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

Optimal. Leaf size=27 \[ \frac{x^2}{2 a}-\frac{b \log \left (a x^2+b\right )}{2 a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0182291, antiderivative size = 27, 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} \[ \frac{x^2}{2 a}-\frac{b \log \left (a x^2+b\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(b + a*x^2),x]

[Out]

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

Mathematica [A]  time = 0.0042536, size = 27, normalized size = 1. \[ \frac{x^2}{2 a}-\frac{b \log \left (a x^2+b\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(b + a*x^2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.002, size = 24, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2\,a}}-{\frac{b\ln \left ( a{x}^{2}+b \right ) }{2\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.938308, size = 31, normalized size = 1.15 \begin{align*} \frac{x^{2}}{2 \, a} - \frac{b \log \left (a x^{2} + b\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2/a - 1/2*b*log(a*x^2 + b)/a^2

________________________________________________________________________________________

Fricas [A]  time = 1.56157, size = 49, normalized size = 1.81 \begin{align*} \frac{a x^{2} - b \log \left (a x^{2} + b\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*x^2 - b*log(a*x^2 + b))/a^2

________________________________________________________________________________________

Sympy [A]  time = 0.293708, size = 20, normalized size = 0.74 \begin{align*} \frac{x^{2}}{2 a} - \frac{b \log{\left (a x^{2} + b \right )}}{2 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.09707, size = 32, normalized size = 1.19 \begin{align*} \frac{x^{2}}{2 \, a} - \frac{b \log \left ({\left | a x^{2} + b \right |}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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