\(\int \frac {x^5}{a+\frac {b}{x^3}} \, dx\) [1965]

Optimal result

Integrand size = 13, antiderivative size = 40 \[ \int \frac {x^5}{a+\frac {b}{x^3}} \, dx=-\frac {b x^3}{3 a^2}+\frac {x^6}{6 a}+\frac {b^2 \log \left (b+a x^3\right )}{3 a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {269, 272, 45} \[ \int \frac {x^5}{a+\frac {b}{x^3}} \, dx=\frac {b^2 \log \left (a x^3+b\right )}{3 a^3}-\frac {b x^3}{3 a^2}+\frac {x^6}{6 a} \]

[In]

Int[x^5/(a + b/x^3),x]

[Out]

-1/3*(b*x^3)/a^2 + x^6/(6*a) + (b^2*Log[b + a*x^3])/(3*a^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 269

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{a+\frac {b}{x^3}} \, dx=-\frac {b x^3}{3 a^2}+\frac {x^6}{6 a}+\frac {b^2 \log \left (b+a x^3\right )}{3 a^3} \]

[In]

Integrate[x^5/(a + b/x^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85

method	result	size
parallelrisch	\(\frac {a^{2} x^{6}-2 a b \,x^{3}+2 b^{2} \ln \left (a \,x^{3}+b \right )}{6 a^{3}}\)	\(34\)
norman	\(-\frac {b \,x^{3}}{3 a^{2}}+\frac {x^{6}}{6 a}+\frac {b^{2} \ln \left (a \,x^{3}+b \right )}{3 a^{3}}\)	\(35\)
default	\(\frac {\frac {1}{2} x^{6} a -b \,x^{3}}{3 a^{2}}+\frac {b^{2} \ln \left (a \,x^{3}+b \right )}{3 a^{3}}\)	\(36\)
risch	\(\frac {x^{6}}{6 a}-\frac {b \,x^{3}}{3 a^{2}}+\frac {b^{2}}{6 a^{3}}+\frac {b^{2} \ln \left (a \,x^{3}+b \right )}{3 a^{3}}\)	\(43\)

[In]

int(x^5/(a+b/x^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {x^5}{a+\frac {b}{x^3}} \, dx=\frac {a^{2} x^{6} - 2 \, a b x^{3} + 2 \, b^{2} \log \left (a x^{3} + b\right )}{6 \, a^{3}} \]

[In]

integrate(x^5/(a+b/x^3),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{a+\frac {b}{x^3}} \, dx=\frac {x^{6}}{6 a} - \frac {b x^{3}}{3 a^{2}} + \frac {b^{2} \log {\left (a x^{3} + b \right )}}{3 a^{3}} \]

[In]

integrate(x**5/(a+b/x**3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {x^5}{a+\frac {b}{x^3}} \, dx=\frac {b^{2} \log \left (a x^{3} + b\right )}{3 \, a^{3}} + \frac {a x^{6} - 2 \, b x^{3}}{6 \, a^{2}} \]

[In]

integrate(x^5/(a+b/x^3),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int \frac {x^5}{a+\frac {b}{x^3}} \, dx=\frac {b^{2} \log \left ({\left | a x^{3} + b \right |}\right )}{3 \, a^{3}} + \frac {a x^{6} - 2 \, b x^{3}}{6 \, a^{2}} \]

[In]

integrate(x^5/(a+b/x^3),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {x^5}{a+\frac {b}{x^3}} \, dx=\frac {2\,b^2\,\ln \left (a\,x^3+b\right )+a^2\,x^6-2\,a\,b\,x^3}{6\,a^3} \]

[In]

int(x^5/(a + b/x^3),x)

[Out]

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