\(\int x^3 \log ^3(c (a+b x^2)^p) \, dx\) [92]

Optimal result

Integrand size = 18, antiderivative size = 211 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {3 a p^3 x^2}{b}-\frac {3 p^3 \left (a+b x^2\right )^2}{16 b^2}-\frac {3 a p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {3 p^2 \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{8 b^2}+\frac {3 a p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {3 p \left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{8 b^2}-\frac {a \left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {\left (a+b x^2\right )^2 \log ^3\left (c \left (a+b x^2\right )^p\right )}{4 b^2} \]

[Out]

3*a*p^3*x^2/b-3/16*p^3*(b*x^2+a)^2/b^2-3*a*p^2*(b*x^2+a)*ln(c*(b*x^2+a)^p)/b^2+3/8*p^2*(b*x^2+a)^2*ln(c*(b*x^2
+a)^p)/b^2+3/2*a*p*(b*x^2+a)*ln(c*(b*x^2+a)^p)^2/b^2-3/8*p*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)^2/b^2-1/2*a*(b*x^2+a)
*ln(c*(b*x^2+a)^p)^3/b^2+1/4*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)^3/b^2

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2504, 2448, 2436, 2333, 2332, 2437, 2342, 2341} \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {3 p^2 \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{8 b^2}-\frac {3 a p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {\left (a+b x^2\right )^2 \log ^3\left (c \left (a+b x^2\right )^p\right )}{4 b^2}-\frac {a \left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {3 p \left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{8 b^2}+\frac {3 a p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {3 p^3 \left (a+b x^2\right )^2}{16 b^2}+\frac {3 a p^3 x^2}{b} \]

[In]

Int[x^3*Log[c*(a + b*x^2)^p]^3,x]

[Out]

(3*a*p^3*x^2)/b - (3*p^3*(a + b*x^2)^2)/(16*b^2) - (3*a*p^2*(a + b*x^2)*Log[c*(a + b*x^2)^p])/b^2 + (3*p^2*(a
+ b*x^2)^2*Log[c*(a + b*x^2)^p])/(8*b^2) + (3*a*p*(a + b*x^2)*Log[c*(a + b*x^2)^p]^2)/(2*b^2) - (3*p*(a + b*x^
2)^2*Log[c*(a + b*x^2)^p]^2)/(8*b^2) - (a*(a + b*x^2)*Log[c*(a + b*x^2)^p]^3)/(2*b^2) + ((a + b*x^2)^2*Log[c*(
a + b*x^2)^p]^3)/(4*b^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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x \log ^3\left (c (a+b x)^p\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a \log ^3\left (c (a+b x)^p\right )}{b}+\frac {(a+b x) \log ^3\left (c (a+b x)^p\right )}{b}\right ) \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int (a+b x) \log ^3\left (c (a+b x)^p\right ) \, dx,x,x^2\right )}{2 b}-\frac {a \text {Subst}\left (\int \log ^3\left (c (a+b x)^p\right ) \, dx,x,x^2\right )}{2 b} \\ & = \frac {\text {Subst}\left (\int x \log ^3\left (c x^p\right ) \, dx,x,a+b x^2\right )}{2 b^2}-\frac {a \text {Subst}\left (\int \log ^3\left (c x^p\right ) \, dx,x,a+b x^2\right )}{2 b^2} \\ & = -\frac {a \left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {\left (a+b x^2\right )^2 \log ^3\left (c \left (a+b x^2\right )^p\right )}{4 b^2}-\frac {(3 p) \text {Subst}\left (\int x \log ^2\left (c x^p\right ) \, dx,x,a+b x^2\right )}{4 b^2}+\frac {(3 a p) \text {Subst}\left (\int \log ^2\left (c x^p\right ) \, dx,x,a+b x^2\right )}{2 b^2} \\ & = \frac {3 a p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {3 p \left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{8 b^2}-\frac {a \left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {\left (a+b x^2\right )^2 \log ^3\left (c \left (a+b x^2\right )^p\right )}{4 b^2}+\frac {\left (3 p^2\right ) \text {Subst}\left (\int x \log \left (c x^p\right ) \, dx,x,a+b x^2\right )}{4 b^2}-\frac {\left (3 a p^2\right ) \text {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,a+b x^2\right )}{b^2} \\ & = \frac {3 a p^3 x^2}{b}-\frac {3 p^3 \left (a+b x^2\right )^2}{16 b^2}-\frac {3 a p^2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {3 p^2 \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{8 b^2}+\frac {3 a p \left (a+b x^2\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{2 b^2}-\frac {3 p \left (a+b x^2\right )^2 \log ^2\left (c \left (a+b x^2\right )^p\right )}{8 b^2}-\frac {a \left (a+b x^2\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{2 b^2}+\frac {\left (a+b x^2\right )^2 \log ^3\left (c \left (a+b x^2\right )^p\right )}{4 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.69 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {3 b p^3 x^2 \left (-14 a+b x^2\right )+6 a^2 p^3 \log \left (a+b x^2\right )+6 p^2 \left (6 a^2+6 a b x^2-b^2 x^4\right ) \log \left (c \left (a+b x^2\right )^p\right )-6 p \left (3 a^2+2 a b x^2-b^2 x^4\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )+4 \left (a^2-b^2 x^4\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )}{16 b^2} \]

