\(\int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx\) [1646]

Optimal result

Integrand size = 28, antiderivative size = 137 \[ \int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {7 e (b d-a e)^2 \sqrt {d+e x}}{b^4}+\frac {7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {7 e (d+e x)^{5/2}}{5 b^2}-\frac {(d+e x)^{7/2}}{b (a+b x)}-\frac {7 e (b d-a e)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}} \]

[Out]

7/3*e*(-a*e+b*d)*(e*x+d)^(3/2)/b^3+7/5*e*(e*x+d)^(5/2)/b^2-(e*x+d)^(7/2)/b/(b*x+a)-7*e*(-a*e+b*d)^(5/2)*arctan
h(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(9/2)+7*e*(-a*e+b*d)^2*(e*x+d)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 43, 52, 65, 214} \[ \int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {7 e (b d-a e)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}}+\frac {7 e \sqrt {d+e x} (b d-a e)^2}{b^4}+\frac {7 e (d+e x)^{3/2} (b d-a e)}{3 b^3}-\frac {(d+e x)^{7/2}}{b (a+b x)}+\frac {7 e (d+e x)^{5/2}}{5 b^2} \]

[In]

Int[(d + e*x)^(7/2)/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(7*e*(b*d - a*e)^2*Sqrt[d + e*x])/b^4 + (7*e*(b*d - a*e)*(d + e*x)^(3/2))/(3*b^3) + (7*e*(d + e*x)^(5/2))/(5*b
^2) - (d + e*x)^(7/2)/(b*(a + b*x)) - (7*e*(b*d - a*e)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])
/b^(9/2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

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] - Dist[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] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^{7/2}}{(a+b x)^2} \, dx \\ & = -\frac {(d+e x)^{7/2}}{b (a+b x)}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{a+b x} \, dx}{2 b} \\ & = \frac {7 e (d+e x)^{5/2}}{5 b^2}-\frac {(d+e x)^{7/2}}{b (a+b x)}+\frac {(7 e (b d-a e)) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{2 b^2} \\ & = \frac {7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {7 e (d+e x)^{5/2}}{5 b^2}-\frac {(d+e x)^{7/2}}{b (a+b x)}+\frac {\left (7 e (b d-a e)^2\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{2 b^3} \\ & = \frac {7 e (b d-a e)^2 \sqrt {d+e x}}{b^4}+\frac {7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {7 e (d+e x)^{5/2}}{5 b^2}-\frac {(d+e x)^{7/2}}{b (a+b x)}+\frac {\left (7 e (b d-a e)^3\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 b^4} \\ & = \frac {7 e (b d-a e)^2 \sqrt {d+e x}}{b^4}+\frac {7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {7 e (d+e x)^{5/2}}{5 b^2}-\frac {(d+e x)^{7/2}}{b (a+b x)}+\frac {\left (7 (b d-a e)^3\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{b^4} \\ & = \frac {7 e (b d-a e)^2 \sqrt {d+e x}}{b^4}+\frac {7 e (b d-a e) (d+e x)^{3/2}}{3 b^3}+\frac {7 e (d+e x)^{5/2}}{5 b^2}-\frac {(d+e x)^{7/2}}{b (a+b x)}-\frac {7 e (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {\sqrt {d+e x} \left (-105 a^3 e^3+35 a^2 b e^2 (7 d-2 e x)+7 a b^2 e \left (-23 d^2+24 d e x+2 e^2 x^2\right )+b^3 \left (15 d^3-116 d^2 e x-32 d e^2 x^2-6 e^3 x^3\right )\right )}{15 b^4 (a+b x)}-\frac {7 e (-b d+a e)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{9/2}} \]

[In]

Integrate[(d + e*x)^(7/2)/(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

-1/15*(Sqrt[d + e*x]*(-105*a^3*e^3 + 35*a^2*b*e^2*(7*d - 2*e*x) + 7*a*b^2*e*(-23*d^2 + 24*d*e*x + 2*e^2*x^2) +
b^3*(15*d^3 - 116*d^2*e*x - 32*d*e^2*x^2 - 6*e^3*x^3)))/(b^4*(a + b*x)) - (7*e*(-(b*d) + a*e)^(5/2)*ArcTan[(S
qrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/b^(9/2)

