\(\int \frac {(f+g x) (c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}}{(d+e x)^{11/2}} \, dx\) [212]

Optimal result

Integrand size = 46, antiderivative size = 285 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\frac {(c e f-5 c d g+2 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 e^2 (d+e x)^{5/2}}-\frac {c (c e f-21 c d g+10 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{8 e^2 (2 c d-b e) (d+e x)^{3/2}}-\frac {(e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (d+e x)^{9/2}}-\frac {c^2 (c e f+11 c d g-6 b e g) \text {arctanh}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {2 c d-b e} \sqrt {d+e x}}\right )}{8 e^2 (2 c d-b e)^{3/2}} \] Output:

1/4*(2*b*e*g-5*c*d*g+c*e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/e^2/(e* 
x+d)^(5/2)-1/8*c*(10*b*e*g-21*c*d*g+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)/(e*x+d)^(3/2)-1/3*(-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)^(9/2)-1/8*c^2*(-6*b*e*g+11*c*d*g+c*e*f)*arc 
tanh((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/(-b*e+2*c*d)^(3/2)
 

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.89 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\frac {c^2 ((d+e x) (-b e+c (d-e x)))^{3/2} \left (\frac {-4 b^2 e^2 (2 e f+d g+3 e g x)+2 b c e^2 (d (9 f+g x)-e x (7 f+15 g x))+c^2 \left (19 d^3 g-3 e^3 f x^2+d^2 e (-7 f+50 g x)+d e^2 x (22 f+63 g x)\right )}{c^2 (2 c d-b e) (d+e x)^3 (-b e+c (d-e x))}-\frac {3 (c e f+11 c d g-6 b e g) \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {-2 c d+b e}}\right )}{(-2 c d+b e)^{3/2} (-b e+c (d-e x))^{3/2}}\right )}{24 e^2 (d+e x)^{3/2}} \] 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 
)^(11/2),x]
 

Output:

(c^2*((d + e*x)*(-(b*e) + c*(d - e*x)))^(3/2)*((-4*b^2*e^2*(2*e*f + d*g + 
3*e*g*x) + 2*b*c*e^2*(d*(9*f + g*x) - e*x*(7*f + 15*g*x)) + c^2*(19*d^3*g 
- 3*e^3*f*x^2 + d^2*e*(-7*f + 50*g*x) + d*e^2*x*(22*f + 63*g*x)))/(c^2*(2* 
c*d - b*e)*(d + e*x)^3*(-(b*e) + c*(d - e*x))) - (3*(c*e*f + 11*c*d*g - 6* 
b*e*g)*ArcTan[Sqrt[c*d - b*e - c*e*x]/Sqrt[-2*c*d + b*e]])/((-2*c*d + b*e) 
^(3/2)*(-(b*e) + c*(d - e*x))^(3/2))))/(24*e^2*(d + e*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {1220, 1130, 1130, 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.

\(\displaystyle \int \frac {(f+g x) \left (-b d e-b e^2 x+c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(-6 b e g+11 c d g+c e f) \int \frac {\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{3/2}}{(d+e x)^{9/2}}dx}{6 e (2 c d-b e)}-\frac {(e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{3 e^2 (d+e x)^{11/2} (2 c d-b e)}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(-6 b e g+11 c d g+c e f) \left (-\frac {3}{4} c \int \frac {\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}{(d+e x)^{5/2}}dx-\frac {\left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}\right )}{6 e (2 c d-b e)}-\frac {(e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{3 e^2 (d+e x)^{11/2} (2 c d-b e)}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(-6 b e g+11 c d g+c e f) \left (-\frac {3}{4} c \left (-\frac {1}{2} c \int \frac {1}{\sqrt {d+e x} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2}}\right )-\frac {\left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}\right )}{6 e (2 c d-b e)}-\frac {(e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{3 e^2 (d+e x)^{11/2} (2 c d-b e)}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(-6 b e g+11 c d g+c e f) \left (-\frac {3}{4} c \left (-c e \int \frac {1}{\frac {e^2 \left (-c x^2 e^2-b x e^2+d (c d-b e)\right )}{d+e x}-e^2 (2 c d-b e)}d\frac {\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}{\sqrt {d+e x}}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2}}\right )-\frac {\left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}\right )}{6 e (2 c d-b e)}-\frac {(e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{3 e^2 (d+e x)^{11/2} (2 c d-b e)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\left (-\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e \sqrt {2 c d-b e}}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2}}\right )-\frac {\left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}\right ) (-6 b e g+11 c d g+c e f)}{6 e (2 c d-b e)}-\frac {(e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{3 e^2 (d+e x)^{11/2} (2 c d-b e)}\)

