3.7.15 \(\int \frac {1}{\sqrt {d+e x} (a-c x^2)} \, dx\) [615]

Optimal. Leaf size=134 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d-\sqrt {a} e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d+\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d+\sqrt {a} e}} \]

[Out]

-arctanh(c^(1/4)*(e*x+d)^(1/2)/(-e*a^(1/2)+d*c^(1/2))^(1/2))/c^(1/4)/a^(1/2)/(-e*a^(1/2)+d*c^(1/2))^(1/2)+arct
anh(c^(1/4)*(e*x+d)^(1/2)/(e*a^(1/2)+d*c^(1/2))^(1/2))/c^(1/4)/a^(1/2)/(e*a^(1/2)+d*c^(1/2))^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {722, 1107, 214} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {a} e+\sqrt {c} d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d-\sqrt {a} e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d + e*x]*(a - c*x^2)),x]

[Out]

-(ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]]/(Sqrt[a]*c^(1/4)*Sqrt[Sqrt[c]*d - Sqrt[a]*e]))
+ ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]]/(Sqrt[a]*c^(1/4)*Sqrt[Sqrt[c]*d + Sqrt[a]*e])

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]

Rule 722

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(c*d^2 + a*e^2 - 2*c
*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1107

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && PosQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d+e x} \left (a-c x^2\right )} \, dx &=(2 e) \text {Subst}\left (\int \frac {1}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt {d+e x}\right )\\ &=-\frac {\sqrt {c} \text {Subst}\left (\int \frac {1}{c d-\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )}{\sqrt {a}}+\frac {\sqrt {c} \text {Subst}\left (\int \frac {1}{c d+\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )}{\sqrt {a}}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d-\sqrt {a} e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d+\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d+\sqrt {a} e}}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 153, normalized size = 1.14 \begin {gather*} \frac {\frac {\tan ^{-1}\left (\frac {\sqrt {-c d-\sqrt {a} \sqrt {c} e} \sqrt {d+e x}}{\sqrt {c} d+\sqrt {a} e}\right )}{\sqrt {-c d-\sqrt {a} \sqrt {c} e}}-\frac {\tan ^{-1}\left (\frac {\sqrt {-c d+\sqrt {a} \sqrt {c} e} \sqrt {d+e x}}{\sqrt {c} d-\sqrt {a} e}\right )}{\sqrt {-c d+\sqrt {a} \sqrt {c} e}}}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[(Sqrt[-(c*d) - Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]*d + Sqrt[a]*e)]/Sqrt[-(c*d) - Sqrt[a]*Sqrt[c
]*e] - ArcTan[(Sqrt[-(c*d) + Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]*d - Sqrt[a]*e)]/Sqrt[-(c*d) + Sqrt[a]*
Sqrt[c]*e])/Sqrt[a]

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Maple [A]
time = 0.44, size = 112, normalized size = 0.84