Maple [A] (verified)

Time = 2.35 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.22

method	result	size
pseudoelliptic	\(-\frac {7 \left (e \left (a e -b d \right )^{3} \left (b x +a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )-\sqrt {e x +d}\, \left (\left (\frac {2}{35} e^{3} x^{3}+\frac {32}{105} d \,e^{2} x^{2}+\frac {116}{105} d^{2} e x -\frac {1}{7} d^{3}\right ) b^{3}+\frac {23 \left (-\frac {2}{23} x^{2} e^{2}-\frac {24}{23} d e x +d^{2}\right ) e a \,b^{2}}{15}-\frac {7 \left (-\frac {2 e x}{7}+d \right ) e^{2} a^{2} b}{3}+a^{3} e^{3}\right ) \sqrt {\left (a e -b d \right ) b}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{4} \left (b x +a \right )}\)	\(167\)
risch	\(\frac {2 e \left (3 x^{2} b^{2} e^{2}-10 x a b \,e^{2}+16 b^{2} d e x +45 a^{2} e^{2}-100 a b d e +58 b^{2} d^{2}\right ) \sqrt {e x +d}}{15 b^{4}}-\frac {\left (2 a^{3} e^{3}-6 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) e \left (-\frac {\sqrt {e x +d}}{2 \left (b \left (e x +d \right )+a e -b d \right )}+\frac {7 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{4}}\)	\(171\)
derivativedivides	\(2 e \left (\frac {\frac {b^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 a b e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+3 a^{2} e^{2} \sqrt {e x +d}-6 a b d e \sqrt {e x +d}+3 b^{2} d^{2} \sqrt {e x +d}}{b^{4}}-\frac {\frac {\left (-\frac {1}{2} a^{3} e^{3}+\frac {3}{2} a^{2} b d \,e^{2}-\frac {3}{2} a \,b^{2} d^{2} e +\frac {1}{2} b^{3} d^{3}\right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {7 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}}{b^{4}}\right )\)	\(230\)
default	\(2 e \left (\frac {\frac {b^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 a b e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+3 a^{2} e^{2} \sqrt {e x +d}-6 a b d e \sqrt {e x +d}+3 b^{2} d^{2} \sqrt {e x +d}}{b^{4}}-\frac {\frac {\left (-\frac {1}{2} a^{3} e^{3}+\frac {3}{2} a^{2} b d \,e^{2}-\frac {3}{2} a \,b^{2} d^{2} e +\frac {1}{2} b^{3} d^{3}\right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {7 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}}{b^{4}}\right )\)	\(230\)

[In]

int((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)

[Out]

-7*(e*(a*e-b*d)^3*(b*x+a)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))-(e*x+d)^(1/2)*((2/35*e^3*x^3+32/105*d*e^
2*x^2+116/105*d^2*e*x-1/7*d^3)*b^3+23/15*(-2/23*x^2*e^2-24/23*d*e*x+d^2)*e*a*b^2-7/3*(-2/7*e*x+d)*e^2*a^2*b+a^
3*e^3)*((a*e-b*d)*b)^(1/2))/((a*e-b*d)*b)^(1/2)/b^4/(b*x+a)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (115) = 230\).

Time = 0.28 (sec) , antiderivative size = 486, normalized size of antiderivative = 3.55 \[ \int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx=\left [\frac {105 \, {\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \, {\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {105 \, {\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \, {\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]

[In]

integrate((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

[1/30*(105*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a^3*e^3 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x)*sqrt((b*d - a*e
)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(6*b^3*e^3*x^3 - 15*b^3*
d^3 + 161*a*b^2*d^2*e - 245*a^2*b*d*e^2 + 105*a^3*e^3 + 2*(16*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 + 2*(58*b^3*d^2*e -
84*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^5*x + a*b^4), -1/15*(105*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a
^3*e^3 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d -
a*e)/b)/(b*d - a*e)) - (6*b^3*e^3*x^3 - 15*b^3*d^3 + 161*a*b^2*d^2*e - 245*a^2*b*d*e^2 + 105*a^3*e^3 + 2*(16*b
^3*d*e^2 - 7*a*b^2*e^3)*x^2 + 2*(58*b^3*d^2*e - 84*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d))/(b^5*x + a*b^
4)]

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx=\text {Timed out} \]

[In]

integrate((e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (115) = 230\).

