\(\int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\) [231]

Optimal result

Integrand size = 46, antiderivative size = 233 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (2 c d-b e) (d+e x)^{5/2}}-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 e^2 (2 c d-b e)^2 (d+e x)^{3/2}}-\frac {c (3 c e f+5 c d g-4 b e g) \text {arctanh}\left (\frac {\sqrt {2 c d-b e} \sqrt {d+e x}}{\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{4 e^2 (2 c d-b e)^{5/2}} \] Output:

-1/2*(-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)^(5/2)-1/4*(-4*b*e*g+5*c*d*g+3*c*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)^2/(e*x+d)^(3/2)-1/4*c*(-4*b*e*g+5*c*d*g+3*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)^(5/2)
 

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.84 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {c \sqrt {d+e x} \left (-\frac {(-b e+c (d-e x)) \left (-2 b e (d g+e (f+2 g x))+c \left (d^2 g+3 e^2 f x+d e (7 f+5 g x)\right )\right )}{c (-2 c d+b e)^2 (d+e x)^2}+\frac {(3 c e f+5 c d g-4 b e g) \sqrt {c d-b e-c e x} \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {-2 c d+b e}}\right )}{(-2 c d+b e)^{5/2}}\right )}{4 e^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \] Input:

Integrate[(f + g*x)/((d + e*x)^(5/2)*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2* 
x^2]),x]
 

Output:

(c*Sqrt[d + e*x]*(-(((-(b*e) + c*(d - e*x))*(-2*b*e*(d*g + e*(f + 2*g*x)) 
+ c*(d^2*g + 3*e^2*f*x + d*e*(7*f + 5*g*x))))/(c*(-2*c*d + b*e)^2*(d + e*x 
)^2)) + ((3*c*e*f + 5*c*d*g - 4*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)^(5/2)))/(4*e^2*Sqr 
t[(d + e*x)*(-(b*e) + c*(d - e*x))])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1220, 1135, 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.

\(\displaystyle \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(-4 b e g+5 c d g+3 c e f) \int \frac {1}{(d+e x)^{3/2} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{4 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {(-4 b e g+5 c d g+3 c e f) \left (\frac {c \int \frac {1}{\sqrt {d+e x} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{2 (2 c d-b e)}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2} (2 c d-b e)}\right )}{4 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(-4 b e g+5 c d g+3 c e f) \left (\frac {c e \int \frac {1}{\frac {e^2 \left (-c x^2 e^2-b x e^2+d (c d-b e)\right )}{d+e x}-e^2 (2 c d-b e)}d\frac {\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}{\sqrt {d+e x}}}{2 c d-b e}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2} (2 c d-b e)}\right )}{4 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\left (-\frac {c \text {arctanh}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e (2 c d-b e)^{3/2}}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2} (2 c d-b e)}\right ) (-4 b e g+5 c d g+3 c e f)}{4 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}\)

Input:

Int[(f + g*x)/((d + e*x)^(5/2)*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2]), 
x]
 

Output:

-1/2*((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)^(5/2)) + ((3*c*e*f + 5*c*d*g - 4*b*e*g)*(-(Sqrt[d*(c*d - b 
*e) - b*e^2*x - c*e^2*x^2]/(e*(2*c*d - b*e)*(d + e*x)^(3/2))) - (c*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*c*d - b*e)^(3/2))))/(4*e*(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 1135

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* 
c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e)))   Int 
[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I 
ntegerQ[2*p]
 

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. \(621\) vs. \(2(211)=422\).

Time = 1.56 (sec) , antiderivative size = 622, normalized size of antiderivative = 2.67

method	result	size
default	\(-\frac {\sqrt {-\left (e x +d \right ) \left (c e x +b e -c d \right )}\, \left (4 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c \,e^{3} g \,x^{2}-5 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d \,e^{2} g \,x^{2}-3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} e^{3} f \,x^{2}+8 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c d \,e^{2} g x -10 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{2} e g x -6 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d \,e^{2} f x +4 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c \,d^{2} e g -5 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{3} g -3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{2} e f -4 b \,e^{2} g x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+5 c d e g x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+3 c \,e^{2} f x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-2 b d e g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-2 b \,e^{2} f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+c \,d^{2} g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+7 c d e f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\right )}{4 \left (e x +d \right )^{\frac {5}{2}} \left (b e -2 c d \right )^{\frac {5}{2}} e^{2} \sqrt {-c e x -b e +c d}}\)	\(622\)

Input:

int((g*x+f)/(e*x+d)^(5/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x,method= 
_RETURNVERBOSE)
 

Output:

-1/4*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)*(4*arctan((-c*e*x-b*e+c*d)^(1/2)/(b* 
e-2*c*d)^(1/2))*b*c*e^3*g*x^2-5*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^ 
(1/2))*c^2*d*e^2*g*x^2-3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))* 
c^2*e^3*f*x^2+8*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*d*e^2 
*g*x-10*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d^2*e*g*x-6*a 
rctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d*e^2*f*x+4*arctan((-c 
*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*d^2*e*g-5*arctan((-c*e*x-b*e+c* 
d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d^3*g-3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e 
-2*c*d)^(1/2))*c^2*d^2*e*f-4*b*e^2*g*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^ 
(1/2)+5*c*d*e*g*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+3*c*e^2*f*x*(-c 
*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-2*b*d*e*g*(-c*e*x-b*e+c*d)^(1/2)*(b* 
e-2*c*d)^(1/2)-2*b*e^2*f*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+c*d^2*g* 
(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+7*c*d*e*f*(-c*e*x-b*e+c*d)^(1/2)* 
(b*e-2*c*d)^(1/2))/(e*x+d)^(5/2)/(b*e-2*c*d)^(5/2)/e^2/(-c*e*x-b*e+c*d)^(1 
/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 570 vs. \(2 (211) = 422\).

