\(\int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx\) [3]

Optimal result

Integrand size = 28, antiderivative size = 135 \[ \int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx=\frac {b \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {e+f x^2}}\right )}{\sqrt {d} \sqrt {f}}+\frac {a \sqrt {e} \sqrt {c+d x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ),1-\frac {d e}{c f}\right )}{c \sqrt {f} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}} \sqrt {e+f x^2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.34 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01 \[ \int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx=\frac {b \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x^2}}{\sqrt {f} \sqrt {c+d x^2}}\right )}{\sqrt {d} \sqrt {f}}-\frac {i a \sqrt {1+\frac {d x^2}{c}} \sqrt {1+\frac {f x^2}{e}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {d}{c}} x\right ),\frac {c f}{d e}\right )}{\sqrt {\frac {d}{c}} \sqrt {c+d x^2} \sqrt {e+f x^2}} \] Input:

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

Output:

(b*ArcTanh[(Sqrt[d]*Sqrt[e + f*x^2])/(Sqrt[f]*Sqrt[c + d*x^2])])/(Sqrt[d]* 
Sqrt[f]) - (I*a*Sqrt[1 + (d*x^2)/c]*Sqrt[1 + (f*x^2)/e]*EllipticF[I*ArcSin 
h[Sqrt[d/c]*x], (c*f)/(d*e)])/(Sqrt[d/c]*Sqrt[c + d*x^2]*Sqrt[e + f*x^2])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1343, 320, 353, 66, 221}

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.

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

\(\Big \downarrow \) 1343

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

\(\Big \downarrow \) 320

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {a \sqrt {e} \sqrt {c+d x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ),1-\frac {d e}{c f}\right )}{c \sqrt {f} \sqrt {e+f x^2} \sqrt {\frac {e \left (c+d x^2\right )}{c \left (e+f x^2\right )}}}+\frac {b \text {arctanh}\left (\frac {\sqrt {f} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {e+f x^2}}\right )}{\sqrt {d} \sqrt {f}}\)

Input:

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

Output:

(b*ArcTanh[(Sqrt[f]*Sqrt[c + d*x^2])/(Sqrt[d]*Sqrt[e + f*x^2])])/(Sqrt[d]* 
Sqrt[f]) + (a*Sqrt[e]*Sqrt[c + d*x^2]*EllipticF[ArcTan[(Sqrt[f]*x)/Sqrt[e] 
], 1 - (d*e)/(c*f)])/(c*Sqrt[f]*Sqrt[(e*(c + d*x^2))/(c*(e + f*x^2))]*Sqrt 
[e + f*x^2])
 

Defintions of rubi rules used

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 221

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

rule 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( 
c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 353

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

rule 1343

Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q 
_), x_Symbol] :> Simp[g   Int[(a + c*x^2)^p*(d + f*x^2)^q, x], x] + Simp[h 
  Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h, p, 
q}, x]
 

Maple [A] (verified)

Time = 5.89 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.34

method	result	size
elliptic	\(\frac {\sqrt {\left (d \,x^{2}+c \right ) \left (f \,x^{2}+e \right )}\, \left (\frac {a \sqrt {1+\frac {x^{2} d}{c}}\, \sqrt {1+\frac {f \,x^{2}}{e}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {d}{c}}, \sqrt {-1+\frac {c f +d e}{e d}}\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}}+\frac {b \ln \left (\frac {2 d f \,x^{2}+c f +d e}{\sqrt {d f}}+2 \sqrt {d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}\right )}{2 \sqrt {d f}}\right )}{\sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}\)	\(181\)
default	\(\frac {\left (2 \sqrt {\frac {d \,x^{2}+c}{c}}\, \sqrt {\frac {f \,x^{2}+e}{e}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {d}{c}}, \sqrt {\frac {c f}{d e}}\right ) \sqrt {d f}\, a +b \ln \left (\frac {2 d f \,x^{2}+2 \sqrt {\left (d \,x^{2}+c \right ) \left (f \,x^{2}+e \right )}\, \sqrt {d f}+c f +d e}{\sqrt {d f}}\right ) \sqrt {-\frac {d}{c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (f \,x^{2}+e \right )}\right ) \sqrt {d \,x^{2}+c}\, \sqrt {f \,x^{2}+e}}{2 \sqrt {d f}\, \sqrt {-\frac {d}{c}}\, \left (d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e \right )}\)	\(187\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13 \[ \int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx=\frac {4 \, \sqrt {d f} a f \sqrt {-\frac {e}{f}} F(\arcsin \left (\frac {\sqrt {-\frac {e}{f}}}{x}\right )\,|\,\frac {c f}{d e}) + \sqrt {d f} b e \log \left (-8 \, d^{2} f^{2} x^{4} - d^{2} e^{2} - 6 \, c d e f - c^{2} f^{2} - 8 \, {\left (d^{2} e f + c d f^{2}\right )} x^{2} - 4 \, {\left (2 \, d f x^{2} + d e + c f\right )} \sqrt {d x^{2} + c} \sqrt {f x^{2} + e} \sqrt {d f}\right )}{4 \, d e f} \] Input:

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

Output:

1/4*(4*sqrt(d*f)*a*f*sqrt(-e/f)*elliptic_f(arcsin(sqrt(-e/f)/x), c*f/(d*e) 
) + sqrt(d*f)*b*e*log(-8*d^2*f^2*x^4 - d^2*e^2 - 6*c*d*e*f - c^2*f^2 - 8*( 
d^2*e*f + c*d*f^2)*x^2 - 4*(2*d*f*x^2 + d*e + c*f)*sqrt(d*x^2 + c)*sqrt(f* 
x^2 + e)*sqrt(d*f)))/(d*e*f)
 

Sympy [F]

\[ \int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx=\int \frac {a + b x}{\sqrt {c + d x^{2}} \sqrt {e + f x^{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx=\int { \frac {b x + a}{\sqrt {d x^{2} + c} \sqrt {f x^{2} + e}} \,d x } \] Input:

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

Output:

integrate((b*x + a)/(sqrt(d*x^2 + c)*sqrt(f*x^2 + e)), x)
 

Giac [F]

\[ \int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx=\int { \frac {b x + a}{\sqrt {d x^{2} + c} \sqrt {f x^{2} + e}} \,d x } \] Input:

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

Output:

integrate((b*x + a)/(sqrt(d*x^2 + c)*sqrt(f*x^2 + e)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx=\int \frac {a+b\,x}{\sqrt {d\,x^2+c}\,\sqrt {f\,x^2+e}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {a+b x}{\sqrt {c+d x^2} \sqrt {e+f x^2}} \, dx=\frac {\sqrt {f}\, \sqrt {d}\, \mathrm {log}\left (-\sqrt {d}\, \sqrt {d \,x^{2}+c}\, f -\sqrt {f}\, \sqrt {f \,x^{2}+e}\, d \right ) b +\left (\int \frac {\sqrt {f \,x^{2}+e}\, \sqrt {d \,x^{2}+c}}{d f \,x^{4}+c f \,x^{2}+d e \,x^{2}+c e}d x \right ) a d f}{d f} \] Input:

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

Output:

(sqrt(f)*sqrt(d)*log( - sqrt(d)*sqrt(c + d*x**2)*f - sqrt(f)*sqrt(e + f*x* 
*2)*d)*b + int((sqrt(e + f*x**2)*sqrt(c + d*x**2))/(c*e + c*f*x**2 + d*e*x 
**2 + d*f*x**4),x)*a*d*f)/(d*f)