3.81 \(\int \frac{x^{17/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=51 \[ \frac{2 x^{15/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

[Out]

x^(17/2)/(7*a*(a*x + b*x^3)^(7/2)) + (2*x^(15/2))/(35*a^2*(a*x + b*x^3)^(5/2))

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Rubi [A]  time = 0.0747788, 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} \[ \frac{2 x^{15/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(17/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(17/2)/(7*a*(a*x + b*x^3)^(7/2)) + (2*x^(15/2))/(35*a^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^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac{x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{2 \int \frac{x^{15/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac{x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{2 x^{15/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0210368, size = 44, normalized size = 0.86 \[ \frac{x^{9/2} \sqrt{x \left (a+b x^2\right )} \left (7 a+2 b x^2\right )}{35 a^2 \left (a+b x^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(17/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(9/2)*Sqrt[x*(a + b*x^2)]*(7*a + 2*b*x^2))/(35*a^2*(a + b*x^2)^4)

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Maple [A]  time = 0.004, size = 37, normalized size = 0.7 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \left ( 2\,b{x}^{2}+7\,a \right ) }{35\,{a}^{2}}{x}^{{\frac{19}{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^(17/2)/(b*x^3+a*x)^(9/2),x)

[Out]

1/35*(b*x^2+a)*x^(19/2)*(2*b*x^2+7*a)/a^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{17}{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^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(17/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.43253, size = 159, normalized size = 3.12 \begin{align*} \frac{{\left (2 \, b x^{6} + 7 \, a x^{4}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{35 \,{\left (a^{2} b^{4} x^{8} + 4 \, a^{3} b^{3} x^{6} + 6 \, a^{4} b^{2} x^{4} + 4 \, a^{5} b x^{2} + a^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

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

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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**(17/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33704, size = 39, normalized size = 0.76 \begin{align*} \frac{x^{5}{\left (\frac{2 \, b x^{2}}{a^{2}} + \frac{7}{a}\right )}}{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^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

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