3.13.86 \(\int \frac {x^9}{2 b+b x^5} \, dx\) [1286]

Optimal. Leaf size=24 \[ \frac {x^5}{5 b}-\frac {2 \log \left (2+x^5\right )}{5 b} \]

[Out]

1/5*x^5/b-2/5*ln(x^5+2)/b

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Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \begin {gather*} \frac {x^5}{5 b}-\frac {2 \log \left (x^5+2\right )}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(2*b + b*x^5),x]

[Out]

x^5/(5*b) - (2*Log[2 + x^5])/(5*b)

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^9}{2 b+b x^5} \, dx &=\frac {1}{5} \text {Subst}\left (\int \frac {x}{2 b+b x} \, dx,x,x^5\right )\\ &=\frac {1}{5} \text {Subst}\left (\int \left (\frac {1}{b}-\frac {2}{b (2+x)}\right ) \, dx,x,x^5\right )\\ &=\frac {x^5}{5 b}-\frac {2 \log \left (2+x^5\right )}{5 b}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 24, normalized size = 1.00 \begin {gather*} \frac {x^5}{5 b}-\frac {2 \log \left (2+x^5\right )}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(2*b + b*x^5),x]

[Out]

x^5/(5*b) - (2*Log[2 + x^5])/(5*b)

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Maple [A]
time = 0.18, size = 19, normalized size = 0.79

method	result	size
default	\(\frac {\frac {x^{5}}{5}-\frac {2 \ln \left (x^{5}+2\right )}{5}}{b}\)	\(19\)
norman	\(\frac {x^{5}}{5 b}-\frac {2 \ln \left (x^{5}+2\right )}{5 b}\)	\(21\)
risch	\(\frac {x^{5}}{5 b}-\frac {2 \ln \left (x^{5}+2\right )}{5 b}\)	\(21\)
meijerg	\(\frac {\frac {x^{5}}{5}-\frac {2 \ln \left (1+\frac {x^{5}}{2}\right )}{5}}{b}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^5+2*b),x,method=_RETURNVERBOSE)

[Out]

1/b*(1/5*x^5-2/5*ln(x^5+2))

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Maxima [A]
time = 0.29, size = 20, normalized size = 0.83 \begin {gather*} \frac {x^{5}}{5 \, b} - \frac {2 \, \log \left (x^{5} + 2\right )}{5 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+2*b),x, algorithm="maxima")

[Out]

1/5*x^5/b - 2/5*log(x^5 + 2)/b

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Fricas [A]
time = 0.36, size = 17, normalized size = 0.71 \begin {gather*} \frac {x^{5} - 2 \, \log \left (x^{5} + 2\right )}{5 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+2*b),x, algorithm="fricas")

[Out]

1/5*(x^5 - 2*log(x^5 + 2))/b

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Sympy [A]
time = 0.08, size = 17, normalized size = 0.71 \begin {gather*} \frac {x^{5}}{5 b} - \frac {2 \log {\left (x^{5} + 2 \right )}}{5 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**5+2*b),x)

[Out]

x**5/(5*b) - 2*log(x**5 + 2)/(5*b)

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Giac [A]
time = 1.69, size = 21, normalized size = 0.88 \begin {gather*} \frac {x^{5}}{5 \, b} - \frac {2 \, \log \left ({\left | x^{5} + 2 \right |}\right )}{5 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+2*b),x, algorithm="giac")

[Out]

1/5*x^5/b - 2/5*log(abs(x^5 + 2))/b

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Mupad [B]
time = 0.04, size = 19, normalized size = 0.79 \begin {gather*} -\frac {2\,\ln \left (x^5+2\right )-x^5}{5\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(2*b + b*x^5),x)

[Out]

-(2*log(x^5 + 2) - x^5)/(5*b)

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