\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/4}}{(d+e x)^5} \, dx\) [359]

Optimal result

Integrand size = 37, antiderivative size = 290 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^5} \, dx=-\frac {20 c d \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{77 e^2 (d+e x)^2}+\frac {20 c^2 d^2 \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{231 e^2 \left (c d^2-a e^2\right ) (d+e x)}-\frac {4 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{11 e (d+e x)^4}-\frac {40 c^2 d^2 (a e+c d x)^{3/2} \left (\frac {c d (d+e x)}{e (a e+c d x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {c d^2-a e^2}}{\sqrt {e} \sqrt {a e+c d x}}\right ),2\right )}{231 e^{3/2} \left (c d^2-a e^2\right )^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \] Output:

-20/77*c*d*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/4)/e^2/(e*x+d)^2+20/231*c^ 
2*d^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/4)/e^2/(-a*e^2+c*d^2)/(e*x+d)-4 
/11*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/e/(e*x+d)^4-40/231*c^2*d^2*(c* 
d*x+a*e)^(3/2)*(c*d*(e*x+d)/e/(c*d*x+a*e))^(3/4)*InverseJacobiAM(1/2*arcta 
n((-a*e^2+c*d^2)^(1/2)/e^(1/2)/(c*d*x+a*e)^(1/2)),2^(1/2))/e^(3/2)/(-a*e^2 
+c*d^2)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.39 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^5} \, dx=\frac {4 c^3 d^3 (a e+c d x)^2 \sqrt [4]{(a e+c d x) (d+e x)} \operatorname {Hypergeometric2F1}\left (\frac {9}{4},\frac {15}{4},\frac {13}{4},\frac {e (a e+c d x)}{-c d^2+a e^2}\right )}{9 \left (c d^2-a e^2\right )^4 \sqrt [4]{\frac {c d (d+e x)}{c d^2-a e^2}}} \] Input:

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/4)/(d + e*x)^5,x]
 

Output:

(4*c^3*d^3*(a*e + c*d*x)^2*((a*e + c*d*x)*(d + e*x))^(1/4)*Hypergeometric2 
F1[9/4, 15/4, 13/4, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(9*(c*d^2 - a*e 
^2)^4*((c*d*(d + e*x))/(c*d^2 - a*e^2))^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.82 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {1138, 57, 57, 61, 73, 768, 858, 807, 229}

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.

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

\(\Big \downarrow \) 1138

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \int \frac {(a e+c d x)^{5/4}}{\left (\frac {e x}{d}+1\right )^{15/4}}dx}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {5 c d^2 \int \frac {\sqrt [4]{a e+c d x}}{\left (\frac {e x}{d}+1\right )^{11/4}}dx}{11 e}-\frac {4 d (a e+c d x)^{5/4}}{11 e \left (\frac {e x}{d}+1\right )^{11/4}}\right )}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {5 c d^2 \left (\frac {c d^2 \int \frac {1}{(a e+c d x)^{3/4} \left (\frac {e x}{d}+1\right )^{7/4}}dx}{7 e}-\frac {4 d \sqrt [4]{a e+c d x}}{7 e \left (\frac {e x}{d}+1\right )^{7/4}}\right )}{11 e}-\frac {4 d (a e+c d x)^{5/4}}{11 e \left (\frac {e x}{d}+1\right )^{11/4}}\right )}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {5 c d^2 \left (\frac {c d^2 \left (\frac {2 c d \int \frac {1}{(a e+c d x)^{3/4} \left (\frac {e x}{d}+1\right )^{3/4}}dx}{3 \left (c d-\frac {a e^2}{d}\right )}+\frac {4 \sqrt [4]{a e+c d x}}{3 \left (\frac {e x}{d}+1\right )^{3/4} \left (c d-\frac {a e^2}{d}\right )}\right )}{7 e}-\frac {4 d \sqrt [4]{a e+c d x}}{7 e \left (\frac {e x}{d}+1\right )^{7/4}}\right )}{11 e}-\frac {4 d (a e+c d x)^{5/4}}{11 e \left (\frac {e x}{d}+1\right )^{11/4}}\right )}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {5 c d^2 \left (\frac {c d^2 \left (\frac {8 \int \frac {1}{\left (-\frac {a e^2}{c d^2}+\frac {(a e+c d x) e}{c d^2}+1\right )^{3/4}}d\sqrt [4]{a e+c d x}}{3 \left (c d-\frac {a e^2}{d}\right )}+\frac {4 \sqrt [4]{a e+c d x}}{3 \left (\frac {e x}{d}+1\right )^{3/4} \left (c d-\frac {a e^2}{d}\right )}\right )}{7 e}-\frac {4 d \sqrt [4]{a e+c d x}}{7 e \left (\frac {e x}{d}+1\right )^{7/4}}\right )}{11 e}-\frac {4 d (a e+c d x)^{5/4}}{11 e \left (\frac {e x}{d}+1\right )^{11/4}}\right )}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