method	result	size
derivativedivides	\(-2 e c \left (-\frac {\arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}-\frac {\arctanh \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )\)	\(112\)
default	\(-2 e c \left (-\frac {\arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}-\frac {\arctanh \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-c*x^2+a)/(e*x+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2*e*c*(-1/2/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan(c*(e*x+d)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c
)^(1/2))-1/2/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh(c*(e*x+d)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)
^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(1/((c*x^2 - a)*sqrt(x*e + d)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 861 vs. \(2 (100) = 200\).
time = 1.12, size = 861, normalized size = 6.43 \begin {gather*} \frac {1}{2} \, \sqrt {\frac {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} + d}{a c d^{2} - a^{2} e^{2}}} \log \left ({\left (a e^{2} - \frac {{\left (a c^{2} d^{3} - a^{2} c d e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}}\right )} \sqrt {\frac {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} + d}{a c d^{2} - a^{2} e^{2}}} + \sqrt {x e + d} e\right ) - \frac {1}{2} \, \sqrt {\frac {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} + d}{a c d^{2} - a^{2} e^{2}}} \log \left (-{\left (a e^{2} - \frac {{\left (a c^{2} d^{3} - a^{2} c d e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}}\right )} \sqrt {\frac {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} + d}{a c d^{2} - a^{2} e^{2}}} + \sqrt {x e + d} e\right ) + \frac {1}{2} \, \sqrt {-\frac {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} - d}{a c d^{2} - a^{2} e^{2}}} \log \left ({\left (a e^{2} + \frac {{\left (a c^{2} d^{3} - a^{2} c d e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}}\right )} \sqrt {-\frac {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} - d}{a c d^{2} - a^{2} e^{2}}} + \sqrt {x e + d} e\right ) - \frac {1}{2} \, \sqrt {-\frac {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} - d}{a c d^{2} - a^{2} e^{2}}} \log \left (-{\left (a e^{2} + \frac {{\left (a c^{2} d^{3} - a^{2} c d e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}}\right )} \sqrt {-\frac {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} e}{\sqrt {a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} - d}{a c d^{2} - a^{2} e^{2}}} + \sqrt {x e + d} e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(((a*c*d^2 - a^2*e^2)*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4) + d)/(a*c*d^2 - a^2*e^2))*log(
(a*e^2 - (a*c^2*d^3 - a^2*c*d*e^2)*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4))*sqrt(((a*c*d^2 - a^2*e^2
)*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4) + d)/(a*c*d^2 - a^2*e^2)) + sqrt(x*e + d)*e) - 1/2*sqrt(((
a*c*d^2 - a^2*e^2)*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4) + d)/(a*c*d^2 - a^2*e^2))*log(-(a*e^2 - (
a*c^2*d^3 - a^2*c*d*e^2)*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4))*sqrt(((a*c*d^2 - a^2*e^2)*e/sqrt(a
*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4) + d)/(a*c*d^2 - a^2*e^2)) + sqrt(x*e + d)*e) + 1/2*sqrt(-((a*c*d^2 -
a^2*e^2)*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4) - d)/(a*c*d^2 - a^2*e^2))*log((a*e^2 + (a*c^2*d^3
- a^2*c*d*e^2)*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4))*sqrt(-((a*c*d^2 - a^2*e^2)*e/sqrt(a*c^3*d^4
- 2*a^2*c^2*d^2*e^2 + a^3*c*e^4) - d)/(a*c*d^2 - a^2*e^2)) + sqrt(x*e + d)*e) - 1/2*sqrt(-((a*c*d^2 - a^2*e^2)
*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4) - d)/(a*c*d^2 - a^2*e^2))*log(-(a*e^2 + (a*c^2*d^3 - a^2*c*
d*e^2)*e/sqrt(a*c^3*d^4 - 2*a^2*c^2*d^2*e^2 + a^3*c*e^4))*sqrt(-((a*c*d^2 - a^2*e^2)*e/sqrt(a*c^3*d^4 - 2*a^2*
c^2*d^2*e^2 + a^3*c*e^4) - d)/(a*c*d^2 - a^2*e^2)) + sqrt(x*e + d)*e)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{- a \sqrt {d + e x} + c x^{2} \sqrt {d + e x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(-a*sqrt(d + e*x) + c*x**2*sqrt(d + e*x)), x)

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Giac [A]
time = 2.16, size = 159, normalized size = 1.19 \begin {gather*} -\frac {\sqrt {a c} {\left | c \right |} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {c d + \sqrt {c^{2} d^{2} - {\left (c d^{2} - a e^{2}\right )} c}}{c}}}\right )}{\sqrt {-c^{2} d - \sqrt {a c} c e} a c} + \frac {\sqrt {a c} {\left | c \right |} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {c d - \sqrt {c^{2} d^{2} - {\left (c d^{2} - a e^{2}\right )} c}}{c}}}\right )}{\sqrt {-c^{2} d + \sqrt {a c} c e} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a*c)*abs(c)*arctan(sqrt(x*e + d)/sqrt(-(c*d + sqrt(c^2*d^2 - (c*d^2 - a*e^2)*c))/c))/(sqrt(-c^2*d - sqrt
(a*c)*c*e)*a*c) + sqrt(a*c)*abs(c)*arctan(sqrt(x*e + d)/sqrt(-(c*d - sqrt(c^2*d^2 - (c*d^2 - a*e^2)*c))/c))/(s
qrt(-c^2*d + sqrt(a*c)*c*e)*a*c)

