3.21.75 \(\int \frac {(a+b x) (d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [2075]

Optimal. Leaf size=146 \[ \frac {35 e^2 (b d-a e) \sqrt {d+e x}}{4 b^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}-\frac {35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2}} \]

[Out]

35/12*e^2*(e*x+d)^(3/2)/b^3-7/4*e*(e*x+d)^(5/2)/b^2/(b*x+a)-1/2*(e*x+d)^(7/2)/b/(b*x+a)^2-35/4*e^2*(-a*e+b*d)^
(3/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(9/2)+35/4*e^2*(-a*e+b*d)*(e*x+d)^(1/2)/b^4

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Rubi [A]
time = 0.06, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 43, 52, 65, 214} \begin {gather*} -\frac {35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2}}+\frac {35 e^2 \sqrt {d+e x} (b d-a e)}{4 b^4}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(35*e^2*(b*d - a*e)*Sqrt[d + e*x])/(4*b^4) + (35*e^2*(d + e*x)^(3/2))/(12*b^3) - (7*e*(d + e*x)^(5/2))/(4*b^2*
(a + b*x)) - (d + e*x)^(7/2)/(2*b*(a + b*x)^2) - (35*e^2*(b*d - a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqr
t[b*d - a*e]])/(4*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*} \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^{7/2}}{(a+b x)^3} \, dx\\ &=-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {\left (35 e^2\right ) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{8 b^2}\\ &=\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {\left (35 e^2 (b d-a e)\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{8 b^3}\\ &=\frac {35 e^2 (b d-a e) \sqrt {d+e x}}{4 b^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {\left (35 e^2 (b d-a e)^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 b^4}\\ &=\frac {35 e^2 (b d-a e) \sqrt {d+e x}}{4 b^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}+\frac {\left (35 e (b d-a e)^2\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 b^4}\\ &=\frac {35 e^2 (b d-a e) \sqrt {d+e x}}{4 b^4}+\frac {35 e^2 (d+e x)^{3/2}}{12 b^3}-\frac {7 e (d+e x)^{5/2}}{4 b^2 (a+b x)}-\frac {(d+e x)^{7/2}}{2 b (a+b x)^2}-\frac {35 e^2 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 162, normalized size = 1.11 \begin {gather*} -\frac {\sqrt {d+e x} \left (105 a^3 e^3+35 a^2 b e^2 (-4 d+5 e x)+7 a b^2 e \left (3 d^2-34 d e x+8 e^2 x^2\right )+b^3 \left (6 d^3+39 d^2 e x-80 d e^2 x^2-8 e^3 x^3\right )\right )}{12 b^4 (a+b x)^2}+\frac {35 e^2 (-b d+a e)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(Sqrt[d + e*x]*(105*a^3*e^3 + 35*a^2*b*e^2*(-4*d + 5*e*x) + 7*a*b^2*e*(3*d^2 - 34*d*e*x + 8*e^2*x^2) + b
^3*(6*d^3 + 39*d^2*e*x - 80*d*e^2*x^2 - 8*e^3*x^3)))/(b^4*(a + b*x)^2) + (35*e^2*(-(b*d) + a*e)^(3/2)*ArcTan[(
Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(4*b^(9/2))

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Maple [A]
time = 0.07, size = 205, normalized size = 1.40

