3.9 \(\int \frac{\sqrt{b x+c x^2}}{x^5} \, dx\)

Optimal. Leaf size=74 \[ -\frac{16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^3}+\frac{8 c \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}-\frac{2 \left (b x+c x^2\right )^{3/2}}{7 b x^5} \]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(7*b*x^5) + (8*c*(b*x + c*x^2)^(3/2))/(35*b^2*x^4) - (16*c^2*(b*x + c*x^2)^(3/2))/(10
5*b^3*x^3)

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Rubi [A]  time = 0.0282151, antiderivative size = 74, 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 = {658, 650} \[ -\frac{16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^3}+\frac{8 c \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}-\frac{2 \left (b x+c x^2\right )^{3/2}}{7 b x^5} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/x^5,x]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(7*b*x^5) + (8*c*(b*x + c*x^2)^(3/2))/(35*b^2*x^4) - (16*c^2*(b*x + c*x^2)^(3/2))/(10
5*b^3*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^5} \, dx &=-\frac{2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}-\frac{(4 c) \int \frac{\sqrt{b x+c x^2}}{x^4} \, dx}{7 b}\\ &=-\frac{2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}+\frac{8 c \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}+\frac{\left (8 c^2\right ) \int \frac{\sqrt{b x+c x^2}}{x^3} \, dx}{35 b^2}\\ &=-\frac{2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}+\frac{8 c \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}-\frac{16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0108996, size = 51, normalized size = 0.69 \[ -\frac{2 \sqrt{x (b+c x)} \left (3 b^2 c x+15 b^3-4 b c^2 x^2+8 c^3 x^3\right )}{105 b^3 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x + c*x^2]/x^5,x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(15*b^3 + 3*b^2*c*x - 4*b*c^2*x^2 + 8*c^3*x^3))/(105*b^3*x^4)

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Maple [A]  time = 0.049, size = 44, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 8\,{c}^{2}{x}^{2}-12\,bcx+15\,{b}^{2} \right ) }{105\,{b}^{3}{x}^{4}}\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^5,x)

[Out]

-2/105*(c*x+b)*(8*c^2*x^2-12*b*c*x+15*b^2)*(c*x^2+b*x)^(1/2)/b^3/x^4

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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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99353, size = 112, normalized size = 1.51 \begin{align*} -\frac{2 \,{\left (8 \, c^{3} x^{3} - 4 \, b c^{2} x^{2} + 3 \, b^{2} c x + 15 \, b^{3}\right )} \sqrt{c x^{2} + b x}}{105 \, b^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/x^5,x, algorithm="fricas")

[Out]

-2/105*(8*c^3*x^3 - 4*b*c^2*x^2 + 3*b^2*c*x + 15*b^3)*sqrt(c*x^2 + b*x)/(b^3*x^4)

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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^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(1/2)/x**5,x)

[Out]

Integral(sqrt(x*(b + c*x))/x**5, x)

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Giac [B]  time = 1.34038, size = 184, normalized size = 2.49 \begin{align*} \frac{2 \,{\left (140 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} c^{2} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} b c^{\frac{3}{2}} + 273 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} b^{2} c + 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b^{3} \sqrt{c} + 15 \, b^{4}\right )}}{105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/x^5,x, algorithm="giac")

[Out]

2/105*(140*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*c^2 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b*c^(3/2) + 273*(sqrt
(c)*x - sqrt(c*x^2 + b*x))^2*b^2*c + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^3*sqrt(c) + 15*b^4)/(sqrt(c)*x - sq
rt(c*x^2 + b*x))^7