3.87 \(\int \frac{x^{5/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=126 \[ \frac{8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{16 \sqrt{x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{64}{35 a^4 \sqrt{x} \sqrt{a x+b x^3}}-\frac{128 \sqrt{a x+b x^3}}{35 a^5 x^{3/2}}+\frac{x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

[Out]

x^(5/2)/(7*a*(a*x + b*x^3)^(7/2)) + (8*x^(3/2))/(35*a^2*(a*x + b*x^3)^(5/2)) + (16*Sqrt[x])/(35*a^3*(a*x + b*x
^3)^(3/2)) + 64/(35*a^4*Sqrt[x]*Sqrt[a*x + b*x^3]) - (128*Sqrt[a*x + b*x^3])/(35*a^5*x^(3/2))

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Rubi [A]  time = 0.192445, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2015, 2014} \[ \frac{8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{16 \sqrt{x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{64}{35 a^4 \sqrt{x} \sqrt{a x+b x^3}}-\frac{128 \sqrt{a x+b x^3}}{35 a^5 x^{3/2}}+\frac{x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(5/2)/(7*a*(a*x + b*x^3)^(7/2)) + (8*x^(3/2))/(35*a^2*(a*x + b*x^3)^(5/2)) + (16*Sqrt[x])/(35*a^3*(a*x + b*x
^3)^(3/2)) + 64/(35*a^4*Sqrt[x]*Sqrt[a*x + b*x^3]) - (128*Sqrt[a*x + b*x^3])/(35*a^5*x^(3/2))

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^{5/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac{x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{8 \int \frac{x^{3/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac{x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{48 \int \frac{\sqrt{x}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 a^2}\\ &=\frac{x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{16 \sqrt{x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{64 \int \frac{1}{\sqrt{x} \left (a x+b x^3\right )^{3/2}} \, dx}{35 a^3}\\ &=\frac{x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{16 \sqrt{x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{64}{35 a^4 \sqrt{x} \sqrt{a x+b x^3}}+\frac{128 \int \frac{1}{x^{3/2} \sqrt{a x+b x^3}} \, dx}{35 a^4}\\ &=\frac{x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{16 \sqrt{x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{64}{35 a^4 \sqrt{x} \sqrt{a x+b x^3}}-\frac{128 \sqrt{a x+b x^3}}{35 a^5 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0297992, size = 77, normalized size = 0.61 \[ -\frac{\sqrt{x \left (a+b x^2\right )} \left (560 a^2 b^2 x^4+280 a^3 b x^2+35 a^4+448 a b^3 x^6+128 b^4 x^8\right )}{35 a^5 x^{3/2} \left (a+b x^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-(Sqrt[x*(a + b*x^2)]*(35*a^4 + 280*a^3*b*x^2 + 560*a^2*b^2*x^4 + 448*a*b^3*x^6 + 128*b^4*x^8))/(35*a^5*x^(3/2
)*(a + b*x^2)^4)

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Maple [A]  time = 0.004, size = 70, normalized size = 0.6 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( 128\,{b}^{4}{x}^{8}+448\,{b}^{3}{x}^{6}a+560\,{b}^{2}{x}^{4}{a}^{2}+280\,b{x}^{2}{a}^{3}+35\,{a}^{4} \right ) }{35\,{a}^{5}}{x}^{{\frac{7}{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^(5/2)/(b*x^3+a*x)^(9/2),x)

[Out]

-1/35*x^(7/2)*(b*x^2+a)*(128*b^4*x^8+448*a*b^3*x^6+560*a^2*b^2*x^4+280*a^3*b*x^2+35*a^4)/a^5/(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{5}{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^(5/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(5/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.67136, size = 239, normalized size = 1.9 \begin{align*} -\frac{{\left (128 \, b^{4} x^{8} + 448 \, a b^{3} x^{6} + 560 \, a^{2} b^{2} x^{4} + 280 \, a^{3} b x^{2} + 35 \, a^{4}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{35 \,{\left (a^{5} b^{4} x^{10} + 4 \, a^{6} b^{3} x^{8} + 6 \, a^{7} b^{2} x^{6} + 4 \, a^{8} b x^{4} + a^{9} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/35*(128*b^4*x^8 + 448*a*b^3*x^6 + 560*a^2*b^2*x^4 + 280*a^3*b*x^2 + 35*a^4)*sqrt(b*x^3 + a*x)*sqrt(x)/(a^5*
b^4*x^10 + 4*a^6*b^3*x^8 + 6*a^7*b^2*x^6 + 4*a^8*b*x^4 + a^9*x^2)

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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**(5/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.44085, size = 96, normalized size = 0.76 \begin{align*} -\frac{{\left ({\left (x^{2}{\left (\frac{93 \, b^{4} x^{2}}{a^{5}} + \frac{308 \, b^{3}}{a^{4}}\right )} + \frac{350 \, b^{2}}{a^{3}}\right )} x^{2} + \frac{140 \, b}{a^{2}}\right )} x}{35 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} - \frac{\sqrt{b + \frac{a}{x^{2}}}}{a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

-1/35*((x^2*(93*b^4*x^2/a^5 + 308*b^3/a^4) + 350*b^2/a^3)*x^2 + 140*b/a^2)*x/(b*x^2 + a)^(7/2) - sqrt(b + a/x^
2)/a^5