\(\int \frac {f+g x}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\) [229]

Optimal result

Integrand size = 46, antiderivative size = 131 \[ \int \frac {f+g x}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {2 g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e^2 \sqrt {d+e x}}-\frac {2 (e f-d g) \text {arctanh}\left (\frac {\sqrt {2 c d-b e} \sqrt {d+e x}}{\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{e^2 \sqrt {2 c d-b e}} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05 \[ \int \frac {f+g x}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {2 \sqrt {d+e x} \left (\sqrt {-2 c d+b e} g (-c d+b e+c e x)+c (e f-d g) \sqrt {-b e+c (d-e x)} \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {-2 c d+b e}}\right )\right )}{c e^2 \sqrt {-2 c d+b e} \sqrt {(d+e x) (-b e+c (d-e x))}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1221, 1136, 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.

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1136

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 1136

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

rule 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.17

method	result	size
default	\(-\frac {2 \sqrt {-\left (e x +d \right ) \left (c e x +b e -c d \right )}\, \left (\arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c d g -\arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c e f +g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\right )}{\sqrt {e x +d}\, \sqrt {-c e x -b e +c d}\, e^{2} c \sqrt {b e -2 c d}}\)	\(153\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 444, normalized size of antiderivative = 3.39 \[ \int \frac {f+g x}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\left [-\frac {2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c d - b e\right )} \sqrt {e x + d} g + {\left (c d e f - c d^{2} g + {\left (c e^{2} f - c d e g\right )} x\right )} \sqrt {2 \, c d - b e} \log \left (-\frac {c e^{2} x^{2} - 3 \, c d^{2} + 2 \, b d e - 2 \, {\left (c d e - b e^{2}\right )} x - 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {2 \, c d - b e} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right )}{2 \, c^{2} d^{2} e^{2} - b c d e^{3} + {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x}, -\frac {2 \, {\left (\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c d - b e\right )} \sqrt {e x + d} g + {\left (c d e f - c d^{2} g + {\left (c e^{2} f - c d e g\right )} x\right )} \sqrt {-2 \, c d + b e} \arctan \left (-\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {-2 \, c d + b e} \sqrt {e x + d}}{2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x}\right )\right )}}{2 \, c^{2} d^{2} e^{2} - b c d e^{3} + {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x}\right ] \] Input:

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

Output:

[-(2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*d - b*e)*sqrt(e*x + d 
)*g + (c*d*e*f - c*d^2*g + (c*e^2*f - c*d*e*g)*x)*sqrt(2*c*d - b*e)*log(-( 
c*e^2*x^2 - 3*c*d^2 + 2*b*d*e - 2*(c*d*e - b*e^2)*x - 2*sqrt(-c*e^2*x^2 - 
b*e^2*x + c*d^2 - b*d*e)*sqrt(2*c*d - b*e)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e 
*x + d^2)))/(2*c^2*d^2*e^2 - b*c*d*e^3 + (2*c^2*d*e^3 - b*c*e^4)*x), -2*(s 
qrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*d - b*e)*sqrt(e*x + d)*g + 
(c*d*e*f - c*d^2*g + (c*e^2*f - c*d*e*g)*x)*sqrt(-2*c*d + b*e)*arctan(-sqr 
t(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(-2*c*d + b*e)*sqrt(e*x + d)/( 
2*c*d^2 - b*d*e + (2*c*d*e - b*e^2)*x)))/(2*c^2*d^2*e^2 - b*c*d*e^3 + (2*c 
^2*d*e^3 - b*c*e^4)*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.62 \[ \int \frac {f+g x}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {2 \, {\left (\frac {{\left (e f - d g\right )} \arctan \left (\frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{\sqrt {-2 \, c d + b e}} - \frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} g}{c}\right )}}{e^{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05 \[ \int \frac {f+g x}{\sqrt {d+e x} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {-2 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c d g +2 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c e f -2 \sqrt {-c e x -b e +c d}\, b e g +4 \sqrt {-c e x -b e +c d}\, c d g}{c \,e^{2} \left (b e -2 c d \right )} \] Input:

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

Output:

(2*( - sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d) 
)*c*d*g + sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c 
*d))*c*e*f - sqrt( - b*e + c*d - c*e*x)*b*e*g + 2*sqrt( - b*e + c*d - c*e* 
x)*c*d*g))/(c*e**2*(b*e - 2*c*d))