3.80 \(\int \frac{x^{19/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac{4 x^{9/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{8 x^{3/2}}{105 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[Out]

-x^(15/2)/(7*b*(a*x + b*x^3)^(7/2)) - (4*x^(9/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (8*x^(3/2))/(105*b^3*(a*x + b
*x^3)^(3/2))

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Rubi [A]  time = 0.118776, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2015, 2014} \[ -\frac{4 x^{9/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{8 x^{3/2}}{105 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(19/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(15/2)/(7*b*(a*x + b*x^3)^(7/2)) - (4*x^(9/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (8*x^(3/2))/(105*b^3*(a*x + b
*x^3)^(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^{19/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac{x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac{4 \int \frac{x^{13/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b}\\ &=-\frac{x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{4 x^{9/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac{8 \int \frac{x^{7/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 b^2}\\ &=-\frac{x^{15/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{4 x^{9/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{8 x^{3/2}}{105 b^3 \left (a x+b x^3\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0256023, size = 55, normalized size = 0.72 \[ -\frac{\sqrt{x} \left (8 a^2+28 a b x^2+35 b^2 x^4\right )}{105 b^3 \left (a+b x^2\right )^3 \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(19/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-(Sqrt[x]*(8*a^2 + 28*a*b*x^2 + 35*b^2*x^4))/(105*b^3*(a + b*x^2)^3*Sqrt[x*(a + b*x^2)])

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Maple [A]  time = 0.005, size = 48, normalized size = 0.6 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( 35\,{x}^{4}{b}^{2}+28\,a{x}^{2}b+8\,{a}^{2} \right ) }{105\,{b}^{3}}{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^(19/2)/(b*x^3+a*x)^(9/2),x)

[Out]

-1/105*(b*x^2+a)*(35*b^2*x^4+28*a*b*x^2+8*a^2)*x^(9/2)/b^3/(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{19}{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^(19/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(19/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.42644, size = 184, normalized size = 2.42 \begin{align*} -\frac{{\left (35 \, b^{2} x^{4} + 28 \, a b x^{2} + 8 \, a^{2}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{105 \,{\left (b^{7} x^{9} + 4 \, a b^{6} x^{7} + 6 \, a^{2} b^{5} x^{5} + 4 \, a^{3} b^{4} x^{3} + a^{4} b^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/105*(35*b^2*x^4 + 28*a*b*x^2 + 8*a^2)*sqrt(b*x^3 + a*x)*sqrt(x)/(b^7*x^9 + 4*a*b^6*x^7 + 6*a^2*b^5*x^5 + 4*
a^3*b^4*x^3 + a^4*b^3*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**(19/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.24878, size = 68, normalized size = 0.89 \begin{align*} \frac{8}{105 \, a^{\frac{3}{2}} b^{3}} - \frac{35 \,{\left (b x^{2} + a\right )}^{2} - 42 \,{\left (b x^{2} + a\right )} a + 15 \, a^{2}}{105 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

8/105/(a^(3/2)*b^3) - 1/105*(35*(b*x^2 + a)^2 - 42*(b*x^2 + a)*a + 15*a^2)/((b*x^2 + a)^(7/2)*b^3)