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

Optimal. Leaf size=48 \[ \frac{4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 b x^4} \]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(5*b*x^4) + (4*c*(b*x + c*x^2)^(3/2))/(15*b^2*x^3)

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Rubi [A]  time = 0.0165422, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {658, 650} \[ \frac{4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 b x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/x^4,x]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(5*b*x^4) + (4*c*(b*x + c*x^2)^(3/2))/(15*b^2*x^3)

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{b x+c x^2}}{x^4} \, dx &=-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}-\frac{(2 c) \int \frac{\sqrt{b x+c x^2}}{x^3} \, dx}{5 b}\\ &=-\frac{2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}+\frac{4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0110612, size = 29, normalized size = 0.6 \[ \frac{2 (x (b+c x))^{3/2} (2 c x-3 b)}{15 b^2 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x + c*x^2]/x^4,x]

[Out]

(2*(x*(b + c*x))^(3/2)*(-3*b + 2*c*x))/(15*b^2*x^4)

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Maple [A]  time = 0.052, size = 33, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -2\,cx+3\,b \right ) }{15\,{b}^{2}{x}^{3}}\sqrt{c{x}^{2}+bx}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(1/2)/x^4,x)

[Out]

-2/15*(c*x+b)*(-2*c*x+3*b)*(c*x^2+b*x)^(1/2)/b^2/x^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.12743, size = 84, normalized size = 1.75 \begin{align*} \frac{2 \,{\left (2 \, c^{2} x^{2} - b c x - 3 \, b^{2}\right )} \sqrt{c x^{2} + b x}}{15 \, b^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(2*c^2*x^2 - b*c*x - 3*b^2)*sqrt(c*x^2 + b*x)/(b^2*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x \left (b + c x\right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))/x**4, x)

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Giac [B]  time = 1.26466, size = 144, normalized size = 3. \begin{align*} \frac{2 \,{\left (15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} c^{\frac{3}{2}} + 25 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} b c + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b^{2} \sqrt{c} + 3 \, b^{3}\right )}}{15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*c^(3/2) + 25*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b*c + 15*(sqrt(c)*x
- sqrt(c*x^2 + b*x))*b^2*sqrt(c) + 3*b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^5