Time = 0.28 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.97 \[ \int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {7 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{4}} - \frac {\sqrt {e x + d} b^{3} d^{3} e - 3 \, \sqrt {e x + d} a b^{2} d^{2} e^{2} + 3 \, \sqrt {e x + d} a^{2} b d e^{3} - \sqrt {e x + d} a^{3} e^{4}}{{\left ({\left (e x + d\right )} b - b d + a e\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{8} e + 10 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{8} d e + 45 \, \sqrt {e x + d} b^{8} d^{2} e - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{7} e^{2} - 90 \, \sqrt {e x + d} a b^{7} d e^{2} + 45 \, \sqrt {e x + d} a^{2} b^{6} e^{3}\right )}}{15 \, b^{10}} \]

[In]

integrate((e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

7*(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-
b^2*d + a*b*e)*b^4) - (sqrt(e*x + d)*b^3*d^3*e - 3*sqrt(e*x + d)*a*b^2*d^2*e^2 + 3*sqrt(e*x + d)*a^2*b*d*e^3 -
sqrt(e*x + d)*a^3*e^4)/(((e*x + d)*b - b*d + a*e)*b^4) + 2/15*(3*(e*x + d)^(5/2)*b^8*e + 10*(e*x + d)^(3/2)*b
^8*d*e + 45*sqrt(e*x + d)*b^8*d^2*e - 10*(e*x + d)^(3/2)*a*b^7*e^2 - 90*sqrt(e*x + d)*a*b^7*d*e^2 + 45*sqrt(e*
x + d)*a^2*b^6*e^3)/b^10

Mupad [B] (verification not implemented)

Time = 9.41 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.72 \[ \int \frac {(d+e x)^{7/2}}{a^2+2 a b x+b^2 x^2} \, dx=\left (\frac {2\,e\,{\left (2\,b^2\,d-2\,a\,b\,e\right )}^2}{b^6}-\frac {2\,e\,{\left (a\,e-b\,d\right )}^2}{b^4}\right )\,\sqrt {d+e\,x}+\frac {\sqrt {d+e\,x}\,\left (a^3\,e^4-3\,a^2\,b\,d\,e^3+3\,a\,b^2\,d^2\,e^2-b^3\,d^3\,e\right )}{b^5\,\left (d+e\,x\right )-b^5\,d+a\,b^4\,e}+\frac {2\,e\,{\left (d+e\,x\right )}^{5/2}}{5\,b^2}+\frac {2\,e\,\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b^4}-\frac {7\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,{\left (a\,e-b\,d\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^4-3\,a^2\,b\,d\,e^3+3\,a\,b^2\,d^2\,e^2-b^3\,d^3\,e}\right )\,{\left (a\,e-b\,d\right )}^{5/2}}{b^{9/2}} \]

[In]

int((d + e*x)^(7/2)/(a^2 + b^2*x^2 + 2*a*b*x),x)

[Out]

((2*e*(2*b^2*d - 2*a*b*e)^2)/b^6 - (2*e*(a*e - b*d)^2)/b^4)*(d + e*x)^(1/2) + ((d + e*x)^(1/2)*(a^3*e^4 - b^3*
d^3*e + 3*a*b^2*d^2*e^2 - 3*a^2*b*d*e^3))/(b^5*(d + e*x) - b^5*d + a*b^4*e) + (2*e*(d + e*x)^(5/2))/(5*b^2) +
(2*e*(2*b^2*d - 2*a*b*e)*(d + e*x)^(3/2))/(3*b^4) - (7*e*atan((b^(1/2)*e*(a*e - b*d)^(5/2)*(d + e*x)^(1/2))/(a
^3*e^4 - b^3*d^3*e + 3*a*b^2*d^2*e^2 - 3*a^2*b*d*e^3))*(a*e - b*d)^(5/2))/b^(9/2)