3.1.71 \(\int \frac {1}{a-b \cos ^4(x)} \, dx\) [71]

Optimal. Leaf size=101 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \cot (x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \cot (x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}+\sqrt {b}}} \]

[Out]

-1/2*arctan(cot(x)*(a^(1/2)-b^(1/2))^(1/2)/a^(1/4))/a^(3/4)/(a^(1/2)-b^(1/2))^(1/2)-1/2*arctan(cot(x)*(a^(1/2)
+b^(1/2))^(1/2)/a^(1/4))/a^(3/4)/(a^(1/2)+b^(1/2))^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3288, 1180, 211} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \cot (x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\text {ArcTan}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \cot (x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}+\sqrt {b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*Cos[x]^4)^(-1),x]

[Out]

-1/2*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Cot[x])/a^(1/4)]/(a^(3/4)*Sqrt[Sqrt[a] - Sqrt[b]]) - ArcTan[(Sqrt[Sqrt[a]
+ Sqrt[b]]*Cot[x])/a^(1/4)]/(2*a^(3/4)*Sqrt[Sqrt[a] + Sqrt[b]])

Rule 211

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

Rule 1180

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

Rule 3288

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dis
t[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x
]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{a-b \cos ^4(x)} \, dx &=-\text {Subst}\left (\int \frac {1+x^2}{a+2 a x^2+(a-b) x^4} \, dx,x,\cot (x)\right )\\ &=-\left (\frac {1}{2} \left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \text {Subst}\left (\int \frac {1}{a-\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\cot (x)\right )\right )-\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a}}\right ) \text {Subst}\left (\int \frac {1}{a+\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\cot (x)\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \cot (x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}-\sqrt {b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \cot (x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}+\sqrt {b}}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 109, normalized size = 1.08 \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{2 \sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{2 \sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*Cos[x]^4)^(-1),x]

[Out]

ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + Sqrt[a]*Sqrt[b]]]/(2*Sqrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]]) - ArcTanh[(Sqrt[a]*Ta
n[x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]]/(2*Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]])

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Maple [A]
time = 0.14, size = 72, normalized size = 0.71

method	result	size
default	\(a \left (\frac {\arctan \left (\frac {a \tan \left (x \right )}{\sqrt {\left (\sqrt {a b}+a \right ) a}}\right )}{2 a \sqrt {\left (\sqrt {a b}+a \right ) a}}-\frac {\arctanh \left (\frac {a \tan \left (x \right )}{\sqrt {\left (\sqrt {a b}-a \right ) a}}\right )}{2 a \sqrt {\left (\sqrt {a b}-a \right ) a}}\right )\)	\(72\)
risch	\(\munderset {\textit {\_R} =\RootOf \left (1+\left (256 a^{4}-256 a^{3} b \right ) \textit {\_Z}^{4}+32 a^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (-\frac {128 i a^{4}}{b}+128 i a^{3}\right ) \textit {\_R}^{3}+\left (\frac {32 a^{3}}{b}-32 a^{2}\right ) \textit {\_R}^{2}+\left (-\frac {8 i a^{2}}{b}-8 i a \right ) \textit {\_R} +\frac {2 a}{b}+1\right )\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-b*cos(x)^4),x,method=_RETURNVERBOSE)

[Out]

a*(1/2/a/(((a*b)^(1/2)+a)*a)^(1/2)*arctan(a*tan(x)/(((a*b)^(1/2)+a)*a)^(1/2))-1/2/a/(((a*b)^(1/2)-a)*a)^(1/2)*
arctanh(a*tan(x)/(((a*b)^(1/2)-a)*a)^(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/(a-b*cos(x)^4),x, algorithm="maxima")

[Out]

