\(\int \frac {(f+g x) (c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\) [209]

Optimal result

Integrand size = 46, antiderivative size = 250 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 (2 c d-b e) (e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 \sqrt {d+e x}}+\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (d+e x)^{3/2}}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 c e^2 (d+e x)^{5/2}}-\frac {2 (2 c d-b e)^{3/2} (e f-d g) \text {arctanh}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {2 c d-b e} \sqrt {d+e x}}\right )}{e^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.78 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 ((d+e x) (-b e+c (d-e x)))^{3/2} \left (\frac {-3 b^2 e^2 g-2 b c e (10 e f-13 d g+3 e g x)+c^2 \left (-38 d^2 g-e^2 x (5 f+3 g x)+d e (35 f+11 g x)\right )}{c (-b e+c (d-e x))}-\frac {15 (-2 c d+b e)^{3/2} (-e f+d g) \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {-2 c d+b e}}\right )}{(-b e+c (d-e x))^{3/2}}\right )}{15 e^2 (d+e x)^{3/2}} \] Input:

Integrate[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2))/(d + e*x 
)^(5/2),x]
 

Output:

(2*((d + e*x)*(-(b*e) + c*(d - e*x)))^(3/2)*((-3*b^2*e^2*g - 2*b*c*e*(10*e 
*f - 13*d*g + 3*e*g*x) + c^2*(-38*d^2*g - e^2*x*(5*f + 3*g*x) + d*e*(35*f 
+ 11*g*x)))/(c*(-(b*e) + c*(d - e*x))) - (15*(-2*c*d + b*e)^(3/2)*(-(e*f) 
+ d*g)*ArcTan[Sqrt[c*d - b*e - c*e*x]/Sqrt[-2*c*d + b*e]])/(-(b*e) + c*(d 
- e*x))^(3/2)))/(15*e^2*(d + e*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {1221, 1131, 1131, 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.

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1136

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

\(\Big \downarrow \) 221

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

Input:

Int[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(3/2))/(d + e*x)^(5/2 
),x]
 

Output:

(-2*g*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(5/2))/(5*c*e^2*(d + e*x)^(5/2 
)) + ((e*f - d*g)*((2*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(3*e*(d 
 + e*x)^(3/2)) + (2*c*d - b*e)*((2*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^ 
2])/(e*Sqrt[d + e*x]) - (2*Sqrt[2*c*d - b*e]*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)))/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 1131

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x 
] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1)))   Int[(d + e*x)^(m + 1)*(a + 
 b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b 
*d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne 
Q[m + 2*p + 1, 0] && IntegerQ[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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(592\) vs. \(2(226)=452\).

Time = 1.72 (sec) , antiderivative size = 593, normalized size of antiderivative = 2.37

method	result	size
default	\(-\frac {2 \sqrt {-\left (e x +d \right ) \left (c e x +b e -c d \right )}\, \left (3 c^{2} e^{2} g \,x^{2} \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+15 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b^{2} c d \,e^{2} g -15 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b^{2} c \,e^{3} f -60 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,c^{2} d^{2} e g +60 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,c^{2} d \,e^{2} f +60 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{3} g -60 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{2} e f +6 b c \,e^{2} g x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-11 c^{2} d e g x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+5 c^{2} e^{2} f x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+3 b^{2} e^{2} g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-26 b c d e g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+20 b c \,e^{2} f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+38 c^{2} d^{2} g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-35 c^{2} d e f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}\right )}{15 \sqrt {e x +d}\, \sqrt {-c e x -b e +c d}\, e^{2} c \sqrt {b e -2 c d}}\)	\(593\)

Input:

int((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x,method= 
_RETURNVERBOSE)
 

Output:

