3.78 \(\int x^3 \log ^2(c (a+b x^2)^p) \, dx\)

Optimal. Leaf size=145 \[ \frac {\left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{4 b^2}-\frac {a \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{4 b^2}+\frac {a p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {p^2 \left (a+b x^2\right )^2}{8 b^2}-\frac {a p^2 x^2}{b} \]

[Out]

-a*p^2*x^2/b+1/8*p^2*(b*x^2+a)^2/b^2+a*p*(b*x^2+a)*ln(c*(b*x^2+a)^p)/b^2-1/4*p*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)/b
^2-1/2*a*(b*x^2+a)*ln(c*(b*x^2+a)^p)^2/b^2+1/4*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)^2/b^2

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Rubi [A]  time = 0.15, antiderivative size = 145, 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 = {2454, 2401, 2389, 2296, 2295, 2390, 2305, 2304} \[ \frac {\left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{4 b^2}-\frac {a \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{4 b^2}+\frac {a p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {p^2 \left (a+b x^2\right )^2}{8 b^2}-\frac {a p^2 x^2}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Log[c*(a + b*x^2)^p]^2,x]

[Out]

-((a*p^2*x^2)/b) + (p^2*(a + b*x^2)^2)/(8*b^2) + (a*p*(a + b*x^2)*Log[c*(a + b*x^2)^p])/b^2 - (p*(a + b*x^2)^2
*Log[c*(a + b*x^2)^p])/(4*b^2) - (a*(a + b*x^2)*Log[c*(a + b*x^2)^p]^2)/(2*b^2) + ((a + b*x^2)^2*Log[c*(a + b*
x^2)^p]^2)/(4*b^2)

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 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 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
]

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 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 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 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 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])

Rubi steps

\begin {align*} \int x^3 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x \log ^2\left (c (a+b x)^p\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {a \log ^2\left (c (a+b x)^p\right )}{b}+\frac {(a+b x) \log ^2\left (c (a+b x)^p\right )}{b}\right ) \, dx,x,x^2\right )\\ &=\frac {\operatorname {Subst}\left (\int (a+b x) \log ^2\left (c (a+b x)^p\right ) \, dx,x,x^2\right )}{2 b}-\frac {a \operatorname {Subst}\left (\int \log ^2\left (c (a+b x)^p\right ) \, dx,x,x^2\right )}{2 b}\\ &=\frac {\operatorname {Subst}\left (\int x \log ^2\left (c x^p\right ) \, dx,x,a+b x^2\right )}{2 b^2}-\frac {a \operatorname {Subst}\left (\int \log ^2\left (c x^p\right ) \, dx,x,a+b x^2\right )}{2 b^2}\\ &=-\frac {a \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {\left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{4 b^2}-\frac {p \operatorname {Subst}\left (\int x \log \left (c x^p\right ) \, dx,x,a+b x^2\right )}{2 b^2}+\frac {(a p) \operatorname {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,a+b x^2\right )}{b^2}\\ &=-\frac {a p^2 x^2}{b}+\frac {p^2 \left (a+b x^2\right )^2}{8 b^2}+\frac {a p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}-\frac {p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{4 b^2}-\frac {a \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {\left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{4 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 105, normalized size = 0.72 \[ \frac {-2 \left (a^2-b^2 x^4\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )+2 p \left (2 a^2+2 a b x^2-b^2 x^4\right ) \log \left (c \left (a+b x^2\right )^p\right )+2 a^2 p^2 \log \left (a+b x^2\right )+b p^2 x^2 \left (b x^2-6 a\right )}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Log[c*(a + b*x^2)^p]^2,x]

[Out]

(b*p^2*x^2*(-6*a + b*x^2) + 2*a^2*p^2*Log[a + b*x^2] + 2*p*(2*a^2 + 2*a*b*x^2 - b^2*x^4)*Log[c*(a + b*x^2)^p]
- 2*(a^2 - b^2*x^4)*Log[c*(a + b*x^2)^p]^2)/(8*b^2)

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fricas [A]  time = 0.48, size = 148, normalized size = 1.02 \[ \frac {b^{2} p^{2} x^{4} + 2 \, b^{2} x^{4} \log \relax (c)^{2} - 6 \, a b p^{2} x^{2} + 2 \, {\left (b^{2} p^{2} x^{4} - a^{2} p^{2}\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b^{2} p^{2} x^{4} - 2 \, a b p^{2} x^{2} - 3 \, a^{2} p^{2} - 2 \, {\left (b^{2} p x^{4} - a^{2} p\right )} \log \relax (c)\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b^{2} p x^{4} - 2 \, a b p x^{2}\right )} \log \relax (c)}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^2*p^2*x^4 + 2*b^2*x^4*log(c)^2 - 6*a*b*p^2*x^2 + 2*(b^2*p^2*x^4 - a^2*p^2)*log(b*x^2 + a)^2 - 2*(b^2*p^
2*x^4 - 2*a*b*p^2*x^2 - 3*a^2*p^2 - 2*(b^2*p*x^4 - a^2*p)*log(c))*log(b*x^2 + a) - 2*(b^2*p*x^4 - 2*a*b*p*x^2)
*log(c))/b^2

