\(\int \frac {1}{(a+b x+c x^2)^{3/2}} \, dx\) [2386]

Optimal result

Integrand size = 14, antiderivative size = 32 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

[Out]

-2*(2*c*x+b)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {627} \[ \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

[In]

Int[(a + b*x + c*x^2)^(-3/2),x]

[Out]

(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])

Rule 627

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

[In]

Integrate[(a + b*x + c*x^2)^(-3/2),x]

[Out]

(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03

method	result	size
gosper	\(\frac {4 c x +2 b}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\)	\(33\)
default	\(\frac {4 c x +2 b}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\)	\(33\)
trager	\(\frac {4 c x +2 b}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\)	\(33\)

[In]

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

[Out]

2*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (30) = 60\).

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.91 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )}}{a b^{2} - 4 \, a^{2} c + {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} - 4 \, a b c\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*x + c*x**2)**(-3/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (\frac {2 \, c x}{b^{2} - 4 \, a c} + \frac {b}{b^{2} - 4 \, a c}\right )}}{\sqrt {c x^{2} + b x + a}} \]

[In]

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

[Out]

-2*(2*c*x/(b^2 - 4*a*c) + b/(b^2 - 4*a*c))/sqrt(c*x^2 + b*x + a)

Mupad [B] (verification not implemented)

Time = 9.98 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {\frac {b}{2}+c\,x}{\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}} \]

[In]

int(1/(a + b*x + c*x^2)^(3/2),x)

[Out]

(b/2 + c*x)/((a*c - b^2/4)*(a + b*x + c*x^2)^(1/2))