\(\int \frac {(d+e x)^{5/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [196]

Optimal result

Integrand size = 37, antiderivative size = 146 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\sqrt {d+e x}}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac {3 e \sqrt {d+e x}}{4 \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{5/2}} \] Output:

-1/2*(e*x+d)^(1/2)/(-a*e^2+c*d^2)/(c*d*x+a*e)^2+3/4*e*(e*x+d)^(1/2)/(-a*e^ 
2+c*d^2)^2/(c*d*x+a*e)-3/4*e^2*arctanh(c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-a*e 
^2+c*d^2)^(1/2))/c^(1/2)/d^(1/2)/(-a*e^2+c*d^2)^(5/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {1}{4} \left (\frac {\sqrt {d+e x} \left (5 a e^2+c d (-2 d+3 e x)\right )}{\left (c d^2-a e^2\right )^2 (a e+c d x)^2}+\frac {3 e^2 \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\sqrt {c} \sqrt {d} \left (-c d^2+a e^2\right )^{5/2}}\right ) \] Input:

Integrate[(d + e*x)^(5/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1121, 52, 52, 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.

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

\(\Big \downarrow \) 1121

\(\displaystyle \int \frac {1}{\sqrt {d+e x} (a e+c d x)^3}dx\)

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(d + e*x)^(5/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]
 

Output:

-1/2*Sqrt[d + e*x]/((c*d^2 - a*e^2)*(a*e + c*d*x)^2) - (3*e*(-(Sqrt[d + e* 
x]/((c*d^2 - a*e^2)*(a*e + c*d*x))) + (e*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + 
 e*x])/Sqrt[c*d^2 - a*e^2]])/(Sqrt[c]*Sqrt[d]*(c*d^2 - a*e^2)^(3/2))))/(4* 
(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 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 = 15.50 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right ) e^{2} \left (c d x +a e \right )^{2}}{4}+\frac {5 \sqrt {e x +d}\, \left (-\frac {2 \left (-\frac {3 e x}{2}+d \right ) d c}{5}+a \,e^{2}\right ) \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}{4}}{\left (a \,e^{2}-c \,d^{2}\right )^{2} \left (c d x +a e \right )^{2} \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\)	\(129\)
derivativedivides	\(2 e^{2} \left (\frac {\sqrt {e x +d}}{4 \left (a \,e^{2}-c \,d^{2}\right ) \left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {\frac {3 \sqrt {e x +d}}{8 \left (a \,e^{2}-c \,d^{2}\right ) \left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{a \,e^{2}-c \,d^{2}}\right )\)	\(175\)
default	\(2 e^{2} \left (\frac {\sqrt {e x +d}}{4 \left (a \,e^{2}-c \,d^{2}\right ) \left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {\frac {3 \sqrt {e x +d}}{8 \left (a \,e^{2}-c \,d^{2}\right ) \left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{a \,e^{2}-c \,d^{2}}\right )\)	\(175\)

Input:

int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^3,x,method=_RETURNVERB 
OSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (122) = 244\).

Time = 0.11 (sec) , antiderivative size = 654, normalized size of antiderivative = 4.48 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\left [\frac {3 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {c^{2} d^{3} - a c d e^{2}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {e x + d}}{c d x + a e}\right ) - 2 \, {\left (2 \, c^{3} d^{5} - 7 \, a c^{2} d^{3} e^{2} + 5 \, a^{2} c d e^{4} - 3 \, {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} c^{4} d^{7} e^{2} - 3 \, a^{3} c^{3} d^{5} e^{4} + 3 \, a^{4} c^{2} d^{3} e^{6} - a^{5} c d e^{8} + {\left (c^{6} d^{9} - 3 \, a c^{5} d^{7} e^{2} + 3 \, a^{2} c^{4} d^{5} e^{4} - a^{3} c^{3} d^{3} e^{6}\right )} x^{2} + 2 \, {\left (a c^{5} d^{8} e - 3 \, a^{2} c^{4} d^{6} e^{3} + 3 \, a^{3} c^{3} d^{4} e^{5} - a^{4} c^{2} d^{2} e^{7}\right )} x\right )}}, \frac {3 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {e x + d}}{c d e x + c d^{2}}\right ) - {\left (2 \, c^{3} d^{5} - 7 \, a c^{2} d^{3} e^{2} + 5 \, a^{2} c d e^{4} - 3 \, {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} c^{4} d^{7} e^{2} - 3 \, a^{3} c^{3} d^{5} e^{4} + 3 \, a^{4} c^{2} d^{3} e^{6} - a^{5} c d e^{8} + {\left (c^{6} d^{9} - 3 \, a c^{5} d^{7} e^{2} + 3 \, a^{2} c^{4} d^{5} e^{4} - a^{3} c^{3} d^{3} e^{6}\right )} x^{2} + 2 \, {\left (a c^{5} d^{8} e - 3 \, a^{2} c^{4} d^{6} e^{3} + 3 \, a^{3} c^{3} d^{4} e^{5} - a^{4} c^{2} d^{2} e^{7}\right )} x\right )}}\right ] \] Input:

