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

Optimal. Leaf size=101 \[ \frac {16 x^{3/2}}{35 a^4 \sqrt {a x+b x^3}}+\frac {8 x^{5/2}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 101, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {8 x^{5/2}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {16 x^{3/2}}{35 a^4 \sqrt {a x+b x^3}}+\frac {x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(9/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(9/2)/(7*a*(a*x + b*x^3)^(7/2)) + (6*x^(7/2))/(35*a^2*(a*x + b*x^3)^(5/2)) + (8*x^(5/2))/(35*a^3*(a*x + b*x^
3)^(3/2)) + (16*x^(3/2))/(35*a^4*Sqrt[a*x + b*x^3])

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])

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])

Rubi steps

\begin {align*} \int \frac {x^{9/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac {x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {6 \int \frac {x^{7/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac {x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {24 \int \frac {x^{5/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 a^2}\\ &=\frac {x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {8 x^{5/2}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {16 \int \frac {x^{3/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{35 a^3}\\ &=\frac {x^{9/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {6 x^{7/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {8 x^{5/2}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {16 x^{3/2}}{35 a^4 \sqrt {a x+b x^3}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.65 \begin {gather*} \frac {\sqrt {x} \sqrt {x \left (a+b x^2\right )} \left (35 a^3+70 a^2 b x^2+56 a b^2 x^4+16 b^3 x^6\right )}{35 a^4 \left (a+b x^2\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.89, size = 57, normalized size = 0.56 \begin {gather*} \frac {x^{9/2} \left (35 a^3+70 a^2 b x^2+56 a b^2 x^4+16 b^3 x^6\right )}{35 a^4 \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(9/2)*(35*a^3 + 70*a^2*b*x^2 + 56*a*b^2*x^4 + 16*b^3*x^6))/(35*a^4*(a*x + b*x^3)^(7/2))

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fricas [A]  time = 0.40, size = 95, normalized size = 0.94 \begin {gather*} \frac {{\left (16 \, b^{3} x^{6} + 56 \, a b^{2} x^{4} + 70 \, a^{2} b x^{2} + 35 \, a^{3}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 55, normalized size = 0.54 \begin {gather*} \frac {{\left (2 \, {\left (4 \, x^{2} {\left (\frac {2 \, b^{3} x^{2}}{a^{4}} + \frac {7 \, b^{2}}{a^{3}}\right )} + \frac {35 \, b}{a^{2}}\right )} x^{2} + \frac {35}{a}\right )} x}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(2*(4*x^2*(2*b^3*x^2/a^4 + 7*b^2/a^3) + 35*b/a^2)*x^2 + 35/a)*x/(b*x^2 + a)^(7/2)

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maple [A]  time = 0.04, size = 59, normalized size = 0.58 \begin {gather*} \frac {\left (b \,x^{2}+a \right ) \left (16 b^{3} x^{6}+56 a \,b^{2} x^{4}+70 a^{2} b \,x^{2}+35 a^{3}\right ) x^{\frac {11}{2}}}{35 \left (b \,x^{3}+a x \right )^{\frac {9}{2}} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(b*x^2+a)*x^(11/2)*(16*b^3*x^6+56*a*b^2*x^4+70*a^2*b*x^2+35*a^3)/a^4/(b*x^3+a*x)^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {9}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{9/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(9/2)/(a*x + b*x^3)^(9/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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