3.73 \(\int x^4 \log (a^2+x^2) \, dx\)

Optimal. Leaf size=54 \[ \frac {2}{5} a^5 \tan ^{-1}\left (\frac {x}{a}\right )-\frac {2 a^4 x}{5}+\frac {2 a^2 x^3}{15}+\frac {1}{5} x^5 \log \left (a^2+x^2\right )-\frac {2 x^5}{25} \]

[Out]

-2/5*a^4*x+2/15*a^2*x^3-2/25*x^5+2/5*a^5*arctan(x/a)+1/5*x^5*ln(a^2+x^2)

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Rubi [A]  time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2455, 302, 203} \[ \frac {2 a^2 x^3}{15}+\frac {1}{5} x^5 \log \left (a^2+x^2\right )-\frac {2 a^4 x}{5}+\frac {2}{5} a^5 \tan ^{-1}\left (\frac {x}{a}\right )-\frac {2 x^5}{25} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Log[a^2 + x^2],x]

[Out]

(-2*a^4*x)/5 + (2*a^2*x^3)/15 - (2*x^5)/25 + (2*a^5*ArcTan[x/a])/5 + (x^5*Log[a^2 + x^2])/5

Rule 203

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

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^4 \log \left (a^2+x^2\right ) \, dx &=\frac {1}{5} x^5 \log \left (a^2+x^2\right )-\frac {2}{5} \int \frac {x^6}{a^2+x^2} \, dx\\ &=\frac {1}{5} x^5 \log \left (a^2+x^2\right )-\frac {2}{5} \int \left (a^4-a^2 x^2+x^4-\frac {a^6}{a^2+x^2}\right ) \, dx\\ &=-\frac {2 a^4 x}{5}+\frac {2 a^2 x^3}{15}-\frac {2 x^5}{25}+\frac {1}{5} x^5 \log \left (a^2+x^2\right )+\frac {1}{5} \left (2 a^6\right ) \int \frac {1}{a^2+x^2} \, dx\\ &=-\frac {2 a^4 x}{5}+\frac {2 a^2 x^3}{15}-\frac {2 x^5}{25}+\frac {2}{5} a^5 \tan ^{-1}\left (\frac {x}{a}\right )+\frac {1}{5} x^5 \log \left (a^2+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 54, normalized size = 1.00 \[ \frac {2}{5} a^5 \tan ^{-1}\left (\frac {x}{a}\right )-\frac {2 a^4 x}{5}+\frac {2 a^2 x^3}{15}+\frac {1}{5} x^5 \log \left (a^2+x^2\right )-\frac {2 x^5}{25} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Log[a^2 + x^2],x]

[Out]

(-2*a^4*x)/5 + (2*a^2*x^3)/15 - (2*x^5)/25 + (2*a^5*ArcTan[x/a])/5 + (x^5*Log[a^2 + x^2])/5

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fricas [A]  time = 0.40, size = 44, normalized size = 0.81 \[ \frac {2}{5} \, a^{5} \arctan \left (\frac {x}{a}\right ) + \frac {1}{5} \, x^{5} \log \left (a^{2} + x^{2}\right ) - \frac {2}{5} \, a^{4} x + \frac {2}{15} \, a^{2} x^{3} - \frac {2}{25} \, x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(a^2+x^2),x, algorithm="fricas")

[Out]

2/5*a^5*arctan(x/a) + 1/5*x^5*log(a^2 + x^2) - 2/5*a^4*x + 2/15*a^2*x^3 - 2/25*x^5

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giac [A]  time = 1.00, size = 44, normalized size = 0.81 \[ \frac {2}{5} \, a^{5} \arctan \left (\frac {x}{a}\right ) + \frac {1}{5} \, x^{5} \log \left (a^{2} + x^{2}\right ) - \frac {2}{5} \, a^{4} x + \frac {2}{15} \, a^{2} x^{3} - \frac {2}{25} \, x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(a^2+x^2),x, algorithm="giac")

[Out]

2/5*a^5*arctan(x/a) + 1/5*x^5*log(a^2 + x^2) - 2/5*a^4*x + 2/15*a^2*x^3 - 2/25*x^5

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maple [A]  time = 0.00, size = 45, normalized size = 0.83 \[ \frac {2 a^{5} \arctan \left (\frac {x}{a}\right )}{5}+\frac {x^{5} \ln \left (a^{2}+x^{2}\right )}{5}-\frac {2 a^{4} x}{5}+\frac {2 a^{2} x^{3}}{15}-\frac {2 x^{5}}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*ln(a^2+x^2),x)

[Out]

-2/5*a^4*x+2/15*a^2*x^3-2/25*x^5+2/5*a^5*arctan(1/a*x)+1/5*x^5*ln(a^2+x^2)

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maxima [A]  time = 0.96, size = 44, normalized size = 0.81 \[ \frac {2}{5} \, a^{5} \arctan \left (\frac {x}{a}\right ) + \frac {1}{5} \, x^{5} \log \left (a^{2} + x^{2}\right ) - \frac {2}{5} \, a^{4} x + \frac {2}{15} \, a^{2} x^{3} - \frac {2}{25} \, x^{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*log(a^2+x^2),x, algorithm="maxima")

[Out]

2/5*a^5*arctan(x/a) + 1/5*x^5*log(a^2 + x^2) - 2/5*a^4*x + 2/15*a^2*x^3 - 2/25*x^5

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mupad [B]  time = 0.13, size = 73, normalized size = 1.35 \[ \frac {x^5\,\ln \left (a^2+x^2\right )}{5}-\frac {2\,a^4\,x}{5}-\frac {\ln \left (x-\sqrt {-a^2}\right )\,{\left (-a^2\right )}^{5/2}}{5}+\frac {\ln \left (x+\sqrt {-a^2}\right )\,{\left (-a^2\right )}^{5/2}}{5}-\frac {2\,x^5}{25}+\frac {2\,a^2\,x^3}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*log(a^2 + x^2),x)

[Out]

(x^5*log(a^2 + x^2))/5 - (2*a^4*x)/5 - (log(x - (-a^2)^(1/2))*(-a^2)^(5/2))/5 + (log(x + (-a^2)^(1/2))*(-a^2)^
(5/2))/5 - (2*x^5)/25 + (2*a^2*x^3)/15

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sympy [C]  time = 0.21, size = 63, normalized size = 1.17 \[ - 2 a^{5} \left (\frac {i \log {\left (- i a + x \right )}}{10} - \frac {i \log {\left (i a + x \right )}}{10}\right ) - \frac {2 a^{4} x}{5} + \frac {2 a^{2} x^{3}}{15} + \frac {x^{5} \log {\left (a^{2} + x^{2} \right )}}{5} - \frac {2 x^{5}}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*ln(a**2+x**2),x)

[Out]

-2*a**5*(I*log(-I*a + x)/10 - I*log(I*a + x)/10) - 2*a**4*x/5 + 2*a**2*x**3/15 + x**5*log(a**2 + x**2)/5 - 2*x
**5/25

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