3.85 \(\int \frac{x^{9/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=101 \[ \frac{6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{8 x^{5/2}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{16 x^{3/2}}{35 a^4 \sqrt{a x+b x^3}}+\frac{x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

[Out]

x^(9/2)/(7*a*(a*x + b*x^3)^(7/2)) + (6*x^(7/2))/(35*a^2*(a*x + b*x^3)^(5/2)) + (8*x^(5/2))/(35*a^3*(a*x + b*x^
3)^(3/2)) + (16*x^(3/2))/(35*a^4*Sqrt[a*x + b*x^3])

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Rubi [A]  time = 0.157272, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2015, 2014} \[ \frac{6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{8 x^{5/2}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{16 x^{3/2}}{35 a^4 \sqrt{a x+b x^3}}+\frac{x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(9/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(9/2)/(7*a*(a*x + b*x^3)^(7/2)) + (6*x^(7/2))/(35*a^2*(a*x + b*x^3)^(5/2)) + (8*x^(5/2))/(35*a^3*(a*x + b*x^
3)^(3/2)) + (16*x^(3/2))/(35*a^4*Sqrt[a*x + b*x^3])

Rule 2015

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n, j
] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{x^{9/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac{x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{6 \int \frac{x^{7/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac{x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{24 \int \frac{x^{5/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 a^2}\\ &=\frac{x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{8 x^{5/2}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{16 \int \frac{x^{3/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{35 a^3}\\ &=\frac{x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{8 x^{5/2}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{16 x^{3/2}}{35 a^4 \sqrt{a x+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0231574, size = 66, normalized size = 0.65 \[ \frac{\sqrt{x} \sqrt{x \left (a+b x^2\right )} \left (70 a^2 b x^2+35 a^3+56 a b^2 x^4+16 b^3 x^6\right )}{35 a^4 \left (a+b x^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(9/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(Sqrt[x]*Sqrt[x*(a + b*x^2)]*(35*a^3 + 70*a^2*b*x^2 + 56*a*b^2*x^4 + 16*b^3*x^6))/(35*a^4*(a + b*x^2)^4)

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Maple [A]  time = 0.005, size = 59, normalized size = 0.6 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \left ( 16\,{b}^{3}{x}^{6}+56\,{b}^{2}{x}^{4}a+70\,b{x}^{2}{a}^{2}+35\,{a}^{3} \right ) }{35\,{a}^{4}}{x}^{{\frac{11}{2}}} \left ( b{x}^{3}+ax \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)/(b*x^3+a*x)^(9/2),x)

[Out]

1/35*(b*x^2+a)*x^(11/2)*(16*b^3*x^6+56*a*b^2*x^4+70*a^2*b*x^2+35*a^3)/a^4/(b*x^3+a*x)^(9/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{9}{2}}}{{\left (b x^{3} + a x\right )}^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(9/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.50133, size = 203, normalized size = 2.01 \begin{align*} \frac{{\left (16 \, b^{3} x^{6} + 56 \, a b^{2} x^{4} + 70 \, a^{2} b x^{2} + 35 \, a^{3}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{35 \,{\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

1/35*(16*b^3*x^6 + 56*a*b^2*x^4 + 70*a^2*b*x^2 + 35*a^3)*sqrt(b*x^3 + a*x)*sqrt(x)/(a^4*b^4*x^8 + 4*a^5*b^3*x^
6 + 6*a^6*b^2*x^4 + 4*a^7*b*x^2 + a^8)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(9/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.31053, size = 74, normalized size = 0.73 \begin{align*} \frac{{\left (2 \,{\left (4 \, x^{2}{\left (\frac{2 \, b^{3} x^{2}}{a^{4}} + \frac{7 \, b^{2}}{a^{3}}\right )} + \frac{35 \, b}{a^{2}}\right )} x^{2} + \frac{35}{a}\right )} x}{35 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

1/35*(2*(4*x^2*(2*b^3*x^2/a^4 + 7*b^2/a^3) + 35*b/a^2)*x^2 + 35/a)*x/(b*x^2 + a)^(7/2)