3.1.67 \(\int \frac {1}{\sqrt {a+c x^2} (d+e x+f x^2)} \, dx\) [67]

3.1.67.1 Optimal result

Integrand size = 24, antiderivative size = 266 \[ \int \frac {1}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=-\frac {\sqrt {2} f \text {arctanh}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {\sqrt {2} f \text {arctanh}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}} \]

output

-f*arctanh(1/2*(2*a*f-c*x*(e-(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/ 
(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2))*2^(1/2)/(-4*d*f+e^2)^( 
1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)+f*arctanh(1/2*(2*a 
*f-c*x*(e+(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d 
*f+e*(-4*d*f+e^2)^(1/2)))^(1/2))*2^(1/2)/(-4*d*f+e^2)^(1/2)/(2*a*f^2+c*(e^ 
2-2*d*f+e*(-4*d*f+e^2)^(1/2)))^(1/2)
 

3.1.67.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.49 \[ \int \frac {1}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=-2 \sqrt {c} \text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {\log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ] \]

input

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

output

-2*Sqrt[c]*RootSum[a^2*f + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 - 2*a*f*#1^2 - 2* 
Sqrt[c]*e*#1^3 + f*#1^4 & , (Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1)/ 
(a*Sqrt[c]*e + 4*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*#1^3) & ]
 

3.1.67.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1315, 488, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx\)

\(\Big \downarrow \) 1315

\(\displaystyle \frac {2 f \int \frac {1}{\left (e+2 f x-\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+a}}dx}{\sqrt {e^2-4 d f}}-\frac {2 f \int \frac {1}{\left (e+2 f x+\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+a}}dx}{\sqrt {e^2-4 d f}}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {2 f \int \frac {1}{4 a f^2+c \left (e+\sqrt {e^2-4 d f}\right )^2-\frac {\left (2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x\right )^2}{c x^2+a}}d\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {c x^2+a}}}{\sqrt {e^2-4 d f}}-\frac {2 f \int \frac {1}{4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2-\frac {\left (2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x\right )^2}{c x^2+a}}d\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {c x^2+a}}}{\sqrt {e^2-4 d f}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {2} f \text {arctanh}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {\sqrt {2} f \text {arctanh}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\)

input

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

output

-((Sqrt[2]*f*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2 
*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[e 
^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])])) + (Sqr 
t[2]*f*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 
 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[e^2 - 4 
*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])])
 

3.1.67.3.1 Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 488

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

rule 1315

Int[1/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Sym 
bol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int[1/((b - q + 2*c*x 
)*Sqrt[d + f*x^2]), x], x] - Simp[2*(c/q)   Int[1/((b + q + 2*c*x)*Sqrt[d + 
 f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f}, x] && NeQ[b^2 - 4*a*c, 0] && Po 
sQ[b^2 - 4*a*c]
 

3.1.67.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(588\) vs. \(2(232)=464\).

Time = 0.63 (sec) , antiderivative size = 589, normalized size of antiderivative = 2.21

method	result	size
default	\(\frac {\sqrt {2}\, \ln \left (\frac {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2} c -\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}-\frac {\sqrt {2}\, \ln \left (\frac {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2} c -\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}\)	\(589\)

input

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

output

1/(-4*d*f+e^2)^(1/2)*2^(1/2)/(((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^ 
2)/f^2)^(1/2)*ln((((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2-c*(e+ 
(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*(((-4*d 
*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x+1/2*(e+(-4*d*f+e 
^2)^(1/2))/f)^2*c-4*c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2 
))/f)+2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2))/(x+1/2* 
(e+(-4*d*f+e^2)^(1/2))/f))-1/(-4*d*f+e^2)^(1/2)*2^(1/2)/((-(-4*d*f+e^2)^(1 
/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a 
*f^2-2*c*d*f+c*e^2)/f^2-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^ 
2)^(1/2)))+1/2*2^(1/2)*((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^ 
2)^(1/2)*(4*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2*c-4*c*(e-(-4*d*f+e^2)^(1/2 
))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2- 
2*c*d*f+c*e^2)/f^2)^(1/2))/(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2))))
 

3.1.67.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5073 vs. \(2 (230) = 460\).

Time = 1.03 (sec) , antiderivative size = 5073, normalized size of antiderivative = 19.07 \[ \int \frac {1}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.1.67.6 Sympy [F]

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

input

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

output

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

3.1.67.7 Maxima [F(-2)]

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

input

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

output

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

3.1.67.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.1.67.9 Mupad [F(-1)]

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

input

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

output

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