3.84 \(\int x^4 \log ^2(c (a+b x^2)^p) \, dx\)

Optimal. Leaf size=336 \[ \frac {4 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {4 i a^{5/2} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{5 b^{5/2}}+\frac {4 i a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 b^{5/2}}-\frac {184 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{75 b^{5/2}}+\frac {8 a^{5/2} p^2 \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {184 a^2 p^2 x}{75 b^2}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {64 a p^2 x^3}{225 b}+\frac {8 p^2 x^5}{125} \]

[Out]

184/75*a^2*p^2*x/b^2-64/225*a*p^2*x^3/b+8/125*p^2*x^5-184/75*a^(5/2)*p^2*arctan(x*b^(1/2)/a^(1/2))/b^(5/2)+4/5
*I*a^(5/2)*p^2*arctan(x*b^(1/2)/a^(1/2))^2/b^(5/2)-4/5*a^2*p*x*ln(c*(b*x^2+a)^p)/b^2+4/15*a*p*x^3*ln(c*(b*x^2+
a)^p)/b-4/25*p*x^5*ln(c*(b*x^2+a)^p)+4/5*a^(5/2)*p*arctan(x*b^(1/2)/a^(1/2))*ln(c*(b*x^2+a)^p)/b^(5/2)+1/5*x^5
*ln(c*(b*x^2+a)^p)^2+8/5*a^(5/2)*p^2*arctan(x*b^(1/2)/a^(1/2))*ln(2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/b^(5/2)+4/5
*I*a^(5/2)*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/b^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {2457, 2476, 2448, 321, 205, 2455, 302, 2470, 12, 4920, 4854, 2402, 2315} \[ \frac {4 i a^{5/2} p^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 b^{5/2}}-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {4 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {184 a^2 p^2 x}{75 b^2}+\frac {4 i a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 b^{5/2}}-\frac {184 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{75 b^{5/2}}+\frac {8 a^{5/2} p^2 \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {64 a p^2 x^3}{225 b}+\frac {8 p^2 x^5}{125} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Log[c*(a + b*x^2)^p]^2,x]

[Out]

(184*a^2*p^2*x)/(75*b^2) - (64*a*p^2*x^3)/(225*b) + (8*p^2*x^5)/125 - (184*a^(5/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt
[a]])/(75*b^(5/2)) + (((4*I)/5)*a^(5/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2)/b^(5/2) + (8*a^(5/2)*p^2*ArcTan[(Sq
rt[b]*x)/Sqrt[a]]*Log[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/(5*b^(5/2)) - (4*a^2*p*x*Log[c*(a + b*x^2)^p])/(5*
b^2) + (4*a*p*x^3*Log[c*(a + b*x^2)^p])/(15*b) - (4*p*x^5*Log[c*(a + b*x^2)^p])/25 + (4*a^(5/2)*p*ArcTan[(Sqrt
[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/(5*b^(5/2)) + (x^5*Log[c*(a + b*x^2)^p]^2)/5 + (((4*I)/5)*a^(5/2)*p^2*Po
lyLog[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/b^(5/2)

Rule 12

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 2455

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 2457

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x
)^(m + 1)*(a + b*Log[c*(d + e*x^n)^p])^q)/(f*(m + 1)), x] - Dist[(b*e*n*p*q)/(f^n*(m + 1)), Int[((f*x)^(m + n)
*(a + b*Log[c*(d + e*x^n)^p])^(q - 1))/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && IGtQ[q, 1]
&& IntegerQ[n] && NeQ[m, -1]

Rule 2470

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[(u*x^(n - 1))/(d + e*x^n
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

Rule 2476

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] && IntegerQ[s]

