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

Optimal. Leaf size=66 \[ -\frac{a^2}{5 c^2 x^4 \sqrt{c x^2}}-\frac{a b}{2 c^2 x^3 \sqrt{c x^2}}-\frac{b^2}{3 c^2 x^2 \sqrt{c x^2}} \]

[Out]

-a^2/(5*c^2*x^4*Sqrt[c*x^2]) - (a*b)/(2*c^2*x^3*Sqrt[c*x^2]) - b^2/(3*c^2*x^2*Sqrt[c*x^2])

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Rubi [A]  time = 0.0124531, antiderivative size = 66, 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} \[ -\frac{a^2}{5 c^2 x^4 \sqrt{c x^2}}-\frac{a b}{2 c^2 x^3 \sqrt{c x^2}}-\frac{b^2}{3 c^2 x^2 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(x*(c*x^2)^(5/2)),x]

[Out]

-a^2/(5*c^2*x^4*Sqrt[c*x^2]) - (a*b)/(2*c^2*x^3*Sqrt[c*x^2]) - b^2/(3*c^2*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 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 \left (c x^2\right )^{5/2}} \, dx &=\frac{x \int \frac{(a+b x)^2}{x^6} \, dx}{c^2 \sqrt{c x^2}}\\ &=\frac{x \int \left (\frac{a^2}{x^6}+\frac{2 a b}{x^5}+\frac{b^2}{x^4}\right ) \, dx}{c^2 \sqrt{c x^2}}\\ &=-\frac{a^2}{5 c^2 x^4 \sqrt{c x^2}}-\frac{a b}{2 c^2 x^3 \sqrt{c x^2}}-\frac{b^2}{3 c^2 x^2 \sqrt{c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0122521, size = 38, normalized size = 0.58 \[ -\frac{\sqrt{c x^2} \left (6 a^2+15 a b x+10 b^2 x^2\right )}{30 c^3 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c*x^2]*(6*a^2 + 15*a*b*x + 10*b^2*x^2))/(30*c^3*x^6)

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Maple [A]  time = 0.005, size = 29, normalized size = 0.4 \begin{align*} -{\frac{10\,{b}^{2}{x}^{2}+15\,abx+6\,{a}^{2}}{30} \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/x/(c*x^2)^(5/2),x)

[Out]

-1/30*(10*b^2*x^2+15*a*b*x+6*a^2)/(c*x^2)^(5/2)

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Maxima [A]  time = 1.08766, size = 50, normalized size = 0.76 \begin{align*} -\frac{b^{2}}{3 \, \left (c x^{2}\right )^{\frac{3}{2}} c} - \frac{a b}{2 \, c^{\frac{5}{2}} x^{4}} - \frac{a^{2}}{5 \, c^{\frac{5}{2}} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.47218, size = 82, normalized size = 1.24 \begin{align*} -\frac{{\left (10 \, b^{2} x^{2} + 15 \, a b x + 6 \, a^{2}\right )} \sqrt{c x^{2}}}{30 \, c^{3} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(10*b^2*x^2 + 15*a*b*x + 6*a^2)*sqrt(c*x^2)/(c^3*x^6)

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Sympy [A]  time = 1.03065, size = 56, normalized size = 0.85 \begin{align*} - \frac{a^{2}}{5 c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}} - \frac{a b x}{2 c^{\frac{5}{2}} \left (x^{2}\right )^{\frac{5}{2}}} - \frac{b^{2} x^{2}}{3 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/x/(c*x**2)**(5/2),x)

[Out]

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

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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