3.835 \(\int \frac{(a+b x)^2}{x^4 \sqrt{c x^2}} \, dx\)

Optimal. Leaf size=57 \[ -\frac{a^2}{4 x^3 \sqrt{c x^2}}-\frac{2 a b}{3 x^2 \sqrt{c x^2}}-\frac{b^2}{2 x \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0128204, antiderivative size = 57, 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} \[ -\frac{a^2}{4 x^3 \sqrt{c x^2}}-\frac{2 a b}{3 x^2 \sqrt{c x^2}}-\frac{b^2}{2 x \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x^4*Sqrt[c*x^2]),x]

[Out]

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

Mathematica [A]  time = 0.0080514, size = 35, normalized size = 0.61 \[ \frac{-3 a^2-8 a b x-6 b^2 x^2}{12 x^3 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/(x^4*Sqrt[c*x^2]),x]

[Out]

(-3*a^2 - 8*a*b*x - 6*b^2*x^2)/(12*x^3*Sqrt[c*x^2])

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Maple [A]  time = 0.003, size = 32, normalized size = 0.6 \begin{align*} -{\frac{6\,{b}^{2}{x}^{2}+8\,abx+3\,{a}^{2}}{12\,{x}^{3}}{\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^4/(c*x^2)^(1/2),x)

[Out]

-1/12*(6*b^2*x^2+8*a*b*x+3*a^2)/x^3/(c*x^2)^(1/2)

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Maxima [A]  time = 1.08714, size = 45, normalized size = 0.79 \begin{align*} -\frac{b^{2}}{2 \, \sqrt{c} x^{2}} - \frac{2 \, a b}{3 \, \sqrt{c} x^{3}} - \frac{a^{2}}{4 \, \sqrt{c} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.60211, size = 77, normalized size = 1.35 \begin{align*} -\frac{{\left (6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}\right )} \sqrt{c x^{2}}}{12 \, c x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/x^4/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

-1/12*(6*b^2*x^2 + 8*a*b*x + 3*a^2)*sqrt(c*x^2)/(c*x^5)

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Sympy [A]  time = 0.75472, size = 61, normalized size = 1.07 \begin{align*} - \frac{a^{2}}{4 \sqrt{c} x^{3} \sqrt{x^{2}}} - \frac{2 a b}{3 \sqrt{c} x^{2} \sqrt{x^{2}}} - \frac{b^{2}}{2 \sqrt{c} x \sqrt{x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/x**4/(c*x**2)**(1/2),x)

[Out]

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

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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^4/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x