3.82 \(\int \frac{x^{15/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=51 \[ -\frac{2 x^{5/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{x^{11/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[Out]

-x^(11/2)/(7*b*(a*x + b*x^3)^(7/2)) - (2*x^(5/2))/(35*b^2*(a*x + b*x^3)^(5/2))

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Rubi [A]  time = 0.0736555, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2015, 2014} \[ -\frac{2 x^{5/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{x^{11/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(15/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(11/2)/(7*b*(a*x + b*x^3)^(7/2)) - (2*x^(5/2))/(35*b^2*(a*x + b*x^3)^(5/2))

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^{15/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac{x^{11/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac{2 \int \frac{x^{9/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b}\\ &=-\frac{x^{11/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{2 x^{5/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0195845, size = 44, normalized size = 0.86 \[ -\frac{\sqrt{x} \left (2 a+7 b x^2\right )}{35 b^2 \left (a+b x^2\right )^3 \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(15/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-(Sqrt[x]*(2*a + 7*b*x^2))/(35*b^2*(a + b*x^2)^3*Sqrt[x*(a + b*x^2)])

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Maple [A]  time = 0.006, size = 37, normalized size = 0.7 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( 7\,b{x}^{2}+2\,a \right ) }{35\,{b}^{2}}{x}^{{\frac{9}{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^(15/2)/(b*x^3+a*x)^(9/2),x)

[Out]

-1/35*(b*x^2+a)*(7*b*x^2+2*a)*x^(9/2)/b^2/(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{15}{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^(15/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(15/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.59191, size = 158, normalized size = 3.1 \begin{align*} -\frac{\sqrt{b x^{3} + a x}{\left (7 \, b x^{2} + 2 \, a\right )} \sqrt{x}}{35 \,{\left (b^{6} x^{9} + 4 \, a b^{5} x^{7} + 6 \, a^{2} b^{4} x^{5} + 4 \, a^{3} b^{3} x^{3} + a^{4} b^{2} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/35*sqrt(b*x^3 + a*x)*(7*b*x^2 + 2*a)*sqrt(x)/(b^6*x^9 + 4*a*b^5*x^7 + 6*a^2*b^4*x^5 + 4*a^3*b^3*x^3 + a^4*b
^2*x)

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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**(15/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.26349, size = 45, normalized size = 0.88 \begin{align*} -\frac{7 \, b x^{2} + 2 \, a}{35 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} b^{2}} + \frac{2}{35 \, a^{\frac{5}{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

-1/35*(7*b*x^2 + 2*a)/((b*x^2 + a)^(7/2)*b^2) + 2/35/(a^(5/2)*b^2)