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[c*x])^(p + 1))/(b*e*(p + 1)), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b,
c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^4 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{5} (4 b p) \int \frac {x^6 \log \left (c \left (a+b x^2\right )^p\right )}{a+b x^2} \, dx\\ &=\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{5} (4 b p) \int \left (\frac {a^2 \log \left (c \left (a+b x^2\right )^p\right )}{b^3}-\frac {a x^2 \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {x^4 \log \left (c \left (a+b x^2\right )^p\right )}{b}-\frac {a^3 \log \left (c \left (a+b x^2\right )^p\right )}{b^3 \left (a+b x^2\right )}\right ) \, dx\\ &=\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{5} (4 p) \int x^4 \log \left (c \left (a+b x^2\right )^p\right ) \, dx-\frac {\left (4 a^2 p\right ) \int \log \left (c \left (a+b x^2\right )^p\right ) \, dx}{5 b^2}+\frac {\left (4 a^3 p\right ) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{a+b x^2} \, dx}{5 b^2}+\frac {(4 a p) \int x^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx}{5 b}\\ &=-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{15} \left (8 a p^2\right ) \int \frac {x^4}{a+b x^2} \, dx+\frac {\left (8 a^2 p^2\right ) \int \frac {x^2}{a+b x^2} \, dx}{5 b}-\frac {\left (8 a^3 p^2\right ) \int \frac {x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} \left (a+b x^2\right )} \, dx}{5 b}+\frac {1}{25} \left (8 b p^2\right ) \int \frac {x^6}{a+b x^2} \, dx\\ &=\frac {8 a^2 p^2 x}{5 b^2}-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{15} \left (8 a p^2\right ) \int \left (-\frac {a}{b^2}+\frac {x^2}{b}+\frac {a^2}{b^2 \left (a+b x^2\right )}\right ) \, dx-\frac {\left (8 a^3 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{5 b^2}-\frac {\left (8 a^{5/2} p^2\right ) \int \frac {x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a+b x^2} \, dx}{5 b^{3/2}}+\frac {1}{25} \left (8 b p^2\right ) \int \left (\frac {a^2}{b^3}-\frac {a x^2}{b^2}+\frac {x^4}{b}-\frac {a^3}{b^3 \left (a+b x^2\right )}\right ) \, dx\\ &=\frac {184 a^2 p^2 x}{75 b^2}-\frac {64 a p^2 x^3}{225 b}+\frac {8 p^2 x^5}{125}-\frac {8 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{5 b^{5/2}}+\frac {4 i a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 b^{5/2}}-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {\left (8 a^2 p^2\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{i-\frac {\sqrt {b} x}{\sqrt {a}}} \, dx}{5 b^2}-\frac {\left (8 a^3 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{25 b^2}-\frac {\left (8 a^3 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{15 b^2}\\ &=\frac {184 a^2 p^2 x}{75 b^2}-\frac {64 a p^2 x^3}{225 b}+\frac {8 p^2 x^5}{125}-\frac {184 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{75 b^{5/2}}+\frac {4 i a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 b^{5/2}}+\frac {8 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 b^{5/2}}-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {\left (8 a^2 p^2\right ) \int \frac {\log \left (\frac {2}{1+\frac {i \sqrt {b} x}{\sqrt {a}}}\right )}{1+\frac {b x^2}{a}} \, dx}{5 b^2}\\ &=\frac {184 a^2 p^2 x}{75 b^2}-\frac {64 a p^2 x^3}{225 b}+\frac {8 p^2 x^5}{125}-\frac {184 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{75 b^{5/2}}+\frac {4 i a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 b^{5/2}}+\frac {8 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 b^{5/2}}-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {\left (8 i a^{5/2} p^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+\frac {i \sqrt {b} x}{\sqrt {a}}}\right )}{5 b^{5/2}}\\ &=\frac {184 a^2 p^2 x}{75 b^2}-\frac {64 a p^2 x^3}{225 b}+\frac {8 p^2 x^5}{125}-\frac {184 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{75 b^{5/2}}+\frac {4 i a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{5 b^{5/2}}+\frac {8 a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 b^{5/2}}-\frac {4 a^2 p x \log \left (c \left (a+b x^2\right )^p\right )}{5 b^2}+\frac {4 a p x^3 \log \left (c \left (a+b x^2\right )^p\right )}{15 b}-\frac {4}{25} p x^5 \log \left (c \left (a+b x^2\right )^p\right )+\frac {4 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{5 b^{5/2}}+\frac {1}{5} x^5 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {4 i a^{5/2} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{5 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 248, normalized size = 0.74 \[ \frac {60 a^{5/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (15 \log \left (c \left (a+b x^2\right )^p\right )+30 p \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )-46 p\right )+900 i a^{5/2} p^2 \text {Li}_2\left (\frac {\sqrt {b} x+i \sqrt {a}}{\sqrt {b} x-i \sqrt {a}}\right )+900 i a^{5/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2+\sqrt {b} x \left (-60 p \left (15 a^2-5 a b x^2+3 b^2 x^4\right ) \log \left (c \left (a+b x^2\right )^p\right )+8 p^2 \left (345 a^2-40 a b x^2+9 b^2 x^4\right )+225 b^2 x^4 \log ^2\left (c \left (a+b x^2\right )^p\right )\right )}{1125 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Log[c*(a + b*x^2)^p]^2,x]

[Out]

((900*I)*a^(5/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2 + 60*a^(5/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*(-46*p + 30*p*Lo
g[(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)] + 15*Log[c*(a + b*x^2)^p]) + Sqrt[b]*x*(8*p^2*(345*a^2 - 40*a*b*x^2 + 9
*b^2*x^4) - 60*p*(15*a^2 - 5*a*b*x^2 + 3*b^2*x^4)*Log[c*(a + b*x^2)^p] + 225*b^2*x^4*Log[c*(a + b*x^2)^p]^2) +
(900*I)*a^(5/2)*p^2*PolyLog[2, (I*Sqrt[a] + Sqrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])/(1125*b^(5/2))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(c*(b*x^2+a)^p)^2,x, algorithm="fricas")

[Out]

integral(x^4*log((b*x^2 + a)^p*c)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(c*(b*x^2+a)^p)^2,x, algorithm="giac")

[Out]

integrate(x^4*log((b*x^2 + a)^p*c)^2, x)

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int x^{4} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*ln(c*(b*x^2+a)^p)^2,x)

[Out]

int(x^4*ln(c*(b*x^2+a)^p)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{5} \, p^{2} x^{5} \log \left (b x^{2} + a\right )^{2} + \int \frac {5 \, b x^{6} \log \relax (c)^{2} + 5 \, a x^{4} \log \relax (c)^{2} - 2 \, {\left ({\left (2 \, p^{2} - 5 \, p \log \relax (c)\right )} b x^{6} - 5 \, a p x^{4} \log \relax (c)\right )} \log \left (b x^{2} + a\right )}{5 \, {\left (b x^{2} + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(c*(b*x^2+a)^p)^2,x, algorithm="maxima")

[Out]

1/5*p^2*x^5*log(b*x^2 + a)^2 + integrate(1/5*(5*b*x^6*log(c)^2 + 5*a*x^4*log(c)^2 - 2*((2*p^2 - 5*p*log(c))*b*
x^6 - 5*a*p*x^4*log(c))*log(b*x^2 + a))/(b*x^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*log(c*(a + b*x^2)^p)^2,x)

[Out]

int(x^4*log(c*(a + b*x^2)^p)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*ln(c*(b*x**2+a)**p)**2,x)

[Out]

Integral(x**4*log(c*(a + b*x**2)**p)**2, x)

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