\(\Big \downarrow \) 768

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {5 c d^2 \left (\frac {c d^2 \left (\frac {8 (a e+c d x)^{3/4} \left (\frac {c d^2-a e^2}{e (a e+c d x)}+1\right )^{3/4} \int \frac {1}{(a e+c d x)^{3/4} \left (\frac {c d^2-a e^2}{e (a e+c d x)}+1\right )^{3/4}}d\sqrt [4]{a e+c d x}}{3 \left (c d-\frac {a e^2}{d}\right ) \left (-\frac {a e^2}{c d^2}+\frac {e (a e+c d x)}{c d^2}+1\right )^{3/4}}+\frac {4 \sqrt [4]{a e+c d x}}{3 \left (\frac {e x}{d}+1\right )^{3/4} \left (c d-\frac {a e^2}{d}\right )}\right )}{7 e}-\frac {4 d \sqrt [4]{a e+c d x}}{7 e \left (\frac {e x}{d}+1\right )^{7/4}}\right )}{11 e}-\frac {4 d (a e+c d x)^{5/4}}{11 e \left (\frac {e x}{d}+1\right )^{11/4}}\right )}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {5 c d^2 \left (\frac {c d^2 \left (\frac {4 \sqrt [4]{a e+c d x}}{3 \left (\frac {e x}{d}+1\right )^{3/4} \left (c d-\frac {a e^2}{d}\right )}-\frac {8 (a e+c d x)^{3/4} \left (\frac {c d^2-a e^2}{e (a e+c d x)}+1\right )^{3/4} \int \frac {1}{\sqrt [4]{a e+c d x} \left (\frac {\left (c d^2-a e^2\right ) (a e+c d x)}{e}+1\right )^{3/4}}d\frac {1}{\sqrt [4]{a e+c d x}}}{3 \left (c d-\frac {a e^2}{d}\right ) \left (-\frac {a e^2}{c d^2}+\frac {e (a e+c d x)}{c d^2}+1\right )^{3/4}}\right )}{7 e}-\frac {4 d \sqrt [4]{a e+c d x}}{7 e \left (\frac {e x}{d}+1\right )^{7/4}}\right )}{11 e}-\frac {4 d (a e+c d x)^{5/4}}{11 e \left (\frac {e x}{d}+1\right )^{11/4}}\right )}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {5 c d^2 \left (\frac {c d^2 \left (\frac {4 \sqrt [4]{a e+c d x}}{3 \left (\frac {e x}{d}+1\right )^{3/4} \left (c d-\frac {a e^2}{d}\right )}-\frac {4 (a e+c d x)^{3/4} \left (\frac {c d^2-a e^2}{e (a e+c d x)}+1\right )^{3/4} \int \frac {1}{\left (\frac {\sqrt {a e+c d x} \left (c d^2-a e^2\right )}{e}+1\right )^{3/4}}d\sqrt {a e+c d x}}{3 \left (c d-\frac {a e^2}{d}\right ) \left (-\frac {a e^2}{c d^2}+\frac {e (a e+c d x)}{c d^2}+1\right )^{3/4}}\right )}{7 e}-\frac {4 d \sqrt [4]{a e+c d x}}{7 e \left (\frac {e x}{d}+1\right )^{7/4}}\right )}{11 e}-\frac {4 d (a e+c d x)^{5/4}}{11 e \left (\frac {e x}{d}+1\right )^{11/4}}\right )}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {5 c d^2 \left (\frac {c d^2 \left (\frac {4 \sqrt [4]{a e+c d x}}{3 \left (\frac {e x}{d}+1\right )^{3/4} \left (c d-\frac {a e^2}{d}\right )}-\frac {8 \sqrt {e} (a e+c d x)^{3/4} \left (\frac {c d^2-a e^2}{e (a e+c d x)}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {c d^2-a e^2} \sqrt {a e+c d x}}{\sqrt {e}}\right ),2\right )}{3 \sqrt {c d^2-a e^2} \left (c d-\frac {a e^2}{d}\right ) \left (-\frac {a e^2}{c d^2}+\frac {e (a e+c d x)}{c d^2}+1\right )^{3/4}}\right )}{7 e}-\frac {4 d \sqrt [4]{a e+c d x}}{7 e \left (\frac {e x}{d}+1\right )^{7/4}}\right )}{11 e}-\frac {4 d (a e+c d x)^{5/4}}{11 e \left (\frac {e x}{d}+1\right )^{11/4}}\right )}{d^5 \sqrt [4]{\frac {e x}{d}+1} \sqrt [4]{a e+c d x}}\)

