3.2.30 \(\int x^2 \log ^2(c (d+e x^3)^p) \, dx\) [130]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 = {2504, 2436, 2333, 2332} \begin {gather*} \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} \end {gather*}

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

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Mathematica [A]
time = 0.01, size = 63, normalized size = 0.95 \begin {gather*} \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 (-p x^3+\frac {\left (d+e x^3\right ) \log \left (c \left (d+e x^3\right )^p\right )}{e}\right )\right ) \end {gather*}

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.76, size = 1036, normalized size = 15.70

method	result	size
risch	\(\text {Expression too large to display}\)	\(1036\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*ln(c)^2*x^3-2/3*d*p^2/e*ln(e*x^3+d)+1/3*x^3*ln((e*x
^3+d)^p)^2-1/3*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*c
sgn(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*csg
n(I*c*(e*x^3+d)^p)^2*csgn(I*c)+2/3*p^2*x^3+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)-1/12*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*csg
n(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^3*csgn(I*c)^2+2/3/e*ln(c)*ln(e*x^3+d)*d*p-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*ln(c)*Pi*x^3*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2+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)-1/3/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*(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)

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Maxima [A]
time = 0.28, size = 97, normalized size = 1.47 \begin {gather*} \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 {gather*}

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

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Fricas [A]
time = 0.35, size = 103, normalized size = 1.56 \begin {gather*} \frac {1}{3} \, {\left (2 \, p^{2} x^{3} e - 2 \, p x^{3} e \log \left (c\right ) + x^{3} e \log \left (c\right )^{2} + {\left (p^{2} x^{3} e + d p^{2}\right )} \log \left (x^{3} e + d\right )^{2} - 2 \, {\left (p^{2} x^{3} e + d p^{2} - {\left (p x^{3} e + d p\right )} \log \left (c\right )\right )} \log \left (x^{3} e + d\right )\right )} e^{\left (-1\right )} \end {gather*}

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

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Sympy [A]
time = 1.28, size = 100, normalized size = 1.52 \begin {gather*} \begin {cases} - \frac {2 d p \log {\left (c \left (d + e x^{3}\right )^{p} \right )}}{3 e} + \frac {d \log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}{3 e} + \frac {2 p^{2} x^{3}}{3} - \frac {2 p x^{3} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}}{3} + \frac {x^{3} \log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}{3} & \text {for}\: e \neq 0 \\\frac {x^{3} \log {\left (c d^{p} \right )}^{2}}{3} & \text {otherwise} \end {cases} \end {gather*}

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

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Giac [A]
time = 4.73, size = 104, normalized size = 1.58 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.23, size = 71, normalized size = 1.08 \begin {gather*} \frac {2\,p^2\,x^3}{3}+{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2\,\left (\frac {d}{3\,e}+\frac {x^3}{3}\right )-\frac {2\,p\,x^3\,\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}{3}-\frac {2\,d\,p^2\,\ln \left (e\,x^3+d\right )}{3\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*log(c*(d + e*x^3)^p)^2,x)

[Out]

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

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