3.318 \(\int \frac{x^8}{a+b x^3} \, dx\)

Optimal. Leaf size=40 \[ \frac{a^2 \log \left (a+b x^3\right )}{3 b^3}-\frac{a x^3}{3 b^2}+\frac{x^6}{6 b} \]

[Out]

-(a*x^3)/(3*b^2) + x^6/(6*b) + (a^2*Log[a + b*x^3])/(3*b^3)

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Rubi [A]  time = 0.0255807, antiderivative size = 40, 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} \[ \frac{a^2 \log \left (a+b x^3\right )}{3 b^3}-\frac{a x^3}{3 b^2}+\frac{x^6}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^3),x]

[Out]

-(a*x^3)/(3*b^2) + x^6/(6*b) + (a^2*Log[a + b*x^3])/(3*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^8}{a+b x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2}{a+b x} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{a}{b^2}+\frac{x}{b}+\frac{a^2}{b^2 (a+b x)}\right ) \, dx,x,x^3\right )\\ &=-\frac{a x^3}{3 b^2}+\frac{x^6}{6 b}+\frac{a^2 \log \left (a+b x^3\right )}{3 b^3}\\ \end{align*}

Mathematica [A]  time = 0.005954, size = 40, normalized size = 1. \[ \frac{a^2 \log \left (a+b x^3\right )}{3 b^3}-\frac{a x^3}{3 b^2}+\frac{x^6}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^3),x]

[Out]

-(a*x^3)/(3*b^2) + x^6/(6*b) + (a^2*Log[a + b*x^3])/(3*b^3)

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Maple [A]  time = 0.003, size = 35, normalized size = 0.9 \begin{align*} -{\frac{a{x}^{3}}{3\,{b}^{2}}}+{\frac{{x}^{6}}{6\,b}}+{\frac{{a}^{2}\ln \left ( b{x}^{3}+a \right ) }{3\,{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a),x)

[Out]

-1/3*a*x^3/b^2+1/6*x^6/b+1/3*a^2*ln(b*x^3+a)/b^3

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Maxima [A]  time = 1.09444, size = 46, normalized size = 1.15 \begin{align*} \frac{a^{2} \log \left (b x^{3} + a\right )}{3 \, b^{3}} + \frac{b x^{6} - 2 \, a x^{3}}{6 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a),x, algorithm="maxima")

[Out]

1/3*a^2*log(b*x^3 + a)/b^3 + 1/6*(b*x^6 - 2*a*x^3)/b^2

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Fricas [A]  time = 1.5901, size = 73, normalized size = 1.82 \begin{align*} \frac{b^{2} x^{6} - 2 \, a b x^{3} + 2 \, a^{2} \log \left (b x^{3} + a\right )}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a),x, algorithm="fricas")

[Out]

1/6*(b^2*x^6 - 2*a*b*x^3 + 2*a^2*log(b*x^3 + a))/b^3

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Sympy [A]  time = 0.429362, size = 32, normalized size = 0.8 \begin{align*} \frac{a^{2} \log{\left (a + b x^{3} \right )}}{3 b^{3}} - \frac{a x^{3}}{3 b^{2}} + \frac{x^{6}}{6 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**3+a),x)

[Out]

a**2*log(a + b*x**3)/(3*b**3) - a*x**3/(3*b**2) + x**6/(6*b)

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Giac [A]  time = 1.13711, size = 47, normalized size = 1.18 \begin{align*} \frac{a^{2} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{3}} + \frac{b x^{6} - 2 \, a x^{3}}{6 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a),x, algorithm="giac")

[Out]

1/3*a^2*log(abs(b*x^3 + a))/b^3 + 1/6*(b*x^6 - 2*a*x^3)/b^2