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

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

[Out]

-1/7*x^(23/2)/b/(b*x^3+a*x)^(7/2)-8/35*x^(17/2)/b^2/(b*x^3+a*x)^(5/2)-16/35*x^(11/2)/b^3/(b*x^3+a*x)^(3/2)-64/
35*x^(5/2)/b^4/(b*x^3+a*x)^(1/2)+128/35*(b*x^3+a*x)^(1/2)/b^5/x^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 126, normalized size of antiderivative = 1.00, 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 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^{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*}

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Mathematica [A]  time = 0.04, size = 77, normalized size = 0.61 \[ \frac {\sqrt {x} \left (128 a^4+448 a^3 b x^2+560 a^2 b^2 x^4+280 a b^3 x^6+35 b^4 x^8\right )}{35 b^5 \left (a+b x^2\right )^3 \sqrt {x \left (a+b x^2\right )}} \]

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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fricas [A]  time = 0.65, size = 108, normalized size = 0.86 \[ \frac {{\left (35 \, b^{4} x^{8} + 280 \, a b^{3} x^{6} + 560 \, a^{2} b^{2} x^{4} + 448 \, a^{3} b x^{2} + 128 \, a^{4}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (b^{9} x^{9} + 4 \, a b^{8} x^{7} + 6 \, a^{2} b^{7} x^{5} + 4 \, a^{3} b^{6} x^{3} + a^{4} b^{5} x\right )}} \]

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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giac [A]  time = 0.21, size = 80, normalized size = 0.63 \[ \frac {\sqrt {b x^{2} + a}}{b^{5}} - \frac {128 \, \sqrt {a}}{35 \, b^{5}} + \frac {140 \, {\left (b x^{2} + a\right )}^{3} a - 70 \, {\left (b x^{2} + a\right )}^{2} a^{2} + 28 \, {\left (b x^{2} + a\right )} a^{3} - 5 \, a^{4}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{5}} \]

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]

sqrt(b*x^2 + a)/b^5 - 128/35*sqrt(a)/b^5 + 1/35*(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)

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maple [A]  time = 0.04, size = 70, normalized size = 0.56 \[ \frac {\left (b \,x^{2}+a \right ) \left (35 x^{8} b^{4}+280 a \,x^{6} b^{3}+560 a^{2} x^{4} b^{2}+448 a^{3} x^{2} b +128 a^{4}\right ) x^{\frac {9}{2}}}{35 \left (b \,x^{3}+a x \right )^{\frac {9}{2}} b^{5}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {27}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{27/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(27/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(27/2)/(a*x + b*x^3)^(9/2), x)

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

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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