3.49 \(\int \frac{1}{x^3 \sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{16 c^2 \sqrt{b x+c x^2}}{15 b^3 x}+\frac{8 c \sqrt{b x+c x^2}}{15 b^2 x^2}-\frac{2 \sqrt{b x+c x^2}}{5 b x^3} \]

[Out]

(-2*Sqrt[b*x + c*x^2])/(5*b*x^3) + (8*c*Sqrt[b*x + c*x^2])/(15*b^2*x^2) - (16*c^2*Sqrt[b*x + c*x^2])/(15*b^3*x
)

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Rubi [A]  time = 0.0263216, 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} \[ -\frac{16 c^2 \sqrt{b x+c x^2}}{15 b^3 x}+\frac{8 c \sqrt{b x+c x^2}}{15 b^2 x^2}-\frac{2 \sqrt{b x+c x^2}}{5 b x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*Sqrt[b*x + c*x^2])/(5*b*x^3) + (8*c*Sqrt[b*x + c*x^2])/(15*b^2*x^2) - (16*c^2*Sqrt[b*x + c*x^2])/(15*b^3*x
)

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{1}{x^3 \sqrt{b x+c x^2}} \, dx &=-\frac{2 \sqrt{b x+c x^2}}{5 b x^3}-\frac{(4 c) \int \frac{1}{x^2 \sqrt{b x+c x^2}} \, dx}{5 b}\\ &=-\frac{2 \sqrt{b x+c x^2}}{5 b x^3}+\frac{8 c \sqrt{b x+c x^2}}{15 b^2 x^2}+\frac{\left (8 c^2\right ) \int \frac{1}{x \sqrt{b x+c x^2}} \, dx}{15 b^2}\\ &=-\frac{2 \sqrt{b x+c x^2}}{5 b x^3}+\frac{8 c \sqrt{b x+c x^2}}{15 b^2 x^2}-\frac{16 c^2 \sqrt{b x+c x^2}}{15 b^3 x}\\ \end{align*}

Mathematica [A]  time = 0.011888, size = 40, normalized size = 0.54 \[ -\frac{2 \sqrt{x (b+c x)} \left (3 b^2-4 b c x+8 c^2 x^2\right )}{15 b^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[b*x + c*x^2]),x]

[Out]

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

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Maple [A]  time = 0.046, size = 44, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 8\,{c}^{2}{x}^{2}-4\,bcx+3\,{b}^{2} \right ) }{15\,{x}^{2}{b}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(c*x^2+b*x)^(1/2),x)

[Out]

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

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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(1/x^3/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88494, size = 88, normalized size = 1.19 \begin{align*} -\frac{2 \,{\left (8 \, c^{2} x^{2} - 4 \, b c x + 3 \, b^{2}\right )} \sqrt{c x^{2} + b x}}{15 \, b^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{x \left (b + c x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(c*x**2+b*x)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(x*(b + c*x))), x)

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Giac [A]  time = 1.20529, size = 105, normalized size = 1.42 \begin{align*} \frac{2 \,{\left (20 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} c + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b \sqrt{c} + 3 \, b^{2}\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(1/x^3/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

2/15*(20*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*c + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b*sqrt(c) + 3*b^2)/(sqrt(c)*
x - sqrt(c*x^2 + b*x))^5