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

Optimal. Leaf size=29 \[ -\frac{(a+b x)^3}{3 a c x^2 \sqrt{c x^2}} \]

[Out]

-(a + b*x)^3/(3*a*c*x^2*Sqrt[c*x^2])

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Rubi [A]  time = 0.0043833, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 37} \[ -\frac{(a+b x)^3}{3 a c x^2 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x*(c*x^2)^(3/2)),x]

[Out]

-(a + b*x)^3/(3*a*c*x^2*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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0117217, size = 36, normalized size = 1.24 \[ \frac{c x^2 \left (-a^2-3 a b x-3 b^2 x^2\right )}{3 \left (c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^2*(-a^2 - 3*a*b*x - 3*b^2*x^2))/(3*(c*x^2)^(5/2))

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Maple [A]  time = 0.003, size = 27, normalized size = 0.9 \begin{align*} -{\frac{3\,{b}^{2}{x}^{2}+3\,abx+{a}^{2}}{3} \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/x/(c*x^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.05268, size = 50, normalized size = 1.72 \begin{align*} -\frac{b^{2}}{\sqrt{c x^{2}} c} - \frac{a b}{c^{\frac{3}{2}} x^{2}} - \frac{a^{2}}{3 \, c^{\frac{3}{2}} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.48277, size = 76, normalized size = 2.62 \begin{align*} -\frac{{\left (3 \, b^{2} x^{2} + 3 \, a b x + a^{2}\right )} \sqrt{c x^{2}}}{3 \, c^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.610677, size = 53, normalized size = 1.83 \begin{align*} - \frac{a^{2}}{3 c^{\frac{3}{2}} \left (x^{2}\right )^{\frac{3}{2}}} - \frac{a b x}{c^{\frac{3}{2}} \left (x^{2}\right )^{\frac{3}{2}}} - \frac{b^{2} x^{2}}{c^{\frac{3}{2}} \left (x^{2}\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{2}}{\left (c x^{2}\right )^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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