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]

-2/5*(c*x^2+b*x)^(1/2)/b/x^3+8/15*c*(c*x^2+b*x)^(1/2)/b^2/x^2-16/15*c^2*(c*x^2+b*x)^(1/2)/b^3/x

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, 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 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]

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]

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 = 0.01, 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)

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 38, normalized size = 0.51 \[ -\frac {2 \, {\left (8 \, c^{2} x^{2} - 4 \, b c x + 3 \, b^{2}\right )} \sqrt {c x^{2} + b x}}{15 \, b^{3} x^{3}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.23, size = 78, normalized size = 1.05 \[ \frac {2 \, {\left (20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b \sqrt {c} + 3 \, b^{2}\right )}}{15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5}} \]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 44, normalized size = 0.59 \[ -\frac {2 \left (c x +b \right ) \left (8 c^{2} x^{2}-4 b c x +3 b^{2}\right )}{15 \sqrt {c \,x^{2}+b x}\, b^{3} x^{2}} \]

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)

________________________________________________________________________________________

maxima [A]  time = 1.35, size = 62, normalized size = 0.84 \[ -\frac {16 \, \sqrt {c x^{2} + b x} c^{2}}{15 \, b^{3} x} + \frac {8 \, \sqrt {c x^{2} + b x} c}{15 \, b^{2} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x}}{5 \, b x^{3}} \]

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]

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

________________________________________________________________________________________

mupad [B]  time = 0.19, size = 38, normalized size = 0.51 \[ -\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (3\,b^2-4\,b\,c\,x+8\,c^2\,x^2\right )}{15\,b^3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(b*x + c*x^2)^(1/2)),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \sqrt {x \left (b + c x\right )}}\, dx \]

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)

________________________________________________________________________________________