3.1.87 \(\int \frac {x^{5/2}}{(a x+b x^3)^{9/2}} \, dx\) [87]

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

[Out]

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

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Rubi [A]
time = 0.13, 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 = {2040, 2039} \begin {gather*} -\frac {128 \sqrt {a x+b x^3}}{35 a^5 x^{3/2}}+\frac {64}{35 a^4 \sqrt {x} \sqrt {a x+b x^3}}+\frac {16 \sqrt {x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

Rule 2039

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] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2040

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^{5/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac {x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {8 \int \frac {x^{3/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac {x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {48 \int \frac {\sqrt {x}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 a^2}\\ &=\frac {x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {16 \sqrt {x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {64 \int \frac {1}{\sqrt {x} \left (a x+b x^3\right )^{3/2}} \, dx}{35 a^3}\\ &=\frac {x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {16 \sqrt {x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {64}{35 a^4 \sqrt {x} \sqrt {a x+b x^3}}+\frac {128 \int \frac {1}{x^{3/2} \sqrt {a x+b x^3}} \, dx}{35 a^4}\\ &=\frac {x^{5/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {8 x^{3/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {16 \sqrt {x}}{35 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {64}{35 a^4 \sqrt {x} \sqrt {a x+b x^3}}-\frac {128 \sqrt {a x+b x^3}}{35 a^5 x^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 68, normalized size = 0.54 \begin {gather*} \frac {x^{5/2} \left (-35 a^4-280 a^3 b x^2-560 a^2 b^2 x^4-448 a b^3 x^6-128 b^4 x^8\right )}{35 a^5 \left (x \left (a+b x^2\right )\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(5/2)*(-35*a^4 - 280*a^3*b*x^2 - 560*a^2*b^2*x^4 - 448*a*b^3*x^6 - 128*b^4*x^8))/(35*a^5*(x*(a + b*x^2))^(7
/2))

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Maple [A]
time = 0.46, size = 72, normalized size = 0.57

method	result	size
gosper	\(-\frac {x^{\frac {7}{2}} \left (b \,x^{2}+a \right ) \left (128 b^{4} x^{8}+448 a \,b^{3} x^{6}+560 a^{2} b^{2} x^{4}+280 a^{3} b \,x^{2}+35 a^{4}\right )}{35 a^{5} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\)	\(70\)
default	\(-\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (128 b^{4} x^{8}+448 a \,b^{3} x^{6}+560 a^{2} b^{2} x^{4}+280 a^{3} b \,x^{2}+35 a^{4}\right )}{35 x^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{4} a^{5}}\)	\(72\)
risch	\(-\frac {b \,x^{2}+a}{a^{5} \sqrt {x}\, \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {\left (b \,x^{2}+a \right ) x^{\frac {3}{2}} \left (93 b^{3} x^{6}+308 a \,b^{2} x^{4}+350 a^{2} b \,x^{2}+140 a^{3}\right ) b}{35 \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right ) a^{5} \sqrt {x \left (b \,x^{2}+a \right )}}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/35/x^(3/2)*(x*(b*x^2+a))^(1/2)*(128*b^4*x^8+448*a*b^3*x^6+560*a^2*b^2*x^4+280*a^3*b*x^2+35*a^4)/(b*x^2+a)^4
/a^5

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(5/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]
time = 1.57, size = 110, normalized size = 0.87 \begin {gather*} -\frac {{\left (128 \, b^{4} x^{8} + 448 \, a b^{3} x^{6} + 560 \, a^{2} b^{2} x^{4} + 280 \, a^{3} b x^{2} + 35 \, a^{4}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (a^{5} b^{4} x^{10} + 4 \, a^{6} b^{3} x^{8} + 6 \, a^{7} b^{2} x^{6} + 4 \, a^{8} b x^{4} + a^{9} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/35*(128*b^4*x^8 + 448*a*b^3*x^6 + 560*a^2*b^2*x^4 + 280*a^3*b*x^2 + 35*a^4)*sqrt(b*x^3 + a*x)*sqrt(x)/(a^5*
b^4*x^10 + 4*a^6*b^3*x^8 + 6*a^7*b^2*x^6 + 4*a^8*b*x^4 + a^9*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {5}{2}}}{\left (x \left (a + b x^{2}\right )\right )^{\frac {9}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Integral(x**(5/2)/(x*(a + b*x**2))**(9/2), x)

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Giac [A]
time = 0.68, size = 90, normalized size = 0.71 \begin {gather*} -\frac {{\left ({\left (x^{2} {\left (\frac {93 \, b^{4} x^{2}}{a^{5}} + \frac {308 \, b^{3}}{a^{4}}\right )} + \frac {350 \, b^{2}}{a^{3}}\right )} x^{2} + \frac {140 \, b}{a^{2}}\right )} x}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} + \frac {2 \, \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

-1/35*((x^2*(93*b^4*x^2/a^5 + 308*b^3/a^4) + 350*b^2/a^3)*x^2 + 140*b/a^2)*x/(b*x^2 + a)^(7/2) + 2*sqrt(b)/(((
sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)*a^4)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}}{{\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^(5/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(5/2)/(a*x + b*x^3)^(9/2), x)

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