3.38 \(\int x^2 \log (c (a+\frac{b}{x^2})^p) \, dx\)

Optimal. Leaf size=58 \[ -\frac{2 b^{3/2} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{3 a^{3/2}}+\frac{1}{3} x^3 \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )+\frac{2 b p x}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.0270836, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2455, 193, 321, 205} \[ -\frac{2 b^{3/2} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{3 a^{3/2}}+\frac{1}{3} x^3 \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )+\frac{2 b p x}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Log[c*(a + b/x^2)^p],x]

[Out]

(2*b*p*x)/(3*a) - (2*b^(3/2)*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(3*a^(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 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

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

Mathematica [C]  time = 0.0024465, size = 47, normalized size = 0.81 \[ \frac{1}{3} x^3 \log \left (c \left (a+\frac{b}{x^2}\right )^p\right )+\frac{2 b p x \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{b}{a x^2}\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[c*(a + b/x^2)^p],x]

[Out]

(2*b*p*x*Hypergeometric2F1[-1/2, 1, 1/2, -(b/(a*x^2))])/(3*a) + (x^3*Log[c*(a + b/x^2)^p])/3

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Maple [F]  time = 0.247, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}\ln \left ( c \left ( a+{\frac{b}{{x}^{2}}} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*(a+b/x^2)^p),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.21636, size = 320, normalized size = 5.52 \begin{align*} \left [\frac{a p x^{3} \log \left (\frac{a x^{2} + b}{x^{2}}\right ) + a x^{3} \log \left (c\right ) + b p \sqrt{-\frac{b}{a}} \log \left (\frac{a x^{2} - 2 \, a x \sqrt{-\frac{b}{a}} - b}{a x^{2} + b}\right ) + 2 \, b p x}{3 \, a}, \frac{a p x^{3} \log \left (\frac{a x^{2} + b}{x^{2}}\right ) + a x^{3} \log \left (c\right ) - 2 \, b p \sqrt{\frac{b}{a}} \arctan \left (\frac{a x \sqrt{\frac{b}{a}}}{b}\right ) + 2 \, b p x}{3 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 85.1369, size = 146, normalized size = 2.52 \begin{align*} \begin{cases} \frac{p x^{3} \log{\left (a + \frac{b}{x^{2}} \right )}}{3} + \frac{x^{3} \log{\left (c \right )}}{3} + \frac{2 b p x}{3 a} + \frac{i b^{\frac{3}{2}} p \log{\left (- i \sqrt{b} \sqrt{\frac{1}{a}} + x \right )}}{3 a^{2} \sqrt{\frac{1}{a}}} - \frac{i b^{\frac{3}{2}} p \log{\left (i \sqrt{b} \sqrt{\frac{1}{a}} + x \right )}}{3 a^{2} \sqrt{\frac{1}{a}}} & \text{for}\: a \neq 0 \\\frac{p x^{3} \log{\left (b \right )}}{3} - \frac{2 p x^{3} \log{\left (x \right )}}{3} + \frac{2 p x^{3}}{9} + \frac{x^{3} \log{\left (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*(a+b/x**2)**p),x)

[Out]

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

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Giac [A]  time = 1.17803, size = 85, normalized size = 1.47 \begin{align*} \frac{1}{3} \, p x^{3} \log \left (a x^{2} + b\right ) - \frac{1}{3} \, p x^{3} \log \left (x^{2}\right ) + \frac{1}{3} \, x^{3} \log \left (c\right ) - \frac{2 \, b^{2} p \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{3 \, \sqrt{a b} a} + \frac{2 \, b p x}{3 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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