-2/15*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)*(3*c^2*e^2*g*x^2*(-c*e*x-b*e+c*d)^( 
1/2)*(b*e-2*c*d)^(1/2)+15*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2)) 
*b^2*c*d*e^2*g-15*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b^2*c*e 
^3*f-60*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c^2*d^2*e*g+60* 
arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c^2*d*e^2*f+60*arctan(( 
-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*d^3*g-60*arctan((-c*e*x-b*e+c 
*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*d^2*e*f+6*b*c*e^2*g*x*(-c*e*x-b*e+c*d)^(1 
/2)*(b*e-2*c*d)^(1/2)-11*c^2*d*e*g*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1 
/2)+5*c^2*e^2*f*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+3*b^2*e^2*g*(-c 
*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-26*b*c*d*e*g*(-c*e*x-b*e+c*d)^(1/2)* 
(b*e-2*c*d)^(1/2)+20*b*c*e^2*f*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+38 
*c^2*d^2*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-35*c^2*d*e*f*(-c*e*x-b 
*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2))/(e*x+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 = 0.11 (sec) , antiderivative size = 658, normalized size of antiderivative = 2.63 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\left [\frac {15 \, \sqrt {2 \, c d - b e} {\left ({\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} f - {\left (2 \, c^{2} d^{3} - b c d^{2} e\right )} g + {\left ({\left (2 \, c^{2} d e^{2} - b c e^{3}\right )} f - {\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} g\right )} x\right )} \log \left (-\frac {c e^{2} x^{2} - 3 \, c d^{2} + 2 \, b d e - 2 \, {\left (c d e - b e^{2}\right )} x + 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {2 \, c d - b e} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, {\left (3 \, c^{2} e^{2} g x^{2} - 5 \, {\left (7 \, c^{2} d e - 4 \, b c e^{2}\right )} f + {\left (38 \, c^{2} d^{2} - 26 \, b c d e + 3 \, b^{2} e^{2}\right )} g + {\left (5 \, c^{2} e^{2} f - {\left (11 \, c^{2} d e - 6 \, b c e^{2}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}{15 \, {\left (c e^{3} x + c d e^{2}\right )}}, -\frac {2 \, {\left (15 \, \sqrt {-2 \, c d + b e} {\left ({\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} f - {\left (2 \, c^{2} d^{3} - b c d^{2} e\right )} g + {\left ({\left (2 \, c^{2} d e^{2} - b c e^{3}\right )} f - {\left (2 \, c^{2} d^{2} e - b c d e^{2}\right )} g\right )} x\right )} \arctan \left (-\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {-2 \, c d + b e} \sqrt {e x + d}}{2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x}\right ) + {\left (3 \, c^{2} e^{2} g x^{2} - 5 \, {\left (7 \, c^{2} d e - 4 \, b c e^{2}\right )} f + {\left (38 \, c^{2} d^{2} - 26 \, b c d e + 3 \, b^{2} e^{2}\right )} g + {\left (5 \, c^{2} e^{2} f - {\left (11 \, c^{2} d e - 6 \, b c e^{2}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}\right )}}{15 \, {\left (c e^{3} x + c d e^{2}\right )}}\right ] \] Input:

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x, 
algorithm="fricas")
 

Output:

[1/15*(15*sqrt(2*c*d - b*e)*((2*c^2*d^2*e - b*c*d*e^2)*f - (2*c^2*d^3 - b* 
c*d^2*e)*g + ((2*c^2*d*e^2 - b*c*e^3)*f - (2*c^2*d^2*e - b*c*d*e^2)*g)*x)* 
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*(3*c^2*e^2*g*x^2 - 5*(7*c^2*d*e - 4*b*c*e^2)*f + (38* 
c^2*d^2 - 26*b*c*d*e + 3*b^2*e^2)*g + (5*c^2*e^2*f - (11*c^2*d*e - 6*b*c*e 
^2)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d))/(c*e^3 
*x + c*d*e^2), -2/15*(15*sqrt(-2*c*d + b*e)*((2*c^2*d^2*e - b*c*d*e^2)*f - 
 (2*c^2*d^3 - b*c*d^2*e)*g + ((2*c^2*d*e^2 - b*c*e^3)*f - (2*c^2*d^2*e - b 
*c*d*e^2)*g)*x)*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)) + (3*c^ 
2*e^2*g*x^2 - 5*(7*c^2*d*e - 4*b*c*e^2)*f + (38*c^2*d^2 - 26*b*c*d*e + 3*b 
^2*e^2)*g + (5*c^2*e^2*f - (11*c^2*d*e - 6*b*c*e^2)*g)*x)*sqrt(-c*e^2*x^2 
- b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d))/(c*e^3*x + c*d*e^2)]
 

Sympy [F]

\[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}} \left (f + g x\right )}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((g*x+f)*(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(3/2)/(e*x+d)**(5/ 
2),x)
 

Output:

Integral((-(d + e*x)*(b*e - c*d + c*e*x))**(3/2)*(f + g*x)/(d + e*x)**(5/2 
), x)
 

