3.478 \(\int \frac {x^5}{a^2+2 a b x^2+b^2 x^4} \, dx\)

Optimal. Leaf size=44 \[ -\frac {a^2}{2 b^3 \left (a+b x^2\right )}-\frac {a \log \left (a+b x^2\right )}{b^3}+\frac {x^2}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 266, 43} \[ -\frac {a^2}{2 b^3 \left (a+b x^2\right )}-\frac {a \log \left (a+b x^2\right )}{b^3}+\frac {x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

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

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^5}{a^2+2 a b x^2+b^2 x^4} \, dx &=b^2 \int \frac {x^5}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=\frac {1}{2} b^2 \operatorname {Subst}\left (\int \frac {x^2}{\left (a b+b^2 x\right )^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} b^2 \operatorname {Subst}\left (\int \left (\frac {1}{b^4}+\frac {a^2}{b^4 (a+b x)^2}-\frac {2 a}{b^4 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac {x^2}{2 b^2}-\frac {a^2}{2 b^3 \left (a+b x^2\right )}-\frac {a \log \left (a+b x^2\right )}{b^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 38, normalized size = 0.86 \[ \frac {-\frac {a^2}{a+b x^2}-2 a \log \left (a+b x^2\right )+b x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

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

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fricas [A]  time = 0.81, size = 56, normalized size = 1.27 \[ \frac {b^{2} x^{4} + a b x^{2} - a^{2} - 2 \, {\left (a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

1/2*(b^2*x^4 + a*b*x^2 - a^2 - 2*(a*b*x^2 + a^2)*log(b*x^2 + a))/(b^4*x^2 + a*b^3)

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giac [A]  time = 0.17, size = 49, normalized size = 1.11 \[ \frac {x^{2}}{2 \, b^{2}} - \frac {a \log \left ({\left | b x^{2} + a \right |}\right )}{b^{3}} + \frac {2 \, a b x^{2} + a^{2}}{2 \, {\left (b x^{2} + a\right )} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 41, normalized size = 0.93 \[ \frac {x^{2}}{2 b^{2}}-\frac {a^{2}}{2 \left (b \,x^{2}+a \right ) b^{3}}-\frac {a \ln \left (b \,x^{2}+a \right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

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

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maxima [A]  time = 1.30, size = 43, normalized size = 0.98 \[ -\frac {a^{2}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} + \frac {x^{2}}{2 \, b^{2}} - \frac {a \log \left (b x^{2} + a\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

-1/2*a^2/(b^4*x^2 + a*b^3) + 1/2*x^2/b^2 - a*log(b*x^2 + a)/b^3

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mupad [B]  time = 0.05, size = 45, normalized size = 1.02 \[ \frac {x^2}{2\,b^2}-\frac {a^2}{2\,\left (b^4\,x^2+a\,b^3\right )}-\frac {a\,\ln \left (b\,x^2+a\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a^2 + b^2*x^4 + 2*a*b*x^2),x)

[Out]

x^2/(2*b^2) - a^2/(2*(a*b^3 + b^4*x^2)) - (a*log(a + b*x^2))/b^3

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sympy [A]  time = 0.25, size = 39, normalized size = 0.89 \[ - \frac {a^{2}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {a \log {\left (a + b x^{2} \right )}}{b^{3}} + \frac {x^{2}}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

-a**2/(2*a*b**3 + 2*b**4*x**2) - a*log(a + b*x**2)/b**3 + x**2/(2*b**2)

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