3.78 \(\int \frac{x^{23/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=101 \[ -\frac{6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{16 \sqrt{x}}{35 b^4 \sqrt{a x+b x^3}}-\frac{x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[Out]

-x^(19/2)/(7*b*(a*x + b*x^3)^(7/2)) - (6*x^(13/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (8*x^(7/2))/(35*b^3*(a*x + b
*x^3)^(3/2)) - (16*Sqrt[x])/(35*b^4*Sqrt[a*x + b*x^3])

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Rubi [A]  time = 0.161451, 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} \[ -\frac{6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{16 \sqrt{x}}{35 b^4 \sqrt{a x+b x^3}}-\frac{x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(23/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(19/2)/(7*b*(a*x + b*x^3)^(7/2)) - (6*x^(13/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (8*x^(7/2))/(35*b^3*(a*x + b
*x^3)^(3/2)) - (16*Sqrt[x])/(35*b^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^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac{x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac{6 \int \frac{x^{17/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b}\\ &=-\frac{x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac{24 \int \frac{x^{11/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 b^2}\\ &=-\frac{x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}+\frac{16 \int \frac{x^{5/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{35 b^3}\\ &=-\frac{x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{16 \sqrt{x}}{35 b^4 \sqrt{a x+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0302392, size = 66, normalized size = 0.65 \[ -\frac{\sqrt{x} \left (56 a^2 b x^2+16 a^3+70 a b^2 x^4+35 b^3 x^6\right )}{35 b^4 \left (a+b x^2\right )^3 \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(23/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

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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 ( 35\,{x}^{6}{b}^{3}+70\,a{x}^{4}{b}^{2}+56\,{a}^{2}{x}^{2}b+16\,{a}^{3} \right ) }{35\,{b}^{4}}{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^(23/2)/(b*x^3+a*x)^(9/2),x)

[Out]

-1/35*(b*x^2+a)*(35*b^3*x^6+70*a*b^2*x^4+56*a^2*b*x^2+16*a^3)*x^(9/2)/b^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{23}{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^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(23/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.4131, size = 207, normalized size = 2.05 \begin{align*} -\frac{{\left (35 \, b^{3} x^{6} + 70 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} + 16 \, a^{3}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{35 \,{\left (b^{8} x^{9} + 4 \, a b^{7} x^{7} + 6 \, a^{2} b^{6} x^{5} + 4 \, a^{3} b^{5} x^{3} + a^{4} b^{4} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/35*(35*b^3*x^6 + 70*a*b^2*x^4 + 56*a^2*b*x^2 + 16*a^3)*sqrt(b*x^3 + a*x)*sqrt(x)/(b^8*x^9 + 4*a*b^7*x^7 + 6
*a^2*b^6*x^5 + 4*a^3*b^5*x^3 + a^4*b^4*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**(23/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27062, size = 86, normalized size = 0.85 \begin{align*} \frac{16}{35 \, \sqrt{a} b^{4}} - \frac{35 \,{\left (b x^{2} + a\right )}^{3} - 35 \,{\left (b x^{2} + a\right )}^{2} a + 21 \,{\left (b x^{2} + a\right )} a^{2} - 5 \, a^{3}}{35 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

16/35/(sqrt(a)*b^4) - 1/35*(35*(b*x^2 + a)^3 - 35*(b*x^2 + a)^2*a + 21*(b*x^2 + a)*a^2 - 5*a^3)/((b*x^2 + a)^(
7/2)*b^4)