*x^6*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^5-1/2*d*p^2*x^3/e+1/2*d^2*p^2/e^2*ln(e*x^3+d)-1/6*I/e*P
i*d*p*x^3*csgn(I*c*(e*x^3+d)^p)^3+1/6*I/e^2*Pi*ln(e*x^3+d)*d^2*p*csgn(I*c*(e*x^3+d)^p)^3-1/6*I*ln(c)*Pi*x^6*cs
gn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)+1/12*I*Pi*p*x^6*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*cs
gn(I*c)-1/24*Pi^2*x^6*csgn(I*c*(e*x^3+d)^p)^6+1/6*I*ln(c)*Pi*x^6*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2+1
/3/e*ln(c)*d*p*x^3-1/3/e^2*ln(c)*ln(e*x^3+d)*d^2*p-1/24*Pi^2*x^6*csgn(I*(e*x^3+d)^p)^2*csgn(I*c*(e*x^3+d)^p)^2
*csgn(I*c)^2+1/12*Pi^2*x^6*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^3*csgn(I*c)^2+1/12*Pi^2*x^6*csgn(I*(e*x^3
+d)^p)^2*csgn(I*c*(e*x^3+d)^p)^3*csgn(I*c)-1/6*Pi^2*x^6*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^4*csgn(I*c)-
1/6*I*ln(c)*Pi*x^6*csgn(I*c*(e*x^3+d)^p)^3+1/12*I*Pi*p*x^6*csgn(I*c*(e*x^3+d)^p)^3+1/6*I/e*Pi*d*p*x^3*csgn(I*c
*(e*x^3+d)^p)^2*csgn(I*c)+1/6*I/e*Pi*d*p*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2-1/6*I/e^2*Pi*ln(e*x^3
+d)*d^2*p*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)-1/6*I/e^2*Pi*ln(e*x^3+d)*d^2*p*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3
+d)^p)^2-1/12*I*Pi*p*x^6*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)-1/12*I*Pi*p*x^6*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3
+d)^p)^2+1/6*I*ln(c)*Pi*x^6*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)-1/6*I/e*Pi*d*p*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*
(e*x^3+d)^p)*csgn(I*c)+1/6*I/e^2*Pi*ln(e*x^3+d)*d^2*p*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)

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Maxima [A]  time = 1.07146, size = 162, normalized size = 1.08 \begin{align*} \frac{1}{6} \, x^{6} \log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2} - \frac{1}{6} \, e p{\left (\frac{2 \, d^{2} \log \left (e x^{3} + d\right )}{e^{3}} + \frac{e x^{6} - 2 \, d x^{3}}{e^{2}}\right )} \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) + \frac{{\left (e^{2} x^{6} - 6 \, d e x^{3} + 2 \, d^{2} \log \left (e x^{3} + d\right )^{2} + 6 \, d^{2} \log \left (e x^{3} + d\right )\right )} p^{2}}{12 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*log(c*(e*x^3+d)^p)^2,x, algorithm="maxima")

[Out]

1/6*x^6*log((e*x^3 + d)^p*c)^2 - 1/6*e*p*(2*d^2*log(e*x^3 + d)/e^3 + (e*x^6 - 2*d*x^3)/e^2)*log((e*x^3 + d)^p*
c) + 1/12*(e^2*x^6 - 6*d*e*x^3 + 2*d^2*log(e*x^3 + d)^2 + 6*d^2*log(e*x^3 + d))*p^2/e^2

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Fricas [A]  time = 2.04078, size = 317, normalized size = 2.11 \begin{align*} \frac{e^{2} p^{2} x^{6} + 2 \, e^{2} x^{6} \log \left (c\right )^{2} - 6 \, d e p^{2} x^{3} + 2 \,{\left (e^{2} p^{2} x^{6} - d^{2} p^{2}\right )} \log \left (e x^{3} + d\right )^{2} - 2 \,{\left (e^{2} p^{2} x^{6} - 2 \, d e p^{2} x^{3} - 3 \, d^{2} p^{2} - 2 \,{\left (e^{2} p x^{6} - d^{2} p\right )} \log \left (c\right )\right )} \log \left (e x^{3} + d\right ) - 2 \,{\left (e^{2} p x^{6} - 2 \, d e p x^{3}\right )} \log \left (c\right )}{12 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*log(c*(e*x^3+d)^p)^2,x, algorithm="fricas")

[Out]