Input:

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/4)/(d + e*x)^5,x]
 

Output:

((a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(1/4)*((-4*d*(a*e + c*d*x)^(5/4)) 
/(11*e*(1 + (e*x)/d)^(11/4)) + (5*c*d^2*((-4*d*(a*e + c*d*x)^(1/4))/(7*e*( 
1 + (e*x)/d)^(7/4)) + (c*d^2*((4*(a*e + c*d*x)^(1/4))/(3*(c*d - (a*e^2)/d) 
*(1 + (e*x)/d)^(3/4)) - (8*Sqrt[e]*(a*e + c*d*x)^(3/4)*(1 + (c*d^2 - a*e^2 
)/(e*(a*e + c*d*x)))^(3/4)*EllipticF[ArcTan[(Sqrt[c*d^2 - a*e^2]*Sqrt[a*e 
+ c*d*x])/Sqrt[e]]/2, 2])/(3*Sqrt[c*d^2 - a*e^2]*(c*d - (a*e^2)/d)*(1 - (a 
*e^2)/(c*d^2) + (e*(a*e + c*d*x))/(c*d^2))^(3/4))))/(7*e)))/(11*e)))/(d^5* 
(a*e + c*d*x)^(1/4)*(1 + (e*x)/d)^(1/4))
 

Defintions of rubi rules used

rule 57

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

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 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 1138

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d^m*((a + b*x + c*x^2)^FracPart[p]/((1 + e*(x/d))^FracPart[p] 
*(a/d + (c*x)/e)^FracPart[p]))   Int[(1 + e*(x/d))^(m + p)*(a/d + (c/e)*x)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
&& (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (IntegerQ[3*p] || Integer 
Q[4*p]))
 

Maple [F]

\[\int \frac {{\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e \right )}^{\frac {5}{4}}}{\left (e x +d \right )^{5}}d x\]

Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/4)/(e*x+d)^5,x)
 

Output:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/4)/(e*x+d)^5,x)
 

Fricas [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^5} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}}}{{\left (e x + d\right )}^{5}} \,d x } \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/(e*x+d)^5,x, algorithm=" 
fricas")
 