[In]

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

[Out]

-1/16*(3*b*p^3*x^2*(-14*a + b*x^2) + 6*a^2*p^3*Log[a + b*x^2] + 6*p^2*(6*a^2 + 6*a*b*x^2 - b^2*x^4)*Log[c*(a +
b*x^2)^p] - 6*p*(3*a^2 + 2*a*b*x^2 - b^2*x^4)*Log[c*(a + b*x^2)^p]^2 + 4*(a^2 - b^2*x^4)*Log[c*(a + b*x^2)^p]
^3)/b^2

Maple [A] (verified)

Time = 11.03 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(-\frac {-4 x^{4} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{3} b^{2}+6 x^{4} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} b^{2} p -6 x^{4} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b^{2} p^{2}+3 b^{2} p^{3} x^{4}-12 x^{2} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a b p +36 x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a b \,p^{2}-42 x^{2} a b \,p^{3}+78 \ln \left (b \,x^{2}+a \right ) a^{2} p^{3}+4 {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{3} a^{2}-18 {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a^{2} p -36 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{2} p^{2}+42 a^{2} p^{3}}{16 b^{2}}\)	\(223\)
risch	\(\text {Expression too large to display}\)	\(241126\)

[In]

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

[Out]

-1/16*(-4*x^4*ln(c*(b*x^2+a)^p)^3*b^2+6*x^4*ln(c*(b*x^2+a)^p)^2*b^2*p-6*x^4*ln(c*(b*x^2+a)^p)*b^2*p^2+3*b^2*p^
3*x^4-12*x^2*ln(c*(b*x^2+a)^p)^2*a*b*p+36*x^2*ln(c*(b*x^2+a)^p)*a*b*p^2-42*x^2*a*b*p^3+78*ln(b*x^2+a)*a^2*p^3+
4*ln(c*(b*x^2+a)^p)^3*a^2-18*ln(c*(b*x^2+a)^p)^2*a^2*p-36*ln(c*(b*x^2+a)^p)*a^2*p^2+42*a^2*p^3)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.30 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=-\frac {3 \, b^{2} p^{3} x^{4} - 4 \, b^{2} x^{4} \log \left (c\right )^{3} - 42 \, a b p^{3} x^{2} - 4 \, {\left (b^{2} p^{3} x^{4} - a^{2} p^{3}\right )} \log \left (b x^{2} + a\right )^{3} + 6 \, {\left (b^{2} p^{3} x^{4} - 2 \, a b p^{3} x^{2} - 3 \, a^{2} p^{3} - 2 \, {\left (b^{2} p^{2} x^{4} - a^{2} p^{2}\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right )^{2} + 6 \, {\left (b^{2} p x^{4} - 2 \, a b p x^{2}\right )} \log \left (c\right )^{2} - 6 \, {\left (b^{2} p^{3} x^{4} - 6 \, a b p^{3} x^{2} - 7 \, a^{2} p^{3} + 2 \, {\left (b^{2} p x^{4} - a^{2} p\right )} \log \left (c\right )^{2} - 2 \, {\left (b^{2} p^{2} x^{4} - 2 \, a b p^{2} x^{2} - 3 \, a^{2} p^{2}\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (b^{2} p^{2} x^{4} - 6 \, a b p^{2} x^{2}\right )} \log \left (c\right )}{16 \, b^{2}} \]

