3.2.29 \(\int x^5 \log ^2(c (d+e x^3)^p) \, dx\) [129]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 150, normalized size of antiderivative = 1.00, 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 = {2504, 2448, 2436, 2333, 2332, 2437, 2342, 2341} \begin {gather*} \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} \end {gather*}

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 2332

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

Rule 2333

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 2341

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
]

Rule 2342

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 2436

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 2437

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 2448

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 2504

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])

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 105, normalized size = 0.70 \begin {gather*} \frac {e p^2 x^3 \left (-6 d+e x^3\right )+2 d^2 p^2 \log \left (d+e x^3\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 \left (d^2-e^2 x^6\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{12 e^2} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 5.14, size = 24313, normalized size = 162.09

method	result	size
risch	\(\text {Expression too large to display}\)	\(24313\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [A]
time = 0.28, size = 120, normalized size = 0.80 \begin {gather*} \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 {gather*}

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 = 0.36, size = 146, normalized size = 0.97 \begin {gather*} \frac {1}{12} \, {\left (p^{2} x^{6} e^{2} + 2 \, x^{6} e^{2} \log \left (c\right )^{2} - 6 \, d p^{2} x^{3} e + 2 \, {\left (p^{2} x^{6} e^{2} - d^{2} p^{2}\right )} \log \left (x^{3} e + d\right )^{2} - 2 \, {\left (p^{2} x^{6} e^{2} - 2 \, d p^{2} x^{3} e - 3 \, d^{2} p^{2} - 2 \, {\left (p x^{6} e^{2} - d^{2} p\right )} \log \left (c\right )\right )} \log \left (x^{3} e + d\right ) - 2 \, {\left (p x^{6} e^{2} - 2 \, d p x^{3} e\right )} \log \left (c\right )\right )} e^{\left (-2\right )} \end {gather*}

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

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Sympy [A]
time = 4.71, size = 136, normalized size = 0.91 \begin {gather*} \begin {cases} \frac {d^{2} p \log {\left (c \left (d + e x^{3}\right )^{p} \right )}}{2 e^{2}} - \frac {d^{2} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}{6 e^{2}} - \frac {d p^{2} x^{3}}{2 e} + \frac {d p x^{3} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}}{3 e} + \frac {p^{2} x^{6}}{12} - \frac {p x^{6} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}}{6} + \frac {x^{6} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}{6} & \text {for}\: e \neq 0 \\\frac {x^{6} \log {\left (c d^{p} \right )}^{2}}{6} & \text {otherwise} \end {cases} \end {gather*}

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

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Giac [A]
time = 3.84, size = 232, normalized size = 1.55 \begin {gather*} \frac {1}{12} \, {\left (2 \, {\left (x^{3} e + d\right )}^{2} p^{2} \log \left (x^{3} e + d\right )^{2} - 2 \, {\left (x^{3} e + d\right )}^{2} p^{2} \log \left (x^{3} e + d\right ) + 4 \, {\left (x^{3} e + d\right )}^{2} p \log \left (x^{3} e + d\right ) \log \left (c\right ) + {\left (x^{3} e + d\right )}^{2} p^{2} - 2 \, {\left (x^{3} e + d\right )}^{2} p \log \left (c\right ) + 2 \, {\left (x^{3} e + d\right )}^{2} \log \left (c\right )^{2}\right )} e^{\left (-2\right )} - \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 )} d p^{2} - 2 \, {\left (x^{3} e - {\left (x^{3} e + d\right )} \log \left (x^{3} e + d\right ) + d\right )} d p \log \left (c\right ) + {\left (x^{3} e + d\right )} d \log \left (c\right )^{2}\right )} e^{\left (-2\right )} \end {gather*}

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*p^2*log(x^3*e + d)^2 - 2*(x^3*e + d)^2*p^2*log(x^3*e + d) + 4*(x^3*e + d)^2*p*log(x^3*e
+ d)*log(c) + (x^3*e + d)^2*p^2 - 2*(x^3*e + d)^2*p*log(c) + 2*(x^3*e + d)^2*log(c)^2)*e^(-2) - 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)*d*p^2 - 2*(x^3*e - (x^3*e + d)*log(x^3*e
+ d) + d)*d*p*log(c) + (x^3*e + d)*d*log(c)^2)*e^(-2)

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Mupad [B]
time = 0.28, size = 100, normalized size = 0.67 \begin {gather*} \frac {p^2\,x^6}{12}-\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )\,\left (\frac {p\,x^6}{6}-\frac {d\,p\,x^3}{3\,e}\right )+{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2\,\left (\frac {x^6}{6}-\frac {d^2}{6\,e^2}\right )-\frac {d\,p^2\,x^3}{2\,e}+\frac {d^2\,p^2\,\ln \left (e\,x^3+d\right )}{2\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*log(c*(d + e*x^3)^p)^2,x)

[Out]

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

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