\(\int \frac {1}{(\frac {b e}{2 c}+e x) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx\) [2410]

Optimal result

Integrand size = 39, antiderivative size = 27 \[ \int \frac {1}{\left (\frac {b e}{2 c}+e x\right ) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx=-\frac {2}{e \sqrt {\frac {b^2}{c}+4 b x+4 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {657, 643} \[ \int \frac {1}{\left (\frac {b e}{2 c}+e x\right ) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx=-\frac {2}{e \sqrt {\frac {b^2}{c}+4 b x+4 c x^2}} \]

[In]

Int[1/(((b*e)/(2*c) + e*x)*Sqrt[b^2/(4*c) + b*x + c*x^2]),x]

[Out]

-2/(e*Sqrt[b^2/c + 4*b*x + 4*c*x^2])

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c,
0] &&  !IntegerQ[p] && EqQ[2*c*d - b*e, 0] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {c \int \frac {\frac {b e}{2 c}+e x}{\left (\frac {b^2}{4 c}+b x+c x^2\right )^{3/2}} \, dx}{e^2} \\ & = -\frac {2}{e \sqrt {\frac {b^2}{c}+4 b x+4 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (\frac {b e}{2 c}+e x\right ) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx=-\frac {2}{e \sqrt {\frac {(b+2 c x)^2}{c}}} \]

[In]

Integrate[1/(((b*e)/(2*c) + e*x)*Sqrt[b^2/(4*c) + b*x + c*x^2]),x]

[Out]

-2/(e*Sqrt[(b + 2*c*x)^2/c])

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {2}{e \sqrt {\frac {\left (2 c x +b \right )^{2}}{c}}}\)	\(20\)
pseudoelliptic	\(-\frac {2}{e \sqrt {\frac {\left (2 c x +b \right )^{2}}{c}}}\)	\(20\)
gosper	\(-\frac {2}{\sqrt {\frac {4 c^{2} x^{2}+4 b x c +b^{2}}{c}}\, e}\)	\(29\)
default	\(-\frac {2}{\sqrt {\frac {4 c^{2} x^{2}+4 b x c +b^{2}}{c}}\, e}\)	\(29\)
trager	\(\frac {4 c^{2} x \sqrt {-\frac {-4 c^{2} x^{2}-4 b x c -b^{2}}{c}}}{b e \left (2 c x +b \right )^{2}}\)	\(47\)

[In]

int(2/(1/2*b*e/c+e*x)/(b^2/c+4*b*x+4*c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {1}{\left (\frac {b e}{2 c}+e x\right ) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx=-\frac {2 \, c \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{c}}}{4 \, c^{2} e x^{2} + 4 \, b c e x + b^{2} e} \]

[In]

integrate(2/(1/2*b*e/c+e*x)/(b^2/c+4*b*x+4*c*x^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {1}{\left (\frac {b e}{2 c}+e x\right ) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx=2 \left (\begin {cases} - \frac {1}{e \sqrt {\frac {b^{2}}{c} + 4 b x + 4 c x^{2}}} & \text {for}\: e \neq 0 \\\frac {\tilde {\infty } \left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate(2/(1/2*b*e/c+e*x)/(b**2/c+4*b*x+4*c*x**2)**(1/2),x)

[Out]

2*Piecewise((-1/(e*sqrt(b**2/c + 4*b*x + 4*c*x**2)), Ne(e, 0)), (zoo*(b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/
(2*c) + x)**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\left (\frac {b e}{2 c}+e x\right ) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx=-\frac {2}{2 \, e^{2} x \sqrt {\frac {c}{e^{2}}} + \frac {b e^{2} \sqrt {\frac {c}{e^{2}}}}{c}} \]

[In]

integrate(2/(1/2*b*e/c+e*x)/(b^2/c+4*b*x+4*c*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (\frac {b e}{2 c}+e x\right ) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx=-\frac {2 \, c^{\frac {3}{2}}}{{\left (2 \, c x + b\right )} e {\left | c \right |} \mathrm {sgn}\left (2 \, c x + b\right )} \]

[In]

integrate(2/(1/2*b*e/c+e*x)/(b^2/c+4*b*x+4*c*x^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (\frac {b e}{2 c}+e x\right ) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx=-\frac {2}{e\,\sqrt {4\,b\,x+4\,c\,x^2+\frac {b^2}{c}}} \]

[In]

int(2/((e*x + (b*e)/(2*c))*(4*b*x + 4*c*x^2 + b^2/c)^(1/2)),x)

[Out]

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