3.18.9 \(\int \frac {\sqrt {d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\)

Optimal. Leaf size=208 \[ -\frac {35 c^{3/2} d^{3/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}+\frac {35 c d e^2}{4 \sqrt {d+e x} \left (c d^2-a e^2\right )^4}+\frac {35 e^2}{12 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}+\frac {7 e}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {1}{2 (d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)^2} \]

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Rubi [A]  time = 0.14, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {626, 51, 63, 208} \begin {gather*} -\frac {35 c^{3/2} d^{3/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}+\frac {35 c d e^2}{4 \sqrt {d+e x} \left (c d^2-a e^2\right )^4}+\frac {35 e^2}{12 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}+\frac {7 e}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {1}{2 (d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(35*e^2)/(12*(c*d^2 - a*e^2)^3*(d + e*x)^(3/2)) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*(d + e*x)^(3/2)) + (7*e
)/(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)*(d + e*x)^(3/2)) + (35*c*d*e^2)/(4*(c*d^2 - a*e^2)^4*Sqrt[d + e*x]) - (35
*c^(3/2)*d^(3/2)*e^2*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(4*(c*d^2 - a*e^2)^(9/2))

Rule 51

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] - Dist[(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] && NeQ[b*c - a*d, 0] && 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 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac {1}{(a e+c d x)^3 (d+e x)^{5/2}} \, dx\\ &=-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}-\frac {(7 e) \int \frac {1}{(a e+c d x)^2 (d+e x)^{5/2}} \, dx}{4 \left (c d^2-a e^2\right )}\\ &=-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {\left (35 e^2\right ) \int \frac {1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{8 \left (c d^2-a e^2\right )^2}\\ &=\frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {\left (35 c d e^2\right ) \int \frac {1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{8 \left (c d^2-a e^2\right )^3}\\ &=\frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt {d+e x}}+\frac {\left (35 c^2 d^2 e^2\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 \left (c d^2-a e^2\right )^4}\\ &=\frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt {d+e x}}+\frac {\left (35 c^2 d^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c d^2}{e}+a e+\frac {c d x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 \left (c d^2-a e^2\right )^4}\\ &=\frac {35 e^2}{12 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{3/2}}+\frac {7 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{3/2}}+\frac {35 c d e^2}{4 \left (c d^2-a e^2\right )^4 \sqrt {d+e x}}-\frac {35 c^{3/2} d^{3/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.29 \begin {gather*} -\frac {2 e^2 \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{3 (d+e x)^{3/2} \left (a e^2-c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e^2*Hypergeometric2F1[-3/2, 3, -1/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(3*(-(c*d^2) + a*e^2)^3*(d +
e*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.82, size = 282, normalized size = 1.36 \begin {gather*} \frac {-8 a^3 e^8+24 a^2 c d^2 e^6+56 a^2 c d e^6 (d+e x)-24 a c^2 d^4 e^4-112 a c^2 d^3 e^4 (d+e x)+175 a c^2 d^2 e^4 (d+e x)^2+8 c^3 d^6 e^2+56 c^3 d^5 e^2 (d+e x)-175 c^3 d^4 e^2 (d+e x)^2+105 c^3 d^3 e^2 (d+e x)^3}{12 (d+e x)^{3/2} \left (c d^2-a e^2\right )^4 \left (-a e^2+c d^2-c d (d+e x)\right )^2}-\frac {35 c^{3/2} d^{3/2} e^2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x} \sqrt {a e^2-c d^2}}{c d^2-a e^2}\right )}{4 \left (a e^2-c d^2\right )^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*c^3*d^6*e^2 - 24*a*c^2*d^4*e^4 + 24*a^2*c*d^2*e^6 - 8*a^3*e^8 + 56*c^3*d^5*e^2*(d + e*x) - 112*a*c^2*d^3*e^
4*(d + e*x) + 56*a^2*c*d*e^6*(d + e*x) - 175*c^3*d^4*e^2*(d + e*x)^2 + 175*a*c^2*d^2*e^4*(d + e*x)^2 + 105*c^3
*d^3*e^2*(d + e*x)^3)/(12*(c*d^2 - a*e^2)^4*(d + e*x)^(3/2)*(c*d^2 - a*e^2 - c*d*(d + e*x))^2) - (35*c^(3/2)*d
^(3/2)*e^2*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[-(c*d^2) + a*e^2]*Sqrt[d + e*x])/(c*d^2 - a*e^2)])/(4*(-(c*d^2) + a*e^
2)^(9/2))

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fricas [B]  time = 0.45, size = 1353, normalized size = 6.50

