3.5.39 \(\int \frac {(a+b x)^2}{\sqrt {x}} \, dx\)

Optimal. Leaf size=34 \[ 2 a^2 \sqrt {x}+\frac {4}{3} a b x^{3/2}+\frac {2}{5} b^2 x^{5/2} \]

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Rubi [A]  time = 0.01, antiderivative size = 34, 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*} 2 a^2 \sqrt {x}+\frac {4}{3} a b x^{3/2}+\frac {2}{5} b^2 x^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/Sqrt[x],x]

[Out]

2*a^2*Sqrt[x] + (4*a*b*x^(3/2))/3 + (2*b^2*x^(5/2))/5

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.82 \begin {gather*} \frac {2}{15} \sqrt {x} \left (15 a^2+10 a b x+3 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/Sqrt[x],x]

[Out]

(2*Sqrt[x]*(15*a^2 + 10*a*b*x + 3*b^2*x^2))/15

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IntegrateAlgebraic [A]  time = 0.01, size = 34, normalized size = 1.00 \begin {gather*} \frac {2}{15} \left (15 a^2 \sqrt {x}+10 a b x^{3/2}+3 b^2 x^{5/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(15*a^2*Sqrt[x] + 10*a*b*x^(3/2) + 3*b^2*x^(5/2)))/15

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fricas [A]  time = 0.84, size = 24, normalized size = 0.71 \begin {gather*} \frac {2}{15} \, {\left (3 \, b^{2} x^{2} + 10 \, a b x + 15 \, a^{2}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*b^2*x^2 + 10*a*b*x + 15*a^2)*sqrt(x)

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giac [A]  time = 0.92, size = 24, normalized size = 0.71 \begin {gather*} \frac {2}{5} \, b^{2} x^{\frac {5}{2}} + \frac {4}{3} \, a b x^{\frac {3}{2}} + 2 \, a^{2} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^(1/2),x, algorithm="giac")

[Out]

2/5*b^2*x^(5/2) + 4/3*a*b*x^(3/2) + 2*a^2*sqrt(x)

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maple [A]  time = 0.00, size = 25, normalized size = 0.74 \begin {gather*} \frac {2 \left (3 b^{2} x^{2}+10 a b x +15 a^{2}\right ) \sqrt {x}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/x^(1/2),x)

[Out]

2/15*x^(1/2)*(3*b^2*x^2+10*a*b*x+15*a^2)

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maxima [A]  time = 1.38, size = 24, normalized size = 0.71 \begin {gather*} \frac {2}{5} \, b^{2} x^{\frac {5}{2}} + \frac {4}{3} \, a b x^{\frac {3}{2}} + 2 \, a^{2} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^(1/2),x, algorithm="maxima")

[Out]

2/5*b^2*x^(5/2) + 4/3*a*b*x^(3/2) + 2*a^2*sqrt(x)

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mupad [B]  time = 0.04, size = 24, normalized size = 0.71 \begin {gather*} \frac {2\,\sqrt {x}\,\left (15\,a^2+10\,a\,b\,x+3\,b^2\,x^2\right )}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/x^(1/2),x)

[Out]

(2*x^(1/2)*(15*a^2 + 3*b^2*x^2 + 10*a*b*x))/15

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sympy [A]  time = 0.26, size = 32, normalized size = 0.94 \begin {gather*} 2 a^{2} \sqrt {x} + \frac {4 a b x^{\frac {3}{2}}}{3} + \frac {2 b^{2} x^{\frac {5}{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/x**(1/2),x)

[Out]

2*a**2*sqrt(x) + 4*a*b*x**(3/2)/3 + 2*b**2*x**(5/2)/5

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