\(\int x^5 \log ^2(c (a+b x^2)^p) \, dx\) [77]

Optimal result

Integrand size = 18, antiderivative size = 215 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {a^2 p^2 x^2}{b^2}-\frac {a p^2 \left (a+b x^2\right )^2}{4 b^3}+\frac {p^2 \left (a+b x^2\right )^3}{27 b^3}-\frac {a^3 p^2 \log ^2\left (a+b x^2\right )}{6 b^3}-\frac {a^2 p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^3}+\frac {a p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^3}-\frac {p \left (a+b x^2\right )^3 \log \left (c \left (a+b x^2\right )^p\right )}{9 b^3}+\frac {a^3 p \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^3}+\frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right ) \]

[Out]

a^2*p^2*x^2/b^2-1/4*a*p^2*(b*x^2+a)^2/b^3+1/27*p^2*(b*x^2+a)^3/b^3-1/6*a^3*p^2*ln(b*x^2+a)^2/b^3-a^2*p*(b*x^2+
a)*ln(c*(b*x^2+a)^p)/b^3+1/2*a*p*(b*x^2+a)^2*ln(c*(b*x^2+a)^p)/b^3-1/9*p*(b*x^2+a)^3*ln(c*(b*x^2+a)^p)/b^3+1/3
*a^3*p*ln(b*x^2+a)*ln(c*(b*x^2+a)^p)/b^3+1/6*x^6*ln(c*(b*x^2+a)^p)^2

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2504, 2445, 2458, 45, 2372, 12, 14, 2338} \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {a^3 p \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^3}-\frac {a^3 p^2 \log ^2\left (a+b x^2\right )}{6 b^3}-\frac {a^2 p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^3}+\frac {a^2 p^2 x^2}{b^2}-\frac {p \left (a+b x^2\right )^3 \log \left (c \left (a+b x^2\right )^p\right )}{9 b^3}+\frac {a p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^3}+\frac {p^2 \left (a+b x^2\right )^3}{27 b^3}-\frac {a p^2 \left (a+b x^2\right )^2}{4 b^3}+\frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right ) \]

[In]

Int[x^5*Log[c*(a + b*x^2)^p]^2,x]

[Out]

(a^2*p^2*x^2)/b^2 - (a*p^2*(a + b*x^2)^2)/(4*b^3) + (p^2*(a + b*x^2)^3)/(27*b^3) - (a^3*p^2*Log[a + b*x^2]^2)/
(6*b^3) - (a^2*p*(a + b*x^2)*Log[c*(a + b*x^2)^p])/b^3 + (a*p*(a + b*x^2)^2*Log[c*(a + b*x^2)^p])/(2*b^3) - (p
*(a + b*x^2)^3*Log[c*(a + b*x^2)^p])/(9*b^3) + (a^3*p*Log[a + b*x^2]*Log[c*(a + b*x^2)^p])/(3*b^3) + (x^6*Log[
c*(a + b*x^2)^p]^2)/6

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2372

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]]
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2445

