\(\int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\) [230]

Optimal result

Integrand size = 46, antiderivative size = 153 \[ \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (2 c d-b e) (d+e x)^{3/2}}-\frac {(c e f+3 c d g-2 b e g) \text {arctanh}\left (\frac {\sqrt {2 c d-b e} \sqrt {d+e x}}{\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{e^2 (2 c d-b e)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00 \[ \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {\sqrt {d+e x} \left (\frac {(e f-d g) (-c d+b e+c e x)}{(2 c d-b e) (d+e x)}-\frac {(c e f+3 c d g-2 b e g) \sqrt {c d-b e-c e x} \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {-2 c d+b e}}\right )}{(-2 c d+b e)^{3/2}}\right )}{e^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \] Input:

Integrate[(f + g*x)/((d + e*x)^(3/2)*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2* 
x^2]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 153, 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 = {1220, 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.

\(\displaystyle \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(-2 b e g+3 c d g+c e f) \int \frac {1}{\sqrt {d+e x} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{2 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (d+e x)^{3/2} (2 c d-b e)}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(-2 b e g+3 c d g+c e f) \int \frac {1}{\frac {e^2 \left (-c x^2 e^2-b x e^2+d (c d-b e)\right )}{d+e x}-e^2 (2 c d-b e)}d\frac {\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}{\sqrt {d+e x}}}{2 c d-b e}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (d+e x)^{3/2} (2 c d-b e)}\)

\(\Big \downarrow \) 221

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

Input:

Int[(f + g*x)/((d + e*x)^(3/2)*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2]), 
x]
 

Output:

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

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 1220

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(319\) vs. \(2(141)=282\).

Time = 1.68 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.09

method	result	size
default	\(\frac {\left (2 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,e^{2} g x -3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c d e g x -\arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c \,e^{2} f x +2 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b d e g -3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c \,d^{2} g -\arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c d e f -\sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, d g +\sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\, e f \right ) \sqrt {-\left (e x +d \right ) \left (c e x +b e -c d \right )}}{\left (b e -2 c d \right )^{\frac {3}{2}} e^{2} \sqrt {-c e x -b e +c d}\, \left (e x +d \right )^{\frac {3}{2}}}\)	\(320\)

Input:

int((g*x+f)/(e*x+d)^(3/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x,method= 
_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (141) = 282\).

Time = 0.10 (sec) , antiderivative size = 700, normalized size of antiderivative = 4.58 \[ \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\left [\frac {{\left (c d^{2} e f + {\left (c e^{3} f + {\left (3 \, c d e^{2} - 2 \, b e^{3}\right )} g\right )} x^{2} + {\left (3 \, c d^{3} - 2 \, b d^{2} e\right )} g + 2 \, {\left (c d e^{2} f + {\left (3 \, c d^{2} e - 2 \, b d e^{2}\right )} g\right )} x\right )} \sqrt {2 \, c d - b e} \log \left (-\frac {c e^{2} x^{2} - 3 \, c d^{2} + 2 \, b d e - 2 \, {\left (c d e - b e^{2}\right )} x + 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {2 \, c d - b e} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left ({\left (2 \, c d e - b e^{2}\right )} f - {\left (2 \, c d^{2} - b d e\right )} g\right )} \sqrt {e x + d}}{2 \, {\left (4 \, c^{2} d^{4} e^{2} - 4 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4} + {\left (4 \, c^{2} d^{2} e^{4} - 4 \, b c d e^{5} + b^{2} e^{6}\right )} x^{2} + 2 \, {\left (4 \, c^{2} d^{3} e^{3} - 4 \, b c d^{2} e^{4} + b^{2} d e^{5}\right )} x\right )}}, -\frac {{\left (c d^{2} e f + {\left (c e^{3} f + {\left (3 \, c d e^{2} - 2 \, b e^{3}\right )} g\right )} x^{2} + {\left (3 \, c d^{3} - 2 \, b d^{2} e\right )} g + 2 \, {\left (c d e^{2} f + {\left (3 \, c d^{2} e - 2 \, b d e^{2}\right )} g\right )} x\right )} \sqrt {-2 \, c d + b e} \arctan \left (-\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {-2 \, c d + b e} \sqrt {e x + d}}{2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x}\right ) + \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left ({\left (2 \, c d e - b e^{2}\right )} f - {\left (2 \, c d^{2} - b d e\right )} g\right )} \sqrt {e x + d}}{4 \, c^{2} d^{4} e^{2} - 4 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4} + {\left (4 \, c^{2} d^{2} e^{4} - 4 \, b c d e^{5} + b^{2} e^{6}\right )} x^{2} + 2 \, {\left (4 \, c^{2} d^{3} e^{3} - 4 \, b c d^{2} e^{4} + b^{2} d e^{5}\right )} x}\right ] \] Input:

integrate((g*x+f)/(e*x+d)^(3/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, 
algorithm="fricas")
 

