\(\int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx\) [139]

Optimal result

Integrand size = 44, antiderivative size = 143 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx=\frac {(4 c e f-4 c d g+b e g+2 c e g x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c e^2}+\frac {(2 c d-b e) (4 c e f-2 c d g-b e g) \arctan \left (\frac {\sqrt {c} (d+e x)}{\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{4 c^{3/2} e^2} \] Output:

1/4*(2*c*e*g*x+b*e*g-4*c*d*g+4*c*e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/ 
2)/c/e^2+1/4*(-b*e+2*c*d)*(-b*e*g-2*c*d*g+4*c*e*f)*arctan(c^(1/2)*(e*x+d)/ 
(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2))/c^(3/2)/e^2
 

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.27 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx=\frac {\sqrt {d+e x} \sqrt {c d-b e-c e x} \left (\sqrt {d+e x} \sqrt {c d-b e-c e x} (b e g+2 c (2 e f-2 d g+e g x))+\sqrt {-\frac {1}{c}} (2 c d-b e) (-4 c e f+2 c d g+b e g) \log \left (\sqrt {d+e x}+\left (-\frac {1}{c}\right )^{3/2} c \sqrt {c d-b e-c e x}\right )\right )}{4 c e^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \] Input:

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

Output:

(Sqrt[d + e*x]*Sqrt[c*d - b*e - c*e*x]*(Sqrt[d + e*x]*Sqrt[c*d - b*e - c*e 
*x]*(b*e*g + 2*c*(2*e*f - 2*d*g + e*g*x)) + Sqrt[-c^(-1)]*(2*c*d - b*e)*(- 
4*c*e*f + 2*c*d*g + b*e*g)*Log[Sqrt[d + e*x] + (-c^(-1))^(3/2)*c*Sqrt[c*d 
- b*e - c*e*x]]))/(4*c*e^2*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1215, 1225, 1092, 217}

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.

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

\(\Big \downarrow \) 1215

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

\(\Big \downarrow \) 1225

\(\displaystyle \frac {(2 c d-b e) (-b e g-2 c d g+4 c e f) \int \frac {1}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{8 c e}+\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (b e g-4 c d g+4 c e f+2 c e g x)}{4 c e^2}\)

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {(2 c d-b e) \arctan \left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right ) (-b e g-2 c d g+4 c e f)}{8 c^{3/2} e^2}+\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (b e g-4 c d g+4 c e f+2 c e g x)}{4 c e^2}\)

Input:

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

Output:

((4*c*e*f - 4*c*d*g + b*e*g + 2*c*e*g*x)*Sqrt[d*(c*d - b*e) - b*e^2*x - c* 
e^2*x^2])/(4*c*e^2) + ((2*c*d - b*e)*(4*c*e*f - 2*c*d*g - b*e*g)*ArcTan[(e 
*(b + 2*c*x))/(2*Sqrt[c]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])])/(8*c 
^(3/2)*e^2)
 

Defintions of rubi rules used

rule 217

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

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1215

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

rule 1225

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 
 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), 
 x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p 
+ 3))/(2*c^2*(2*p + 3))   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, p}, x] &&  !LeQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(292\) vs. \(2(131)=262\).

Time = 2.45 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.05

method	result	size
default	\(\frac {g \left (-\frac {\left (-2 c \,e^{2} x -b \,e^{2}\right ) \sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}{4 c \,e^{2}}-\frac {\left (-4 c \,e^{2} \left (-b d e +c \,d^{2}\right )-b^{2} e^{4}\right ) \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {b}{2 c}\right )}{\sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}\right )}{8 c \,e^{2} \sqrt {c \,e^{2}}}\right )}{e}-\frac {\left (d g -e f \right ) \left (\sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )}+\frac {\left (-b \,e^{2}+2 d e c \right ) \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {d}{e}-\frac {-b \,e^{2}+2 d e c}{2 c \,e^{2}}\right )}{\sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {c \,e^{2}}}\right )}{e^{2}}\)	\(293\)

Input:

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

Output:

g/e*(-1/4*(-2*c*e^2*x-b*e^2)/c/e^2*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)- 
1/8*(-4*c*e^2*(-b*d*e+c*d^2)-b^2*e^4)/c/e^2/(c*e^2)^(1/2)*arctan((c*e^2)^( 
1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)))-(d*g-e*f)/e^2*(( 
-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)+1/2*(-b*e^2+2*c*d*e)/(c*e 
^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-c*e^2* 
(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.76 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx=\left [-\frac {{\left (4 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} f - {\left (4 \, c^{2} d^{2} - b^{2} e^{2}\right )} g\right )} \sqrt {-c} \log \left (8 \, c^{2} e^{2} x^{2} + 8 \, b c e^{2} x - 4 \, c^{2} d^{2} + 4 \, b c d e + b^{2} e^{2} - 4 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {-c}\right ) - 4 \, {\left (2 \, c^{2} e g x + 4 \, c^{2} e f - {\left (4 \, c^{2} d - b c e\right )} g\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}{16 \, c^{2} e^{2}}, -\frac {{\left (4 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} f - {\left (4 \, c^{2} d^{2} - b^{2} e^{2}\right )} g\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {c}}{2 \, {\left (c^{2} e^{2} x^{2} + b c e^{2} x - c^{2} d^{2} + b c d e\right )}}\right ) - 2 \, {\left (2 \, c^{2} e g x + 4 \, c^{2} e f - {\left (4 \, c^{2} d - b c e\right )} g\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}}{8 \, c^{2} e^{2}}\right ] \] Input:

