\(\int x^2 \log (a^2+x^2) \, dx\) [72]

Optimal result

Integrand size = 12, antiderivative size = 44 \[ \int x^2 \log \left (a^2+x^2\right ) \, dx=\frac {2 a^2 x}{3}-\frac {2 x^3}{9}-\frac {2}{3} a^3 \arctan \left (\frac {x}{a}\right )+\frac {1}{3} x^3 \log \left (a^2+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2505, 308, 209} \[ \int x^2 \log \left (a^2+x^2\right ) \, dx=-\frac {2}{3} a^3 \arctan \left (\frac {x}{a}\right )+\frac {1}{3} x^3 \log \left (a^2+x^2\right )+\frac {2 a^2 x}{3}-\frac {2 x^3}{9} \]

[In]

Int[x^2*Log[a^2 + x^2],x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \log \left (a^2+x^2\right )-\frac {2}{3} \int \frac {x^4}{a^2+x^2} \, dx \\ & = \frac {1}{3} x^3 \log \left (a^2+x^2\right )-\frac {2}{3} \int \left (-a^2+x^2+\frac {a^4}{a^2+x^2}\right ) \, dx \\ & = \frac {2 a^2 x}{3}-\frac {2 x^3}{9}+\frac {1}{3} x^3 \log \left (a^2+x^2\right )-\frac {1}{3} \left (2 a^4\right ) \int \frac {1}{a^2+x^2} \, dx \\ & = \frac {2 a^2 x}{3}-\frac {2 x^3}{9}-\frac {2}{3} a^3 \arctan \left (\frac {x}{a}\right )+\frac {1}{3} x^3 \log \left (a^2+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int x^2 \log \left (a^2+x^2\right ) \, dx=\frac {2 a^2 x}{3}-\frac {2 x^3}{9}-\frac {2}{3} a^3 \arctan \left (\frac {x}{a}\right )+\frac {1}{3} x^3 \log \left (a^2+x^2\right ) \]

[In]

Integrate[x^2*Log[a^2 + x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {2 a^{2} x}{3}-\frac {2 x^{3}}{9}-\frac {2 a^{3} \arctan \left (\frac {x}{a}\right )}{3}+\frac {x^{3} \ln \left (a^{2}+x^{2}\right )}{3}\)	\(37\)
risch	\(\frac {2 a^{2} x}{3}-\frac {2 x^{3}}{9}-\frac {2 a^{3} \arctan \left (\frac {x}{a}\right )}{3}+\frac {x^{3} \ln \left (a^{2}+x^{2}\right )}{3}\)	\(37\)
parts	\(\frac {2 a^{2} x}{3}-\frac {2 x^{3}}{9}-\frac {2 a^{3} \arctan \left (\frac {x}{a}\right )}{3}+\frac {x^{3} \ln \left (a^{2}+x^{2}\right )}{3}\)	\(37\)
parallelrisch	\(\frac {2 i \ln \left (-i a +x \right ) a^{3}}{3}-\frac {i \ln \left (a^{2}+x^{2}\right ) a^{3}}{3}+\frac {x^{3} \ln \left (a^{2}+x^{2}\right )}{3}-\frac {2 x^{3}}{9}+\frac {2 a^{2} x}{3}\)	\(53\)

[In]

int(x^2*ln(a^2+x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int x^2 \log \left (a^2+x^2\right ) \, dx=-\frac {2}{3} \, a^{3} \arctan \left (\frac {x}{a}\right ) + \frac {1}{3} \, x^{3} \log \left (a^{2} + x^{2}\right ) + \frac {2}{3} \, a^{2} x - \frac {2}{9} \, x^{3} \]

[In]

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

[Out]

-2/3*a^3*arctan(x/a) + 1/3*x^3*log(a^2 + x^2) + 2/3*a^2*x - 2/9*x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20 \[ \int x^2 \log \left (a^2+x^2\right ) \, dx=- 2 a^{3} \left (- \frac {i \log {\left (- i a + x \right )}}{6} + \frac {i \log {\left (i a + x \right )}}{6}\right ) + \frac {2 a^{2} x}{3} + \frac {x^{3} \log {\left (a^{2} + x^{2} \right )}}{3} - \frac {2 x^{3}}{9} \]

[In]

integrate(x**2*ln(a**2+x**2),x)

[Out]

-2*a**3*(-I*log(-I*a + x)/6 + I*log(I*a + x)/6) + 2*a**2*x/3 + x**3*log(a**2 + x**2)/3 - 2*x**3/9

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int x^2 \log \left (a^2+x^2\right ) \, dx=-\frac {2}{3} \, a^{3} \arctan \left (\frac {x}{a}\right ) + \frac {1}{3} \, x^{3} \log \left (a^{2} + x^{2}\right ) + \frac {2}{3} \, a^{2} x - \frac {2}{9} \, x^{3} \]

[In]

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

[Out]

-2/3*a^3*arctan(x/a) + 1/3*x^3*log(a^2 + x^2) + 2/3*a^2*x - 2/9*x^3

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int x^2 \log \left (a^2+x^2\right ) \, dx=-\frac {2}{3} \, a^{3} \arctan \left (\frac {x}{a}\right ) + \frac {1}{3} \, x^{3} \log \left (a^{2} + x^{2}\right ) + \frac {2}{3} \, a^{2} x - \frac {2}{9} \, x^{3} \]

[In]

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

[Out]

-2/3*a^3*arctan(x/a) + 1/3*x^3*log(a^2 + x^2) + 2/3*a^2*x - 2/9*x^3

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.48 \[ \int x^2 \log \left (a^2+x^2\right ) \, dx=\frac {2\,a^2\,x}{3}-\frac {\ln \left (x-\sqrt {-a^2}\right )\,{\left (-a^2\right )}^{3/2}}{3}+\frac {x^3\,\ln \left (a^2+x^2\right )}{3}+\frac {\ln \left (x+\sqrt {-a^2}\right )\,{\left (-a^2\right )}^{3/2}}{3}-\frac {2\,x^3}{9} \]

[In]

int(x^2*log(a^2 + x^2),x)

[Out]

(2*a^2*x)/3 - (log(x - (-a^2)^(1/2))*(-a^2)^(3/2))/3 + (x^3*log(a^2 + x^2))/3 + (log(x + (-a^2)^(1/2))*(-a^2)^
(3/2))/3 - (2*x^3)/9