\(\int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx\) [49]

Optimal result

Integrand size = 17, antiderivative size = 74 \[ \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx=-\frac {2 \sqrt {b x+c x^2}}{5 b x^3}+\frac {8 c \sqrt {b x+c x^2}}{15 b^2 x^2}-\frac {16 c^2 \sqrt {b x+c x^2}}{15 b^3 x} \]

[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] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 664} \[ \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx=-\frac {16 c^2 \sqrt {b x+c x^2}}{15 b^3 x}+\frac {8 c \sqrt {b x+c x^2}}{15 b^2 x^2}-\frac {2 \sqrt {b x+c x^2}}{5 b x^3} \]

[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 664

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 672

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*} \text {integral}& = -\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] (verified)

Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (3 b^2-4 b c x+8 c^2 x^2\right )}{15 b^3 x^3} \]

[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)

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.50

method	result	size
pseudoelliptic	\(-\frac {2 \left (8 c^{2} x^{2}-4 b c x +3 b^{2}\right ) \sqrt {x \left (c x +b \right )}}{15 b^{3} x^{3}}\)	\(37\)
trager	\(-\frac {2 \left (8 c^{2} x^{2}-4 b c x +3 b^{2}\right ) \sqrt {c \,x^{2}+b x}}{15 b^{3} x^{3}}\)	\(39\)
risch	\(-\frac {2 \left (c x +b \right ) \left (8 c^{2} x^{2}-4 b c x +3 b^{2}\right )}{15 b^{3} x^{2} \sqrt {x \left (c x +b \right )}}\)	\(42\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (8 c^{2} x^{2}-4 b c x +3 b^{2}\right )}{15 x^{2} b^{3} \sqrt {c \,x^{2}+b x}}\)	\(44\)
default	\(-\frac {2 \sqrt {c \,x^{2}+b x}}{5 b \,x^{3}}-\frac {4 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{3 b \,x^{2}}+\frac {4 c \sqrt {c \,x^{2}+b x}}{3 b^{2} x}\right )}{5 b}\)	\(67\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx=-\frac {2 \, {\left (8 \, c^{2} x^{2} - 4 \, b c x + 3 \, b^{2}\right )} \sqrt {c x^{2} + b x}}{15 \, b^{3} x^{3}} \]

[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)

Sympy [F]

\[ \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx=\int \frac {1}{x^{3} \sqrt {x \left (b + c x\right )}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx=-\frac {16 \, \sqrt {c x^{2} + b x} c^{2}}{15 \, b^{3} x} + \frac {8 \, \sqrt {c x^{2} + b x} c}{15 \, b^{2} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x}}{5 \, b x^{3}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b \sqrt {c} + 3 \, b^{2}\right )}}{15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5}} \]

[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

Mupad [B] (verification not implemented)

Time = 9.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (3\,b^2-4\,b\,c\,x+8\,c^2\,x^2\right )}{15\,b^3\,x^3} \]

[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)