3.76 \(\int \frac{x^{27/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=126 \[ -\frac{8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{64 x^{5/2}}{35 b^4 \sqrt{a x+b x^3}}+\frac{128 \sqrt{a x+b x^3}}{35 b^5 \sqrt{x}}-\frac{x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[Out]

-x^(23/2)/(7*b*(a*x + b*x^3)^(7/2)) - (8*x^(17/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (16*x^(11/2))/(35*b^3*(a*x +
b*x^3)^(3/2)) - (64*x^(5/2))/(35*b^4*Sqrt[a*x + b*x^3]) + (128*Sqrt[a*x + b*x^3])/(35*b^5*Sqrt[x])

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Rubi [A]  time = 0.200279, 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} \[ -\frac{8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{64 x^{5/2}}{35 b^4 \sqrt{a x+b x^3}}+\frac{128 \sqrt{a x+b x^3}}{35 b^5 \sqrt{x}}-\frac{x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(27/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(23/2)/(7*b*(a*x + b*x^3)^(7/2)) - (8*x^(17/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (16*x^(11/2))/(35*b^3*(a*x +
b*x^3)^(3/2)) - (64*x^(5/2))/(35*b^4*Sqrt[a*x + b*x^3]) + (128*Sqrt[a*x + b*x^3])/(35*b^5*Sqrt[x])

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^{27/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac{x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac{8 \int \frac{x^{21/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b}\\ &=-\frac{x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac{48 \int \frac{x^{15/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 b^2}\\ &=-\frac{x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}+\frac{64 \int \frac{x^{9/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{35 b^3}\\ &=-\frac{x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{64 x^{5/2}}{35 b^4 \sqrt{a x+b x^3}}+\frac{128 \int \frac{x^{3/2}}{\sqrt{a x+b x^3}} \, dx}{35 b^4}\\ &=-\frac{x^{23/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{8 x^{17/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{16 x^{11/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{64 x^{5/2}}{35 b^4 \sqrt{a x+b x^3}}+\frac{128 \sqrt{a x+b x^3}}{35 b^5 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0366427, size = 77, normalized size = 0.61 \[ \frac{\sqrt{x} \left (560 a^2 b^2 x^4+448 a^3 b x^2+128 a^4+280 a b^3 x^6+35 b^4 x^8\right )}{35 b^5 \left (a+b x^2\right )^3 \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(27/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

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Maple [A]  time = 0.007, size = 70, normalized size = 0.6 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \left ( 35\,{x}^{8}{b}^{4}+280\,a{x}^{6}{b}^{3}+560\,{a}^{2}{x}^{4}{b}^{2}+448\,{a}^{3}{x}^{2}b+128\,{a}^{4} \right ) }{35\,{b}^{5}}{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^(27/2)/(b*x^3+a*x)^(9/2),x)

[Out]

1/35*(b*x^2+a)*(35*b^4*x^8+280*a*b^3*x^6+560*a^2*b^2*x^4+448*a^3*b*x^2+128*a^4)*x^(9/2)/b^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{27}{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^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(27/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.46616, size = 234, normalized size = 1.86 \begin{align*} \frac{{\left (35 \, b^{4} x^{8} + 280 \, a b^{3} x^{6} + 560 \, a^{2} b^{2} x^{4} + 448 \, a^{3} b x^{2} + 128 \, a^{4}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{35 \,{\left (b^{9} x^{9} + 4 \, a b^{8} x^{7} + 6 \, a^{2} b^{7} x^{5} + 4 \, a^{3} b^{6} x^{3} + a^{4} b^{5} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

1/35*(35*b^4*x^8 + 280*a*b^3*x^6 + 560*a^2*b^2*x^4 + 448*a^3*b*x^2 + 128*a^4)*sqrt(b*x^3 + a*x)*sqrt(x)/(b^9*x
^9 + 4*a*b^8*x^7 + 6*a^2*b^7*x^5 + 4*a^3*b^6*x^3 + a^4*b^5*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**(27/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.28052, size = 108, normalized size = 0.86 \begin{align*} \frac{35 \, \sqrt{b x^{2} + a} + \frac{140 \,{\left (b x^{2} + a\right )}^{3} a - 70 \,{\left (b x^{2} + a\right )}^{2} a^{2} + 28 \,{\left (b x^{2} + a\right )} a^{3} - 5 \, a^{4}}{{\left (b x^{2} + a\right )}^{\frac{7}{2}}}}{35 \, b^{5}} - \frac{128 \, \sqrt{a}}{35 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(27/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

1/35*(35*sqrt(b*x^2 + a) + (140*(b*x^2 + a)^3*a - 70*(b*x^2 + a)^2*a^2 + 28*(b*x^2 + a)*a^3 - 5*a^4)/(b*x^2 +
a)^(7/2))/b^5 - 128/35*sqrt(a)/b^5