\(\int \frac {\sqrt {d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [325]

Optimal result

Integrand size = 39, antiderivative size = 255 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {5 c d (d+e x)^{3/2}}{3 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {5 c d e \sqrt {d+e x}}{\left (c d^2-a e^2\right )^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {5 c d e^{3/2} \arctan \left (\frac {\sqrt {c d^2-a e^2} \sqrt {d+e x}}{\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {\sqrt {d+e x} \left (\sqrt {c d^2-a e^2} \left (3 a^2 e^4+2 a c d e^2 (7 d+10 e x)+c^2 d^2 \left (-2 d^2+10 d e x+15 e^2 x^2\right )\right )+15 c d e^{3/2} (a e+c d x)^{3/2} (d+e x) \arctan \left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )\right )}{3 \left (c d^2-a e^2\right )^{7/2} ((a e+c d x) (d+e x))^{3/2}} \] Input:

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

Output:

(Sqrt[d + e*x]*(Sqrt[c*d^2 - a*e^2]*(3*a^2*e^4 + 2*a*c*d*e^2*(7*d + 10*e*x 
) + c^2*d^2*(-2*d^2 + 10*d*e*x + 15*e^2*x^2)) + 15*c*d*e^(3/2)*(a*e + c*d* 
x)^(3/2)*(d + e*x)*ArcTan[(Sqrt[e]*Sqrt[a*e + c*d*x])/Sqrt[c*d^2 - a*e^2]] 
))/(3*(c*d^2 - a*e^2)^(7/2)*((a*e + c*d*x)*(d + e*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {1132, 1135, 1132, 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.

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

\(\Big \downarrow \) 1132

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

\(\Big \downarrow \) 1135

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

\(\Big \downarrow \) 1132

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

\(\Big \downarrow \) 1136

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

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 1132

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(2*c*d - b*e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 
 1)*(b^2 - 4*a*c))), x] - Simp[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(b^2 - 
 4*a*c)))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Free 
Q[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ 
[0, m, 1] && IntegerQ[2*p]
 

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]
 

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.66

method	result	size
default	\(\frac {\sqrt {\left (e x +d \right ) \left (c d x +a e \right )}\, \left (15 \sqrt {c d x +a e}\, \operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) c^{2} d^{2} e^{3} x^{2}+15 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) a c d \,e^{4} x \sqrt {c d x +a e}+15 \sqrt {c d x +a e}\, \operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) c^{2} d^{3} e^{2} x +15 \,\operatorname {arctanh}\left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right ) a c \,d^{2} e^{3} \sqrt {c d x +a e}-15 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{2} e^{2} x^{2}-20 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, a c d \,e^{3} x -10 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{3} e x -3 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, a^{2} e^{4}-14 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, a c \,d^{2} e^{2}+2 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c^{2} d^{4}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} \left (c d x +a e \right )^{2} \left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\)	\(423\)

Input:

int((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/2),x,method=_RETURN 
VERBOSE)
 

Output:

1/3*((e*x+d)*(c*d*x+a*e))^(1/2)*(15*(c*d*x+a*e)^(1/2)*arctanh(e*(c*d*x+a*e 
)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*c^2*d^2*e^3*x^2+15*arctanh(e*(c*d*x+a*e)^ 
(1/2)/((a*e^2-c*d^2)*e)^(1/2))*a*c*d*e^4*x*(c*d*x+a*e)^(1/2)+15*(c*d*x+a*e 
)^(1/2)*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*c^2*d^3*e^2*x 
+15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*a*c*d^2*e^3*(c*d* 
x+a*e)^(1/2)-15*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^2*e^2*x^2-20*((a*e^2-c*d^2)* 
e)^(1/2)*a*c*d*e^3*x-10*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^3*e*x-3*((a*e^2-c*d^ 
2)*e)^(1/2)*a^2*e^4-14*((a*e^2-c*d^2)*e)^(1/2)*a*c*d^2*e^2+2*((a*e^2-c*d^2 
)*e)^(1/2)*c^2*d^4)/(e*x+d)^(3/2)/(c*d*x+a*e)^2/(a*e^2-c*d^2)^3/((a*e^2-c* 
d^2)*e)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 607 vs. \(2 (229) = 458\).