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)^(11/ 
2),x]
 

Output:

-1/3*((e*f - d*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(5/2))/(e^2*(2*c*d 
 - b*e)*(d + e*x)^(11/2)) + ((c*e*f + 11*c*d*g - 6*b*e*g)*(-1/2*(d*(c*d - 
b*e) - b*e^2*x - c*e^2*x^2)^(3/2)/(e*(d + e*x)^(7/2)) - (3*c*(-(Sqrt[d*(c* 
d - b*e) - b*e^2*x - c*e^2*x^2]/(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*S 
qrt[2*c*d - b*e])))/4))/(6*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 1130

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 + p + 1))), x] 
- Simp[c*(p/(e^2*(m + p + 1)))   Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + 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 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. \(990\) vs. \(2(257)=514\).

Time = 1.56 (sec) , antiderivative size = 991, normalized size of antiderivative = 3.48

method	result	size
default	\(\frac {\left (30 b c \,e^{3} g \,x^{2} \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-63 c^{2} d \,e^{2} g \,x^{2} \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-2 b c d \,e^{2} g x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-19 c^{2} d^{3} g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} e^{4} f \,x^{3}-3 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{3} e f +54 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,c^{2} d \,e^{3} g \,x^{2}+54 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,c^{2} d^{2} e^{2} g x -18 b c d \,e^{2} f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-22 c^{2} d \,e^{2} f x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-50 c^{2} d^{2} e g x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+14 b c \,e^{3} f x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-33 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{4} g +18 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,c^{2} e^{4} g \,x^{3}-33 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d \,e^{3} g \,x^{3}+8 b^{2} e^{3} f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}-9 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d \,e^{3} f \,x^{2}-99 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{3} e g x -9 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{2} e^{2} f x +12 b^{2} e^{3} g x \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+4 b^{2} d \,e^{2} g \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+7 c^{2} d^{2} e f \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+3 c^{2} e^{3} f \,x^{2} \sqrt {-c e x -b e +c d}\, \sqrt {b e -2 c d}+18 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) b \,c^{2} d^{3} e g -99 \arctan \left (\frac {\sqrt {-c e x -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{3} d^{2} e^{2} g \,x^{2}\right ) \sqrt {-\left (e x +d \right ) \left (c e x +b e -c d \right )}}{24 \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 {7}{2}}}\)	\(991\)

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)^(11/2),x,method 
=_RETURNVERBOSE)
 

Output:

1/24*(30*b*c*e^3*g*x^2*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-63*c^2*d*e 
^2*g*x^2*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-2*b*c*d*e^2*g*x*(-c*e*x- 
b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-19*c^2*d^3*g*(-c*e*x-b*e+c*d)^(1/2)*(b*e- 
2*c*d)^(1/2)-3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*e^4*f* 
x^3-3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*d^3*e*f+54*arct 
an((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c^2*d*e^3*g*x^2+54*arctan(( 
-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c^2*d^2*e^2*g*x-18*b*c*d*e^2*f* 
(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-22*c^2*d*e^2*f*x*(-c*e*x-b*e+c*d) 
^(1/2)*(b*e-2*c*d)^(1/2)-50*c^2*d^2*e*g*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c* 
d)^(1/2)+14*b*c*e^3*f*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-33*arctan 
((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*d^4*g+18*arctan((-c*e*x-b*e 
+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c^2*e^4*g*x^3-33*arctan((-c*e*x-b*e+c*d)^ 
(1/2)/(b*e-2*c*d)^(1/2))*c^3*d*e^3*g*x^3+8*b^2*e^3*f*(-c*e*x-b*e+c*d)^(1/2 
)*(b*e-2*c*d)^(1/2)-9*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3 
*d*e^3*f*x^2-99*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*d^3*e 
*g*x-9*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^3*d^2*e^2*f*x+12 
*b^2*e^3*g*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+4*b^2*d*e^2*g*(-c*e* 
x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+7*c^2*d^2*e*f*(-c*e*x-b*e+c*d)^(1/2)*(b 
*e-2*c*d)^(1/2)+3*c^2*e^3*f*x^2*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+1 
8*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c^2*d^3*e*g-99*arc...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 713 vs. \(2 (257) = 514\).

