3.287 \(\int \frac {(a+b x^2)^3}{x^{7/2}} \, dx\)

Optimal. Leaf size=47 \[ -\frac {2 a^3}{5 x^{5/2}}-\frac {6 a^2 b}{\sqrt {x}}+2 a b^2 x^{3/2}+\frac {2}{7} b^3 x^{7/2} \]

[Out]

-2/5*a^3/x^(5/2)+2*a*b^2*x^(3/2)+2/7*b^3*x^(7/2)-6*a^2*b/x^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {270} \[ -\frac {6 a^2 b}{\sqrt {x}}-\frac {2 a^3}{5 x^{5/2}}+2 a b^2 x^{3/2}+\frac {2}{7} b^3 x^{7/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^3/x^(7/2),x]

[Out]

(-2*a^3)/(5*x^(5/2)) - (6*a^2*b)/Sqrt[x] + 2*a*b^2*x^(3/2) + (2*b^3*x^(7/2))/7

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.87 \[ \frac {2 \left (-7 a^3-105 a^2 b x^2+35 a b^2 x^4+5 b^3 x^6\right )}{35 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^3/x^(7/2),x]

[Out]

(2*(-7*a^3 - 105*a^2*b*x^2 + 35*a*b^2*x^4 + 5*b^3*x^6))/(35*x^(5/2))

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fricas [A]  time = 0.75, size = 37, normalized size = 0.79 \[ \frac {2 \, {\left (5 \, b^{3} x^{6} + 35 \, a b^{2} x^{4} - 105 \, a^{2} b x^{2} - 7 \, a^{3}\right )}}{35 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^(7/2),x, algorithm="fricas")

[Out]

2/35*(5*b^3*x^6 + 35*a*b^2*x^4 - 105*a^2*b*x^2 - 7*a^3)/x^(5/2)

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giac [A]  time = 0.61, size = 36, normalized size = 0.77 \[ \frac {2}{7} \, b^{3} x^{\frac {7}{2}} + 2 \, a b^{2} x^{\frac {3}{2}} - \frac {2 \, {\left (15 \, a^{2} b x^{2} + a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^(7/2),x, algorithm="giac")

[Out]

2/7*b^3*x^(7/2) + 2*a*b^2*x^(3/2) - 2/5*(15*a^2*b*x^2 + a^3)/x^(5/2)

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maple [A]  time = 0.00, size = 38, normalized size = 0.81 \[ -\frac {2 \left (-5 b^{3} x^{6}-35 a \,b^{2} x^{4}+105 a^{2} b \,x^{2}+7 a^{3}\right )}{35 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^3/x^(7/2),x)

[Out]

-2/35*(-5*b^3*x^6-35*a*b^2*x^4+105*a^2*b*x^2+7*a^3)/x^(5/2)

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maxima [A]  time = 1.35, size = 36, normalized size = 0.77 \[ \frac {2}{7} \, b^{3} x^{\frac {7}{2}} + 2 \, a b^{2} x^{\frac {3}{2}} - \frac {2 \, {\left (15 \, a^{2} b x^{2} + a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/x^(7/2),x, algorithm="maxima")

[Out]

2/7*b^3*x^(7/2) + 2*a*b^2*x^(3/2) - 2/5*(15*a^2*b*x^2 + a^3)/x^(5/2)

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mupad [B]  time = 0.04, size = 37, normalized size = 0.79 \[ -\frac {14\,a^3+210\,a^2\,b\,x^2-70\,a\,b^2\,x^4-10\,b^3\,x^6}{35\,x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^3/x^(7/2),x)

[Out]

-(14*a^3 - 10*b^3*x^6 + 210*a^2*b*x^2 - 70*a*b^2*x^4)/(35*x^(5/2))

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sympy [A]  time = 3.86, size = 46, normalized size = 0.98 \[ - \frac {2 a^{3}}{5 x^{\frac {5}{2}}} - \frac {6 a^{2} b}{\sqrt {x}} + 2 a b^{2} x^{\frac {3}{2}} + \frac {2 b^{3} x^{\frac {7}{2}}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**3/x**(7/2),x)

[Out]

-2*a**3/(5*x**(5/2)) - 6*a**2*b/sqrt(x) + 2*a*b**2*x**(3/2) + 2*b**3*x**(7/2)/7

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