3.1.85 \(\int x^2 \log ^2(c (a+b x^2)^p) \, dx\) [85]

Optimal. Leaf size=294 \[ -\frac {32 a p^2 x}{9 b}+\frac {8 p^2 x^3}{27}+\frac {32 a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{9 b^{3/2}}-\frac {4 i a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{3 b^{3/2}}-\frac {8 a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 b^{3/2}}+\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {4 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4 i a^{3/2} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 b^{3/2}} \]

[Out]

-32/9*a*p^2*x/b+8/27*p^2*x^3+32/9*a^(3/2)*p^2*arctan(x*b^(1/2)/a^(1/2))/b^(3/2)-4/3*I*a^(3/2)*p^2*arctan(x*b^(
1/2)/a^(1/2))^2/b^(3/2)+4/3*a*p*x*ln(c*(b*x^2+a)^p)/b-4/9*p*x^3*ln(c*(b*x^2+a)^p)-4/3*a^(3/2)*p*arctan(x*b^(1/
2)/a^(1/2))*ln(c*(b*x^2+a)^p)/b^(3/2)+1/3*x^3*ln(c*(b*x^2+a)^p)^2-8/3*a^(3/2)*p^2*arctan(x*b^(1/2)/a^(1/2))*ln
(2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/b^(3/2)-4/3*I*a^(3/2)*p^2*polylog(2,1-2*a^(1/2)/(a^(1/2)+I*x*b^(1/2)))/b^(3/
2)

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Rubi [A]
time = 0.22, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {2507, 2526, 2498, 327, 211, 2505, 308, 2520, 12, 5040, 4964, 2449, 2352} \begin {gather*} -\frac {4 i a^{3/2} p^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 b^{3/2}}-\frac {4 a^{3/2} p \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}-\frac {4 i a^{3/2} p^2 \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{3 b^{3/2}}+\frac {32 a^{3/2} p^2 \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{9 b^{3/2}}-\frac {8 a^{3/2} p^2 \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 b^{3/2}}+\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {32 a p^2 x}{9 b}+\frac {8 p^2 x^3}{27} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Log[c*(a + b*x^2)^p]^2,x]

[Out]

(-32*a*p^2*x)/(9*b) + (8*p^2*x^3)/27 + (32*a^(3/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(9*b^(3/2)) - (((4*I)/3)*a
^(3/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2)/b^(3/2) - (8*a^(3/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[(2*Sqrt[a])
/(Sqrt[a] + I*Sqrt[b]*x)])/(3*b^(3/2)) + (4*a*p*x*Log[c*(a + b*x^2)^p])/(3*b) - (4*p*x^3*Log[c*(a + b*x^2)^p])
/9 - (4*a^(3/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^p])/(3*b^(3/2)) + (x^3*Log[c*(a + b*x^2)^p]^2)
/3 - (((4*I)/3)*a^(3/2)*p^2*PolyLog[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/b^(3/2)

Rule 12

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

Rule 211

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

Rule 308

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 327

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 2352

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

Rule 2449

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 2498

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 2505

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 2507

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 2520

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 2526

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 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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^2 \log ^2\left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} (4 b p) \int \frac {x^4 \log \left (c \left (a+b x^2\right )^p\right )}{a+b x^2} \, dx\\ &=\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} (4 b p) \int \left (-\frac {a \log \left (c \left (a+b x^2\right )^p\right )}{b^2}+\frac {x^2 \log \left (c \left (a+b x^2\right )^p\right )}{b}+\frac {a^2 \log \left (c \left (a+b x^2\right )^p\right )}{b^2 \left (a+b x^2\right )}\right ) \, dx\\ &=\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} (4 p) \int x^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx+\frac {(4 a p) \int \log \left (c \left (a+b x^2\right )^p\right ) \, dx}{3 b}-\frac {\left (4 a^2 p\right ) \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{a+b x^2} \, dx}{3 b}\\ &=\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {4 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} \left (8 a p^2\right ) \int \frac {x^2}{a+b x^2} \, dx+\frac {1}{3} \left (8 a^2 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+\frac {1}{9} \left (8 b p^2\right ) \int \frac {x^4}{a+b x^2} \, dx\\ &=-\frac {8 a p^2 x}{3 b}+\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {4 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {\left (8 a^2 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{3 b}+\frac {\left (8 a^{3/2} p^2\right ) \int \frac {x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a+b x^2} \, dx}{3 \sqrt {b}}+\frac {1}{9} \left (8 b 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 {32 a p^2 x}{9 b}+\frac {8 p^2 x^3}{27}+\frac {8 a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}-\frac {4 i a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{3 b^{3/2}}+\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {4 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {\left (8 a p^2\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{i-\frac {\sqrt {b} x}{\sqrt {a}}} \, dx}{3 b}+\frac {\left (8 a^2 p^2\right ) \int \frac {1}{a+b x^2} \, dx}{9 b}\\ &=-\frac {32 a p^2 x}{9 b}+\frac {8 p^2 x^3}{27}+\frac {32 a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{9 b^{3/2}}-\frac {4 i a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{3 b^{3/2}}-\frac {8 a^{3/2} 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 )}{3 b^{3/2}}+\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {4 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {\left (8 a 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}{3 b}\\ &=-\frac {32 a p^2 x}{9 b}+\frac {8 p^2 x^3}{27}+\frac {32 a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{9 b^{3/2}}-\frac {4 i a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{3 b^{3/2}}-\frac {8 a^{3/2} 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 )}{3 b^{3/2}}+\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {4 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {\left (8 i a^{3/2} p^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+\frac {i \sqrt {b} x}{\sqrt {a}}}\right )}{3 b^{3/2}}\\ &=-\frac {32 a p^2 x}{9 b}+\frac {8 p^2 x^3}{27}+\frac {32 a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{9 b^{3/2}}-\frac {4 i a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{3 b^{3/2}}-\frac {8 a^{3/2} 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 )}{3 b^{3/2}}+\frac {4 a p x \log \left (c \left (a+b x^2\right )^p\right )}{3 b}-\frac {4}{9} p x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {4 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log ^2\left (c \left (a+b x^2\right )^p\right )-\frac {4 i a^{3/2} p^2 \text {Li}_2\left (1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{3 b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 223, normalized size = 0.76 \begin {gather*} \frac {-36 i a^{3/2} p^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2-12 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-8 p+6 p \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )+3 \log \left (c \left (a+b x^2\right )^p\right )\right )+\sqrt {b} x \left (8 p^2 \left (-12 a+b x^2\right )+12 p \left (3 a-b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right )+9 b x^2 \log ^2\left (c \left (a+b x^2\right )^p\right )\right )-36 i a^{3/2} p^2 \text {Li}_2\left (\frac {i \sqrt {a}+\sqrt {b} x}{-i \sqrt {a}+\sqrt {b} x}\right )}{27 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-36*I)*a^(3/2)*p^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2 - 12*a^(3/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*(-8*p + 6*p*Log[
(2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)] + 3*Log[c*(a + b*x^2)^p]) + Sqrt[b]*x*(8*p^2*(-12*a + b*x^2) + 12*p*(3*a
- b*x^2)*Log[c*(a + b*x^2)^p] + 9*b*x^2*Log[c*(a + b*x^2)^p]^2) - (36*I)*a^(3/2)*p^2*PolyLog[2, (I*Sqrt[a] + S
qrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])/(27*b^(3/2))

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int x^{2} \ln \left (c \left (b \,x^{2}+a \right )^{p}\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*log(c*(a + b*x^2)^p)^2,x)

[Out]

int(x^2*log(c*(a + b*x^2)^p)^2, x)

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