Time = 0.20 (sec) , antiderivative size = 1236, normalized size of antiderivative = 4.85 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorit 
hm="fricas")
 

Output:

[-1/6*(15*(c^3*d^3*e^3*x^4 + a^2*c*d^3*e^3 + 2*(c^3*d^4*e^2 + a*c^2*d^2*e^ 
4)*x^3 + (c^3*d^5*e + 4*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x^2 + 2*(a*c^2*d^4*e^ 
2 + a^2*c*d^2*e^4)*x)*sqrt(-e/(c*d^2 - a*e^2))*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)*(c* 
d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-e/(c*d^2 - a*e^2)))/(e^2*x^2 + 2*d*e*x + 
d^2)) - 2*(15*c^2*d^2*e^2*x^2 - 2*c^2*d^4 + 14*a*c*d^2*e^2 + 3*a^2*e^4 + 1 
0*(c^2*d^3*e + 2*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))/(a^2*c^3*d^8*e^2 - 3*a^3*c^2*d^6*e^4 + 3*a^4*c*d^4*e^6 - a 
^5*d^2*e^8 + (c^5*d^8*e^2 - 3*a*c^4*d^6*e^4 + 3*a^2*c^3*d^4*e^6 - a^3*c^2* 
d^2*e^8)*x^4 + 2*(c^5*d^9*e - 2*a*c^4*d^7*e^3 + 2*a^3*c^2*d^3*e^7 - a^4*c* 
d*e^9)*x^3 + (c^5*d^10 + a*c^4*d^8*e^2 - 8*a^2*c^3*d^6*e^4 + 8*a^3*c^2*d^4 
*e^6 - a^4*c*d^2*e^8 - a^5*e^10)*x^2 + 2*(a*c^4*d^9*e - 2*a^2*c^3*d^7*e^3 
+ 2*a^4*c*d^3*e^7 - a^5*d*e^9)*x), 1/3*(15*(c^3*d^3*e^3*x^4 + a^2*c*d^3*e^ 
3 + 2*(c^3*d^4*e^2 + a*c^2*d^2*e^4)*x^3 + (c^3*d^5*e + 4*a*c^2*d^3*e^3 + a 
^2*c*d*e^5)*x^2 + 2*(a*c^2*d^4*e^2 + a^2*c*d^2*e^4)*x)*sqrt(e/(c*d^2 - a*e 
^2))*arctan(-sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d^2 - a*e^2)*s 
qrt(e*x + d)*sqrt(e/(c*d^2 - a*e^2))/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a 
*e^3)*x)) + (15*c^2*d^2*e^2*x^2 - 2*c^2*d^4 + 14*a*c*d^2*e^2 + 3*a^2*e^4 + 
 10*(c^2*d^3*e + 2*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))/(a^2*c^3*d^8*e^2 - 3*a^3*c^2*d^6*e^4 + 3*a^4*c*d^4*e^...
 

Sympy [F]

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

integrate((e*x+d)**(1/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2),x)
 

Output:

Integral(sqrt(d + e*x)/((d + e*x)*(a*e + c*d*x))**(5/2), x)
 

Maxima [F]

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

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorit 
hm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {1}{3} \, {\left (\frac {15 \, c d e \arctan \left (\frac {\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right )}{{\left (c^{3} d^{6} {\left | e \right |} - 3 \, a c^{2} d^{4} e^{2} {\left | e \right |} + 3 \, a^{2} c d^{2} e^{4} {\left | e \right |} - a^{3} e^{6} {\left | e \right |}\right )} \sqrt {c d^{2} e - a e^{3}}} - \frac {2 \, {\left (c^{2} d^{3} e^{2} - a c d e^{4} - 6 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} c d e\right )}}{{\left (c^{3} d^{6} {\left | e \right |} - 3 \, a c^{2} d^{4} e^{2} {\left | e \right |} + 3 \, a^{2} c d^{2} e^{4} {\left | e \right |} - a^{3} e^{6} {\left | e \right |}\right )} {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}}} + \frac {3 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{{\left (c^{3} d^{6} {\left | e \right |} - 3 \, a c^{2} d^{4} e^{2} {\left | e \right |} + 3 \, a^{2} c d^{2} e^{4} {\left | e \right |} - a^{3} e^{6} {\left | e \right |}\right )} {\left (e x + d\right )}}\right )} e^{2} \] Input:

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorit 
hm="giac")
 

Output:

1/3*(15*c*d*e*arctan(sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)/sqrt(c*d^2*e 
- a*e^3))/((c^3*d^6*abs(e) - 3*a*c^2*d^4*e^2*abs(e) + 3*a^2*c*d^2*e^4*abs( 
e) - a^3*e^6*abs(e))*sqrt(c*d^2*e - a*e^3)) - 2*(c^2*d^3*e^2 - a*c*d*e^4 - 
 6*((e*x + d)*c*d*e - c*d^2*e + a*e^3)*c*d*e)/((c^3*d^6*abs(e) - 3*a*c^2*d 
^4*e^2*abs(e) + 3*a^2*c*d^2*e^4*abs(e) - a^3*e^6*abs(e))*((e*x + d)*c*d*e 
- c*d^2*e + a*e^3)^(3/2)) + 3*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)/((c^ 
3*d^6*abs(e) - 3*a*c^2*d^4*e^2*abs(e) + 3*a^2*c*d^2*e^4*abs(e) - a^3*e^6*a 
bs(e))*(e*x + d)))*e^2
 