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f
+ g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])^p/(g*(q + 1))), x] - Dist[b*e*n*(p/(g*(q + 1))), Int[(f + g*x)^(q
+ 1)*((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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^2 \log ^2\left (c (a+b x)^p\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} (b p) \text {Subst}\left (\int \frac {x^3 \log \left (c (a+b x)^p\right )}{a+b x} \, dx,x,x^2\right ) \\ & = \frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} p \text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^3 \log \left (c x^p\right )}{x} \, dx,x,a+b x^2\right ) \\ & = -\frac {a^2 p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^3}+\frac {a p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^3}-\frac {p \left (a+b x^2\right )^3 \log \left (c \left (a+b x^2\right )^p\right )}{9 b^3}+\frac {a^3 p \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^3}+\frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {1}{3} p^2 \text {Subst}\left (\int \frac {18 a^2 x-9 a x^2+2 x^3-6 a^3 \log (x)}{6 b^3 x} \, dx,x,a+b x^2\right ) \\ & = -\frac {a^2 p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^3}+\frac {a p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^3}-\frac {p \left (a+b x^2\right )^3 \log \left (c \left (a+b x^2\right )^p\right )}{9 b^3}+\frac {a^3 p \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^3}+\frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {p^2 \text {Subst}\left (\int \frac {18 a^2 x-9 a x^2+2 x^3-6 a^3 \log (x)}{x} \, dx,x,a+b x^2\right )}{18 b^3} \\ & = -\frac {a^2 p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^3}+\frac {a p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^3}-\frac {p \left (a+b x^2\right )^3 \log \left (c \left (a+b x^2\right )^p\right )}{9 b^3}+\frac {a^3 p \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^3}+\frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {p^2 \text {Subst}\left (\int \left (18 a^2-9 a x+2 x^2-\frac {6 a^3 \log (x)}{x}\right ) \, dx,x,a+b x^2\right )}{18 b^3} \\ & = \frac {a^2 p^2 x^2}{b^2}-\frac {a p^2 \left (a+b x^2\right )^2}{4 b^3}+\frac {p^2 \left (a+b x^2\right )^3}{27 b^3}-\frac {a^2 p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^3}+\frac {a p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^3}-\frac {p \left (a+b x^2\right )^3 \log \left (c \left (a+b x^2\right )^p\right )}{9 b^3}+\frac {a^3 p \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^3}+\frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {\left (a^3 p^2\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x^2\right )}{3 b^3} \\ & = \frac {a^2 p^2 x^2}{b^2}-\frac {a p^2 \left (a+b x^2\right )^2}{4 b^3}+\frac {p^2 \left (a+b x^2\right )^3}{27 b^3}-\frac {a^3 p^2 \log ^2\left (a+b x^2\right )}{6 b^3}-\frac {a^2 p \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{b^3}+\frac {a p \left (a+b x^2\right )^2 \log \left (c \left (a+b x^2\right )^p\right )}{2 b^3}-\frac {p \left (a+b x^2\right )^3 \log \left (c \left (a+b x^2\right )^p\right )}{9 b^3}+\frac {a^3 p \log \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^3}+\frac {1}{6} x^6 \log ^2\left (c \left (a+b x^2\right )^p\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.59 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {b p^2 x^2 \left (66 a^2-15 a b x^2+4 b^2 x^4\right )-30 a^3 p^2 \log \left (a+b x^2\right )-6 p \left (6 a^3+6 a^2 b x^2-3 a b^2 x^4+2 b^3 x^6\right ) \log \left (c \left (a+b x^2\right )^p\right )+18 \left (a^3+b^3 x^6\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )}{108 b^3} \]

[In]

Integrate[x^5*Log[c*(a + b*x^2)^p]^2,x]

[Out]

(b*p^2*x^2*(66*a^2 - 15*a*b*x^2 + 4*b^2*x^4) - 30*a^3*p^2*Log[a + b*x^2] - 6*p*(6*a^3 + 6*a^2*b*x^2 - 3*a*b^2*
x^4 + 2*b^3*x^6)*Log[c*(a + b*x^2)^p] + 18*(a^3 + b^3*x^6)*Log[c*(a + b*x^2)^p]^2)/(108*b^3)

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(-\frac {-18 x^{6} {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} b^{3}+12 x^{6} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) b^{3} p -4 x^{6} b^{3} p^{2}-18 x^{4} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a \,b^{2} p +15 x^{4} a \,b^{2} p^{2}+36 x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{2} b p -66 x^{2} a^{2} b \,p^{2}+102 \ln \left (b \,x^{2}+a \right ) a^{3} p^{2}-18 {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}^{2} a^{3}-36 \ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) a^{3} p +66 a^{3} p^{2}}{108 b^{3}}\)	\(190\)
risch	\(\text {Expression too large to display}\)	\(1436\)

[In]

int(x^5*ln(c*(b*x^2+a)^p)^2,x,method=_RETURNVERBOSE)

[Out]

