3.13.91 \(\int \frac {x^9}{3+b x^5} \, dx\) [1291]

Optimal. Leaf size=26 \[ \frac {x^5}{5 b}-\frac {3 \log \left (3+b x^5\right )}{5 b^2} \]

[Out]

1/5*x^5/b-3/5*ln(b*x^5+3)/b^2

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Rubi [A]
time = 0.01, antiderivative size = 26, 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 {x^5}{5 b}-\frac {3 \log \left (b x^5+3\right )}{5 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(3 + b*x^5),x]

[Out]

x^5/(5*b) - (3*Log[3 + b*x^5])/(5*b^2)

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}{3+b x^5} \, dx &=\frac {1}{5} \text {Subst}\left (\int \frac {x}{3+b x} \, dx,x,x^5\right )\\ &=\frac {1}{5} \text {Subst}\left (\int \left (\frac {1}{b}-\frac {3}{b (3+b x)}\right ) \, dx,x,x^5\right )\\ &=\frac {x^5}{5 b}-\frac {3 \log \left (3+b x^5\right )}{5 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 26, normalized size = 1.00 \begin {gather*} \frac {x^5}{5 b}-\frac {3 \log \left (3+b x^5\right )}{5 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(3 + b*x^5),x]

[Out]

x^5/(5*b) - (3*Log[3 + b*x^5])/(5*b^2)

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Maple [A]
time = 0.16, size = 23, normalized size = 0.88

method	result	size
default	\(\frac {x^{5}}{5 b}-\frac {3 \ln \left (b \,x^{5}+3\right )}{5 b^{2}}\)	\(23\)
norman	\(\frac {x^{5}}{5 b}-\frac {3 \ln \left (b \,x^{5}+3\right )}{5 b^{2}}\)	\(23\)
risch	\(\frac {x^{5}}{5 b}-\frac {3 \ln \left (b \,x^{5}+3\right )}{5 b^{2}}\)	\(23\)
meijerg	\(\frac {\frac {b \,x^{5}}{5}-\frac {3 \ln \left (1+\frac {b \,x^{5}}{3}\right )}{5}}{b^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^5+3),x,method=_RETURNVERBOSE)

[Out]

1/5*x^5/b-3/5*ln(b*x^5+3)/b^2

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Maxima [A]
time = 0.29, size = 22, normalized size = 0.85 \begin {gather*} \frac {x^{5}}{5 \, b} - \frac {3 \, \log \left (b x^{5} + 3\right )}{5 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+3),x, algorithm="maxima")

[Out]

1/5*x^5/b - 3/5*log(b*x^5 + 3)/b^2

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Fricas [A]
time = 0.36, size = 21, normalized size = 0.81 \begin {gather*} \frac {b x^{5} - 3 \, \log \left (b x^{5} + 3\right )}{5 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+3),x, algorithm="fricas")

[Out]

1/5*(b*x^5 - 3*log(b*x^5 + 3))/b^2

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Sympy [A]
time = 0.08, size = 20, normalized size = 0.77 \begin {gather*} \frac {x^{5}}{5 b} - \frac {3 \log {\left (b x^{5} + 3 \right )}}{5 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**5+3),x)

[Out]

x**5/(5*b) - 3*log(b*x**5 + 3)/(5*b**2)

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Giac [A]
time = 2.09, size = 23, normalized size = 0.88 \begin {gather*} \frac {x^{5}}{5 \, b} - \frac {3 \, \log \left ({\left | b x^{5} + 3 \right |}\right )}{5 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^5+3),x, algorithm="giac")

[Out]

1/5*x^5/b - 3/5*log(abs(b*x^5 + 3))/b^2

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Mupad [B]
time = 1.02, size = 22, normalized size = 0.85 \begin {gather*} -\frac {3\,\ln \left (b\,x^5+3\right )-b\,x^5}{5\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^5 + 3),x)

[Out]

-(3*log(b*x^5 + 3) - b*x^5)/(5*b^2)

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