\(\int \frac {1}{(d+e x) \sqrt {\frac {-c d^2+b d e}{e^2}+b x+c x^2}} \, dx\) [2409]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 36, antiderivative size = 48 \[ \int \frac {1}{(d+e x) \sqrt {\frac {-c d^2+b d e}{e^2}+b x+c x^2}} \, dx=\frac {2 e \sqrt {-\frac {d (c d-b e)}{e^2}+b x+c x^2}}{(2 c d-b e) (d+e x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {664} \[ \int \frac {1}{(d+e x) \sqrt {\frac {-c d^2+b d e}{e^2}+b x+c x^2}} \, dx=\frac {2 e \sqrt {-\frac {d (c d-b e)}{e^2}+b x+c x^2}}{(d+e x) (2 c d-b e)} \]

[In]

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

[Out]

(2*e*Sqrt[-((d*(c*d - b*e))/e^2) + b*x + c*x^2])/((2*c*d - b*e)*(d + e*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]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(d+e x) \sqrt {\frac {-c d^2+b d e}{e^2}+b x+c x^2}} \, dx=-\frac {2 e \sqrt {\frac {(d+e x) (-c d+b e+c e x)}{e^2}}}{(-2 c d+b e) (d+e x)} \]

[In]

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

[Out]

(-2*e*Sqrt[((d + e*x)*(-(c*d) + b*e + c*e*x))/e^2])/((-2*c*d + b*e)*(d + e*x))

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.15

method result size
default \(-\frac {2 \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}}}{\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}\) \(55\)
trager \(-\frac {2 e \sqrt {-\frac {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}{e^{2}}}}{\left (b e -2 c d \right ) \left (e x +d \right )}\) \(55\)
gosper \(-\frac {2 \left (c e x +b e -c d \right )}{e \left (b e -2 c d \right ) \sqrt {\frac {c \,e^{2} x^{2}+b \,e^{2} x +b d e -c \,d^{2}}{e^{2}}}}\) \(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(d+e x) \sqrt {\frac {-c d^2+b d e}{e^2}+b x+c x^2}} \, dx=\frac {2 \, e \sqrt {\frac {c e^{2} x^{2} + b e^{2} x - c d^{2} + b d e}{e^{2}}}}{2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(1/(sqrt((d/e + x)*(b - c*d/e + c*x))*(d + e*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x) \sqrt {\frac {-c d^2+b d e}{e^2}+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(e*x+d)/((b*d*e-c*d^2)/e^2+b*x+c*x^2)^(1/2),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(b*e-2*c*d>0)', see `assume?` f
or more deta

Giac [F]

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

[In]

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

[Out]

undef

Mupad [B] (verification not implemented)

Time = 10.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(d+e x) \sqrt {\frac {-c d^2+b d e}{e^2}+b x+c x^2}} \, dx=-\frac {2\,e\,\sqrt {b\,x-\frac {c\,d^2-b\,d\,e}{e^2}+c\,x^2}}{\left (b\,e-2\,c\,d\right )\,\left (d+e\,x\right )} \]

[In]

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

[Out]

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