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

Optimal. Leaf size=66 \[ -\frac {2 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log \left (c \left (a+b x^2\right )^p\right )+\frac {2 a p x}{3 b}-\frac {2 p x^3}{9} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2455, 302, 205} \[ -\frac {2 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log \left (c \left (a+b x^2\right )^p\right )+\frac {2 a p x}{3 b}-\frac {2 p x^3}{9} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Log[c*(a + b*x^2)^p],x]

[Out]

(2*a*p*x)/(3*b) - (2*p*x^3)/9 - (2*a^(3/2)*p*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(3*b^(3/2)) + (x^3*Log[c*(a + b*x^2)
^p])/3

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 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]

Rubi steps

\begin {align*} \int x^2 \log \left (c \left (a+b x^2\right )^p\right ) \, dx &=\frac {1}{3} x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} (2 b p) \int \frac {x^4}{a+b x^2} \, dx\\ &=\frac {1}{3} x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {1}{3} (2 b p) \int \left (-\frac {a}{b^2}+\frac {x^2}{b}+\frac {a^2}{b^2 \left (a+b x^2\right )}\right ) \, dx\\ &=\frac {2 a p x}{3 b}-\frac {2 p x^3}{9}+\frac {1}{3} x^3 \log \left (c \left (a+b x^2\right )^p\right )-\frac {\left (2 a^2 p\right ) \int \frac {1}{a+b x^2} \, dx}{3 b}\\ &=\frac {2 a p x}{3 b}-\frac {2 p x^3}{9}-\frac {2 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{3 b^{3/2}}+\frac {1}{3} x^3 \log \left (c \left (a+b x^2\right )^p\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 62, normalized size = 0.94 \[ \frac {1}{9} \left (-\frac {6 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}+3 x^3 \log \left (c \left (a+b x^2\right )^p\right )+\frac {6 a p x}{b}-2 p x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[c*(a + b*x^2)^p],x]

[Out]

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

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fricas [A]  time = 0.48, size = 152, normalized size = 2.30 \[ \left [\frac {3 \, b p x^{3} \log \left (b x^{2} + a\right ) - 2 \, b p x^{3} + 3 \, b x^{3} \log \relax (c) + 3 \, a p \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 6 \, a p x}{9 \, b}, \frac {3 \, b p x^{3} \log \left (b x^{2} + a\right ) - 2 \, b p x^{3} + 3 \, b x^{3} \log \relax (c) - 6 \, a p \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 6 \, a p x}{9 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/9*(3*b*p*x^3*log(b*x^2 + a) - 2*b*p*x^3 + 3*b*x^3*log(c) + 3*a*p*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) -
a)/(b*x^2 + a)) + 6*a*p*x)/b, 1/9*(3*b*p*x^3*log(b*x^2 + a) - 2*b*p*x^3 + 3*b*x^3*log(c) - 6*a*p*sqrt(a/b)*ar
ctan(b*x*sqrt(a/b)/a) + 6*a*p*x)/b]

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giac [A]  time = 0.20, size = 59, normalized size = 0.89 \[ \frac {1}{3} \, p x^{3} \log \left (b x^{2} + a\right ) - \frac {1}{9} \, {\left (2 \, p - 3 \, \log \relax (c)\right )} x^{3} - \frac {2 \, a^{2} p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b} + \frac {2 \, a p x}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*p*x^3*log(b*x^2 + a) - 1/9*(2*p - 3*log(c))*x^3 - 2/3*a^2*p*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b) + 2/3*a*p*
x/b

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maple [C]  time = 0.35, size = 217, normalized size = 3.29 \[ -\frac {i \pi \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}{6}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{6}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}}{6}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}}{6}-\frac {2 p \,x^{3}}{9}+\frac {x^{3} \ln \relax (c )}{3}+\frac {x^{3} \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{3}+\frac {2 a p x}{3 b}+\frac {\sqrt {-a b}\, a p \ln \left (-a -\sqrt {-a b}\, x \right )}{3 b^{2}}-\frac {\sqrt {-a b}\, a p \ln \left (-a +\sqrt {-a b}\, x \right )}{3 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*ln((b*x^2+a)^p)+1/6*I*Pi*x^3*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/6*I*Pi*x^3*csgn(I*(b*x^2+a)
^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-1/6*I*Pi*x^3*csgn(I*c*(b*x^2+a)^p)^3+1/6*I*Pi*x^3*csgn(I*c*(b*x^2+a)^p)^2*
csgn(I*c)+1/3*ln(c)*x^3-2/9*p*x^3+1/3/b^2*(-a*b)^(1/2)*a*p*ln(-(-a*b)^(1/2)*x-a)-1/3/b^2*(-a*b)^(1/2)*a*p*ln((
-a*b)^(1/2)*x-a)+2/3*a*p*x/b

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maxima [A]  time = 1.57, size = 59, normalized size = 0.89 \[ \frac {1}{3} \, x^{3} \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) - \frac {2}{9} \, b p {\left (\frac {3 \, a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {b x^{3} - 3 \, a x}{b^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log((b*x^2 + a)^p*c) - 2/9*b*p*(3*a^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2) + (b*x^3 - 3*a*x)/b^2)

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mupad [B]  time = 0.23, size = 50, normalized size = 0.76 \[ \frac {x^3\,\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{3}-\frac {2\,p\,x^3}{9}-\frac {2\,a^{3/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{3\,b^{3/2}}+\frac {2\,a\,p\,x}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*log(c*(a + b*x^2)^p),x)

[Out]

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

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sympy [A]  time = 28.76, size = 121, normalized size = 1.83 \[ \begin {cases} - \frac {i a^{\frac {3}{2}} p \log {\left (a + b x^{2} \right )}}{3 b^{2} \sqrt {\frac {1}{b}}} + \frac {2 i a^{\frac {3}{2}} p \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{3 b^{2} \sqrt {\frac {1}{b}}} + \frac {2 a p x}{3 b} + \frac {p x^{3} \log {\left (a + b x^{2} \right )}}{3} - \frac {2 p x^{3}}{9} + \frac {x^{3} \log {\relax (c )}}{3} & \text {for}\: b \neq 0 \\\frac {x^{3} \log {\left (a^{p} c \right )}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*a**(3/2)*p*log(a + b*x**2)/(3*b**2*sqrt(1/b)) + 2*I*a**(3/2)*p*log(-I*sqrt(a)*sqrt(1/b) + x)/(3*
b**2*sqrt(1/b)) + 2*a*p*x/(3*b) + p*x**3*log(a + b*x**2)/3 - 2*p*x**3/9 + x**3*log(c)/3, Ne(b, 0)), (x**3*log(
a**p*c)/3, True))

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