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

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]

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

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Rubi [A]  time = 0.031901, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {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 + b*x^2)^2,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 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]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0224456, 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 + b*x^2)^2,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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Maple [A]  time = 0.009, size = 41, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2\,{b}^{2}}}-{\frac{{a}^{2}}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}-{\frac{a\ln \left ( b{x}^{2}+a \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(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 = 1.41121, size = 58, normalized size = 1.32 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(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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Fricas [A]  time = 1.22029, size = 113, normalized size = 2.57 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(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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Sympy [A]  time = 0.368169, size = 39, normalized size = 0.89 \begin{align*} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(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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Giac [A]  time = 3.29296, size = 66, normalized size = 1.5 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(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)