\(\int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx\) [2378]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 79 \[ \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\frac {\text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c d^2-b d e+a e^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {738, 212} \[ \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\frac {\text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{\sqrt {a e^2-b d e+c d^2}} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )\right ) \\ & = \frac {\tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c d^2-b d e+a e^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {-c d^2+b d e-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{c d^2+e (-b d+a e)} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(156\) vs. \(2(71)=142\).

Time = 0.34 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.99

method result size
default \(-\frac {\ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\) \(157\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (71) = 142\).

Time = 0.32 (sec) , antiderivative size = 343, normalized size of antiderivative = 4.34 \[ \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\left [\frac {\log \left (\frac {8 \, a b d e - 8 \, a^{2} e^{2} - {\left (b^{2} + 4 \, a c\right )} d^{2} - {\left (8 \, c^{2} d^{2} - 8 \, b c d e + {\left (b^{2} + 4 \, a c\right )} e^{2}\right )} x^{2} - 4 \, \sqrt {c d^{2} - b d e + a e^{2}} \sqrt {c x^{2} + b x + a} {\left (b d - 2 \, a e + {\left (2 \, c d - b e\right )} x\right )} - 2 \, {\left (4 \, b c d^{2} + 4 \, a b e^{2} - {\left (3 \, b^{2} + 4 \, a c\right )} d e\right )} x}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right )}{2 \, \sqrt {c d^{2} - b d e + a e^{2}}}, \frac {\sqrt {-c d^{2} + b d e - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e - a e^{2}} \sqrt {c x^{2} + b x + a} {\left (b d - 2 \, a e + {\left (2 \, c d - b e\right )} x\right )}}{2 \, {\left (a c d^{2} - a b d e + a^{2} e^{2} + {\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} x^{2} + {\left (b c d^{2} - b^{2} d e + a b e^{2}\right )} x\right )}}\right )}{c d^{2} - b d e + a e^{2}}\right ] \]

[In]

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

[Out]

[1/2*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt
(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*
b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2))/sqrt(c*d^2 - b*d*e + a*e^2), sqrt(-c*d^2 + b*d*e - a*e^2)*arct
an(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e
+ a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x))/(c*d^2 - b*d*e + a*e^2)]

Sympy [F]

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

[In]

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

[Out]

Integral(1/((d + e*x)*sqrt(a + b*x + c*x**2)), x)

Maxima [F(-2)]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\frac {2 \, \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e - a e^{2}}}\right )}{\sqrt {-c d^{2} + b d e - a e^{2}}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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