[In]

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

[Out]

-1/16*(3*b^2*p^3*x^4 - 4*b^2*x^4*log(c)^3 - 42*a*b*p^3*x^2 - 4*(b^2*p^3*x^4 - a^2*p^3)*log(b*x^2 + a)^3 + 6*(b
^2*p^3*x^4 - 2*a*b*p^3*x^2 - 3*a^2*p^3 - 2*(b^2*p^2*x^4 - a^2*p^2)*log(c))*log(b*x^2 + a)^2 + 6*(b^2*p*x^4 - 2
*a*b*p*x^2)*log(c)^2 - 6*(b^2*p^3*x^4 - 6*a*b*p^3*x^2 - 7*a^2*p^3 + 2*(b^2*p*x^4 - a^2*p)*log(c)^2 - 2*(b^2*p^
2*x^4 - 2*a*b*p^2*x^2 - 3*a^2*p^2)*log(c))*log(b*x^2 + a) - 6*(b^2*p^2*x^4 - 6*a*b*p^2*x^2)*log(c))/b^2

Sympy [A] (verification not implemented)

Time = 2.08 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.06 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} - \frac {21 a^{2} p^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{8 b^{2}} + \frac {9 a^{2} p \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{8 b^{2}} - \frac {a^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{4 b^{2}} + \frac {21 a p^{3} x^{2}}{8 b} - \frac {9 a p^{2} x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{4 b} + \frac {3 a p x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{4 b} - \frac {3 p^{3} x^{4}}{16} + \frac {3 p^{2} x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{8} - \frac {3 p x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{8} + \frac {x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{4} & \text {for}\: b \neq 0 \\\frac {x^{4} \log {\left (a^{p} c \right )}^{3}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-21*a**2*p**2*log(c*(a + b*x**2)**p)/(8*b**2) + 9*a**2*p*log(c*(a + b*x**2)**p)**2/(8*b**2) - a**2*
log(c*(a + b*x**2)**p)**3/(4*b**2) + 21*a*p**3*x**2/(8*b) - 9*a*p**2*x**2*log(c*(a + b*x**2)**p)/(4*b) + 3*a*p
*x**2*log(c*(a + b*x**2)**p)**2/(4*b) - 3*p**3*x**4/16 + 3*p**2*x**4*log(c*(a + b*x**2)**p)/8 - 3*p*x**4*log(c
*(a + b*x**2)**p)**2/8 + x**4*log(c*(a + b*x**2)**p)**3/4, Ne(b, 0)), (x**4*log(a**p*c)**3/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.96 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{4} \, x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3} - \frac {3}{8} \, b p {\left (\frac {2 \, a^{2} \log \left (b x^{2} + a\right )}{b^{3}} + \frac {b x^{4} - 2 \, a x^{2}}{b^{2}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} - \frac {1}{16} \, b p {\left (\frac {{\left (3 \, b^{2} x^{4} + 4 \, a^{2} \log \left (b x^{2} + a\right )^{3} - 42 \, a b x^{2} + 18 \, a^{2} \log \left (b x^{2} + a\right )^{2} + 42 \, a^{2} \log \left (b x^{2} + a\right )\right )} p^{2}}{b^{3}} - \frac {6 \, {\left (b^{2} x^{4} - 6 \, a b x^{2} + 2 \, a^{2} \log \left (b x^{2} + a\right )^{2} + 6 \, a^{2} \log \left (b x^{2} + a\right )\right )} p \log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{b^{3}}\right )} \]

[In]

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

[Out]