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Mupad [B]
time = 0.65, size = 1366, normalized size = 10.19 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {32\,a^2\,c^5\,d^2\,e^2\,\sqrt {-\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^4\,c^6\,d^3\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^4\,c^4\,e^6\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^5\,c^5\,d\,e^5}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a^3\,c^5\,d^2\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}-\frac {32\,c^3\,e^2\,\sqrt {-\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^2\,c^4\,d\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a\,c^3\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}+\frac {32\,a\,c^4\,d\,e^3\,\sqrt {a^3\,c}\,\sqrt {-\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^4\,c^6\,d^3\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^4\,c^4\,e^6\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^5\,c^5\,d\,e^5}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a^3\,c^5\,d^2\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}\right )\,\sqrt {-\frac {e\,\sqrt {a^3\,c}+a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}-2\,\mathrm {atanh}\left (\frac {32\,c^3\,e^2\,\sqrt {\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^2\,c^4\,d\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a\,c^3\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}-\frac {32\,a^2\,c^5\,d^2\,e^2\,\sqrt {\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^4\,c^6\,d^3\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a^4\,c^4\,e^6\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^5\,c^5\,d\,e^5}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^3\,c^5\,d^2\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}+\frac {32\,a\,c^4\,d\,e^3\,\sqrt {a^3\,c}\,\sqrt {\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^4\,c^6\,d^3\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a^4\,c^4\,e^6\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^5\,c^5\,d\,e^5}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^3\,c^5\,d^2\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}\right )\,\sqrt {\frac {e\,\sqrt {a^3\,c}-a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*atanh((32*a^2*c^5*d^2*e^2*(- (e*(a^3*c)^(1/2))/(4*(a^3*c*e^2 - a^2*c^2*d^2)) - (a*c*d)/(4*(a^3*c*e^2 - a^2*c
^2*d^2)))^(1/2)*(d + e*x)^(1/2))/((16*a^4*c^6*d^3*e^3)/(a^3*c*e^2 - a^2*c^2*d^2) - (16*a^4*c^4*e^6*(a^3*c)^(1/
2))/(a^3*c*e^2 - a^2*c^2*d^2) - (16*a^5*c^5*d*e^5)/(a^3*c*e^2 - a^2*c^2*d^2) + (16*a^3*c^5*d^2*e^4*(a^3*c)^(1/
2))/(a^3*c*e^2 - a^2*c^2*d^2)) - (32*c^3*e^2*(- (e*(a^3*c)^(1/2))/(4*(a^3*c*e^2 - a^2*c^2*d^2)) - (a*c*d)/(4*(
a^3*c*e^2 - a^2*c^2*d^2)))^(1/2)*(d + e*x)^(1/2))/((16*a^2*c^4*d*e^3)/(a^3*c*e^2 - a^2*c^2*d^2) + (16*a*c^3*e^
4*(a^3*c)^(1/2))/(a^3*c*e^2 - a^2*c^2*d^2)) + (32*a*c^4*d*e^3*(a^3*c)^(1/2)*(- (e*(a^3*c)^(1/2))/(4*(a^3*c*e^2
- a^2*c^2*d^2)) - (a*c*d)/(4*(a^3*c*e^2 - a^2*c^2*d^2)))^(1/2)*(d + e*x)^(1/2))/((16*a^4*c^6*d^3*e^3)/(a^3*c*
e^2 - a^2*c^2*d^2) - (16*a^4*c^4*e^6*(a^3*c)^(1/2))/(a^3*c*e^2 - a^2*c^2*d^2) - (16*a^5*c^5*d*e^5)/(a^3*c*e^2
- a^2*c^2*d^2) + (16*a^3*c^5*d^2*e^4*(a^3*c)^(1/2))/(a^3*c*e^2 - a^2*c^2*d^2)))*(-(e*(a^3*c)^(1/2) + a*c*d)/(4
*(a^3*c*e^2 - a^2*c^2*d^2)))^(1/2) - 2*atanh((32*c^3*e^2*((e*(a^3*c)^(1/2))/(4*(a^3*c*e^2 - a^2*c^2*d^2)) - (a
*c*d)/(4*(a^3*c*e^2 - a^2*c^2*d^2)))^(1/2)*(d + e*x)^(1/2))/((16*a^2*c^4*d*e^3)/(a^3*c*e^2 - a^2*c^2*d^2) - (1
6*a*c^3*e^4*(a^3*c)^(1/2))/(a^3*c*e^2 - a^2*c^2*d^2)) - (32*a^2*c^5*d^2*e^2*((e*(a^3*c)^(1/2))/(4*(a^3*c*e^2 -
a^2*c^2*d^2)) - (a*c*d)/(4*(a^3*c*e^2 - a^2*c^2*d^2)))^(1/2)*(d + e*x)^(1/2))/((16*a^4*c^6*d^3*e^3)/(a^3*c*e^
2 - a^2*c^2*d^2) + (16*a^4*c^4*e^6*(a^3*c)^(1/2))/(a^3*c*e^2 - a^2*c^2*d^2) - (16*a^5*c^5*d*e^5)/(a^3*c*e^2 -
a^2*c^2*d^2) - (16*a^3*c^5*d^2*e^4*(a^3*c)^(1/2))/(a^3*c*e^2 - a^2*c^2*d^2)) + (32*a*c^4*d*e^3*(a^3*c)^(1/2)*(
(e*(a^3*c)^(1/2))/(4*(a^3*c*e^2 - a^2*c^2*d^2)) - (a*c*d)/(4*(a^3*c*e^2 - a^2*c^2*d^2)))^(1/2)*(d + e*x)^(1/2)
)/((16*a^4*c^6*d^3*e^3)/(a^3*c*e^2 - a^2*c^2*d^2) + (16*a^4*c^4*e^6*(a^3*c)^(1/2))/(a^3*c*e^2 - a^2*c^2*d^2) -
(16*a^5*c^5*d*e^5)/(a^3*c*e^2 - a^2*c^2*d^2) - (16*a^3*c^5*d^2*e^4*(a^3*c)^(1/2))/(a^3*c*e^2 - a^2*c^2*d^2)))
*((e*(a^3*c)^(1/2) - a*c*d)/(4*(a^3*c*e^2 - a^2*c^2*d^2)))^(1/2)

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