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/24*(105*(c^3*d^3*e^4*x^4 + a^2*c*d^3*e^4 + 2*(c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + (c^3*d^5*e^2 + 4*a*c^2*d^3
*e^4 + a^2*c*d*e^6)*x^2 + 2*(a*c^2*d^4*e^3 + a^2*c*d^2*e^5)*x)*sqrt(c*d/(c*d^2 - a*e^2))*log((c*d*e*x + 2*c*d^
2 - a*e^2 - 2*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(c*d/(c*d^2 - a*e^2)))/(c*d*x + a*e)) + 2*(105*c^3*d^3*e^3*x^3
- 6*c^3*d^6 + 39*a*c^2*d^4*e^2 + 80*a^2*c*d^2*e^4 - 8*a^3*e^6 + 35*(4*c^3*d^4*e^2 + 5*a*c^2*d^2*e^4)*x^2 + 7*
(3*c^3*d^5*e + 34*a*c^2*d^3*e^3 + 8*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a^2*c^4*d^10*e^2 - 4*a^3*c^3*d^8*e^4 + 6*a
^4*c^2*d^6*e^6 - 4*a^5*c*d^4*e^8 + a^6*d^2*e^10 + (c^6*d^10*e^2 - 4*a*c^5*d^8*e^4 + 6*a^2*c^4*d^6*e^6 - 4*a^3*
c^3*d^4*e^8 + a^4*c^2*d^2*e^10)*x^4 + 2*(c^6*d^11*e - 3*a*c^5*d^9*e^3 + 2*a^2*c^4*d^7*e^5 + 2*a^3*c^3*d^5*e^7
- 3*a^4*c^2*d^3*e^9 + a^5*c*d*e^11)*x^3 + (c^6*d^12 - 9*a^2*c^4*d^8*e^4 + 16*a^3*c^3*d^6*e^6 - 9*a^4*c^2*d^4*e
^8 + a^6*e^12)*x^2 + 2*(a*c^5*d^11*e - 3*a^2*c^4*d^9*e^3 + 2*a^3*c^3*d^7*e^5 + 2*a^4*c^2*d^5*e^7 - 3*a^5*c*d^3
*e^9 + a^6*d*e^11)*x), -1/12*(105*(c^3*d^3*e^4*x^4 + a^2*c*d^3*e^4 + 2*(c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + (c^
3*d^5*e^2 + 4*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x^2 + 2*(a*c^2*d^4*e^3 + a^2*c*d^2*e^5)*x)*sqrt(-c*d/(c*d^2 - a*e^2
))*arctan(-(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c*d*e*x + c*d^2)) - (105*c^3*d^3*e^3*x^3
- 6*c^3*d^6 + 39*a*c^2*d^4*e^2 + 80*a^2*c*d^2*e^4 - 8*a^3*e^6 + 35*(4*c^3*d^4*e^2 + 5*a*c^2*d^2*e^4)*x^2 + 7*(
3*c^3*d^5*e + 34*a*c^2*d^3*e^3 + 8*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a^2*c^4*d^10*e^2 - 4*a^3*c^3*d^8*e^4 + 6*a^
4*c^2*d^6*e^6 - 4*a^5*c*d^4*e^8 + a^6*d^2*e^10 + (c^6*d^10*e^2 - 4*a*c^5*d^8*e^4 + 6*a^2*c^4*d^6*e^6 - 4*a^3*c
^3*d^4*e^8 + a^4*c^2*d^2*e^10)*x^4 + 2*(c^6*d^11*e - 3*a*c^5*d^9*e^3 + 2*a^2*c^4*d^7*e^5 + 2*a^3*c^3*d^5*e^7 -
3*a^4*c^2*d^3*e^9 + a^5*c*d*e^11)*x^3 + (c^6*d^12 - 9*a^2*c^4*d^8*e^4 + 16*a^3*c^3*d^6*e^6 - 9*a^4*c^2*d^4*e^
8 + a^6*e^12)*x^2 + 2*(a*c^5*d^11*e - 3*a^2*c^4*d^9*e^3 + 2*a^3*c^3*d^7*e^5 + 2*a^4*c^2*d^5*e^7 - 3*a^5*c*d^3*
e^9 + a^6*d*e^11)*x)]

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giac [F(-1)]  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((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.07, size = 262, normalized size = 1.26 \begin {gather*} \frac {13 \sqrt {e x +d}\, a \,c^{2} d^{2} e^{4}}{4 \left (a \,e^{2}-c \,d^{2}\right )^{4} \left (c d e x +a \,e^{2}\right )^{2}}-\frac {13 \sqrt {e x +d}\, c^{3} d^{4} e^{2}}{4 \left (a \,e^{2}-c \,d^{2}\right )^{4} \left (c d e x +a \,e^{2}\right )^{2}}+\frac {11 \left (e x +d \right )^{\frac {3}{2}} c^{3} d^{3} e^{2}}{4 \left (a \,e^{2}-c \,d^{2}\right )^{4} \left (c d e x +a \,e^{2}\right )^{2}}+\frac {35 c^{2} d^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{4 \left (a \,e^{2}-c \,d^{2}\right )^{4} \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}+\frac {6 c d \,e^{2}}{\left (a \,e^{2}-c \,d^{2}\right )^{4} \sqrt {e x +d}}-\frac {2 e^{2}}{3 \left (a \,e^{2}-c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^3,x)