Time = 0.14 (sec) , antiderivative size = 1456, normalized size of antiderivative = 5.11 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\text {Too large to display} \] 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)^(11/2),x, 
 algorithm="fricas")
 

Output:

[1/48*(3*(c^3*d^4*e*f + (c^3*e^5*f + (11*c^3*d*e^4 - 6*b*c^2*e^5)*g)*x^4 + 
 4*(c^3*d*e^4*f + (11*c^3*d^2*e^3 - 6*b*c^2*d*e^4)*g)*x^3 + 6*(c^3*d^2*e^3 
*f + (11*c^3*d^3*e^2 - 6*b*c^2*d^2*e^3)*g)*x^2 + (11*c^3*d^5 - 6*b*c^2*d^4 
*e)*g + 4*(c^3*d^3*e^2*f + (11*c^3*d^4*e - 6*b*c^2*d^3*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*sq 
rt(-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)* 
(3*((2*c^3*d*e^3 - b*c^2*e^4)*f - (42*c^3*d^2*e^2 - 41*b*c^2*d*e^3 + 10*b^ 
2*c*e^4)*g)*x^2 + (14*c^3*d^3*e - 43*b*c^2*d^2*e^2 + 34*b^2*c*d*e^3 - 8*b^ 
3*e^4)*f - (38*c^3*d^4 - 19*b*c^2*d^3*e - 8*b^2*c*d^2*e^2 + 4*b^3*d*e^3)*g 
 - 2*((22*c^3*d^2*e^2 - 25*b*c^2*d*e^3 + 7*b^2*c*e^4)*f + (50*c^3*d^3*e - 
23*b*c^2*d^2*e^2 - 13*b^2*c*d*e^3 + 6*b^3*e^4)*g)*x)*sqrt(e*x + d))/(4*c^2 
*d^6*e^2 - 4*b*c*d^5*e^3 + b^2*d^4*e^4 + (4*c^2*d^2*e^6 - 4*b*c*d*e^7 + b^ 
2*e^8)*x^4 + 4*(4*c^2*d^3*e^5 - 4*b*c*d^2*e^6 + b^2*d*e^7)*x^3 + 6*(4*c^2* 
d^4*e^4 - 4*b*c*d^3*e^5 + b^2*d^2*e^6)*x^2 + 4*(4*c^2*d^5*e^3 - 4*b*c*d^4* 
e^4 + b^2*d^3*e^5)*x), -1/24*(3*(c^3*d^4*e*f + (c^3*e^5*f + (11*c^3*d*e^4 
- 6*b*c^2*e^5)*g)*x^4 + 4*(c^3*d*e^4*f + (11*c^3*d^2*e^3 - 6*b*c^2*d*e^4)* 
g)*x^3 + 6*(c^3*d^2*e^3*f + (11*c^3*d^3*e^2 - 6*b*c^2*d^2*e^3)*g)*x^2 + (1 
1*c^3*d^5 - 6*b*c^2*d^4*e)*g + 4*(c^3*d^3*e^2*f + (11*c^3*d^4*e - 6*b*c^2* 
d^3*e^2)*g)*x)*sqrt(-2*c*d + b*e)*arctan(-sqrt(-c*e^2*x^2 - b*e^2*x + c...
 

Sympy [F]

\[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\int \frac {\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}} \left (f + g x\right )}{\left (d + e x\right )^{\frac {11}{2}}}\, dx \] 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)**(11 
/2),x)
 

Output:

Integral((-(d + e*x)*(b*e - c*d + c*e*x))**(3/2)*(f + g*x)/(d + e*x)**(11/ 
2), x)
 

Maxima [F]

\[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\int { \frac {{\left (-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e\right )}^{\frac {3}{2}} {\left (g x + f\right )}}{{\left (e x + d\right )}^{\frac {11}{2}}} \,d x } \] 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)^(11/2),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) 
^(11/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 592 vs. \(2 (257) = 514\).