-integrate(1/(b*cos(x)^4 - a), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 817 vs. \(2 (65) = 130\).
time = 0.50, size = 817, normalized size = 8.09 \begin {gather*} -\frac {1}{8} \, \sqrt {-\frac {{\left (a^{2} - a b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} + 1}{a^{2} - a b}} \log \left (b \cos \left (x\right )^{2} + 2 \, {\left (a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{4} - a^{3} b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {-\frac {{\left (a^{2} - a b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} + 1}{a^{2} - a b}} + {\left (a^{3} - a^{2} b - 2 \, {\left (a^{3} - a^{2} b\right )} \cos \left (x\right )^{2}\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}}\right ) + \frac {1}{8} \, \sqrt {-\frac {{\left (a^{2} - a b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} + 1}{a^{2} - a b}} \log \left (b \cos \left (x\right )^{2} - 2 \, {\left (a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{4} - a^{3} b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {-\frac {{\left (a^{2} - a b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} + 1}{a^{2} - a b}} + {\left (a^{3} - a^{2} b - 2 \, {\left (a^{3} - a^{2} b\right )} \cos \left (x\right )^{2}\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}}\right ) + \frac {1}{8} \, \sqrt {\frac {{\left (a^{2} - a b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} - 1}{a^{2} - a b}} \log \left (-b \cos \left (x\right )^{2} + 2 \, {\left (a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{4} - a^{3} b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {\frac {{\left (a^{2} - a b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} - 1}{a^{2} - a b}} + {\left (a^{3} - a^{2} b - 2 \, {\left (a^{3} - a^{2} b\right )} \cos \left (x\right )^{2}\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}}\right ) - \frac {1}{8} \, \sqrt {\frac {{\left (a^{2} - a b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} - 1}{a^{2} - a b}} \log \left (-b \cos \left (x\right )^{2} - 2 \, {\left (a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{4} - a^{3} b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {\frac {{\left (a^{2} - a b\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}} - 1}{a^{2} - a b}} + {\left (a^{3} - a^{2} b - 2 \, {\left (a^{3} - a^{2} b\right )} \cos \left (x\right )^{2}\right )} \sqrt {\frac {b}{a^{5} - 2 \, a^{4} b + a^{3} b^{2}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-b*cos(x)^4),x, algorithm="fricas")

[Out]

-1/8*sqrt(-((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) + 1)/(a^2 - a*b))*log(b*cos(x)^2 + 2*(a*b*cos(x)*sin
(x) - (a^4 - a^3*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(-((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b
+ a^3*b^2)) + 1)/(a^2 - a*b)) + (a^3 - a^2*b - 2*(a^3 - a^2*b)*cos(x)^2)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))) +
1/8*sqrt(-((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) + 1)/(a^2 - a*b))*log(b*cos(x)^2 - 2*(a*b*cos(x)*sin(
x) - (a^4 - a^3*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(-((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b +
a^3*b^2)) + 1)/(a^2 - a*b)) + (a^3 - a^2*b - 2*(a^3 - a^2*b)*cos(x)^2)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))) + 1
/8*sqrt(((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) - 1)/(a^2 - a*b))*log(-b*cos(x)^2 + 2*(a*b*cos(x)*sin(x
) + (a^4 - a^3*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a
^3*b^2)) - 1)/(a^2 - a*b)) + (a^3 - a^2*b - 2*(a^3 - a^2*b)*cos(x)^2)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))) - 1/8
*sqrt(((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)) - 1)/(a^2 - a*b))*log(-b*cos(x)^2 - 2*(a*b*cos(x)*sin(x)
+ (a^4 - a^3*b)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(((a^2 - a*b)*sqrt(b/(a^5 - 2*a^4*b + a^3
*b^2)) - 1)/(a^2 - a*b)) + (a^3 - a^2*b - 2*(a^3 - a^2*b)*cos(x)^2)*sqrt(b/(a^5 - 2*a^4*b + a^3*b^2)))

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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(1/(a-b*cos(x)**4),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (65) = 130\).
time = 0.44, size = 299, normalized size = 2.96 \begin {gather*} \frac {{\left (3 \, \sqrt {a^{2} + \sqrt {a b} a} a^{2} - 4 \, \sqrt {a^{2} + \sqrt {a b} a} a b - 3 \, \sqrt {a^{2} + \sqrt {a b} a} \sqrt {a b} a + 4 \, \sqrt {a^{2} + \sqrt {a b} a} \sqrt {a b} b\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (x\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a}}}\right )\right )} {\left | a \right |}}{2 \, {\left (3 \, a^{5} - 7 \, a^{4} b + 4 \, a^{3} b^{2}\right )}} + \frac {{\left (3 \, \sqrt {a^{2} - \sqrt {a b} a} a^{2} - 4 \, \sqrt {a^{2} - \sqrt {a b} a} a b + 3 \, \sqrt {a^{2} - \sqrt {a b} a} \sqrt {a b} a - 4 \, \sqrt {a^{2} - \sqrt {a b} a} \sqrt {a b} b\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (x\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a}}}\right )\right )} {\left | a \right |}}{2 \, {\left (3 \, a^{5} - 7 \, a^{4} b + 4 \, a^{3} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-b*cos(x)^4),x, algorithm="giac")

