\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{11/2}} \, dx\) [304]

Optimal result

Integrand size = 39, antiderivative size = 236 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\frac {15 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3 \sqrt {d+e x}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 e^2 (d+e x)^{5/2}}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}-\frac {15 c^2 d^2 \sqrt {c d^2-a e^2} \arctan \left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{4 e^{7/2}} \] Output:

15/4*c^2*d^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/e^3/(e*x+d)^(1/2)-5/4 
*c*d*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/e^2/(e*x+d)^(5/2)-1/2*(a*d*e+ 
(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/e/(e*x+d)^(9/2)-15/4*c^2*d^2*(-a*e^2+c*d^ 
2)^(1/2)*arctan(e^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(-a*e^2+c* 
d^2)^(1/2)/(e*x+d)^(1/2))/e^(7/2)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\frac {\sqrt {(a e+c d x) (d+e x)} \left (\sqrt {e} \sqrt {a e+c d x} \left (-2 a^2 e^4-a c d e^2 (5 d+9 e x)+c^2 d^2 \left (15 d^2+25 d e x+8 e^2 x^2\right )\right )-15 c^2 d^2 \sqrt {c d^2-a e^2} (d+e x)^2 \arctan \left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )\right )}{4 e^{7/2} \sqrt {a e+c d x} (d+e x)^{5/2}} \] Input:

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(11/2),x 
]
 

Output:

(Sqrt[(a*e + c*d*x)*(d + e*x)]*(Sqrt[e]*Sqrt[a*e + c*d*x]*(-2*a^2*e^4 - a* 
c*d*e^2*(5*d + 9*e*x) + c^2*d^2*(15*d^2 + 25*d*e*x + 8*e^2*x^2)) - 15*c^2* 
d^2*Sqrt[c*d^2 - a*e^2]*(d + e*x)^2*ArcTan[(Sqrt[e]*Sqrt[a*e + c*d*x])/Sqr 
t[c*d^2 - a*e^2]]))/(4*e^(7/2)*Sqrt[a*e + c*d*x]*(d + e*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {1130, 1130, 1131, 1136, 218}

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.

\(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {5 c d \int \frac {\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/2}}{(d+e x)^{7/2}}dx}{4 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {5 c d \left (\frac {3 c d \int \frac {\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{(d+e x)^{3/2}}dx}{2 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}\right )}{4 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}\)

\(\Big \downarrow \) 1131

\(\displaystyle \frac {5 c d \left (\frac {3 c d \left (\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e \sqrt {d+e x}}-\frac {\left (c d^2-a e^2\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{e}\right )}{2 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}\right )}{4 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {5 c d \left (\frac {3 c d \left (\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e \sqrt {d+e x}}-2 \left (c d^2-a e^2\right ) \int \frac {1}{\frac {\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right ) e^2}{d+e x}+\left (c d^2-a e^2\right ) e}d\frac {\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{\sqrt {d+e x}}\right )}{2 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}\right )}{4 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {5 c d \left (\frac {3 c d \left (\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e \sqrt {d+e x}}-\frac {2 \sqrt {c d^2-a e^2} \arctan \left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{e^{3/2}}\right )}{2 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{e (d+e x)^{5/2}}\right )}{4 e}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 e (d+e x)^{9/2}}\)

Input:

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(11/2),x]
 

Output:

-1/2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(e*(d + e*x)^(9/2)) + ( 
5*c*d*(-((a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(e*(d + e*x)^(5/2)) 
) + (3*c*d*((2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(e*Sqrt[d + e* 
x]) - (2*Sqrt[c*d^2 - a*e^2]*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)* 
x + c*d*e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/e^(3/2)))/(2*e)))/(4 
*e)
 

Defintions of rubi rules used

rule 218

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(516\) vs. \(2(204)=408\).

