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

Optimal. Leaf size=49 \[ -\frac{a^2}{2 x \sqrt{c x^2}}-\frac{2 a b}{\sqrt{c x^2}}+\frac{b^2 x \log (x)}{\sqrt{c x^2}} \]

[Out]

(-2*a*b)/Sqrt[c*x^2] - a^2/(2*x*Sqrt[c*x^2]) + (b^2*x*Log[x])/Sqrt[c*x^2]

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Rubi [A]  time = 0.0113112, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 43} \[ -\frac{a^2}{2 x \sqrt{c x^2}}-\frac{2 a b}{\sqrt{c x^2}}+\frac{b^2 x \log (x)}{\sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x^2*Sqrt[c*x^2]),x]

[Out]

(-2*a*b)/Sqrt[c*x^2] - a^2/(2*x*Sqrt[c*x^2]) + (b^2*x*Log[x])/Sqrt[c*x^2]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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

Mathematica [A]  time = 0.0080317, size = 35, normalized size = 0.71 \[ \frac{c x \left (2 b^2 x^2 \log (x)-a (a+4 b x)\right )}{2 \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 34, normalized size = 0.7 \begin{align*}{\frac{2\,{b}^{2}\ln \left ( x \right ){x}^{2}-4\,abx-{a}^{2}}{2\,x}{\frac{1}{\sqrt{c{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.08382, size = 42, normalized size = 0.86 \begin{align*} \frac{b^{2} \log \left (x\right )}{\sqrt{c}} - \frac{2 \, a b}{\sqrt{c} x} - \frac{a^{2}}{2 \, \sqrt{c} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*log(x)/sqrt(c) - 2*a*b/(sqrt(c)*x) - 1/2*a^2/(sqrt(c)*x^2)

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Fricas [A]  time = 1.46495, size = 81, normalized size = 1.65 \begin{align*} \frac{{\left (2 \, b^{2} x^{2} \log \left (x\right ) - 4 \, a b x - a^{2}\right )} \sqrt{c x^{2}}}{2 \, c x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*x^2*log(x) - 4*a*b*x - a^2)*sqrt(c*x^2)/(c*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{2}}{x^{2} \sqrt{c x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**2/(x**2*sqrt(c*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x