Time = 0.44 (sec) , antiderivative size = 592, normalized size of antiderivative = 2.08 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\frac {\frac {3 \, {\left (c^{4} e f + 11 \, c^{4} d g - 6 \, b c^{3} e g\right )} \arctan \left (\frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{{\left (2 \, c d - b e\right )} \sqrt {-2 \, c d + b e}} + \frac {12 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{6} d^{2} e f - 12 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{5} d e^{2} f + 3 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b^{2} c^{4} e^{3} f + 132 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{6} d^{3} g - 204 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{5} d^{2} e g + 105 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b^{2} c^{4} d e^{2} g - 18 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b^{3} c^{3} e^{3} g - 16 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{5} d e f + 8 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} b c^{4} e^{2} f - 176 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{5} d^{2} g + 184 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} b c^{4} d e g - 48 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} b^{2} c^{3} e^{2} g - 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{4} e f + 63 \, {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{4} d g - 30 \, {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{3} e g}{{\left (2 \, c d - b e\right )} {\left (e x + d\right )}^{3} c^{3}}}{24 \, c e^{2}} \] 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)^(11/2),x, 
 algorithm="giac")
 

Output:

1/24*(3*(c^4*e*f + 11*c^4*d*g - 6*b*c^3*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)) + (12*sq 
rt(-(e*x + d)*c + 2*c*d - b*e)*c^6*d^2*e*f - 12*sqrt(-(e*x + d)*c + 2*c*d 
- b*e)*b*c^5*d*e^2*f + 3*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b^2*c^4*e^3*f + 
132*sqrt(-(e*x + d)*c + 2*c*d - b*e)*c^6*d^3*g - 204*sqrt(-(e*x + d)*c + 2 
*c*d - b*e)*b*c^5*d^2*e*g + 105*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b^2*c^4*d 
*e^2*g - 18*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b^3*c^3*e^3*g - 16*(-(e*x + d 
)*c + 2*c*d - b*e)^(3/2)*c^5*d*e*f + 8*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)* 
b*c^4*e^2*f - 176*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c^5*d^2*g + 184*(-(e* 
x + d)*c + 2*c*d - b*e)^(3/2)*b*c^4*d*e*g - 48*(-(e*x + d)*c + 2*c*d - b*e 
)^(3/2)*b^2*c^3*e^2*g - 3*((e*x + d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d)*c 
+ 2*c*d - b*e)*c^4*e*f + 63*((e*x + d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d)* 
c + 2*c*d - b*e)*c^4*d*g - 30*((e*x + d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d 
)*c + 2*c*d - b*e)*b*c^3*e*g)/((2*c*d - b*e)*(e*x + d)^3*c^3))/(c*e^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\int \frac {\left (f+g\,x\right )\,{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{3/2}}{{\left (d+e\,x\right )}^{11/2}} \,d x \] 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)^(11/ 
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)^(11/ 
2), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 1239, normalized size of antiderivative = 4.35 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx =\text {Too large to display} \] 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)^(11/2),x)
 

Output:

(18*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*b 
*c**2*d**3*e*g + 54*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt 
(b*e - 2*c*d))*b*c**2*d**2*e**2*g*x + 54*sqrt(b*e - 2*c*d)*atan(sqrt( - b* 
e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*b*c**2*d*e**3*g*x**2 + 18*sqrt(b*e - 2 
*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*b*c**2*e**4*g*x** 
3 - 33*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d) 
)*c**3*d**4*g - 3*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b 
*e - 2*c*d))*c**3*d**3*e*f - 99*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - 
 c*e*x)/sqrt(b*e - 2*c*d))*c**3*d**3*e*g*x - 9*sqrt(b*e - 2*c*d)*atan(sqrt 
( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**3*d**2*e**2*f*x - 99*sqrt(b*e 
 - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**3*d**2*e** 
2*g*x**2 - 9*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - c*e*x)/sqrt(b*e - 
2*c*d))*c**3*d*e**3*f*x**2 - 33*sqrt(b*e - 2*c*d)*atan(sqrt( - b*e + c*d - 
 c*e*x)/sqrt(b*e - 2*c*d))*c**3*d*e**3*g*x**3 - 3*sqrt(b*e - 2*c*d)*atan(s 
qrt( - b*e + c*d - c*e*x)/sqrt(b*e - 2*c*d))*c**3*e**4*f*x**3 + 4*sqrt( - 
b*e + c*d - c*e*x)*b**3*d*e**3*g + 8*sqrt( - b*e + c*d - c*e*x)*b**3*e**4* 
f + 12*sqrt( - b*e + c*d - c*e*x)*b**3*e**4*g*x - 8*sqrt( - b*e + c*d - c* 
e*x)*b**2*c*d**2*e**2*g - 34*sqrt( - b*e + c*d - c*e*x)*b**2*c*d*e**3*f - 
26*sqrt( - b*e + c*d - c*e*x)*b**2*c*d*e**3*g*x + 14*sqrt( - b*e + c*d - c 
*e*x)*b**2*c*e**4*f*x + 30*sqrt( - b*e + c*d - c*e*x)*b**2*c*e**4*g*x**...