[Out]

1/2*(3*sqrt(a^2 + sqrt(a*b)*a)*a^2 - 4*sqrt(a^2 + sqrt(a*b)*a)*a*b - 3*sqrt(a^2 + sqrt(a*b)*a)*sqrt(a*b)*a + 4
*sqrt(a^2 + sqrt(a*b)*a)*sqrt(a*b)*b)*(pi*floor(x/pi + 1/2) + arctan(2*tan(x)/sqrt((4*a + sqrt(-16*(a - b)*a +
16*a^2))/a)))*abs(a)/(3*a^5 - 7*a^4*b + 4*a^3*b^2) + 1/2*(3*sqrt(a^2 - sqrt(a*b)*a)*a^2 - 4*sqrt(a^2 - sqrt(a
*b)*a)*a*b + 3*sqrt(a^2 - sqrt(a*b)*a)*sqrt(a*b)*a - 4*sqrt(a^2 - sqrt(a*b)*a)*sqrt(a*b)*b)*(pi*floor(x/pi + 1
/2) + arctan(2*tan(x)/sqrt((4*a - sqrt(-16*(a - b)*a + 16*a^2))/a)))*abs(a)/(3*a^5 - 7*a^4*b + 4*a^3*b^2)

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Mupad [B]
time = 2.60, size = 938, normalized size = 9.29 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {8\,a^6\,b\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {a^2}{16\,\left (a^3\,b-a^4\right )}+\frac {\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^4\right )}}}{2\,a^5\,b-2\,a^4\,b^2-\frac {2\,a^8\,b^2}{a^3\,b-a^4}+\frac {2\,a^9\,b}{a^3\,b-a^4}-\frac {2\,a^6\,b^2\,\sqrt {a^3\,b}}{a^3\,b-a^4}+\frac {2\,a^7\,b\,\sqrt {a^3\,b}}{a^3\,b-a^4}}-\frac {8\,a^2\,b\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {a^2}{16\,\left (a^3\,b-a^4\right )}+\frac {\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^4\right )}}}{2\,a\,b+\frac {2\,a^5\,b}{a^3\,b-a^4}+\frac {2\,a^3\,b\,\sqrt {a^3\,b}}{a^3\,b-a^4}}+\frac {8\,a^4\,b\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {a^2}{16\,\left (a^3\,b-a^4\right )}+\frac {\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^4\right )}}\,\sqrt {a^3\,b}}{2\,a^5\,b-2\,a^4\,b^2-\frac {2\,a^8\,b^2}{a^3\,b-a^4}+\frac {2\,a^9\,b}{a^3\,b-a^4}-\frac {2\,a^6\,b^2\,\sqrt {a^3\,b}}{a^3\,b-a^4}+\frac {2\,a^7\,b\,\sqrt {a^3\,b}}{a^3\,b-a^4}}\right )\,\sqrt {\frac {a^2+\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^4\right )}}-2\,\mathrm {atanh}\left (\frac {8\,a^2\,b\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {a^2}{16\,\left (a^3\,b-a^4\right )}-\frac {\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^4\right )}}}{2\,a\,b+\frac {2\,a^5\,b}{a^3\,b-a^4}-\frac {2\,a^3\,b\,\sqrt {a^3\,b}}{a^3\,b-a^4}}-\frac {8\,a^6\,b\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {a^2}{16\,\left (a^3\,b-a^4\right )}-\frac {\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^4\right )}}}{2\,a^5\,b-2\,a^4\,b^2-\frac {2\,a^8\,b^2}{a^3\,b-a^4}+\frac {2\,a^9\,b}{a^3\,b-a^4}+\frac {2\,a^6\,b^2\,\sqrt {a^3\,b}}{a^3\,b-a^4}-\frac {2\,a^7\,b\,\sqrt {a^3\,b}}{a^3\,b-a^4}}+\frac {8\,a^4\,b\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {a^2}{16\,\left (a^3\,b-a^4\right )}-\frac {\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^4\right )}}\,\sqrt {a^3\,b}}{2\,a^5\,b-2\,a^4\,b^2-\frac {2\,a^8\,b^2}{a^3\,b-a^4}+\frac {2\,a^9\,b}{a^3\,b-a^4}+\frac {2\,a^6\,b^2\,\sqrt {a^3\,b}}{a^3\,b-a^4}-\frac {2\,a^7\,b\,\sqrt {a^3\,b}}{a^3\,b-a^4}}\right )\,\sqrt {\frac {a^2-\sqrt {a^3\,b}}{16\,\left (a^3\,b-a^4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - b*cos(x)^4),x)

