\(\int \frac {(a+b x) (d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [248]

Optimal result

Integrand size = 35, antiderivative size = 204 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {5 e^2 \sqrt {d+e x}}{8 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e^3 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} \sqrt {b d-a e} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

-5/8*e^2*(e*x+d)^(1/2)/b^3/((b*x+a)^2)^(1/2)-5/12*e*(e*x+d)^(3/2)/b^2/(b*x 
+a)/((b*x+a)^2)^(1/2)-1/3*(e*x+d)^(5/2)/b/(b*x+a)^2/((b*x+a)^2)^(1/2)-5/8* 
e^3*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)/(-a*e+ 
b*d)^(1/2)/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3 (a+b x) \left (-\frac {\sqrt {b} \sqrt {d+e x} \left (15 a^2 e^2+10 a b e (d+4 e x)+b^2 \left (8 d^2+26 d e x+33 e^2 x^2\right )\right )}{e^3 (a+b x)^3}+\frac {15 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {-b d+a e}}\right )}{24 b^{7/2} \sqrt {(a+b x)^2}} \] Input:

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

Output:

(e^3*(a + b*x)*(-((Sqrt[b]*Sqrt[d + e*x]*(15*a^2*e^2 + 10*a*b*e*(d + 4*e*x 
) + b^2*(8*d^2 + 26*d*e*x + 33*e^2*x^2)))/(e^3*(a + b*x)^3)) + (15*ArcTan[ 
(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/Sqrt[-(b*d) + a*e]))/(24*b^(7 
/2)*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.78, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1187, 27, 51, 51, 51, 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.

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

\(\Big \downarrow \) 1187

\(\displaystyle \frac {b^5 (a+b x) \int \frac {(d+e x)^{5/2}}{b^5 (a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^{5/2}}{(a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(a+b x) \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{(a+b x)^3}dx}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(a+b x) \left (\frac {5 e \left (\frac {3 e \int \frac {\sqrt {d+e x}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(a+b x) \left (\frac {5 e \left (\frac {3 e \left (\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+b x) \left (\frac {5 e \left (\frac {3 e \left (\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(a+b x) \left (\frac {5 e \left (\frac {3 e \left (-\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

Input:

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

Output:

((a + b*x)*(-1/3*(d + e*x)^(5/2)/(b*(a + b*x)^3) + (5*e*(-1/2*(d + e*x)^(3 
/2)/(b*(a + b*x)^2) + (3*e*(-(Sqrt[d + e*x]/(b*(a + b*x))) - (e*ArcTanh[(S 
qrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b^(3/2)*Sqrt[b*d - a*e])))/(4*b)) 
)/(6*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 51

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, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[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 1187

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ 
IntPart[p]*(b/2 + c*x)^(2*FracPart[p]))   Int[(d + e*x)^m*(f + g*x)^n*(b/2 
+ c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 
 - 4*a*c, 0] &&  !IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(315\) vs. \(2(136)=272\).

Time = 1.46 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.55

method	result	size
default	\(-\frac {\left (-15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{3} e^{3} x^{3}-45 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,b^{2} e^{3} x^{2}+33 \left (e x +d \right )^{\frac {5}{2}} \sqrt {b \left (a e -b d \right )}\, b^{2}-45 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} b \,e^{3} x +40 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, a b e -40 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, b^{2} d -15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{3} e^{3}+15 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a^{2} e^{2}-30 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a b d e +15 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b^{2} d^{2}\right ) \left (b x +a \right )^{2}}{24 \sqrt {b \left (a e -b d \right )}\, b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(316\)

Input:

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

Output:

-1/24*(-15*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*b^3*e^3*x^3-45*arct 
an(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a*b^2*e^3*x^2+33*(e*x+d)^(5/2)*(b* 
(a*e-b*d))^(1/2)*b^2-45*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^2*b* 
e^3*x+40*(e*x+d)^(3/2)*(b*(a*e-b*d))^(1/2)*a*b*e-40*(e*x+d)^(3/2)*(b*(a*e- 
b*d))^(1/2)*b^2*d-15*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^3*e^3+1 
5*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*a^2*e^2-30*(e*x+d)^(1/2)*(b*(a*e-b*d)) 
^(1/2)*a*b*d*e+15*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*b^2*d^2)*(b*x+a)^2/(b* 
(a*e-b*d))^(1/2)/b^3/((b*x+a)^2)^(5/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (136) = 272\).

Time = 0.09 (sec) , antiderivative size = 563, normalized size of antiderivative = 2.76 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (8 \, b^{4} d^{3} + 2 \, a b^{3} d^{2} e + 5 \, a^{2} b^{2} d e^{2} - 15 \, a^{3} b e^{3} + 33 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (13 \, b^{4} d^{2} e + 7 \, a b^{3} d e^{2} - 20 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{5} d - a^{4} b^{4} e + {\left (b^{8} d - a b^{7} e\right )} x^{3} + 3 \, {\left (a b^{7} d - a^{2} b^{6} e\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d - a^{3} b^{5} e\right )} x\right )}}, \frac {15 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (8 \, b^{4} d^{3} + 2 \, a b^{3} d^{2} e + 5 \, a^{2} b^{2} d e^{2} - 15 \, a^{3} b e^{3} + 33 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (13 \, b^{4} d^{2} e + 7 \, a b^{3} d e^{2} - 20 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{5} d - a^{4} b^{4} e + {\left (b^{8} d - a b^{7} e\right )} x^{3} + 3 \, {\left (a b^{7} d - a^{2} b^{6} e\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d - a^{3} b^{5} e\right )} x\right )}}\right ] \] Input:

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

Output:

[1/48*(15*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(b 
^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + 
d))/(b*x + a)) - 2*(8*b^4*d^3 + 2*a*b^3*d^2*e + 5*a^2*b^2*d*e^2 - 15*a^3*b 
*e^3 + 33*(b^4*d*e^2 - a*b^3*e^3)*x^2 + 2*(13*b^4*d^2*e + 7*a*b^3*d*e^2 - 
20*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d - a^4*b^4*e + (b^8*d - a*b^7* 
e)*x^3 + 3*(a*b^7*d - a^2*b^6*e)*x^2 + 3*(a^2*b^6*d - a^3*b^5*e)*x), 1/24* 
(15*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(-b^2*d 
+ a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (8*b^4 
*d^3 + 2*a*b^3*d^2*e + 5*a^2*b^2*d*e^2 - 15*a^3*b*e^3 + 33*(b^4*d*e^2 - a* 
b^3*e^3)*x^2 + 2*(13*b^4*d^2*e + 7*a*b^3*d*e^2 - 20*a^2*b^2*e^3)*x)*sqrt(e 
*x + d))/(a^3*b^5*d - a^4*b^4*e + (b^8*d - a*b^7*e)*x^3 + 3*(a*b^7*d - a^2 
*b^6*e)*x^2 + 3*(a^2*b^6*d - a^3*b^5*e)*x)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {5 \, e^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, \sqrt {-b^{2} d + a b e} b^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {33 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} e^{3} - 40 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} d e^{3} + 15 \, \sqrt {e x + d} b^{2} d^{2} e^{3} + 40 \, {\left (e x + d\right )}^{\frac {3}{2}} a b e^{4} - 30 \, \sqrt {e x + d} a b d e^{4} + 15 \, \sqrt {e x + d} a^{2} e^{5}}{24 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3} b^{3} \mathrm {sgn}\left (b x + a\right )} \] Input:

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

Output:

5/8*e^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e) 
*b^3*sgn(b*x + a)) - 1/24*(33*(e*x + d)^(5/2)*b^2*e^3 - 40*(e*x + d)^(3/2) 
*b^2*d*e^3 + 15*sqrt(e*x + d)*b^2*d^2*e^3 + 40*(e*x + d)^(3/2)*a*b*e^4 - 3 
0*sqrt(e*x + d)*a*b*d*e^4 + 15*sqrt(e*x + d)*a^2*e^5)/(((e*x + d)*b - b*d 
+ a*e)^3*b^3*sgn(b*x + a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.01 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{3} e^{3}+45 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{2} b \,e^{3} x +45 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a \,b^{2} e^{3} x^{2}+15 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{3} e^{3} x^{3}-15 \sqrt {e x +d}\, a^{3} b \,e^{3}+5 \sqrt {e x +d}\, a^{2} b^{2} d \,e^{2}-40 \sqrt {e x +d}\, a^{2} b^{2} e^{3} x +2 \sqrt {e x +d}\, a \,b^{3} d^{2} e +14 \sqrt {e x +d}\, a \,b^{3} d \,e^{2} x -33 \sqrt {e x +d}\, a \,b^{3} e^{3} x^{2}+8 \sqrt {e x +d}\, b^{4} d^{3}+26 \sqrt {e x +d}\, b^{4} d^{2} e x +33 \sqrt {e x +d}\, b^{4} d \,e^{2} x^{2}}{24 b^{4} \left (a \,b^{3} e \,x^{3}-b^{4} d \,x^{3}+3 a^{2} b^{2} e \,x^{2}-3 a \,b^{3} d \,x^{2}+3 a^{3} b e x -3 a^{2} b^{2} d x +a^{4} e -a^{3} b d \right )} \] Input:

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

Output:

(15*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d 
)))*a**3*e**3 + 45*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b) 
*sqrt(a*e - b*d)))*a**2*b*e**3*x + 45*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d 
 + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a*b**2*e**3*x**2 + 15*sqrt(b)*sqrt(a 
*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b**3*e**3*x**3 
 - 15*sqrt(d + e*x)*a**3*b*e**3 + 5*sqrt(d + e*x)*a**2*b**2*d*e**2 - 40*sq 
rt(d + e*x)*a**2*b**2*e**3*x + 2*sqrt(d + e*x)*a*b**3*d**2*e + 14*sqrt(d + 
 e*x)*a*b**3*d*e**2*x - 33*sqrt(d + e*x)*a*b**3*e**3*x**2 + 8*sqrt(d + e*x 
)*b**4*d**3 + 26*sqrt(d + e*x)*b**4*d**2*e*x + 33*sqrt(d + e*x)*b**4*d*e** 
2*x**2)/(24*b**4*(a**4*e - a**3*b*d + 3*a**3*b*e*x - 3*a**2*b**2*d*x + 3*a 
**2*b**2*e*x**2 - 3*a*b**3*d*x**2 + a*b**3*e*x**3 - b**4*d*x**3))