\(\int x \log (a+x^2) \, dx\) [232]

Optimal result

Integrand size = 8, antiderivative size = 23 \[ \int x \log \left (a+x^2\right ) \, dx=-\frac {x^2}{2}+\frac {1}{2} \left (a+x^2\right ) \log \left (a+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2504, 2436, 2332} \[ \int x \log \left (a+x^2\right ) \, dx=\frac {1}{2} \left (a+x^2\right ) \log \left (a+x^2\right )-\frac {x^2}{2} \]

[In]

Int[x*Log[a + x^2],x]

[Out]

-1/2*x^2 + ((a + x^2)*Log[a + x^2])/2

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int x \log \left (a+x^2\right ) \, dx=\frac {1}{2} \left (-x^2+\left (a+x^2\right ) \log \left (a+x^2\right )\right ) \]

[In]

Integrate[x*Log[a + x^2],x]

[Out]

(-x^2 + (a + x^2)*Log[a + x^2])/2

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\left (x^{2}+a \right ) \ln \left (x^{2}+a \right )}{2}-\frac {x^{2}}{2}-\frac {a}{2}\)	\(23\)
default	\(\frac {\left (x^{2}+a \right ) \ln \left (x^{2}+a \right )}{2}-\frac {x^{2}}{2}-\frac {a}{2}\)	\(23\)
norman	\(-\frac {x^{2}}{2}+\frac {\ln \left (x^{2}+a \right ) a}{2}+\frac {\ln \left (x^{2}+a \right ) x^{2}}{2}\)	\(27\)
risch	\(-\frac {x^{2}}{2}+\frac {\ln \left (x^{2}+a \right ) a}{2}+\frac {\ln \left (x^{2}+a \right ) x^{2}}{2}\)	\(27\)
parts	\(-\frac {x^{2}}{2}+\frac {\ln \left (x^{2}+a \right ) a}{2}+\frac {\ln \left (x^{2}+a \right ) x^{2}}{2}\)	\(27\)
parallelrisch	\(\frac {\ln \left (x^{2}+a \right ) x^{2}}{2}-\frac {x^{2}}{2}+\frac {\ln \left (x^{2}+a \right ) a}{2}+\frac {a}{2}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int x \log \left (a+x^2\right ) \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{2} \, {\left (x^{2} + a\right )} \log \left (x^{2} + a\right ) \]

[In]

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

[Out]

-1/2*x^2 + 1/2*(x^2 + a)*log(x^2 + a)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int x \log \left (a+x^2\right ) \, dx=\frac {a \log {\left (a + x^{2} \right )}}{2} + \frac {x^{2} \log {\left (a + x^{2} \right )}}{2} - \frac {x^{2}}{2} \]

[In]

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

[Out]

a*log(a + x**2)/2 + x**2*log(a + x**2)/2 - x**2/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int x \log \left (a+x^2\right ) \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{2} \, {\left (x^{2} + a\right )} \log \left (x^{2} + a\right ) - \frac {1}{2} \, a \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int x \log \left (a+x^2\right ) \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{2} \, {\left (x^{2} + a\right )} \log \left (x^{2} + a\right ) - \frac {1}{2} \, a \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int x \log \left (a+x^2\right ) \, dx=\frac {a\,\ln \left (x+\sqrt {-a}\right )}{2}+\frac {x^2\,\ln \left (x^2+a\right )}{2}+\frac {a\,\ln \left (x-\sqrt {-a}\right )}{2}-\frac {x^2}{2} \]

[In]

int(x*log(a + x^2),x)

[Out]

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