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm=" 
fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((e*x+d)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,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 = 0.18 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.20 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {3 \, e^{2} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {3 \, {\left (e x + d\right )}^{\frac {3}{2}} c d e^{2} - 5 \, \sqrt {e x + d} c d^{2} e^{2} + 5 \, \sqrt {e x + d} a e^{4}}{4 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2}} \] Input:

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm=" 
giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 5.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {\frac {5\,e^2\,\sqrt {d+e\,x}}{4\,\left (a\,e^2-c\,d^2\right )}+\frac {3\,c\,d\,e^2\,{\left (d+e\,x\right )}^{3/2}}{4\,{\left (a\,e^2-c\,d^2\right )}^2}}{a^2\,e^4+c^2\,d^4-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,\left (d+e\,x\right )+c^2\,d^2\,{\left (d+e\,x\right )}^2-2\,a\,c\,d^2\,e^2}+\frac {3\,e^2\,\mathrm {atan}\left (\frac {c\,d\,\sqrt {d+e\,x}}{\sqrt {c\,d}\,\sqrt {a\,e^2-c\,d^2}}\right )}{4\,\sqrt {c\,d}\,{\left (a\,e^2-c\,d^2\right )}^{5/2}} \] Input:

int((d + e*x)^(5/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.97 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {3 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{2} e^{4}+6 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a c d \,e^{3} x +3 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{2} e^{2} x^{2}+5 \sqrt {e x +d}\, a^{2} c d \,e^{4}-7 \sqrt {e x +d}\, a \,c^{2} d^{3} e^{2}+3 \sqrt {e x +d}\, a \,c^{2} d^{2} e^{3} x +2 \sqrt {e x +d}\, c^{3} d^{5}-3 \sqrt {e x +d}\, c^{3} d^{4} e x}{4 c d \left (a^{3} c^{2} d^{2} e^{6} x^{2}-3 a^{2} c^{3} d^{4} e^{4} x^{2}+3 a \,c^{4} d^{6} e^{2} x^{2}-c^{5} d^{8} x^{2}+2 a^{4} c d \,e^{7} x -6 a^{3} c^{2} d^{3} e^{5} x +6 a^{2} c^{3} d^{5} e^{3} x -2 a \,c^{4} d^{7} e x +a^{5} e^{8}-3 a^{4} c \,d^{2} e^{6}+3 a^{3} c^{2} d^{4} e^{4}-a^{2} c^{3} d^{6} e^{2}\right )} \] Input:

int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)
 

Output:

(3*sqrt(d)*sqrt(c)*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**2*e**4 + 6*sqrt(d)*sqrt(c)*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*e**3*x + 3*sqrt(d)*sqrt(c)*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 + 5* 
sqrt(d + e*x)*a**2*c*d*e**4 - 7*sqrt(d + e*x)*a*c**2*d**3*e**2 + 3*sqrt(d 
+ e*x)*a*c**2*d**2*e**3*x + 2*sqrt(d + e*x)*c**3*d**5 - 3*sqrt(d + e*x)*c* 
*3*d**4*e*x)/(4*c*d*(a**5*e**8 - 3*a**4*c*d**2*e**6 + 2*a**4*c*d*e**7*x + 
3*a**3*c**2*d**4*e**4 - 6*a**3*c**2*d**3*e**5*x + a**3*c**2*d**2*e**6*x**2 
 - a**2*c**3*d**6*e**2 + 6*a**2*c**3*d**5*e**3*x - 3*a**2*c**3*d**4*e**4*x 
**2 - 2*a*c**4*d**7*e*x + 3*a*c**4*d**6*e**2*x**2 - c**5*d**8*x**2))