\(\int \frac {x^8}{a+b x^3} \, dx\) [318]

Optimal result

Integrand size = 13, antiderivative size = 40 \[ \int \frac {x^8}{a+b x^3} \, dx=-\frac {a x^3}{3 b^2}+\frac {x^6}{6 b}+\frac {a^2 \log \left (a+b x^3\right )}{3 b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {x^8}{a+b x^3} \, dx=\frac {a^2 \log \left (a+b x^3\right )}{3 b^3}-\frac {a x^3}{3 b^2}+\frac {x^6}{6 b} \]

[In]

Int[x^8/(a + b*x^3),x]

[Out]

-1/3*(a*x^3)/b^2 + x^6/(6*b) + (a^2*Log[a + b*x^3])/(3*b^3)

Rule 45

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])

Rule 272

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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {x^2}{a+b x} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {x^8}{a+b x^3} \, dx=-\frac {a x^3}{3 b^2}+\frac {x^6}{6 b}+\frac {a^2 \log \left (a+b x^3\right )}{3 b^3} \]

[In]

Integrate[x^8/(a + b*x^3),x]

[Out]

-1/3*(a*x^3)/b^2 + x^6/(6*b) + (a^2*Log[a + b*x^3])/(3*b^3)

Maple [A] (verified)

Time = 3.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85

method	result	size
parallelrisch	\(\frac {b^{2} x^{6}-2 a b \,x^{3}+2 a^{2} \ln \left (b \,x^{3}+a \right )}{6 b^{3}}\)	\(34\)
default	\(-\frac {-\frac {1}{2} b \,x^{6}+a \,x^{3}}{3 b^{2}}+\frac {a^{2} \ln \left (b \,x^{3}+a \right )}{3 b^{3}}\)	\(35\)
norman	\(-\frac {a \,x^{3}}{3 b^{2}}+\frac {x^{6}}{6 b}+\frac {a^{2} \ln \left (b \,x^{3}+a \right )}{3 b^{3}}\)	\(35\)
risch	\(\frac {x^{6}}{6 b}-\frac {a \,x^{3}}{3 b^{2}}+\frac {a^{2}}{6 b^{3}}+\frac {a^{2} \ln \left (b \,x^{3}+a \right )}{3 b^{3}}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {x^8}{a+b x^3} \, dx=\frac {b^{2} x^{6} - 2 \, a b x^{3} + 2 \, a^{2} \log \left (b x^{3} + a\right )}{6 \, b^{3}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int \frac {x^8}{a+b x^3} \, dx=\frac {a^{2} \log {\left (a + b x^{3} \right )}}{3 b^{3}} - \frac {a x^{3}}{3 b^{2}} + \frac {x^{6}}{6 b} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {x^8}{a+b x^3} \, dx=\frac {a^{2} \log \left (b x^{3} + a\right )}{3 \, b^{3}} + \frac {b x^{6} - 2 \, a x^{3}}{6 \, b^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int \frac {x^8}{a+b x^3} \, dx=\frac {a^{2} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{3}} + \frac {b x^{6} - 2 \, a x^{3}}{6 \, b^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {x^8}{a+b x^3} \, dx=\frac {2\,a^2\,\ln \left (b\,x^3+a\right )+b^2\,x^6-2\,a\,b\,x^3}{6\,b^3} \]

[In]

int(x^8/(a + b*x^3),x)

[Out]

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