Time = 0.13 (sec) , antiderivative size = 1170, normalized size of antiderivative = 5.02 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\text {Too large to display} \] Input:

integrate((g*x+f)/(e*x+d)^(5/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, 
algorithm="fricas")
 

Output:

[-1/8*((3*c^2*d^3*e*f + (3*c^2*e^4*f + (5*c^2*d*e^3 - 4*b*c*e^4)*g)*x^3 + 
3*(3*c^2*d*e^3*f + (5*c^2*d^2*e^2 - 4*b*c*d*e^3)*g)*x^2 + (5*c^2*d^4 - 4*b 
*c*d^3*e)*g + 3*(3*c^2*d^2*e^2*f + (5*c^2*d^3*e - 4*b*c*d^2*e^2)*g)*x)*sqr 
t(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)*((14*c^2*d^2*e - 11*b*c*d*e^2 + 2*b^2*e^3)*f + (2*c^2*d^3 - 5*b*c*d^2 
*e + 2*b^2*d*e^2)*g + (3*(2*c^2*d*e^2 - b*c*e^3)*f + (10*c^2*d^2*e - 13*b* 
c*d*e^2 + 4*b^2*e^3)*g)*x)*sqrt(e*x + d))/(8*c^3*d^6*e^2 - 12*b*c^2*d^5*e^ 
3 + 6*b^2*c*d^4*e^4 - b^3*d^3*e^5 + (8*c^3*d^3*e^5 - 12*b*c^2*d^2*e^6 + 6* 
b^2*c*d*e^7 - b^3*e^8)*x^3 + 3*(8*c^3*d^4*e^4 - 12*b*c^2*d^3*e^5 + 6*b^2*c 
*d^2*e^6 - b^3*d*e^7)*x^2 + 3*(8*c^3*d^5*e^3 - 12*b*c^2*d^4*e^4 + 6*b^2*c* 
d^3*e^5 - b^3*d^2*e^6)*x), -1/4*((3*c^2*d^3*e*f + (3*c^2*e^4*f + (5*c^2*d* 
e^3 - 4*b*c*e^4)*g)*x^3 + 3*(3*c^2*d*e^3*f + (5*c^2*d^2*e^2 - 4*b*c*d*e^3) 
*g)*x^2 + (5*c^2*d^4 - 4*b*c*d^3*e)*g + 3*(3*c^2*d^2*e^2*f + (5*c^2*d^3*e 
- 4*b*c*d^2*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)*((14*c^2*d 
^2*e - 11*b*c*d*e^2 + 2*b^2*e^3)*f + (2*c^2*d^3 - 5*b*c*d^2*e + 2*b^2*d*e^ 
2)*g + (3*(2*c^2*d*e^2 - b*c*e^3)*f + (10*c^2*d^2*e - 13*b*c*d*e^2 + 4*...
 

Sympy [F]

\[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {f + g x}{\sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )} \left (d + e x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((g*x+f)/(e*x+d)**(5/2)/(-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)**(5/2)) 
, x)
 

Maxima [F]

\[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int { \frac {g x + f}{\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((g*x+f)/(e*x+d)^(5/2)/(-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)^ 
(5/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.53 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {\frac {{\left (3 \, c^{3} e f + 5 \, c^{3} d g - 4 \, b c^{2} e g\right )} \arctan \left (\frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{{\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sqrt {-2 \, c d + b e}} - \frac {10 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{4} d e f - 5 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{3} e^{2} f + 6 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{4} d^{2} g - 11 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{3} d e g + 4 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b^{2} c^{2} e^{2} g - 3 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{3} e f - 5 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{3} d g + 4 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} b c^{2} e g}{{\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} {\left (e x + d\right )}^{2} c^{2}}}{4 \, c e^{2}} \] Input:

integrate((g*x+f)/(e*x+d)^(5/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, 
algorithm="giac")
 

Output:

1/4*((3*c^3*e*f + 5*c^3*d*g - 4*b*c^2*e*g)*arctan(sqrt(-(e*x + d)*c + 2*c* 
d - b*e)/sqrt(-2*c*d + b*e))/((4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*sqrt(-2*c* 
d + b*e)) - (10*sqrt(-(e*x + d)*c + 2*c*d - b*e)*c^4*d*e*f - 5*sqrt(-(e*x 
+ d)*c + 2*c*d - b*e)*b*c^3*e^2*f + 6*sqrt(-(e*x + d)*c + 2*c*d - b*e)*c^4 
*d^2*g - 11*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b*c^3*d*e*g + 4*sqrt(-(e*x + 
d)*c + 2*c*d - b*e)*b^2*c^2*e^2*g - 3*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c 
^3*e*f - 5*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c^3*d*g + 4*(-(e*x + d)*c + 
2*c*d - b*e)^(3/2)*b*c^2*e*g)/((4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*(e*x + d) 
^2*c^2))/(c*e^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {f+g\,x}{{\left (d+e\,x\right )}^{5/2}\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}} \,d x \] Input:

int((f + g*x)/((d + e*x)^(5/2)*(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)^(5/2)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2) 
), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 828, normalized size of antiderivative = 3.55 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {-4 \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 c \,d^{2} e g -8 \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 c d \,e^{2} g x -4 \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 c \,e^{3} g \,x^{2}+5 \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^{2} d^{3} g +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^{2} d^{2} e f +10 \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^{2} d^{2} e g x +6 \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^{2} d \,e^{2} f x +5 \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^{2} d \,e^{2} g \,x^{2}+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^{2} e^{3} f \,x^{2}+2 \sqrt {-c e x -b e +c d}\, b^{2} d \,e^{2} g +2 \sqrt {-c e x -b e +c d}\, b^{2} e^{3} f +4 \sqrt {-c e x -b e +c d}\, b^{2} e^{3} g x -5 \sqrt {-c e x -b e +c d}\, b c \,d^{2} e g -11 \sqrt {-c e x -b e +c d}\, b c d \,e^{2} f -13 \sqrt {-c e x -b e +c d}\, b c d \,e^{2} g x -3 \sqrt {-c e x -b e +c d}\, b c \,e^{3} f x +2 \sqrt {-c e x -b e +c d}\, c^{2} d^{3} g +14 \sqrt {-c e x -b e +c d}\, c^{2} d^{2} e f +10 \sqrt {-c e x -b e +c d}\, c^{2} d^{2} e g x +6 \sqrt {-c e x -b e +c d}\, c^{2} d \,e^{2} f x}{4 e^{2} \left (b^{3} e^{5} x^{2}-6 b^{2} c d \,e^{4} x^{2}+12 b \,c^{2} d^{2} e^{3} x^{2}-8 c^{3} d^{3} e^{2} x^{2}+2 b^{3} d \,e^{4} x -12 b^{2} c \,d^{2} e^{3} x +24 b \,c^{2} d^{3} e^{2} x -16 c^{3} d^{4} e x +b^{3} d^{2} e^{3}-6 b^{2} c \,d^{3} e^{2}+12 b \,c^{2} d^{4} e -8 c^{3} d^{5}\right )} \] Input:

int((g*x+f)/(e*x+d)^(5/2)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x)
 

Output:

( - 4*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d)) 
*b*c*d**2*e*g - 8*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b 
*e - 2*c*d))*b*c*d*e**2*g*x - 4*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - 
 c*e*x)/sqrt(b*e - 2*c*d))*b*c*e**3*g*x**2 + 5*sqrt(b*e - 2*c*d)*atan(sqrt 
( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**2*d**3*g + 3*sqrt(b*e - 2*c*d 
)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**2*d**2*e*f + 10*sq 
rt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**2*d* 
*2*e*g*x + 6*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 
2*c*d))*c**2*d*e**2*f*x + 5*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e 
*x)/sqrt(b*e - 2*c*d))*c**2*d*e**2*g*x**2 + 3*sqrt(b*e - 2*c*d)*atan(sqrt( 
 - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**2*e**3*f*x**2 + 2*sqrt( - b*e 
+ c*d - c*e*x)*b**2*d*e**2*g + 2*sqrt( - b*e + c*d - c*e*x)*b**2*e**3*f + 
4*sqrt( - b*e + c*d - c*e*x)*b**2*e**3*g*x - 5*sqrt( - b*e + c*d - c*e*x)* 
b*c*d**2*e*g - 11*sqrt( - b*e + c*d - c*e*x)*b*c*d*e**2*f - 13*sqrt( - b*e 
 + c*d - c*e*x)*b*c*d*e**2*g*x - 3*sqrt( - b*e + c*d - c*e*x)*b*c*e**3*f*x 
 + 2*sqrt( - b*e + c*d - c*e*x)*c**2*d**3*g + 14*sqrt( - b*e + c*d - c*e*x 
)*c**2*d**2*e*f + 10*sqrt( - b*e + c*d - c*e*x)*c**2*d**2*e*g*x + 6*sqrt( 
- b*e + c*d - c*e*x)*c**2*d*e**2*f*x)/(4*e**2*(b**3*d**2*e**3 + 2*b**3*d*e 
**4*x + b**3*e**5*x**2 - 6*b**2*c*d**3*e**2 - 12*b**2*c*d**2*e**3*x - 6*b* 
*2*c*d*e**4*x**2 + 12*b*c**2*d**4*e + 24*b*c**2*d**3*e**2*x + 12*b*c**2...