Output:

integral((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(1/4)*(c*d*x + a*e)/(e^4* 
x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)
 

Sympy [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^5} \, dx=\int \frac {\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {5}{4}}}{\left (d + e x\right )^{5}}\, dx \] Input:

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/4)/(e*x+d)**5,x)
 

Output:

Integral(((d + e*x)*(a*e + c*d*x))**(5/4)/(d + e*x)**5, x)
 

Maxima [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^5} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}}}{{\left (e x + d\right )}^{5}} \,d x } \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/(e*x+d)^5,x, algorithm=" 
maxima")
 

Output:

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/4)/(e*x + d)^5, x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^5} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}}}{{\left (e x + d\right )}^{5}} \,d x } \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/(e*x+d)^5,x, algorithm=" 
giac")
 

Output:

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/4)/(e*x + d)^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^5} \, dx=\int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/4}}{{\left (d+e\,x\right )}^5} \,d x \] Input:

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/4)/(d + e*x)^5,x)
 

Output:

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/4)/(d + e*x)^5, x)
 

Reduce [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^5} \, dx=\text {too large to display} \] Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/(e*x+d)^5,x)
 

Output:

( - 24*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*e**3 - 16*(a 
*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c*d**2*e - 44*(a*d*e + a 
*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c*d*e**2*x + 4*(a*d*e + a*e**2*x 
 + c*d**2*x + c*d*e*x**2)**(1/4)*c**2*d**3*x - 55*int(((a*d*e + a*e**2*x + 
 c*d**2*x + c*d*e*x**2)**(1/4)*x)/(11*a**2*d**4*e**3 + 44*a**2*d**3*e**4*x 
 + 66*a**2*d**2*e**5*x**2 + 44*a**2*d*e**6*x**3 + 11*a**2*e**7*x**4 - a*c* 
d**6*e + 7*a*c*d**5*e**2*x + 38*a*c*d**4*e**3*x**2 + 62*a*c*d**3*e**4*x**3 
 + 43*a*c*d**2*e**5*x**4 + 11*a*c*d*e**6*x**5 - c**2*d**7*x - 4*c**2*d**6* 
e*x**2 - 6*c**2*d**5*e**2*x**3 - 4*c**2*d**4*e**3*x**4 - c**2*d**3*e**4*x* 
*5),x)*a**3*c*d**4*e**6 - 165*int(((a*d*e + a*e**2*x + c*d**2*x + c*d*e*x* 
*2)**(1/4)*x)/(11*a**2*d**4*e**3 + 44*a**2*d**3*e**4*x + 66*a**2*d**2*e**5 
*x**2 + 44*a**2*d*e**6*x**3 + 11*a**2*e**7*x**4 - a*c*d**6*e + 7*a*c*d**5* 
e**2*x + 38*a*c*d**4*e**3*x**2 + 62*a*c*d**3*e**4*x**3 + 43*a*c*d**2*e**5* 
x**4 + 11*a*c*d*e**6*x**5 - c**2*d**7*x - 4*c**2*d**6*e*x**2 - 6*c**2*d**5 
*e**2*x**3 - 4*c**2*d**4*e**3*x**4 - c**2*d**3*e**4*x**5),x)*a**3*c*d**3*e 
**7*x - 165*int(((a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*x)/(11* 
a**2*d**4*e**3 + 44*a**2*d**3*e**4*x + 66*a**2*d**2*e**5*x**2 + 44*a**2*d* 
e**6*x**3 + 11*a**2*e**7*x**4 - a*c*d**6*e + 7*a*c*d**5*e**2*x + 38*a*c*d* 
*4*e**3*x**2 + 62*a*c*d**3*e**4*x**3 + 43*a*c*d**2*e**5*x**4 + 11*a*c*d*e* 
*6*x**5 - c**2*d**7*x - 4*c**2*d**6*e*x**2 - 6*c**2*d**5*e**2*x**3 - 4*...