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

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

[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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Rubi [A]  time = 0.0121488, antiderivative size = 47, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0108319, size = 41, normalized size = 0.87 \[ \frac{2 \left (-105 a^2 b x^2-7 a^3+35 a b^2 x^4+5 b^3 x^6\right )}{35 x^{5/2}} \]

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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Maple [A]  time = 0.004, size = 38, normalized size = 0.8 \begin{align*} -{\frac{-10\,{b}^{3}{x}^{6}-70\,a{b}^{2}{x}^{4}+210\,{a}^{2}b{x}^{2}+14\,{a}^{3}}{35}{x}^{-{\frac{5}{2}}}} \end{align*}

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 = 2.45956, size = 49, normalized size = 1.04 \begin{align*} \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}}} \end{align*}

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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Fricas [A]  time = 1.22153, size = 88, normalized size = 1.87 \begin{align*} \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}}} \end{align*}

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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Sympy [A]  time = 3.84514, size = 46, normalized size = 0.98 \begin{align*} - \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} \end{align*}

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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Giac [A]  time = 1.61519, size = 49, normalized size = 1.04 \begin{align*} \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}}} \end{align*}

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)