3.5.41 \(\int \frac {(a+b x)^2}{x^{5/2}} \, dx\)

Optimal. Leaf size=32 \[ -\frac {2 a^2}{3 x^{3/2}}-\frac {4 a b}{\sqrt {x}}+2 b^2 \sqrt {x} \]

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Rubi [A]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {43} \begin {gather*} -\frac {2 a^2}{3 x^{3/2}}-\frac {4 a b}{\sqrt {x}}+2 b^2 \sqrt {x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/x^(5/2),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {(a+b x)^2}{x^{5/2}} \, dx &=\int \left (\frac {a^2}{x^{5/2}}+\frac {2 a b}{x^{3/2}}+\frac {b^2}{\sqrt {x}}\right ) \, dx\\ &=-\frac {2 a^2}{3 x^{3/2}}-\frac {4 a b}{\sqrt {x}}+2 b^2 \sqrt {x}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.81 \begin {gather*} -\frac {2 \left (a^2+6 a b x-3 b^2 x^2\right )}{3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/x^(5/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.02, size = 28, normalized size = 0.88 \begin {gather*} \frac {2 \left (-a^2-6 a b x+3 b^2 x^2\right )}{3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^2/x^(5/2),x]

[Out]

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

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fricas [A]  time = 0.81, size = 24, normalized size = 0.75 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} x^{2} - 6 \, a b x - a^{2}\right )}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*b^2*x^2 - 6*a*b*x - a^2)/x^(3/2)

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giac [A]  time = 1.00, size = 23, normalized size = 0.72 \begin {gather*} 2 \, b^{2} \sqrt {x} - \frac {2 \, {\left (6 \, a b x + a^{2}\right )}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 23, normalized size = 0.72 \begin {gather*} -\frac {2 \left (-3 b^{2} x^{2}+6 a b x +a^{2}\right )}{3 x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/x^(5/2),x)

[Out]

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

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maxima [A]  time = 1.29, size = 23, normalized size = 0.72 \begin {gather*} 2 \, b^{2} \sqrt {x} - \frac {2 \, {\left (6 \, a b x + a^{2}\right )}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 24, normalized size = 0.75 \begin {gather*} -\frac {2\,a^2+12\,a\,b\,x-6\,b^2\,x^2}{3\,x^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/x^(5/2),x)

[Out]

-(2*a^2 - 6*b^2*x^2 + 12*a*b*x)/(3*x^(3/2))

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sympy [A]  time = 0.59, size = 31, normalized size = 0.97 \begin {gather*} - \frac {2 a^{2}}{3 x^{\frac {3}{2}}} - \frac {4 a b}{\sqrt {x}} + 2 b^{2} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**2/(3*x**(3/2)) - 4*a*b/sqrt(x) + 2*b**2*sqrt(x)

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