-1/108*(-18*x^6*ln(c*(b*x^2+a)^p)^2*b^3+12*x^6*ln(c*(b*x^2+a)^p)*b^3*p-4*x^6*b^3*p^2-18*x^4*ln(c*(b*x^2+a)^p)*
a*b^2*p+15*x^4*a*b^2*p^2+36*x^2*ln(c*(b*x^2+a)^p)*a^2*b*p-66*x^2*a^2*b*p^2+102*ln(b*x^2+a)*a^3*p^2-18*ln(c*(b*
x^2+a)^p)^2*a^3-36*ln(c*(b*x^2+a)^p)*a^3*p+66*a^3*p^2)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.88 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {4 \, b^{3} p^{2} x^{6} + 18 \, b^{3} x^{6} \log \left (c\right )^{2} - 15 \, a b^{2} p^{2} x^{4} + 66 \, a^{2} b p^{2} x^{2} + 18 \, {\left (b^{3} p^{2} x^{6} + a^{3} p^{2}\right )} \log \left (b x^{2} + a\right )^{2} - 6 \, {\left (2 \, b^{3} p^{2} x^{6} - 3 \, a b^{2} p^{2} x^{4} + 6 \, a^{2} b p^{2} x^{2} + 11 \, a^{3} p^{2} - 6 \, {\left (b^{3} p x^{6} + a^{3} p\right )} \log \left (c\right )\right )} \log \left (b x^{2} + a\right ) - 6 \, {\left (2 \, b^{3} p x^{6} - 3 \, a b^{2} p x^{4} + 6 \, a^{2} b p x^{2}\right )} \log \left (c\right )}{108 \, b^{3}} \]

[In]

integrate(x^5*log(c*(b*x^2+a)^p)^2,x, algorithm="fricas")

[Out]

1/108*(4*b^3*p^2*x^6 + 18*b^3*x^6*log(c)^2 - 15*a*b^2*p^2*x^4 + 66*a^2*b*p^2*x^2 + 18*(b^3*p^2*x^6 + a^3*p^2)*
log(b*x^2 + a)^2 - 6*(2*b^3*p^2*x^6 - 3*a*b^2*p^2*x^4 + 6*a^2*b*p^2*x^2 + 11*a^3*p^2 - 6*(b^3*p*x^6 + a^3*p)*l
og(c))*log(b*x^2 + a) - 6*(2*b^3*p*x^6 - 3*a*b^2*p*x^4 + 6*a^2*b*p*x^2)*log(c))/b^3

Sympy [A] (verification not implemented)

Time = 3.22 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.85 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\begin {cases} - \frac {11 a^{3} p \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{18 b^{3}} + \frac {a^{3} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{6 b^{3}} + \frac {11 a^{2} p^{2} x^{2}}{18 b^{2}} - \frac {a^{2} p x^{2} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{3 b^{2}} - \frac {5 a p^{2} x^{4}}{36 b} + \frac {a p x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{6 b} + \frac {p^{2} x^{6}}{27} - \frac {p x^{6} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{9} + \frac {x^{6} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}{6} & \text {for}\: b \neq 0 \\\frac {x^{6} \log {\left (a^{p} c \right )}^{2}}{6} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5*ln(c*(b*x**2+a)**p)**2,x)

[Out]

Piecewise((-11*a**3*p*log(c*(a + b*x**2)**p)/(18*b**3) + a**3*log(c*(a + b*x**2)**p)**2/(6*b**3) + 11*a**2*p**
2*x**2/(18*b**2) - a**2*p*x**2*log(c*(a + b*x**2)**p)/(3*b**2) - 5*a*p**2*x**4/(36*b) + a*p*x**4*log(c*(a + b*
x**2)**p)/(6*b) + p**2*x**6/27 - p*x**6*log(c*(a + b*x**2)**p)/9 + x**6*log(c*(a + b*x**2)**p)**2/6, Ne(b, 0))
, (x**6*log(a**p*c)**2/6, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.67 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {1}{6} \, x^{6} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2} + \frac {1}{18} \, b p {\left (\frac {6 \, a^{3} \log \left (b x^{2} + a\right )}{b^{4}} - \frac {2 \, b^{2} x^{6} - 3 \, a b x^{4} + 6 \, a^{2} x^{2}}{b^{3}}\right )} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) + \frac {{\left (4 \, b^{3} x^{6} - 15 \, a b^{2} x^{4} + 66 \, a^{2} b x^{2} - 18 \, a^{3} \log \left (b x^{2} + a\right )^{2} - 66 \, a^{3} \log \left (b x^{2} + a\right )\right )} p^{2}}{108 \, b^{3}} \]

[In]

integrate(x^5*log(c*(b*x^2+a)^p)^2,x, algorithm="maxima")

[Out]

