\(\int \frac {\sqrt {b x+c x^2}}{x^3} \, dx\) [7]

Optimal result

Integrand size = 17, antiderivative size = 23 \[ \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx=-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 b x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {664} \[ \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx=-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 b x^3} \]

[In]

Int[Sqrt[b*x + c*x^2]/x^3,x]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(3*b*x^3)

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (b x+c x^2\right )^{3/2}}{3 b x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx=-\frac {2 (x (b+c x))^{3/2}}{3 b x^3} \]

[In]

Integrate[Sqrt[b*x + c*x^2]/x^3,x]

[Out]

(-2*(x*(b + c*x))^(3/2))/(3*b*x^3)

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3 b \,x^{3}}\)	\(20\)
pseudoelliptic	\(-\frac {2 \left (c x +b \right ) \sqrt {x \left (c x +b \right )}}{3 x^{2} b}\)	\(23\)
gosper	\(-\frac {2 \left (c x +b \right ) \sqrt {c \,x^{2}+b x}}{3 x^{2} b}\)	\(25\)
trager	\(-\frac {2 \left (c x +b \right ) \sqrt {c \,x^{2}+b x}}{3 x^{2} b}\)	\(25\)
risch	\(-\frac {2 \left (c x +b \right )^{2}}{3 x \sqrt {x \left (c x +b \right )}\, b}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx=-\frac {2 \, \sqrt {c x^{2} + b x} {\left (c x + b\right )}}{3 \, b x^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))/x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx=-\frac {2 \, \sqrt {c x^{2} + b x} c}{3 \, b x} - \frac {2 \, \sqrt {c x^{2} + b x}}{3 \, x^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.30 \[ \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx=\frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} c + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b \sqrt {c} + b^{2}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (b+c\,x\right )}{3\,b\,x^2} \]

[In]

int((b*x + c*x^2)^(1/2)/x^3,x)

[Out]

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