[Out]

2*atanh((8*a^6*b*tan(x)*(a^2/(16*(a^3*b - a^4)) + (a^3*b)^(1/2)/(16*(a^3*b - a^4)))^(1/2))/(2*a^5*b - 2*a^4*b^
2 - (2*a^8*b^2)/(a^3*b - a^4) + (2*a^9*b)/(a^3*b - a^4) - (2*a^6*b^2*(a^3*b)^(1/2))/(a^3*b - a^4) + (2*a^7*b*(
a^3*b)^(1/2))/(a^3*b - a^4)) - (8*a^2*b*tan(x)*(a^2/(16*(a^3*b - a^4)) + (a^3*b)^(1/2)/(16*(a^3*b - a^4)))^(1/
2))/(2*a*b + (2*a^5*b)/(a^3*b - a^4) + (2*a^3*b*(a^3*b)^(1/2))/(a^3*b - a^4)) + (8*a^4*b*tan(x)*(a^2/(16*(a^3*
b - a^4)) + (a^3*b)^(1/2)/(16*(a^3*b - a^4)))^(1/2)*(a^3*b)^(1/2))/(2*a^5*b - 2*a^4*b^2 - (2*a^8*b^2)/(a^3*b -
a^4) + (2*a^9*b)/(a^3*b - a^4) - (2*a^6*b^2*(a^3*b)^(1/2))/(a^3*b - a^4) + (2*a^7*b*(a^3*b)^(1/2))/(a^3*b - a
^4)))*((a^2 + (a^3*b)^(1/2))/(16*(a^3*b - a^4)))^(1/2) - 2*atanh((8*a^2*b*tan(x)*(a^2/(16*(a^3*b - a^4)) - (a^
3*b)^(1/2)/(16*(a^3*b - a^4)))^(1/2))/(2*a*b + (2*a^5*b)/(a^3*b - a^4) - (2*a^3*b*(a^3*b)^(1/2))/(a^3*b - a^4)
) - (8*a^6*b*tan(x)*(a^2/(16*(a^3*b - a^4)) - (a^3*b)^(1/2)/(16*(a^3*b - a^4)))^(1/2))/(2*a^5*b - 2*a^4*b^2 -
(2*a^8*b^2)/(a^3*b - a^4) + (2*a^9*b)/(a^3*b - a^4) + (2*a^6*b^2*(a^3*b)^(1/2))/(a^3*b - a^4) - (2*a^7*b*(a^3*
b)^(1/2))/(a^3*b - a^4)) + (8*a^4*b*tan(x)*(a^2/(16*(a^3*b - a^4)) - (a^3*b)^(1/2)/(16*(a^3*b - a^4)))^(1/2)*(
a^3*b)^(1/2))/(2*a^5*b - 2*a^4*b^2 - (2*a^8*b^2)/(a^3*b - a^4) + (2*a^9*b)/(a^3*b - a^4) + (2*a^6*b^2*(a^3*b)^
(1/2))/(a^3*b - a^4) - (2*a^7*b*(a^3*b)^(1/2))/(a^3*b - a^4)))*((a^2 - (a^3*b)^(1/2))/(16*(a^3*b - a^4)))^(1/2
)

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