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*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

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Rubi [A]  time = 0.036543, antiderivative size = 66, normalized size of antiderivative = 1., 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 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 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 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]

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*}

Mathematica [A]  time = 0.0234834, 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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Maple [C]  time = 0.421, size = 217, normalized size = 3.3 \begin{align*}{\frac{{x}^{3}\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{3}}-{\frac{i}{6}}\pi \,{x}^{3}{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{6}}\pi \,{x}^{3} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{i}{6}}\pi \,{x}^{3}{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}-{\frac{i}{6}}\pi \,{x}^{3} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+{\frac{\ln \left ( c \right ){x}^{3}}{3}}-{\frac{2\,p{x}^{3}}{9}}+{\frac{ap}{3\,{b}^{2}}\sqrt{-ab}\ln \left ( -\sqrt{-ab}x-a \right ) }-{\frac{ap}{3\,{b}^{2}}\sqrt{-ab}\ln \left ( \sqrt{-ab}x-a \right ) }+{\frac{2\,apx}{3\,b}} \end{align*}

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)*csgn(I*c)+1/6*I*Pi*x^3*csgn(I*c
*(b*x^2+a)^p)^2*csgn(I*c)+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*c*(b*x^
2+a)^p)^3+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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.98627, size = 350, normalized size = 5.3 \begin{align*} \left [\frac{3 \, b p x^{3} \log \left (b x^{2} + a\right ) - 2 \, b p x^{3} + 3 \, b x^{3} \log \left (c\right ) + 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 \left (c\right ) - 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 ] \end{align*}

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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Sympy [A]  time = 56.7637, size = 121, normalized size = 1.83 \begin{align*} \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{\left (c \right )}}{3} & \text{for}\: b \neq 0 \\\frac{x^{3} \log{\left (a^{p} c \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.1917, size = 80, normalized size = 1.21 \begin{align*} \frac{1}{3} \, p x^{3} \log \left (b x^{2} + a\right ) - \frac{1}{9} \,{\left (2 \, p - 3 \, \log \left (c\right )\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} \end{align*}

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