[Out]

11/4*e^2/(a*e^2-c*d^2)^4*c^3*d^3/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)+13/4*e^4/(a*e^2-c*d^2)^4*c^2*d^2/(c*d*e*x+a*e
^2)^2*(e*x+d)^(1/2)*a-13/4*e^2/(a*e^2-c*d^2)^4*c^3*d^4/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)+35/4*e^2/(a*e^2-c*d^2)^
4*c^2*d^2/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)*c*d)-2/3*e^2/(a*e^2-c*d^2)^
3/(e*x+d)^(3/2)+6*e^2/(a*e^2-c*d^2)^4*c*d/(e*x+d)^(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((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,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^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 positive or negative?

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mupad [B]  time = 0.86, size = 296, normalized size = 1.42 \begin {gather*} \frac {\frac {14\,c\,d\,e^2\,\left (d+e\,x\right )}{3\,{\left (a\,e^2-c\,d^2\right )}^2}-\frac {2\,e^2}{3\,\left (a\,e^2-c\,d^2\right )}+\frac {175\,c^2\,d^2\,e^2\,{\left (d+e\,x\right )}^2}{12\,{\left (a\,e^2-c\,d^2\right )}^3}+\frac {35\,c^3\,d^3\,e^2\,{\left (d+e\,x\right )}^3}{4\,{\left (a\,e^2-c\,d^2\right )}^4}}{{\left (d+e\,x\right )}^{3/2}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,{\left (d+e\,x\right )}^{5/2}+c^2\,d^2\,{\left (d+e\,x\right )}^{7/2}}+\frac {35\,c^{3/2}\,d^{3/2}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^4\,e^8-4\,a^3\,c\,d^2\,e^6+6\,a^2\,c^2\,d^4\,e^4-4\,a\,c^3\,d^6\,e^2+c^4\,d^8\right )}{{\left (a\,e^2-c\,d^2\right )}^{9/2}}\right )}{4\,{\left (a\,e^2-c\,d^2\right )}^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(1/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)

[Out]

((14*c*d*e^2*(d + e*x))/(3*(a*e^2 - c*d^2)^2) - (2*e^2)/(3*(a*e^2 - c*d^2)) + (175*c^2*d^2*e^2*(d + e*x)^2)/(1
2*(a*e^2 - c*d^2)^3) + (35*c^3*d^3*e^2*(d + e*x)^3)/(4*(a*e^2 - c*d^2)^4))/((d + e*x)^(3/2)*(a^2*e^4 + c^2*d^4
- 2*a*c*d^2*e^2) - (2*c^2*d^3 - 2*a*c*d*e^2)*(d + e*x)^(5/2) + c^2*d^2*(d + e*x)^(7/2)) + (35*c^(3/2)*d^(3/2)
*e^2*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^4*e^8 + c^4*d^8 - 4*a*c^3*d^6*e^2 - 4*a^3*c*d^2*e^6 + 6*a^2*c^2*
d^4*e^4))/(a*e^2 - c*d^2)^(9/2)))/(4*(a*e^2 - c*d^2)^(9/2))

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sympy [B]  time = 119.85, size = 1826, normalized size = 8.78

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)

[Out]

