3.12 \(\int \frac{x^4}{a x+b x^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0227478, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1584, 266, 43} \[ \frac{x^2}{2 b}-\frac{a \log \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a*x + b*x^3),x]

[Out]

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

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A]  time = 0.0039764, size = 27, normalized size = 1. \[ \frac{x^2}{2 b}-\frac{a \log \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a*x + b*x^3),x]

[Out]

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

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Maple [A]  time = 0.001, size = 24, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2\,b}}-{\frac{a\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^4/(b*x^3+a*x),x)

[Out]

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

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Maxima [A]  time = 1.06432, size = 31, normalized size = 1.15 \begin{align*} \frac{x^{2}}{2 \, b} - \frac{a \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^4/(b*x^3+a*x),x, algorithm="maxima")

[Out]

1/2*x^2/b - 1/2*a*log(b*x^2 + a)/b^2

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Fricas [A]  time = 1.38497, size = 49, normalized size = 1.81 \begin{align*} \frac{b x^{2} - a \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^4/(b*x^3+a*x),x, algorithm="fricas")

[Out]

1/2*(b*x^2 - a*log(b*x^2 + a))/b^2

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Sympy [A]  time = 0.289509, size = 20, normalized size = 0.74 \begin{align*} - \frac{a \log{\left (a + b x^{2} \right )}}{2 b^{2}} + \frac{x^{2}}{2 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**3+a*x),x)

[Out]

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

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Giac [A]  time = 1.20129, size = 32, normalized size = 1.19 \begin{align*} \frac{x^{2}}{2 \, b} - \frac{a \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^3+a*x),x, algorithm="giac")

[Out]

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