Mupad [F(-1)]

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

int((d + e*x)^(1/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2),x)
 

Output:

int((d + e*x)^(1/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 584, normalized size of antiderivative = 2.29 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {15 \sqrt {e}\, \sqrt {c d x +a e}\, \sqrt {-a \,e^{2}+c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}}\right ) a c \,d^{2} e^{2}+15 \sqrt {e}\, \sqrt {c d x +a e}\, \sqrt {-a \,e^{2}+c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}}\right ) a c d \,e^{3} x +15 \sqrt {e}\, \sqrt {c d x +a e}\, \sqrt {-a \,e^{2}+c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}}\right ) c^{2} d^{3} e x +15 \sqrt {e}\, \sqrt {c d x +a e}\, \sqrt {-a \,e^{2}+c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {e}\, \sqrt {-a \,e^{2}+c \,d^{2}}}\right ) c^{2} d^{2} e^{2} x^{2}-3 a^{3} e^{6}-11 a^{2} c \,d^{2} e^{4}-20 a^{2} c d \,e^{5} x +16 a \,c^{2} d^{4} e^{2}+10 a \,c^{2} d^{3} e^{3} x -15 a \,c^{2} d^{2} e^{4} x^{2}-2 c^{3} d^{6}+10 c^{3} d^{5} e x +15 c^{3} d^{4} e^{2} x^{2}}{3 \sqrt {c d x +a e}\, \left (a^{4} c d \,e^{9} x^{2}-4 a^{3} c^{2} d^{3} e^{7} x^{2}+6 a^{2} c^{3} d^{5} e^{5} x^{2}-4 a \,c^{4} d^{7} e^{3} x^{2}+c^{5} d^{9} e \,x^{2}+a^{5} e^{10} x -3 a^{4} c \,d^{2} e^{8} x +2 a^{3} c^{2} d^{4} e^{6} x +2 a^{2} c^{3} d^{6} e^{4} x -3 a \,c^{4} d^{8} e^{2} x +c^{5} d^{10} x +a^{5} d \,e^{9}-4 a^{4} c \,d^{3} e^{7}+6 a^{3} c^{2} d^{5} e^{5}-4 a^{2} c^{3} d^{7} e^{3}+a \,c^{4} d^{9} e \right )} \] Input:

int((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x)
 

Output:

(15*sqrt(e)*sqrt(a*e + c*d*x)*sqrt( - a*e**2 + c*d**2)*atan((sqrt(a*e + c* 
d*x)*e)/(sqrt(e)*sqrt( - a*e**2 + c*d**2)))*a*c*d**2*e**2 + 15*sqrt(e)*sqr 
t(a*e + c*d*x)*sqrt( - a*e**2 + c*d**2)*atan((sqrt(a*e + c*d*x)*e)/(sqrt(e 
)*sqrt( - a*e**2 + c*d**2)))*a*c*d*e**3*x + 15*sqrt(e)*sqrt(a*e + c*d*x)*s 
qrt( - 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 + c*d*x)*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**2*e**2*x**2 - 3*a**3*e**6 - 11*a**2*c*d**2*e**4 - 20*a**2*c*d*e**5*x 
 + 16*a*c**2*d**4*e**2 + 10*a*c**2*d**3*e**3*x - 15*a*c**2*d**2*e**4*x**2 
- 2*c**3*d**6 + 10*c**3*d**5*e*x + 15*c**3*d**4*e**2*x**2)/(3*sqrt(a*e + c 
*d*x)*(a**5*d*e**9 + a**5*e**10*x - 4*a**4*c*d**3*e**7 - 3*a**4*c*d**2*e** 
8*x + a**4*c*d*e**9*x**2 + 6*a**3*c**2*d**5*e**5 + 2*a**3*c**2*d**4*e**6*x 
 - 4*a**3*c**2*d**3*e**7*x**2 - 4*a**2*c**3*d**7*e**3 + 2*a**2*c**3*d**6*e 
**4*x + 6*a**2*c**3*d**5*e**5*x**2 + a*c**4*d**9*e - 3*a*c**4*d**8*e**2*x 
- 4*a*c**4*d**7*e**3*x**2 + c**5*d**10*x + c**5*d**9*e*x**2))