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giac [A]  time = 0.19, size = 207, normalized size = 1.43 \[ \frac {\frac {{\left (2 \, {\left (b x^{2} + a\right )}^{2} \log \left (b x^{2} + a\right )^{2} - 4 \, {\left (b x^{2} + a\right )} a \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )}^{2} \log \left (b x^{2} + a\right ) + 8 \, {\left (b x^{2} + a\right )} a \log \left (b x^{2} + a\right ) + {\left (b x^{2} + a\right )}^{2} - 8 \, {\left (b x^{2} + a\right )} a\right )} p^{2}}{b} + \frac {2 \, {\left (2 \, {\left (b x^{2} + a\right )}^{2} \log \left (b x^{2} + a\right ) - 4 \, {\left (b x^{2} + a\right )} a \log \left (b x^{2} + a\right ) - {\left (b x^{2} + a\right )}^{2} + 4 \, {\left (b x^{2} + a\right )} a\right )} p \log \relax (c)}{b} + \frac {2 \, {\left ({\left (b x^{2} + a\right )}^{2} - 2 \, {\left (b x^{2} + a\right )} a\right )} \log \relax (c)^{2}}{b}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((2*(b*x^2 + a)^2*log(b*x^2 + a)^2 - 4*(b*x^2 + a)*a*log(b*x^2 + a)^2 - 2*(b*x^2 + a)^2*log(b*x^2 + a) + 8
*(b*x^2 + a)*a*log(b*x^2 + a) + (b*x^2 + a)^2 - 8*(b*x^2 + a)*a)*p^2/b + 2*(2*(b*x^2 + a)^2*log(b*x^2 + a) - 4
*(b*x^2 + a)*a*log(b*x^2 + a) - (b*x^2 + a)^2 + 4*(b*x^2 + a)*a)*p*log(c)/b + 2*((b*x^2 + a)^2 - 2*(b*x^2 + a)
*a)*log(c)^2/b)/b

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maple [C]  time = 0.47, size = 1242, normalized size = 8.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*ln(c*(b*x^2+a)^p)^2,x)

[Out]

-1/4*ln(c)*p*x^4-1/16*Pi^2*x^4*csgn(I*c*(b*x^2+a)^p)^6+1/4*(I*Pi*b^2*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a
)^p)^2-I*Pi*b^2*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*b^2*x^4*csgn(I*c*(b*x^2+a)^p)^3+I
*Pi*b^2*x^4*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*ln(c)*b^2*x^4-b^2*p*x^4+2*a*b*p*x^2-2*a^2*p*ln(b*x^2+a))/b^2*l
n((b*x^2+a)^p)+1/8*x^4*p^2+3/4*a^2*p^2/b^2*ln(b*x^2+a)-3/4*a*p^2*x^2/b+1/4/b^2*a^2*p^2*ln(b*x^2+a)^2-1/16*Pi^2
*x^4*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^4+1/8*Pi^2*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^5+1/
8*Pi^2*x^4*csgn(I*c*(b*x^2+a)^p)^5*csgn(I*c)-1/16*Pi^2*x^4*csgn(I*c*(b*x^2+a)^p)^4*csgn(I*c)^2+1/4*x^4*ln((b*x
^2+a)^p)^2+1/4*I/b^2*Pi*ln(b*x^2+a)*a^2*p*csgn(I*c*(b*x^2+a)^p)^3-1/4*I*ln(c)*Pi*x^4*csgn(I*(b*x^2+a)^p)*csgn(
I*c*(b*x^2+a)^p)*csgn(I*c)+1/8*I*Pi*p*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-1/4*I/b*Pi*a*p*x
^2*csgn(I*c*(b*x^2+a)^p)^3+1/4*ln(c)^2*x^4+1/8*Pi^2*x^4*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c
)-1/16*Pi^2*x^4*csgn(I*(b*x^2+a)^p)^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)^2-1/4*Pi^2*x^4*csgn(I*(b*x^2+a)^p)*csg
n(I*c*(b*x^2+a)^p)^4*csgn(I*c)+1/8*Pi^2*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^3*csgn(I*c)^2+1/2/b*ln(c
)*a*p*x^2-1/2/b^2*ln(c)*ln(b*x^2+a)*a^2*p-1/4*I*ln(c)*Pi*x^4*csgn(I*c*(b*x^2+a)^p)^3+1/8*I*Pi*p*x^4*csgn(I*c*(
b*x^2+a)^p)^3+1/4*I*ln(c)*Pi*x^4*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/8*I*Pi*p*x^4*csgn(I*(b*x^2+a)^p
)*csgn(I*c*(b*x^2+a)^p)^2+1/4*I*ln(c)*Pi*x^4*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-1/8*I*Pi*p*x^4*csgn(I*c*(b*x^2+
a)^p)^2*csgn(I*c)+1/4*I/b*Pi*a*p*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+1/4*I/b*Pi*a*p*x^2*csgn(I*c*(
b*x^2+a)^p)^2*csgn(I*c)-1/4*I/b^2*Pi*ln(b*x^2+a)*a^2*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/4*I/b^2*P
i*ln(b*x^2+a)*a^2*p*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)-1/4*I/b*Pi*a*p*x^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a
)^p)*csgn(I*c)+1/4*I/b^2*Pi*ln(b*x^2+a)*a^2*p*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)