10*a*c**2*d**2*e**4*sqrt(d + e*x)/(8*a**4*e**8*(a*e**2 - c*d**2)**2 - 16*a**3*c*d**2*e**6*(a*e**2 - c*d**2)**2
+ 16*a**3*c*d*e**7*x*(a*e**2 - c*d**2)**2 - 48*a**2*c**2*d**3*e**5*x*(a*e**2 - c*d**2)**2 + 8*a**2*c**2*d**2*
e**4*(d + e*x)**2*(a*e**2 - c*d**2)**2 + 16*a*c**3*d**6*e**2*(a*e**2 - c*d**2)**2 + 48*a*c**3*d**5*e**3*x*(a*e
**2 - c*d**2)**2 - 16*a*c**3*d**4*e**2*(d + e*x)**2*(a*e**2 - c*d**2)**2 - 8*c**4*d**8*(a*e**2 - c*d**2)**2 -
16*c**4*d**7*e*x*(a*e**2 - c*d**2)**2 + 8*c**4*d**6*(d + e*x)**2*(a*e**2 - c*d**2)**2) - 10*c**3*d**4*e**2*sqr
t(d + e*x)/(8*a**4*e**8*(a*e**2 - c*d**2)**2 - 16*a**3*c*d**2*e**6*(a*e**2 - c*d**2)**2 + 16*a**3*c*d*e**7*x*(
a*e**2 - c*d**2)**2 - 48*a**2*c**2*d**3*e**5*x*(a*e**2 - c*d**2)**2 + 8*a**2*c**2*d**2*e**4*(d + e*x)**2*(a*e*
*2 - c*d**2)**2 + 16*a*c**3*d**6*e**2*(a*e**2 - c*d**2)**2 + 48*a*c**3*d**5*e**3*x*(a*e**2 - c*d**2)**2 - 16*a
*c**3*d**4*e**2*(d + e*x)**2*(a*e**2 - c*d**2)**2 - 8*c**4*d**8*(a*e**2 - c*d**2)**2 - 16*c**4*d**7*e*x*(a*e**
2 - c*d**2)**2 + 8*c**4*d**6*(d + e*x)**2*(a*e**2 - c*d**2)**2) + 6*c**3*d**3*e**2*(d + e*x)**(3/2)/(8*a**4*e*
*8*(a*e**2 - c*d**2)**2 - 16*a**3*c*d**2*e**6*(a*e**2 - c*d**2)**2 + 16*a**3*c*d*e**7*x*(a*e**2 - c*d**2)**2 -
48*a**2*c**2*d**3*e**5*x*(a*e**2 - c*d**2)**2 + 8*a**2*c**2*d**2*e**4*(d + e*x)**2*(a*e**2 - c*d**2)**2 + 16*
a*c**3*d**6*e**2*(a*e**2 - c*d**2)**2 + 48*a*c**3*d**5*e**3*x*(a*e**2 - c*d**2)**2 - 16*a*c**3*d**4*e**2*(d +
e*x)**2*(a*e**2 - c*d**2)**2 - 8*c**4*d**8*(a*e**2 - c*d**2)**2 - 16*c**4*d**7*e*x*(a*e**2 - c*d**2)**2 + 8*c*
*4*d**6*(d + e*x)**2*(a*e**2 - c*d**2)**2) - 3*c**2*d**2*e**2*sqrt(-1/(c*d*(a*e**2 - c*d**2)**5))*log(-a**3*e*
*6*sqrt(-1/(c*d*(a*e**2 - c*d**2)**5)) + 3*a**2*c*d**2*e**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**5)) - 3*a*c**2*d**
4*e**2*sqrt(-1/(c*d*(a*e**2 - c*d**2)**5)) + c**3*d**6*sqrt(-1/(c*d*(a*e**2 - c*d**2)**5)) + sqrt(d + e*x))/(8
*(a*e**2 - c*d**2)**2) + 3*c**2*d**2*e**2*sqrt(-1/(c*d*(a*e**2 - c*d**2)**5))*log(a**3*e**6*sqrt(-1/(c*d*(a*e*
*2 - c*d**2)**5)) - 3*a**2*c*d**2*e**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**5)) + 3*a*c**2*d**4*e**2*sqrt(-1/(c*d*(
a*e**2 - c*d**2)**5)) - c**3*d**6*sqrt(-1/(c*d*(a*e**2 - c*d**2)**5)) + sqrt(d + e*x))/(8*(a*e**2 - c*d**2)**2
) - c**2*d**2*e**2*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3))*log(-a**2*e**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + 2*
a*c*d**2*e**2*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) - c**2*d**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + sqrt(d + e
*x))/(a*e**2 - c*d**2)**3 + c**2*d**2*e**2*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3))*log(a**2*e**4*sqrt(-1/(c*d*(a*e
**2 - c*d**2)**3)) - 2*a*c*d**2*e**2*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + c**2*d**4*sqrt(-1/(c*d*(a*e**2 - c*
d**2)**3)) + sqrt(d + e*x))/(a*e**2 - c*d**2)**3 + 4*c**2*d**2*e**2*sqrt(d + e*x)/(2*a**2*e**4*(a*e**2 - c*d**
2)**3 - 2*a*c*d**2*e**2*(a*e**2 - c*d**2)**3 + 2*a*c*d*e**3*x*(a*e**2 - c*d**2)**3 - 2*c**2*d**3*e*x*(a*e**2 -
c*d**2)**3) + 6*c*d*e**2*atan(sqrt(d + e*x)/sqrt(a*e**2/(c*d) - d))/((a*e**2 - c*d**2)**4*sqrt(a*e**2/(c*d) -
d)) + 6*c*d*e**2/(sqrt(d + e*x)*(a*e**2 - c*d**2)**4) - 2*e**2/(3*(d + e*x)**(3/2)*(a*e**2 - c*d**2)**3)

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