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

Optimal. Leaf size=47 \[ -\frac {2 a^3}{\sqrt {x}}+2 a^2 b x^{3/2}+\frac {6}{7} a b^2 x^{7/2}+\frac {2}{11} b^3 x^{11/2} \]

[Out]

2*a^2*b*x^(3/2)+6/7*a*b^2*x^(7/2)+2/11*b^3*x^(11/2)-2*a^3/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} \[ 2 a^2 b x^{3/2}-\frac {2 a^3}{\sqrt {x}}+\frac {6}{7} a b^2 x^{7/2}+\frac {2}{11} b^3 x^{11/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^3/x^(3/2),x]

[Out]

(-2*a^3)/Sqrt[x] + 2*a^2*b*x^(3/2) + (6*a*b^2*x^(7/2))/7 + (2*b^3*x^(11/2))/11

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^{3/2}} \, dx &=\int \left (\frac {a^3}{x^{3/2}}+3 a^2 b \sqrt {x}+3 a b^2 x^{5/2}+b^3 x^{9/2}\right ) \, dx\\ &=-\frac {2 a^3}{\sqrt {x}}+2 a^2 b x^{3/2}+\frac {6}{7} a b^2 x^{7/2}+\frac {2}{11} b^3 x^{11/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.87 \[ \frac {2 \left (-77 a^3+77 a^2 b x^2+33 a b^2 x^4+7 b^3 x^6\right )}{77 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^3/x^(3/2),x]

[Out]

(2*(-77*a^3 + 77*a^2*b*x^2 + 33*a*b^2*x^4 + 7*b^3*x^6))/(77*Sqrt[x])

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fricas [A]  time = 0.87, size = 37, normalized size = 0.79 \[ \frac {2 \, {\left (7 \, b^{3} x^{6} + 33 \, a b^{2} x^{4} + 77 \, a^{2} b x^{2} - 77 \, a^{3}\right )}}{77 \, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77*(7*b^3*x^6 + 33*a*b^2*x^4 + 77*a^2*b*x^2 - 77*a^3)/sqrt(x)

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giac [A]  time = 0.58, size = 35, normalized size = 0.74 \[ \frac {2}{11} \, b^{3} x^{\frac {11}{2}} + \frac {6}{7} \, a b^{2} x^{\frac {7}{2}} + 2 \, a^{2} b x^{\frac {3}{2}} - \frac {2 \, a^{3}}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*b^3*x^(11/2) + 6/7*a*b^2*x^(7/2) + 2*a^2*b*x^(3/2) - 2*a^3/sqrt(x)

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maple [A]  time = 0.01, size = 38, normalized size = 0.81 \[ -\frac {2 \left (-7 b^{3} x^{6}-33 a \,b^{2} x^{4}-77 a^{2} b \,x^{2}+77 a^{3}\right )}{77 \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77*(-7*b^3*x^6-33*a*b^2*x^4-77*a^2*b*x^2+77*a^3)/x^(1/2)

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maxima [A]  time = 1.39, size = 35, normalized size = 0.74 \[ \frac {2}{11} \, b^{3} x^{\frac {11}{2}} + \frac {6}{7} \, a b^{2} x^{\frac {7}{2}} + 2 \, a^{2} b x^{\frac {3}{2}} - \frac {2 \, a^{3}}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*b^3*x^(11/2) + 6/7*a*b^2*x^(7/2) + 2*a^2*b*x^(3/2) - 2*a^3/sqrt(x)

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mupad [B]  time = 0.04, size = 35, normalized size = 0.74 \[ \frac {2\,b^3\,x^{11/2}}{11}-\frac {2\,a^3}{\sqrt {x}}+2\,a^2\,b\,x^{3/2}+\frac {6\,a\,b^2\,x^{7/2}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^3/x^(3/2),x)

[Out]

(2*b^3*x^(11/2))/11 - (2*a^3)/x^(1/2) + 2*a^2*b*x^(3/2) + (6*a*b^2*x^(7/2))/7

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sympy [A]  time = 2.31, size = 46, normalized size = 0.98 \[ - \frac {2 a^{3}}{\sqrt {x}} + 2 a^{2} b x^{\frac {3}{2}} + \frac {6 a b^{2} x^{\frac {7}{2}}}{7} + \frac {2 b^{3} x^{\frac {11}{2}}}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**3/sqrt(x) + 2*a**2*b*x**(3/2) + 6*a*b**2*x**(7/2)/7 + 2*b**3*x**(11/2)/11

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