1/12*(e^2*p^2*x^6 + 2*e^2*x^6*log(c)^2 - 6*d*e*p^2*x^3 + 2*(e^2*p^2*x^6 - d^2*p^2)*log(e*x^3 + d)^2 - 2*(e^2*p
^2*x^6 - 2*d*e*p^2*x^3 - 3*d^2*p^2 - 2*(e^2*p*x^6 - d^2*p)*log(c))*log(e*x^3 + d) - 2*(e^2*p*x^6 - 2*d*e*p*x^3
)*log(c))/e^2

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Sympy [A]  time = 54.5267, size = 206, normalized size = 1.37 \begin{align*} \begin{cases} - \frac{d^{2} p^{2} \log{\left (d + e x^{3} \right )}^{2}}{6 e^{2}} + \frac{d^{2} p^{2} \log{\left (d + e x^{3} \right )}}{2 e^{2}} - \frac{d^{2} p \log{\left (c \right )} \log{\left (d + e x^{3} \right )}}{3 e^{2}} + \frac{d p^{2} x^{3} \log{\left (d + e x^{3} \right )}}{3 e} - \frac{d p^{2} x^{3}}{2 e} + \frac{d p x^{3} \log{\left (c \right )}}{3 e} + \frac{p^{2} x^{6} \log{\left (d + e x^{3} \right )}^{2}}{6} - \frac{p^{2} x^{6} \log{\left (d + e x^{3} \right )}}{6} + \frac{p^{2} x^{6}}{12} + \frac{p x^{6} \log{\left (c \right )} \log{\left (d + e x^{3} \right )}}{3} - \frac{p x^{6} \log{\left (c \right )}}{6} + \frac{x^{6} \log{\left (c \right )}^{2}}{6} & \text{for}\: e \neq 0 \\\frac{x^{6} \log{\left (c d^{p} \right )}^{2}}{6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*ln(c*(e*x**3+d)**p)**2,x)

[Out]

Piecewise((-d**2*p**2*log(d + e*x**3)**2/(6*e**2) + d**2*p**2*log(d + e*x**3)/(2*e**2) - d**2*p*log(c)*log(d +
e*x**3)/(3*e**2) + d*p**2*x**3*log(d + e*x**3)/(3*e) - d*p**2*x**3/(2*e) + d*p*x**3*log(c)/(3*e) + p**2*x**6*
log(d + e*x**3)**2/6 - p**2*x**6*log(d + e*x**3)/6 + p**2*x**6/12 + p*x**6*log(c)*log(d + e*x**3)/3 - p*x**6*l
og(c)/6 + x**6*log(c)**2/6, Ne(e, 0)), (x**6*log(c*d**p)**2/6, True))

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Giac [A]  time = 1.22022, size = 298, normalized size = 1.99 \begin{align*} \frac{1}{12} \,{\left ({\left (2 \,{\left (x^{3} e + d\right )}^{2} \log \left (x^{3} e + d\right )^{2} - 4 \,{\left (x^{3} e + d\right )} d \log \left (x^{3} e + d\right )^{2} - 2 \,{\left (x^{3} e + d\right )}^{2} \log \left (x^{3} e + d\right ) + 8 \,{\left (x^{3} e + d\right )} d \log \left (x^{3} e + d\right ) +{\left (x^{3} e + d\right )}^{2} - 8 \,{\left (x^{3} e + d\right )} d\right )} p^{2} e^{\left (-1\right )} + 2 \,{\left (2 \,{\left (x^{3} e + d\right )}^{2} \log \left (x^{3} e + d\right ) - 4 \,{\left (x^{3} e + d\right )} d \log \left (x^{3} e + d\right ) -{\left (x^{3} e + d\right )}^{2} + 4 \,{\left (x^{3} e + d\right )} d\right )} p e^{\left (-1\right )} \log \left (c\right ) + 2 \,{\left ({\left (x^{3} e + d\right )}^{2} - 2 \,{\left (x^{3} e + d\right )} d\right )} e^{\left (-1\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^5*log(c*(e*x^3+d)^p)^2,x, algorithm="giac")

[Out]

1/12*((2*(x^3*e + d)^2*log(x^3*e + d)^2 - 4*(x^3*e + d)*d*log(x^3*e + d)^2 - 2*(x^3*e + d)^2*log(x^3*e + d) +
8*(x^3*e + d)*d*log(x^3*e + d) + (x^3*e + d)^2 - 8*(x^3*e + d)*d)*p^2*e^(-1) + 2*(2*(x^3*e + d)^2*log(x^3*e +
d) - 4*(x^3*e + d)*d*log(x^3*e + d) - (x^3*e + d)^2 + 4*(x^3*e + d)*d)*p*e^(-1)*log(c) + 2*((x^3*e + d)^2 - 2*
(x^3*e + d)*d)*e^(-1)*log(c)^2)*e^(-1)