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maxima [A]  time = 0.77, size = 120, normalized size = 0.83 \[ \frac {1}{4} \, x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} - \frac {1}{4} \, b p {\left (\frac {2 \, a^{2} \log \left (b x^{2} + a\right )}{b^{3}} + \frac {b x^{4} - 2 \, a x^{2}}{b^{2}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) + \frac {{\left (b^{2} x^{4} - 6 \, a b x^{2} + 2 \, a^{2} \log \left (b x^{2} + a\right )^{2} + 6 \, a^{2} \log \left (b x^{2} + a\right )\right )} p^{2}}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*log((b*x^2 + a)^p*c)^2 - 1/4*b*p*(2*a^2*log(b*x^2 + a)/b^3 + (b*x^4 - 2*a*x^2)/b^2)*log((b*x^2 + a)^p*
c) + 1/8*(b^2*x^4 - 6*a*b*x^2 + 2*a^2*log(b*x^2 + a)^2 + 6*a^2*log(b*x^2 + a))*p^2/b^2

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mupad [B]  time = 0.26, size = 100, normalized size = 0.69 \[ \frac {p^2\,x^4}{8}-\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )\,\left (\frac {p\,x^4}{4}-\frac {a\,p\,x^2}{2\,b}\right )+{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {x^4}{4}-\frac {a^2}{4\,b^2}\right )-\frac {3\,a\,p^2\,x^2}{4\,b}+\frac {3\,a^2\,p^2\,\ln \left (b\,x^2+a\right )}{4\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*log(c*(a + b*x^2)^p)^2,x)

[Out]

(p^2*x^4)/8 - log(c*(a + b*x^2)^p)*((p*x^4)/4 - (a*p*x^2)/(2*b)) + log(c*(a + b*x^2)^p)^2*(x^4/4 - a^2/(4*b^2)
) - (3*a*p^2*x^2)/(4*b) + (3*a^2*p^2*log(a + b*x^2))/(4*b^2)

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sympy [A]  time = 8.64, size = 209, normalized size = 1.44 \[ \begin {cases} - \frac {a^{2} p^{2} \log {\left (a + b x^{2} \right )}^{2}}{4 b^{2}} + \frac {3 a^{2} p^{2} \log {\left (a + b x^{2} \right )}}{4 b^{2}} - \frac {a^{2} p \log {\relax (c )} \log {\left (a + b x^{2} \right )}}{2 b^{2}} + \frac {a p^{2} x^{2} \log {\left (a + b x^{2} \right )}}{2 b} - \frac {3 a p^{2} x^{2}}{4 b} + \frac {a p x^{2} \log {\relax (c )}}{2 b} + \frac {p^{2} x^{4} \log {\left (a + b x^{2} \right )}^{2}}{4} - \frac {p^{2} x^{4} \log {\left (a + b x^{2} \right )}}{4} + \frac {p^{2} x^{4}}{8} + \frac {p x^{4} \log {\relax (c )} \log {\left (a + b x^{2} \right )}}{2} - \frac {p x^{4} \log {\relax (c )}}{4} + \frac {x^{4} \log {\relax (c )}^{2}}{4} & \text {for}\: b \neq 0 \\\frac {x^{4} \log {\left (a^{p} c \right )}^{2}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*ln(c*(b*x**2+a)**p)**2,x)

[Out]

Piecewise((-a**2*p**2*log(a + b*x**2)**2/(4*b**2) + 3*a**2*p**2*log(a + b*x**2)/(4*b**2) - a**2*p*log(c)*log(a
+ b*x**2)/(2*b**2) + a*p**2*x**2*log(a + b*x**2)/(2*b) - 3*a*p**2*x**2/(4*b) + a*p*x**2*log(c)/(2*b) + p**2*x
**4*log(a + b*x**2)**2/4 - p**2*x**4*log(a + b*x**2)/4 + p**2*x**4/8 + p*x**4*log(c)*log(a + b*x**2)/2 - p*x**
4*log(c)/4 + x**4*log(c)**2/4, Ne(b, 0)), (x**4*log(a**p*c)**2/4, True))

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