3.4.29 \(\int \frac {x^8}{(a+b x^3)^2} \, dx\) [329]

Optimal. Leaf size=46 \[ \frac {x^3}{3 b^2}-\frac {a^2}{3 b^3 \left (a+b x^3\right )}-\frac {2 a \log \left (a+b x^3\right )}{3 b^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, 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 = {272, 45} \begin {gather*} -\frac {a^2}{3 b^3 \left (a+b x^3\right )}-\frac {2 a \log \left (a+b x^3\right )}{3 b^3}+\frac {x^3}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^3)^2,x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 38, normalized size = 0.83 \begin {gather*} \frac {b x^3-\frac {a^2}{a+b x^3}-2 a \log \left (a+b x^3\right )}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^3)^2,x]

[Out]

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

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Maple [A]
time = 0.13, size = 44, normalized size = 0.96

method	result	size
risch	\(\frac {x^{3}}{3 b^{2}}-\frac {a^{2}}{3 b^{3} \left (b \,x^{3}+a \right )}-\frac {2 a \ln \left (b \,x^{3}+a \right )}{3 b^{3}}\)	\(41\)
norman	\(\frac {\frac {x^{6}}{3 b}-\frac {2 a^{2}}{3 b^{3}}}{b \,x^{3}+a}-\frac {2 a \ln \left (b \,x^{3}+a \right )}{3 b^{3}}\)	\(43\)
default	\(\frac {x^{3}}{3 b^{2}}-\frac {a \left (\frac {a}{b \left (b \,x^{3}+a \right )}+\frac {2 \ln \left (b \,x^{3}+a \right )}{b}\right )}{3 b^{2}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 43, normalized size = 0.93 \begin {gather*} -\frac {a^{2}}{3 \, {\left (b^{4} x^{3} + a b^{3}\right )}} + \frac {x^{3}}{3 \, b^{2}} - \frac {2 \, a \log \left (b x^{3} + a\right )}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

-1/3*a^2/(b^4*x^3 + a*b^3) + 1/3*x^3/b^2 - 2/3*a*log(b*x^3 + a)/b^3

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Fricas [A]
time = 0.34, size = 56, normalized size = 1.22 \begin {gather*} \frac {b^{2} x^{6} + a b x^{3} - a^{2} - 2 \, {\left (a b x^{3} + a^{2}\right )} \log \left (b x^{3} + a\right )}{3 \, {\left (b^{4} x^{3} + a b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

1/3*(b^2*x^6 + a*b*x^3 - a^2 - 2*(a*b*x^3 + a^2)*log(b*x^3 + a))/(b^4*x^3 + a*b^3)

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Sympy [A]
time = 0.13, size = 42, normalized size = 0.91 \begin {gather*} - \frac {a^{2}}{3 a b^{3} + 3 b^{4} x^{3}} - \frac {2 a \log {\left (a + b x^{3} \right )}}{3 b^{3}} + \frac {x^{3}}{3 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**3+a)**2,x)

[Out]

-a**2/(3*a*b**3 + 3*b**4*x**3) - 2*a*log(a + b*x**3)/(3*b**3) + x**3/(3*b**2)

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Giac [A]
time = 1.41, size = 49, normalized size = 1.07 \begin {gather*} \frac {x^{3}}{3 \, b^{2}} - \frac {2 \, a \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b^{3}} + \frac {2 \, a b x^{3} + a^{2}}{3 \, {\left (b x^{3} + a\right )} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.05, size = 45, normalized size = 0.98 \begin {gather*} \frac {x^3}{3\,b^2}-\frac {a^2}{3\,\left (b^4\,x^3+a\,b^3\right )}-\frac {2\,a\,\ln \left (b\,x^3+a\right )}{3\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(a + b*x^3)^2,x)

[Out]

x^3/(3*b^2) - a^2/(3*(a*b^3 + b^4*x^3)) - (2*a*log(a + b*x^3))/(3*b^3)

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