3.847 \(\int \frac{(a+b x)^2}{(c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=66 \[ -\frac{a^2}{4 c^2 x^3 \sqrt{c x^2}}-\frac{2 a b}{3 c^2 x^2 \sqrt{c x^2}}-\frac{b^2}{2 c^2 x \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0125624, antiderivative size = 66, 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} \[ -\frac{a^2}{4 c^2 x^3 \sqrt{c x^2}}-\frac{2 a b}{3 c^2 x^2 \sqrt{c x^2}}-\frac{b^2}{2 c^2 x \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(c*x^2)^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.0037259, size = 38, normalized size = 0.58 \[ -\frac{\sqrt{c x^2} \left (3 a^2+8 a b x+6 b^2 x^2\right )}{12 c^3 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/(c*x^2)^(5/2),x]

[Out]

-(Sqrt[c*x^2]*(3*a^2 + 8*a*b*x + 6*b^2*x^2))/(12*c^3*x^5)

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Maple [A]  time = 0.004, size = 30, normalized size = 0.5 \begin{align*} -{\frac{x \left ( 6\,{b}^{2}{x}^{2}+8\,abx+3\,{a}^{2} \right ) }{12} \left ( c{x}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/(c*x^2)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.06597, size = 50, normalized size = 0.76 \begin{align*} -\frac{2 \, a b}{3 \, \left (c x^{2}\right )^{\frac{3}{2}} c} - \frac{b^{2}}{2 \, c^{\frac{5}{2}} x^{2}} - \frac{a^{2}}{4 \, c^{\frac{5}{2}} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

-2/3*a*b/((c*x^2)^(3/2)*c) - 1/2*b^2/(c^(5/2)*x^2) - 1/4*a^2/(c^(5/2)*x^4)

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Fricas [A]  time = 1.53816, size = 80, normalized size = 1.21 \begin{align*} -\frac{{\left (6 \, b^{2} x^{2} + 8 \, a b x + 3 \, a^{2}\right )} \sqrt{c x^{2}}}{12 \, c^{3} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.901204, size = 61, normalized size = 0.92 \begin{align*} - \frac{a^{2} x}{4 c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}} - \frac{2 a b x^{2}}{3 c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}} - \frac{b^{2} x^{3}}{2 c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/(c*x**2)**(5/2),x)

[Out]

-a**2*x/(4*c**(5/2)*(x**2)**(5/2)) - 2*a*b*x**2/(3*c**(5/2)*(x**2)**(5/2)) - b**2*x**3/(2*c**(5/2)*(x**2)**(5/
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/(c*x^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x