\(\int \frac {1}{\sqrt {d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [189]

Optimal result

Integrand size = 37, antiderivative size = 158 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {5 e}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{3/2}}-\frac {5 c d e}{\left (c d^2-a e^2\right )^3 \sqrt {d+e x}}+\frac {5 c^{3/2} d^{3/2} e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {2 a^2 e^4-2 a c d e^2 (7 d+5 e x)-c^2 d^2 \left (3 d^2+20 d e x+15 e^2 x^2\right )}{3 \left (c d^2-a e^2\right )^3 (a e+c d x) (d+e x)^{3/2}}+\frac {5 c^{3/2} d^{3/2} e \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{7/2}} \] Input:

Integrate[1/(Sqrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2),x]
 

Output:

(2*a^2*e^4 - 2*a*c*d*e^2*(7*d + 5*e*x) - c^2*d^2*(3*d^2 + 20*d*e*x + 15*e^ 
2*x^2))/(3*(c*d^2 - a*e^2)^3*(a*e + c*d*x)*(d + e*x)^(3/2)) + (5*c^(3/2)*d 
^(3/2)*e*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/( 
-(c*d^2) + a*e^2)^(7/2)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1121, 52, 61, 61, 73, 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.

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

\(\Big \downarrow \) 1121

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(Sqrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2),x]
 

Output:

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

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 1121

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] 
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int 
egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
 

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(2 e \left (-\frac {1}{3 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 c d}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {e x +d}}+\frac {c^{2} d^{2} \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 a \,e^{2}-2 c \,d^{2}}+\frac {5 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{3}}\right )\)	\(153\)
default	\(2 e \left (-\frac {1}{3 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 c d}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {e x +d}}+\frac {c^{2} d^{2} \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 a \,e^{2}-2 c \,d^{2}}+\frac {5 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{3}}\right )\)	\(153\)
pseudoelliptic	\(e \left (-\frac {2}{3 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}+\frac {4 c d}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {e x +d}}+\frac {\sqrt {e x +d}\, c^{2} d^{2}}{e \left (c d x +a e \right ) \left (a \,e^{2}-c \,d^{2}\right )^{3}}+\frac {5 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right ) c^{2} d^{2}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, \left (a \,e^{2}-c \,d^{2}\right )^{3}}\right )\)	\(160\)

Input:

int(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^2,x,method=_RETURNVE 
RBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (136) = 272\).

Time = 0.11 (sec) , antiderivative size = 854, normalized size of antiderivative = 5.41 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {5}{2}} \left (a e + c d x\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*e^2-c*d^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {5 \, c^{2} d^{2} e \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} - \frac {\sqrt {e x + d} c^{2} d^{2} e}{{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}} - \frac {2 \, {\left (6 \, {\left (e x + d\right )} c d e + c d^{2} e - a e^{3}\right )}}{3 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} {\left (e x + d\right )}^{\frac {3}{2}}} \] Input:

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

Output:

-5*c^2*d^2*e*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^3*d^ 
6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(-c^2*d^3 + a*c*d*e^2 
)) - sqrt(e*x + d)*c^2*d^2*e/((c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 
 - a^3*e^6)*((e*x + d)*c*d - c*d^2 + a*e^2)) - 2/3*(6*(e*x + d)*c*d*e + c* 
d^2*e - a*e^3)/((c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*(e 
*x + d)^(3/2))
 

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {\frac {10\,c\,d\,e\,\left (d+e\,x\right )}{3\,{\left (a\,e^2-c\,d^2\right )}^2}-\frac {2\,e}{3\,\left (a\,e^2-c\,d^2\right )}+\frac {5\,c^2\,d^2\,e\,{\left (d+e\,x\right )}^2}{{\left (a\,e^2-c\,d^2\right )}^3}}{\left (a\,e^2-c\,d^2\right )\,{\left (d+e\,x\right )}^{3/2}+c\,d\,{\left (d+e\,x\right )}^{5/2}}+\frac {5\,c^{3/2}\,d^{3/2}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^3\,e^6-3\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2-c^3\,d^6\right )}{{\left (a\,e^2-c\,d^2\right )}^{7/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{7/2}} \] Input:

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

Output:

((10*c*d*e*(d + e*x))/(3*(a*e^2 - c*d^2)^2) - (2*e)/(3*(a*e^2 - c*d^2)) + 
(5*c^2*d^2*e*(d + e*x)^2)/(a*e^2 - c*d^2)^3)/((a*e^2 - c*d^2)*(d + e*x)^(3 
/2) + c*d*(d + e*x)^(5/2)) + (5*c^(3/2)*d^(3/2)*e*atan((c^(1/2)*d^(1/2)*(d 
 + e*x)^(1/2)*(a^3*e^6 - c^3*d^6 + 3*a*c^2*d^4*e^2 - 3*a^2*c*d^2*e^4))/(a* 
e^2 - c*d^2)^(7/2)))/(a*e^2 - c*d^2)^(7/2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 585, normalized size of antiderivative = 3.70 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {15 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a c \,d^{2} e^{2}+15 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a c d \,e^{3} x +15 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{3} e x +15 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{2} e^{2} x^{2}-2 a^{3} e^{6}+16 a^{2} c \,d^{2} e^{4}+10 a^{2} c d \,e^{5} x -11 a \,c^{2} d^{4} e^{2}+10 a \,c^{2} d^{3} e^{3} x +15 a \,c^{2} d^{2} e^{4} x^{2}-3 c^{3} d^{6}-20 c^{3} d^{5} e x -15 c^{3} d^{4} e^{2} x^{2}}{3 \sqrt {e x +d}\, \left (a^{4} c d \,e^{9} x^{2}-4 a^{3} c^{2} d^{3} e^{7} x^{2}+6 a^{2} c^{3} d^{5} e^{5} x^{2}-4 a \,c^{4} d^{7} e^{3} x^{2}+c^{5} d^{9} e \,x^{2}+a^{5} e^{10} x -3 a^{4} c \,d^{2} e^{8} x +2 a^{3} c^{2} d^{4} e^{6} x +2 a^{2} c^{3} d^{6} e^{4} x -3 a \,c^{4} d^{8} e^{2} x +c^{5} d^{10} x +a^{5} d \,e^{9}-4 a^{4} c \,d^{3} e^{7}+6 a^{3} c^{2} d^{5} e^{5}-4 a^{2} c^{3} d^{7} e^{3}+a \,c^{4} d^{9} e \right )} \] Input:

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

Output:

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