method	result	size
derivativedivides	\(2 e^{2} \left (-\frac {-\frac {b \left (e x +d \right )^{\frac {3}{2}}}{3}+3 a e \sqrt {e x +d}-3 b d \sqrt {e x +d}}{b^{4}}+\frac {\frac {\left (-\frac {13}{8} a^{2} b \,e^{2}+\frac {13}{4} a \,b^{2} d e -\frac {13}{8} d^{2} b^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {11}{8} a^{3} e^{3}+\frac {33}{8} a^{2} b d \,e^{2}-\frac {33}{8} a \,b^{2} d^{2} e +\frac {11}{8} b^{3} d^{3}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {35 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}}{b^{4}}\right )\)	\(205\)
default	\(2 e^{2} \left (-\frac {-\frac {b \left (e x +d \right )^{\frac {3}{2}}}{3}+3 a e \sqrt {e x +d}-3 b d \sqrt {e x +d}}{b^{4}}+\frac {\frac {\left (-\frac {13}{8} a^{2} b \,e^{2}+\frac {13}{4} a \,b^{2} d e -\frac {13}{8} d^{2} b^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {11}{8} a^{3} e^{3}+\frac {33}{8} a^{2} b d \,e^{2}-\frac {33}{8} a \,b^{2} d^{2} e +\frac {11}{8} b^{3} d^{3}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {35 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}}{b^{4}}\right )\)	\(205\)
risch	\(-\frac {2 e^{2} \left (-b e x +9 a e -10 b d \right ) \sqrt {e x +d}}{3 b^{4}}-\frac {13 e^{4} \left (e x +d \right )^{\frac {3}{2}} a^{2}}{4 b^{3} \left (b e x +a e \right )^{2}}+\frac {13 e^{3} \left (e x +d \right )^{\frac {3}{2}} a d}{2 b^{2} \left (b e x +a e \right )^{2}}-\frac {13 e^{2} \left (e x +d \right )^{\frac {3}{2}} d^{2}}{4 b \left (b e x +a e \right )^{2}}-\frac {11 e^{5} \sqrt {e x +d}\, a^{3}}{4 b^{4} \left (b e x +a e \right )^{2}}+\frac {33 e^{4} \sqrt {e x +d}\, a^{2} d}{4 b^{3} \left (b e x +a e \right )^{2}}-\frac {33 e^{3} \sqrt {e x +d}\, a \,d^{2}}{4 b^{2} \left (b e x +a e \right )^{2}}+\frac {11 e^{2} \sqrt {e x +d}\, d^{3}}{4 b \left (b e x +a e \right )^{2}}+\frac {35 e^{4} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2}}{4 b^{4} \sqrt {\left (a e -b d \right ) b}}-\frac {35 e^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a d}{2 b^{3} \sqrt {\left (a e -b d \right ) b}}+\frac {35 e^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) d^{2}}{4 b^{2} \sqrt {\left (a e -b d \right ) b}}\)	\(362\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e^2*(-1/b^4*(-1/3*b*(e*x+d)^(3/2)+3*a*e*(e*x+d)^(1/2)-3*b*d*(e*x+d)^(1/2))+1/b^4*(((-13/8*a^2*b*e^2+13/4*a*b
^2*d*e-13/8*d^2*b^3)*(e*x+d)^(3/2)+(-11/8*a^3*e^3+33/8*a^2*b*d*e^2-33/8*a*b^2*d^2*e+11/8*b^3*d^3)*(e*x+d)^(1/2
))/(b*(e*x+d)+a*e-b*d)^2+35/8*(a^2*e^2-2*a*b*d*e+b^2*d^2)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d
)*b)^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 2.70, size = 489, normalized size = 3.35 \begin {gather*} \left [\frac {105 \, {\left ({\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} e^{3} - {\left (b^{3} d x^{2} + 2 \, a b^{2} d x + a^{2} b d\right )} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {2 \, b d + 2 \, \sqrt {x e + d} b \sqrt {\frac {b d - a e}{b}} + {\left (b x - a\right )} e}{b x + a}\right ) - 2 \, {\left (6 \, b^{3} d^{3} - {\left (8 \, b^{3} x^{3} - 56 \, a b^{2} x^{2} - 175 \, a^{2} b x - 105 \, a^{3}\right )} e^{3} - 2 \, {\left (40 \, b^{3} d x^{2} + 119 \, a b^{2} d x + 70 \, a^{2} b d\right )} e^{2} + 3 \, {\left (13 \, b^{3} d^{2} x + 7 \, a b^{2} d^{2}\right )} e\right )} \sqrt {x e + d}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {105 \, {\left ({\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} e^{3} - {\left (b^{3} d x^{2} + 2 \, a b^{2} d x + a^{2} b d\right )} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {x e + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (6 \, b^{3} d^{3} - {\left (8 \, b^{3} x^{3} - 56 \, a b^{2} x^{2} - 175 \, a^{2} b x - 105 \, a^{3}\right )} e^{3} - 2 \, {\left (40 \, b^{3} d x^{2} + 119 \, a b^{2} d x + 70 \, a^{2} b d\right )} e^{2} + 3 \, {\left (13 \, b^{3} d^{2} x + 7 \, a b^{2} d^{2}\right )} e\right )} \sqrt {x e + d}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(105*((a*b^2*x^2 + 2*a^2*b*x + a^3)*e^3 - (b^3*d*x^2 + 2*a*b^2*d*x + a^2*b*d)*e^2)*sqrt((b*d - a*e)/b)*l
og((2*b*d + 2*sqrt(x*e + d)*b*sqrt((b*d - a*e)/b) + (b*x - a)*e)/(b*x + a)) - 2*(6*b^3*d^3 - (8*b^3*x^3 - 56*a
*b^2*x^2 - 175*a^2*b*x - 105*a^3)*e^3 - 2*(40*b^3*d*x^2 + 119*a*b^2*d*x + 70*a^2*b*d)*e^2 + 3*(13*b^3*d^2*x +
7*a*b^2*d^2)*e)*sqrt(x*e + d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/12*(105*((a*b^2*x^2 + 2*a^2*b*x + a^3)*e^3 -
(b^3*d*x^2 + 2*a*b^2*d*x + a^2*b*d)*e^2)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(x*e + d)*b*sqrt(-(b*d - a*e)/b)/(b
*d - a*e)) - (6*b^3*d^3 - (8*b^3*x^3 - 56*a*b^2*x^2 - 175*a^2*b*x - 105*a^3)*e^3 - 2*(40*b^3*d*x^2 + 119*a*b^2
*d*x + 70*a^2*b*d)*e^2 + 3*(13*b^3*d^2*x + 7*a*b^2*d^2)*e)*sqrt(x*e + d))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (124) = 248\).
time = 0.92, size = 265, normalized size = 1.82 \begin {gather*} \frac {35 \, {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{4}} - \frac {13 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{2} - 11 \, \sqrt {x e + d} b^{3} d^{3} e^{2} - 26 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} d e^{3} + 33 \, \sqrt {x e + d} a b^{2} d^{2} e^{3} + 13 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b e^{4} - 33 \, \sqrt {x e + d} a^{2} b d e^{4} + 11 \, \sqrt {x e + d} a^{3} e^{5}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{4}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{6} e^{2} + 9 \, \sqrt {x e + d} b^{6} d e^{2} - 9 \, \sqrt {x e + d} a b^{5} e^{3}\right )}}{3 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/4*(b^2*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*
b^4) - 1/4*(13*(x*e + d)^(3/2)*b^3*d^2*e^2 - 11*sqrt(x*e + d)*b^3*d^3*e^2 - 26*(x*e + d)^(3/2)*a*b^2*d*e^3 + 3
3*sqrt(x*e + d)*a*b^2*d^2*e^3 + 13*(x*e + d)^(3/2)*a^2*b*e^4 - 33*sqrt(x*e + d)*a^2*b*d*e^4 + 11*sqrt(x*e + d)
*a^3*e^5)/(((x*e + d)*b - b*d + a*e)^2*b^4) + 2/3*((x*e + d)^(3/2)*b^6*e^2 + 9*sqrt(x*e + d)*b^6*d*e^2 - 9*sqr
t(x*e + d)*a*b^5*e^3)/b^9

