3.2125 \(\int \frac{(a+b \sqrt{x})^2}{x} \, dx\)

Optimal. Leaf size=21 \[ a^2 \log (x)+4 a b \sqrt{x}+b^2 x \]

[Out]

4*a*b*Sqrt[x] + b^2*x + a^2*Log[x]

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Rubi [A]  time = 0.0119502, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ a^2 \log (x)+4 a b \sqrt{x}+b^2 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^2/x,x]

[Out]

4*a*b*Sqrt[x] + b^2*x + a^2*Log[x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A]  time = 0.0079906, size = 21, normalized size = 1. \[ a^2 \log (x)+4 a b \sqrt{x}+b^2 x \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^2/x,x]

[Out]

4*a*b*Sqrt[x] + b^2*x + a^2*Log[x]

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Maple [A]  time = 0.003, size = 20, normalized size = 1. \begin{align*}{b}^{2}x+{a}^{2}\ln \left ( x \right ) +4\,ab\sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x+a^2*ln(x)+4*a*b*x^(1/2)

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Maxima [A]  time = 0.950582, size = 26, normalized size = 1.24 \begin{align*} b^{2} x + a^{2} \log \left (x\right ) + 4 \, a b \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x + a^2*log(x) + 4*a*b*sqrt(x)

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Fricas [A]  time = 1.51504, size = 59, normalized size = 2.81 \begin{align*} b^{2} x + 2 \, a^{2} \log \left (\sqrt{x}\right ) + 4 \, a b \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x + 2*a^2*log(sqrt(x)) + 4*a*b*sqrt(x)

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Sympy [A]  time = 0.17154, size = 20, normalized size = 0.95 \begin{align*} a^{2} \log{\left (x \right )} + 4 a b \sqrt{x} + b^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*log(x) + 4*a*b*sqrt(x) + b**2*x

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Giac [A]  time = 1.10201, size = 27, normalized size = 1.29 \begin{align*} b^{2} x + a^{2} \log \left ({\left | x \right |}\right ) + 4 \, a b \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x + a^2*log(abs(x)) + 4*a*b*sqrt(x)