3.2409 \(\int \frac {1}{(d+e x) \sqrt {\frac {-c d^2+b d e}{e^2}+b x+c x^2}} \, dx\)

Optimal. Leaf size=48 \[ \frac {2 e \sqrt {-\frac {d (c d-b e)}{e^2}+b x+c x^2}}{(d+e x) (2 c d-b e)} \]

[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]  time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {650} \[ \frac {2 e \sqrt {-\frac {d (c d-b e)}{e^2}+b x+c x^2}}{(d+e x) (2 c d-b e)} \]

Antiderivative was successfully verified.

[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 650

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*} \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)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 45, normalized size = 0.94 \[ -\frac {2 e \sqrt {\frac {(d+e x) (b e-c d+c e x)}{e^2}}}{(d+e x) (b e-2 c d)} \]

Antiderivative was successfully verified.

[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))

________________________________________________________________________________________

fricas [A]  time = 1.39, size = 62, normalized size = 1.29 \[ \frac {2 \, e \sqrt {\frac {c e^{2} x^{2} + b e^{2} x - c d^{2} + b d e}{e^{2}}}}{2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

Exception raised: NotImplementedError >> Unable to parse Giac output: 2*exp(2)*2/2/sqrt(-c*d^2*exp(2)^2-b*d*ex
p(1)^3*exp(2)+b*d*exp(1)*exp(2)^2+c*d^2*exp(1)^2*exp(2))*atan((-d*sqrt(c)*exp(2)+(sqrt(c*x^2+b*x-(c*d^2-b*d*ex
p(1))/exp(2))-sqrt(c)*x)*exp(1)*exp(2))/sqrt(-c*d^2*exp(2)^2-b*d*exp(1)^3*exp(2)+b*d*exp(1)*exp(2)^2+c*d^2*exp
(1)^2*exp(2)))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 59, normalized size = 1.23 \[ -\frac {2 \left (c e x +b e -c d \right )}{\left (b e -2 c d \right ) \sqrt {\frac {c \,e^{2} x^{2}+b \,e^{2} x +b d e -c \,d^{2}}{e^{2}}}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is b*e-2*c*d zero or nonzero?

________________________________________________________________________________________

mupad [B]  time = 1.18, size = 47, normalized size = 0.98 \[ -\frac {2\,e\,\sqrt {b\,x-\frac {c\,d^2-b\,d\,e}{e^2}+c\,x^2}}{\left (b\,e-2\,c\,d\right )\,\left (d+e\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\left (\frac {d}{e} + x\right ) \left (b - \frac {c d}{e} + c x\right )} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________