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Mupad [B]
time = 0.14, size = 268, normalized size = 1.84 \begin {gather*} \frac {2\,e^2\,{\left (d+e\,x\right )}^{3/2}}{3\,b^3}-\frac {\sqrt {d+e\,x}\,\left (\frac {11\,a^3\,e^5}{4}-\frac {33\,a^2\,b\,d\,e^4}{4}+\frac {33\,a\,b^2\,d^2\,e^3}{4}-\frac {11\,b^3\,d^3\,e^2}{4}\right )+{\left (d+e\,x\right )}^{3/2}\,\left (\frac {13\,a^2\,b\,e^4}{4}-\frac {13\,a\,b^2\,d\,e^3}{2}+\frac {13\,b^3\,d^2\,e^2}{4}\right )}{b^6\,{\left (d+e\,x\right )}^2-\left (2\,b^6\,d-2\,a\,b^5\,e\right )\,\left (d+e\,x\right )+b^6\,d^2+a^2\,b^4\,e^2-2\,a\,b^5\,d\,e}+\frac {2\,e^2\,\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,\sqrt {d+e\,x}}{b^6}+\frac {35\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^4-2\,a\,b\,d\,e^3+b^2\,d^2\,e^2}\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{4\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*e^2*(d + e*x)^(3/2))/(3*b^3) - ((d + e*x)^(1/2)*((11*a^3*e^5)/4 - (11*b^3*d^3*e^2)/4 + (33*a*b^2*d^2*e^3)/4
- (33*a^2*b*d*e^4)/4) + (d + e*x)^(3/2)*((13*a^2*b*e^4)/4 + (13*b^3*d^2*e^2)/4 - (13*a*b^2*d*e^3)/2))/(b^6*(d
+ e*x)^2 - (2*b^6*d - 2*a*b^5*e)*(d + e*x) + b^6*d^2 + a^2*b^4*e^2 - 2*a*b^5*d*e) + (2*e^2*(3*b^3*d - 3*a*b^2
*e)*(d + e*x)^(1/2))/b^6 + (35*e^2*atan((b^(1/2)*e^2*(a*e - b*d)^(3/2)*(d + e*x)^(1/2))/(a^2*e^4 + b^2*d^2*e^2
- 2*a*b*d*e^3))*(a*e - b*d)^(3/2))/(4*b^(9/2))

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