1/6*x^6*log((b*x^2 + a)^p*c)^2 + 1/18*b*p*(6*a^3*log(b*x^2 + a)/b^4 - (2*b^2*x^6 - 3*a*b*x^4 + 6*a^2*x^2)/b^3)
*log((b*x^2 + a)^p*c) + 1/108*(4*b^3*x^6 - 15*a*b^2*x^4 + 66*a^2*b*x^2 - 18*a^3*log(b*x^2 + a)^2 - 66*a^3*log(
b*x^2 + a))*p^2/b^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.72 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{3} p^{2} \log \left (b x^{2} + a\right )^{2}}{6 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{2} a p^{2} \log \left (b x^{2} + a\right )^{2}}{2 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{3} p^{2} \log \left (b x^{2} + a\right )}{9 \, b^{3}} + \frac {{\left (b x^{2} + a\right )}^{2} a p^{2} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (b x^{2} + a\right )}^{3} p \log \left (b x^{2} + a\right ) \log \left (c\right )}{3 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{2} a p \log \left (b x^{2} + a\right ) \log \left (c\right )}{b^{3}} + \frac {{\left (b x^{2} + a\right )}^{3} p^{2}}{27 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{2} a p^{2}}{4 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{3} p \log \left (c\right )}{9 \, b^{3}} + \frac {{\left (b x^{2} + a\right )}^{2} a p \log \left (c\right )}{2 \, b^{3}} + \frac {{\left (b x^{2} + a\right )}^{3} \log \left (c\right )^{2}}{6 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{2} a \log \left (c\right )^{2}}{2 \, b^{3}} + \frac {{\left (2 \, b x^{2} + {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right )^{2} - 2 \, {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + 2 \, a\right )} a^{2} p^{2} - 2 \, {\left (b x^{2} - {\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a\right )} a^{2} p \log \left (c\right ) + {\left (b x^{2} + a\right )} a^{2} \log \left (c\right )^{2}}{2 \, b^{3}} \]

[In]

integrate(x^5*log(c*(b*x^2+a)^p)^2,x, algorithm="giac")

[Out]

1/6*(b*x^2 + a)^3*p^2*log(b*x^2 + a)^2/b^3 - 1/2*(b*x^2 + a)^2*a*p^2*log(b*x^2 + a)^2/b^3 - 1/9*(b*x^2 + a)^3*
p^2*log(b*x^2 + a)/b^3 + 1/2*(b*x^2 + a)^2*a*p^2*log(b*x^2 + a)/b^3 + 1/3*(b*x^2 + a)^3*p*log(b*x^2 + a)*log(c
)/b^3 - (b*x^2 + a)^2*a*p*log(b*x^2 + a)*log(c)/b^3 + 1/27*(b*x^2 + a)^3*p^2/b^3 - 1/4*(b*x^2 + a)^2*a*p^2/b^3
- 1/9*(b*x^2 + a)^3*p*log(c)/b^3 + 1/2*(b*x^2 + a)^2*a*p*log(c)/b^3 + 1/6*(b*x^2 + a)^3*log(c)^2/b^3 - 1/2*(b
*x^2 + a)^2*a*log(c)^2/b^3 + 1/2*((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^2*p^2 - 2*(b*x^2 - (b*x^2 + a)*log(b*x^2 + a) + a)*a^2*p*log(c) + (b*x^2 + a)*a^2*log(c)^2)/b^3

Mupad [B] (verification not implemented)

Time = 1.41 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.59 \[ \int x^5 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx=\frac {p^2\,x^6}{27}+{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2\,\left (\frac {x^6}{6}+\frac {a^3}{6\,b^3}\right )-\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )\,\left (\frac {p\,x^6}{9}+\frac {a^2\,p\,x^2}{3\,b^2}-\frac {a\,p\,x^4}{6\,b}\right )-\frac {5\,a\,p^2\,x^4}{36\,b}-\frac {11\,a^3\,p^2\,\ln \left (b\,x^2+a\right )}{18\,b^3}+\frac {11\,a^2\,p^2\,x^2}{18\,b^2} \]

[In]

int(x^5*log(c*(a + b*x^2)^p)^2,x)

[Out]

(p^2*x^6)/27 + log(c*(a + b*x^2)^p)^2*(x^6/6 + a^3/(6*b^3)) - log(c*(a + b*x^2)^p)*((p*x^6)/9 + (a^2*p*x^2)/(3
*b^2) - (a*p*x^4)/(6*b)) - (5*a*p^2*x^4)/(36*b) - (11*a^3*p^2*log(a + b*x^2))/(18*b^3) + (11*a^2*p^2*x^2)/(18*
b^2)