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

Output:

[-1/16*((4*(2*c^2*d*e - b*c*e^2)*f - (4*c^2*d^2 - b^2*e^2)*g)*sqrt(-c)*log 
(8*c^2*e^2*x^2 + 8*b*c*e^2*x - 4*c^2*d^2 + 4*b*c*d*e + b^2*e^2 - 4*sqrt(-c 
*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(-c)) - 4*(2*c^2*e 
*g*x + 4*c^2*e*f - (4*c^2*d - b*c*e)*g)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 
- b*d*e))/(c^2*e^2), -1/8*((4*(2*c^2*d*e - b*c*e^2)*f - (4*c^2*d^2 - b^2*e 
^2)*g)*sqrt(c)*arctan(1/2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c* 
e*x + b*e)*sqrt(c)/(c^2*e^2*x^2 + b*c*e^2*x - c^2*d^2 + b*c*d*e)) - 2*(2*c 
^2*e*g*x + 4*c^2*e*f - (4*c^2*d - b*c*e)*g)*sqrt(-c*e^2*x^2 - b*e^2*x + c* 
d^2 - b*d*e))/(c^2*e^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/(e*x+d),x, algori 
thm="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(b*e-2*c*d>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.24 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx=\frac {1}{4} \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (\frac {2 \, g x}{e} + \frac {4 \, c e^{2} f - 4 \, c d e g + b e^{2} g}{c e^{3}}\right )} - \frac {{\left (8 \, c^{2} d e f - 4 \, b c e^{2} f - 4 \, c^{2} d^{2} g + b^{2} e^{2} g\right )} \log \left ({\left | -b e^{2} + 2 \, {\left (\sqrt {-c e^{2}} x - \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}\right )} \sqrt {-c} {\left | e \right |} \right |}\right )}{8 \, \sqrt {-c} c e {\left | e \right |}} \] Input:

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

Output:

1/4*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*g*x/e + (4*c*e^2*f - 4*c 
*d*e*g + b*e^2*g)/(c*e^3)) - 1/8*(8*c^2*d*e*f - 4*b*c*e^2*f - 4*c^2*d^2*g 
+ b^2*e^2*g)*log(abs(-b*e^2 + 2*(sqrt(-c*e^2)*x - sqrt(-c*e^2*x^2 - b*e^2* 
x + c*d^2 - b*d*e))*sqrt(-c)*abs(e)))/(sqrt(-c)*c*e*abs(e))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.43 \[ \int \frac {(f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx=\frac {i \left (\sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) b^{3} e^{3} g -2 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) b^{2} c d \,e^{2} g -4 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) b^{2} c \,e^{3} f -4 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) b \,c^{2} d^{2} e g +16 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) b \,c^{2} d \,e^{2} f +8 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) c^{3} d^{3} g -16 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) c^{3} d^{2} e f +\sqrt {e x +d}\, \sqrt {b e -2 c d}\, \sqrt {-b e +2 c d}\, \sqrt {-c e x -b e +c d}\, b c e g -4 \sqrt {e x +d}\, \sqrt {b e -2 c d}\, \sqrt {-b e +2 c d}\, \sqrt {-c e x -b e +c d}\, c^{2} d g +4 \sqrt {e x +d}\, \sqrt {b e -2 c d}\, \sqrt {-b e +2 c d}\, \sqrt {-c e x -b e +c d}\, c^{2} e f +2 \sqrt {e x +d}\, \sqrt {b e -2 c d}\, \sqrt {-b e +2 c d}\, \sqrt {-c e x -b e +c d}\, c^{2} e g x \right )}{4 c^{2} e^{2} \left (b e -2 c d \right )} \] Input:

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

Output:

(i*(sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b** 
3*e**3*g - 2*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2* 
c*d))*b**2*c*d*e**2*g - 4*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqr 
t( - b*e + 2*c*d))*b**2*c*e**3*f - 4*sqrt(c)*asinh((sqrt( - b*e + c*d - c* 
e*x)*i)/sqrt( - b*e + 2*c*d))*b*c**2*d**2*e*g + 16*sqrt(c)*asinh((sqrt( - 
b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b*c**2*d*e**2*f + 8*sqrt(c)*as 
inh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*c**3*d**3*g - 16* 
sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*c**3*d* 
*2*e*f + sqrt(d + e*x)*sqrt(b*e - 2*c*d)*sqrt( - b*e + 2*c*d)*sqrt( - b*e 
+ c*d - c*e*x)*b*c*e*g - 4*sqrt(d + e*x)*sqrt(b*e - 2*c*d)*sqrt( - b*e + 2 
*c*d)*sqrt( - b*e + c*d - c*e*x)*c**2*d*g + 4*sqrt(d + e*x)*sqrt(b*e - 2*c 
*d)*sqrt( - b*e + 2*c*d)*sqrt( - b*e + c*d - c*e*x)*c**2*e*f + 2*sqrt(d + 
e*x)*sqrt(b*e - 2*c*d)*sqrt( - b*e + 2*c*d)*sqrt( - b*e + c*d - c*e*x)*c** 
2*e*g*x))/(4*c**2*e**2*(b*e - 2*c*d))