3.155 \(\int \frac{x^3}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=33 \[ \frac{a}{2 b^2 \left (a+b x^2\right )}+\frac{\log \left (a+b x^2\right )}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0237993, antiderivative size = 33, 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} \[ \frac{a}{2 b^2 \left (a+b x^2\right )}+\frac{\log \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x^2)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0076065, size = 27, normalized size = 0.82 \[ \frac{\frac{a}{a+b x^2}+\log \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*x^2)^2,x]

[Out]

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

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Maple [A]  time = 0.008, size = 30, normalized size = 0.9 \begin{align*}{\frac{a}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{\ln \left ( b{x}^{2}+a \right ) }{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)^2,x)

[Out]

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

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Maxima [A]  time = 1.95675, size = 43, normalized size = 1.3 \begin{align*} \frac{a}{2 \,{\left (b^{3} x^{2} + a b^{2}\right )}} + \frac{\log \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.29161, size = 76, normalized size = 2.3 \begin{align*} \frac{{\left (b x^{2} + a\right )} \log \left (b x^{2} + a\right ) + a}{2 \,{\left (b^{3} x^{2} + a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.331859, size = 29, normalized size = 0.88 \begin{align*} \frac{a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{\log{\left (a + b x^{2} \right )}}{2 b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**2+a)**2,x)

[Out]

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

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Giac [A]  time = 2.69281, size = 65, normalized size = 1.97 \begin{align*} -\frac{\frac{\log \left (\frac{{\left | b x^{2} + a \right |}}{{\left (b x^{2} + a\right )}^{2}{\left | b \right |}}\right )}{b} - \frac{a}{{\left (b x^{2} + a\right )} b}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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