Output:

[1/2*((c*d^2*e*f + (c*e^3*f + (3*c*d*e^2 - 2*b*e^3)*g)*x^2 + (3*c*d^3 - 2* 
b*d^2*e)*g + 2*(c*d*e^2*f + (3*c*d^2*e - 2*b*d*e^2)*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*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((2*c*d* 
e - b*e^2)*f - (2*c*d^2 - b*d*e)*g)*sqrt(e*x + d))/(4*c^2*d^4*e^2 - 4*b*c* 
d^3*e^3 + b^2*d^2*e^4 + (4*c^2*d^2*e^4 - 4*b*c*d*e^5 + b^2*e^6)*x^2 + 2*(4 
*c^2*d^3*e^3 - 4*b*c*d^2*e^4 + b^2*d*e^5)*x), -((c*d^2*e*f + (c*e^3*f + (3 
*c*d*e^2 - 2*b*e^3)*g)*x^2 + (3*c*d^3 - 2*b*d^2*e)*g + 2*(c*d*e^2*f + (3*c 
*d^2*e - 2*b*d*e^2)*g)*x)*sqrt(-2*c*d + b*e)*arctan(-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 + 
(2*c*d*e - b*e^2)*x)) + sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((2*c*d 
*e - b*e^2)*f - (2*c*d^2 - b*d*e)*g)*sqrt(e*x + d))/(4*c^2*d^4*e^2 - 4*b*c 
*d^3*e^3 + b^2*d^2*e^4 + (4*c^2*d^2*e^4 - 4*b*c*d*e^5 + b^2*e^6)*x^2 + 2*( 
4*c^2*d^3*e^3 - 4*b*c*d^2*e^4 + b^2*d*e^5)*x)]
 

Sympy [F]

\[ \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {f + g x}{\sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )} \left (d + e x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((g*x+f)/(e*x+d)**(3/2)/(-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))*(d + e*x)**(3/2)) 
, x)
 

Maxima [F]

\[ \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int { \frac {g x + f}{\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((g*x+f)/(e*x+d)^(3/2)/(-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)*(e*x + d)^ 
(3/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.02 \[ \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {\frac {{\left (c^{2} e f + 3 \, c^{2} d g - 2 \, b c e g\right )} \arctan \left (\frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{{\left (2 \, c d - b e\right )} \sqrt {-2 \, c d + b e}} - \frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{2} e f - \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{2} d g}{{\left (2 \, c d - b e\right )} {\left (e x + d\right )} c}}{c e^{2}} \] Input:

integrate((g*x+f)/(e*x+d)^(3/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, 
algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {f+g\,x}{{\left (d+e\,x\right )}^{3/2}\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}} \,d x \] Input:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.60 \[ \int \frac {f+g x}{(d+e x)^{3/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {2 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b d e g +2 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,e^{2} g x -3 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c \,d^{2} g -\sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c d e f -3 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c d e g x -\sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c \,e^{2} f x -\sqrt {-c e x -b e +c d}\, b d e g +\sqrt {-c e x -b e +c d}\, b \,e^{2} f +2 \sqrt {-c e x -b e +c d}\, c \,d^{2} g -2 \sqrt {-c e x -b e +c d}\, c d e f}{e^{2} \left (b^{2} e^{3} x -4 b c d \,e^{2} x +4 c^{2} d^{2} e x +b^{2} d \,e^{2}-4 b c \,d^{2} e +4 c^{2} d^{3}\right )} \] Input:

int((g*x+f)/(e*x+d)^(3/2)/(-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))*b* 
d*e*g + 2*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c 
*d))*b*e**2*g*x - 3*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt 
(b*e - 2*c*d))*c*d**2*g - sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x 
)/sqrt(b*e - 2*c*d))*c*d*e*f - 3*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d 
- c*e*x)/sqrt(b*e - 2*c*d))*c*d*e*g*x - sqrt(b*e - 2*c*d)*atan(sqrt( - b*e 
 + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c*e**2*f*x - sqrt( - b*e + c*d - c*e*x) 
*b*d*e*g + sqrt( - b*e + c*d - c*e*x)*b*e**2*f + 2*sqrt( - b*e + c*d - c*e 
*x)*c*d**2*g - 2*sqrt( - b*e + c*d - c*e*x)*c*d*e*f)/(e**2*(b**2*d*e**2 + 
b**2*e**3*x - 4*b*c*d**2*e - 4*b*c*d*e**2*x + 4*c**2*d**3 + 4*c**2*d**2*e* 
x))