3.839 \(\int \frac{(a+b x)^2}{(c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=58 \[ -\frac{a^2}{2 c x \sqrt{c x^2}}-\frac{2 a b}{c \sqrt{c x^2}}+\frac{b^2 x \log (x)}{c \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0122137, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {15, 43} \[ -\frac{a^2}{2 c x \sqrt{c x^2}}-\frac{2 a b}{c \sqrt{c x^2}}+\frac{b^2 x \log (x)}{c \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(c*x^2)^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.0041649, size = 34, normalized size = 0.59 \[ \frac{x \left (2 b^2 x^2 \log (x)-a (a+4 b x)\right )}{2 \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/(c*x^2)^(3/2),x]

[Out]

(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 = 32, normalized size = 0.6 \begin{align*}{\frac{x \left ( 2\,{b}^{2}\ln \left ( x \right ){x}^{2}-4\,abx-{a}^{2} \right ) }{2} \left ( c{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/(c*x^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.05825, size = 47, normalized size = 0.81 \begin{align*} \frac{b^{2} \log \left (x\right )}{c^{\frac{3}{2}}} - \frac{2 \, a b}{\sqrt{c x^{2}} c} - \frac{a^{2}}{2 \, c^{\frac{3}{2}} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(c*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58561, size = 84, normalized size = 1.45 \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^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(c*x^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(2*b^2*x^2*log(x) - 4*a*b*x - a^2)*sqrt(c*x^2)/(c^2*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}}{\left (c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/(c*x**2)**(3/2),x)

[Out]

Integral((a + b*x)**2/(c*x**2)**(3/2), 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/(c*x^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x