Maxima [F]

\[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int { \frac {{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}} {\left (g x + f\right )}}{{\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x, 
algorithm="maxima")
 

Output:

integrate((-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)^(3/2)*(g*x + f)/(e*x + d) 
^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.26 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {15 \, {\left (4 \, c^{2} d^{2} e f - 4 \, b c d e^{2} f + b^{2} e^{3} f - 4 \, c^{2} d^{3} g + 4 \, b c d^{2} e g - b^{2} d e^{2} g\right )} \arctan \left (\frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{\sqrt {-2 \, c d + b e}} + \frac {30 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{6} d e f - 15 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{5} e^{2} f - 30 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{6} d^{2} g + 15 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{5} d e g + 5 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{5} e f - 5 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{5} d g - 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{4} g}{c^{5}}\right )}}{15 \, e^{2}} \] Input:

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x, 
algorithm="giac")
 

Output:

2/15*(15*(4*c^2*d^2*e*f - 4*b*c*d*e^2*f + b^2*e^3*f - 4*c^2*d^3*g + 4*b*c* 
d^2*e*g - b^2*d*e^2*g)*arctan(sqrt(-(e*x + d)*c + 2*c*d - b*e)/sqrt(-2*c*d 
 + b*e))/sqrt(-2*c*d + b*e) + (30*sqrt(-(e*x + d)*c + 2*c*d - b*e)*c^6*d*e 
*f - 15*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b*c^5*e^2*f - 30*sqrt(-(e*x + d)* 
c + 2*c*d - b*e)*c^6*d^2*g + 15*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b*c^5*d*e 
*g + 5*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c^5*e*f - 5*(-(e*x + d)*c + 2*c* 
d - b*e)^(3/2)*c^5*d*g - 3*((e*x + d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d)*c 
 + 2*c*d - b*e)*c^4*g)/c^5)/e^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (f+g\,x\right )\,{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{3/2}}{{\left (d+e\,x\right )}^{5/2}} \,d x \] Input:

int(((f + g*x)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2))/(d + e*x)^(5/2 
),x)
 

Output:

int(((f + g*x)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(3/2))/(d + e*x)^(5/2 
), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.57 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {-2 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c d e g +2 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c \,e^{2} f +4 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{2} g -4 \sqrt {b e -2 c d}\, \mathit {atan} \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d e f -\frac {2 \sqrt {-c e x -b e +c d}\, b^{2} e^{2} g}{5}+\frac {52 \sqrt {-c e x -b e +c d}\, b c d e g}{15}-\frac {8 \sqrt {-c e x -b e +c d}\, b c \,e^{2} f}{3}-\frac {4 \sqrt {-c e x -b e +c d}\, b c \,e^{2} g x}{5}-\frac {76 \sqrt {-c e x -b e +c d}\, c^{2} d^{2} g}{15}+\frac {14 \sqrt {-c e x -b e +c d}\, c^{2} d e f}{3}+\frac {22 \sqrt {-c e x -b e +c d}\, c^{2} d e g x}{15}-\frac {2 \sqrt {-c e x -b e +c d}\, c^{2} e^{2} f x}{3}-\frac {2 \sqrt {-c e x -b e +c d}\, c^{2} e^{2} g \,x^{2}}{5}}{c \,e^{2}} \] Input:

int((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)/(e*x+d)^(5/2),x)
 

Output:

(2*( - 15*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*g + 15*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt 
(b*e - 2*c*d))*b*c*e**2*f + 30*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*g - 30*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*f - 3*sqrt( - b*e + c*d - c 
*e*x)*b**2*e**2*g + 26*sqrt( - b*e + c*d - c*e*x)*b*c*d*e*g - 20*sqrt( - b 
*e + c*d - c*e*x)*b*c*e**2*f - 6*sqrt( - b*e + c*d - c*e*x)*b*c*e**2*g*x - 
 38*sqrt( - b*e + c*d - c*e*x)*c**2*d**2*g + 35*sqrt( - b*e + c*d - c*e*x) 
*c**2*d*e*f + 11*sqrt( - b*e + c*d - c*e*x)*c**2*d*e*g*x - 5*sqrt( - b*e + 
 c*d - c*e*x)*c**2*e**2*f*x - 3*sqrt( - b*e + c*d - c*e*x)*c**2*e**2*g*x** 
2))/(15*c*e**2)