\(\int \frac {x}{(b x+c x^2)^{3/2}} \, dx\) [55]

Optimal result

Integrand size = 15, antiderivative size = 19 \[ \int \frac {x}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 x}{b \sqrt {b x+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {650} \[ \int \frac {x}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 x}{b \sqrt {b x+c x^2}} \]

[In]

Int[x/(b*x + c*x^2)^(3/2),x]

[Out]

(2*x)/(b*Sqrt[b*x + c*x^2])

Rule 650

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*((b*d - 2*a*e + (2*c*
d - b*e)*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x}{b \sqrt {b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 x}{b \sqrt {x (b+c x)}} \]

[In]

Integrate[x/(b*x + c*x^2)^(3/2),x]

[Out]

(2*x)/(b*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {2 x}{b \sqrt {x \left (c x +b \right )}}\)	\(16\)
trager	\(\frac {2 \sqrt {c \,x^{2}+b x}}{\left (c x +b \right ) b}\)	\(24\)
gosper	\(\frac {2 x^{2} \left (c x +b \right )}{b \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(25\)
default	\(-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {x}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, \sqrt {c x^{2} + b x}}{b c x + b^{2}} \]

[In]

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

[Out]

2*sqrt(c*x^2 + b*x)/(b*c*x + b^2)

Sympy [F]

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

[In]

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

[Out]

Integral(x/(x*(b + c*x))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, x}{\sqrt {c x^{2} + b x} b} \]

[In]

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

[Out]

2*x/(sqrt(c*x^2 + b*x)*b)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int \frac {x}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )} \sqrt {c}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2\,x}{b\,\sqrt {x\,\left (b+c\,x\right )}} \]

[In]

int(x/(b*x + c*x^2)^(3/2),x)

[Out]

(2*x)/(b*(x*(b + c*x))^(1/2))