Time = 1.08 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.19

method	result	size
default	\(-\frac {\sqrt {\left (e x +d \right ) \left (c d x +a e \right )}\, \left (15 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) a \,c^{2} d^{2} e^{4} x^{2}-15 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) c^{3} d^{4} e^{2} x^{2}+30 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) a \,c^{2} d^{3} e^{3} x -30 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) c^{3} d^{5} e x +15 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) a \,c^{2} d^{4} e^{2}-15 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) c^{3} d^{6}-8 c^{2} d^{2} e^{2} x^{2} \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}+9 a c d \,e^{3} x \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}-25 c^{2} d^{3} e x \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}+2 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, a^{2} e^{4}+5 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, a c \,d^{2} e^{2}-15 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, \sqrt {c d x +a e}\, c^{2} d^{4}\right )}{4 \left (e x +d \right )^{\frac {5}{2}} \sqrt {c d x +a e}\, e^{3} \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\)	\(517\)

Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/2)/(e*x+d)^(11/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/4*((e*x+d)*(c*d*x+a*e))^(1/2)*(15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c 
*d^2)*e)^(1/2))*a*c^2*d^2*e^4*x^2-15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c 
*d^2)*e)^(1/2))*c^3*d^4*e^2*x^2+30*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d 
^2)*e)^(1/2))*a*c^2*d^3*e^3*x-30*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2 
)*e)^(1/2))*c^3*d^5*e*x+15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^( 
1/2))*a*c^2*d^4*e^2-15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2) 
)*c^3*d^6-8*c^2*d^2*e^2*x^2*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+9*a* 
c*d*e^3*x*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)-25*c^2*d^3*e*x*(c*d*x+ 
a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+2*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^( 
1/2)*a^2*e^4+5*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a*c*d^2*e^2-15*(( 
a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^4)/(e*x+d)^(5/2)/(c*d*x+a*e) 
^(1/2)/e^3/((a*e^2-c*d^2)*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 594, normalized size of antiderivative = 2.52 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\left [\frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{e}} \log \left (-\frac {c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} e \sqrt {-\frac {c d^{2} - a e^{2}}{e}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (8 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} + {\left (25 \, c^{2} d^{3} e - 9 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{8 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}, \frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {\frac {c d^{2} - a e^{2}}{e}} \arctan \left (-\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} e \sqrt {\frac {c d^{2} - a e^{2}}{e}}}{c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x}\right ) + {\left (8 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} + {\left (25 \, c^{2} d^{3} e - 9 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{4 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}\right ] \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(11/2),x, algori 
thm="fricas")
 

Output:

[1/8*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*s 
qrt(-(c*d^2 - a*e^2)/e)*log(-(c*d*e^2*x^2 + 2*a*e^3*x - c*d^3 + 2*a*d*e^2 
- 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*e*sqrt(-(c*d 
^2 - a*e^2)/e))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(8*c^2*d^2*e^2*x^2 + 15*c^2 
*d^4 - 5*a*c*d^2*e^2 - 2*a^2*e^4 + (25*c^2*d^3*e - 9*a*c*d*e^3)*x)*sqrt(c* 
d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(e^6*x^3 + 3*d*e^5*x^2 
 + 3*d^2*e^4*x + d^3*e^3), 1/4*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 
3*c^2*d^4*e*x + c^2*d^5)*sqrt((c*d^2 - a*e^2)/e)*arctan(-sqrt(c*d*e*x^2 + 
a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*e*sqrt((c*d^2 - a*e^2)/e)/(c*d^3 
- a*d*e^2 + (c*d^2*e - a*e^3)*x)) + (8*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 5*a* 
c*d^2*e^2 - 2*a^2*e^4 + (25*c^2*d^3*e - 9*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a 
*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^ 
4*x + d^3*e^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\text {Timed out} \] Input:

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{\frac {11}{2}}} \,d x } \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(11/2),x, algori 
thm="maxima")
 

Output:

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/(e*x + d)^(11/2), 
x)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\frac {8 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{3} d^{3} {\left | e \right |} - \frac {15 \, {\left (c^{4} d^{5} e {\left | e \right |} - a c^{3} d^{3} e^{3} {\left | e \right |}\right )} \arctan \left (\frac {\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right )}{\sqrt {c d^{2} e - a e^{3}}} + \frac {7 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{5} d^{7} e^{2} {\left | e \right |} - 14 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a c^{4} d^{5} e^{4} {\left | e \right |} + 7 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a^{2} c^{3} d^{3} e^{6} {\left | e \right |} + 9 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{4} d^{5} e {\left | e \right |} - 9 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a c^{3} d^{3} e^{3} {\left | e \right |}}{{\left (e x + d\right )}^{2} c^{2} d^{2} e^{2}}}{4 \, c d e^{5}} \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(11/2),x, algori 
thm="giac")
 

Output:

1/4*(8*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*c^3*d^3*abs(e) - 15*(c^4*d^ 
5*e*abs(e) - a*c^3*d^3*e^3*abs(e))*arctan(sqrt((e*x + d)*c*d*e - c*d^2*e + 
 a*e^3)/sqrt(c*d^2*e - a*e^3))/sqrt(c*d^2*e - a*e^3) + (7*sqrt((e*x + d)*c 
*d*e - c*d^2*e + a*e^3)*c^5*d^7*e^2*abs(e) - 14*sqrt((e*x + d)*c*d*e - c*d 
^2*e + a*e^3)*a*c^4*d^5*e^4*abs(e) + 7*sqrt((e*x + d)*c*d*e - c*d^2*e + a* 
e^3)*a^2*c^3*d^3*e^6*abs(e) + 9*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)* 
c^4*d^5*e*abs(e) - 9*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*c^3*d^3*e 
^3*abs(e))/((e*x + d)^2*c^2*d^2*e^2))/(c*d*e^5)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}}{{\left (d+e\,x\right )}^{11/2}} \,d x \] Input:

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^(11/2),x)
 

Output:

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^(11/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\frac {-15 \sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}}\right ) c^{2} d^{4}-30 \sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}}\right ) c^{2} d^{3} e x -15 \sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}}\right ) c^{2} d^{2} e^{2} x^{2}-2 \sqrt {c d x +a e}\, a^{2} e^{5}-5 \sqrt {c d x +a e}\, a c \,d^{2} e^{3}-9 \sqrt {c d x +a e}\, a c d \,e^{4} x +15 \sqrt {c d x +a e}\, c^{2} d^{4} e +25 \sqrt {c d x +a e}\, c^{2} d^{3} e^{2} x +8 \sqrt {c d x +a e}\, c^{2} d^{2} e^{3} x^{2}}{4 e^{4} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(11/2),x)
 

Output:

( - 15*sqrt(e)*sqrt( - a*e**2 + c*d**2)*atan((sqrt(a*e + c*d*x)*e)/(sqrt(e 
)*sqrt( - a*e**2 + c*d**2)))*c**2*d**4 - 30*sqrt(e)*sqrt( - a*e**2 + c*d** 
2)*atan((sqrt(a*e + c*d*x)*e)/(sqrt(e)*sqrt( - a*e**2 + c*d**2)))*c**2*d** 
3*e*x - 15*sqrt(e)*sqrt( - a*e**2 + c*d**2)*atan((sqrt(a*e + c*d*x)*e)/(sq 
rt(e)*sqrt( - a*e**2 + c*d**2)))*c**2*d**2*e**2*x**2 - 2*sqrt(a*e + c*d*x) 
*a**2*e**5 - 5*sqrt(a*e + c*d*x)*a*c*d**2*e**3 - 9*sqrt(a*e + c*d*x)*a*c*d 
*e**4*x + 15*sqrt(a*e + c*d*x)*c**2*d**4*e + 25*sqrt(a*e + c*d*x)*c**2*d** 
3*e**2*x + 8*sqrt(a*e + c*d*x)*c**2*d**2*e**3*x**2)/(4*e**4*(d**2 + 2*d*e* 
x + e**2*x**2))