1/4*x^4*log((b*x^2 + a)^p*c)^3 - 3/8*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)^2 - 1/16*b*p*((3*b^2*x^4 + 4*a^2*log(b*x^2 + a)^3 - 42*a*b*x^2 + 18*a^2*log(b*x^2 + a)^2 + 42*a^2*log(b*x^2
+ a))*p^2/b^3 - 6*(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*log((b*x^2 + a)^p*c
)/b^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.82 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {4 \, {\left (b x^{2} + a\right )}^{2} p^{3} \log \left (b x^{2} + a\right )^{3} - 6 \, {\left (b x^{2} + a\right )}^{2} p^{3} \log \left (b x^{2} + a\right )^{2} + 12 \, {\left (b x^{2} + a\right )}^{2} p^{2} \log \left (b x^{2} + a\right )^{2} \log \left (c\right ) + 6 \, {\left (b x^{2} + a\right )}^{2} p^{3} \log \left (b x^{2} + a\right ) - 12 \, {\left (b x^{2} + a\right )}^{2} p^{2} \log \left (b x^{2} + a\right ) \log \left (c\right ) + 12 \, {\left (b x^{2} + a\right )}^{2} p \log \left (b x^{2} + a\right ) \log \left (c\right )^{2} - 3 \, {\left (b x^{2} + a\right )}^{2} p^{3} + 6 \, {\left (b x^{2} + a\right )}^{2} p^{2} \log \left (c\right ) - 6 \, {\left (b x^{2} + a\right )}^{2} p \log \left (c\right )^{2} + 4 \, {\left (b x^{2} + a\right )}^{2} \log \left (c\right )^{3}}{16 \, b^{2}} - \frac {{\left ({\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{3} - 6 \, b x^{2} - 3 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} + 6 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) - 6 \, a\right )} a p^{3} + 3 \, {\left (2 \, b x^{2} + {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + 2 \, a\right )} a p^{2} \log \left (c\right ) - 3 \, {\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} a p \log \left (c\right )^{2} + {\left (b x^{2} + a\right )} a \log \left (c\right )^{3}}{2 \, b^{2}} \]

[In]

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

[Out]

1/16*(4*(b*x^2 + a)^2*p^3*log(b*x^2 + a)^3 - 6*(b*x^2 + a)^2*p^3*log(b*x^2 + a)^2 + 12*(b*x^2 + a)^2*p^2*log(b
*x^2 + a)^2*log(c) + 6*(b*x^2 + a)^2*p^3*log(b*x^2 + a) - 12*(b*x^2 + a)^2*p^2*log(b*x^2 + a)*log(c) + 12*(b*x
^2 + a)^2*p*log(b*x^2 + a)*log(c)^2 - 3*(b*x^2 + a)^2*p^3 + 6*(b*x^2 + a)^2*p^2*log(c) - 6*(b*x^2 + a)^2*p*log
(c)^2 + 4*(b*x^2 + a)^2*log(c)^3)/b^2 - 1/2*(((b*x^2 + a)*log(b*x^2 + a)^3 - 6*b*x^2 - 3*(b*x^2 + a)*log(b*x^2
+ a)^2 + 6*(b*x^2 + a)*log(b*x^2 + a) - 6*a)*a*p^3 + 3*(2*b*x^2 + (b*x^2 + a)*log(b*x^2 + a)^2 - 2*(b*x^2 + a
)*log(b*x^2 + a) + 2*a)*a*p^2*log(c) - 3*(b*x^2 - (b*x^2 + a)*log(b*x^2 + a) + a)*a*p*log(c)^2 + (b*x^2 + a)*a
*log(c)^3)/b^2

Mupad [B] (verification not implemented)

Time = 1.45 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.68 \[ \int x^3 \log ^3\left (c \left (a+b x^2\right )^p\right ) \, dx={\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {9\,a^2\,p}{8\,b^2}-\frac {3\,p\,x^4}{8}+\frac {3\,a\,p\,x^2}{4\,b}\right )-\frac {3\,p^3\,x^4}{16}+\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )\,\left (\frac {3\,p^2\,x^4}{8}-\frac {9\,a\,p^2\,x^2}{4\,b}\right )+{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^3\,\left (\frac {x^4}{4}-\frac {a^2}{4\,b^2}\right )+\frac {21\,a\,p^3\,x^2}{8\,b}-\frac {21\,a^2\,p^3\,\ln \left (b\,x^2+a\right )}{8\,b^2} \]

[In]

int(x^3*log(c*(a + b*x^2)^p)^3,x)

[Out]

log(c*(a + b*x^2)^p)^2*((9*a^2*p)/(8*b^2) - (3*p*x^4)/8 + (3*a*p*x^2)/(4*b)) - (3*p^3*x^4)/16 + log(c*(a + b*x
^2)^p)*((3*p^2*x^4)/8 - (9*a*p^2*x^2)/(4*b)) + log(c*(a + b*x^2)^p)^3*(x^4/4 - a^2/(4*b^2)) + (21*a*p^3*x^2)/(
8*b) - (21*a^2*p^3*log(a + b*x^2))/(8*b^2)