3.2.98 \(\int \frac {\sin ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [198]

Optimal. Leaf size=115 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}-\sqrt {b}} b^{3/4} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 \sqrt {\sqrt {a}+\sqrt {b}} b^{3/4} d} \]

[Out]

-1/2*arctan(b^(1/4)*cos(d*x+c)/(a^(1/2)-b^(1/2))^(1/2))/b^(3/4)/d/(a^(1/2)-b^(1/2))^(1/2)+1/2*arctanh(b^(1/4)*
cos(d*x+c)/(a^(1/2)+b^(1/2))^(1/2))/b^(3/4)/d/(a^(1/2)+b^(1/2))^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3294, 1180, 211, 214} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}+\sqrt {b}}}\right )}{2 b^{3/4} d \sqrt {\sqrt {a}+\sqrt {b}}}-\frac {\text {ArcTan}\left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt {\sqrt {a}-\sqrt {b}}}\right )}{2 b^{3/4} d \sqrt {\sqrt {a}-\sqrt {b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]

[Out]

-1/2*ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]]/(Sqrt[Sqrt[a] - Sqrt[b]]*b^(3/4)*d) + ArcTanh[(b^(
1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]]/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(3/4)*d)

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 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 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 3294

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, Dist[-ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4
)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.13, size = 285, normalized size = 2.48 \begin {gather*} -\frac {i \text {RootSum}\left [b-4 b \text {$\#$1}^2-16 a \text {$\#$1}^4+6 b \text {$\#$1}^4-4 b \text {$\#$1}^6+b \text {$\#$1}^8\&,\frac {-2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+6 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-3 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-6 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+3 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^6-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{-b \text {$\#$1}-8 a \text {$\#$1}^3+3 b \text {$\#$1}^3-3 b \text {$\#$1}^5+b \text {$\#$1}^7}\&\right ]}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]

[Out]

((-1/8*I)*RootSum[b - 4*b*#1^2 - 16*a*#1^4 + 6*b*#1^4 - 4*b*#1^6 + b*#1^8 & , (-2*ArcTan[Sin[c + d*x]/(Cos[c +
d*x] - #1)] + I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 6*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - (3*I)*L
og[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - 6*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 + (3*I)*Log[1 - 2*Cos[
c + d*x]*#1 + #1^2]*#1^4 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^6 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^
2]*#1^6)/(-(b*#1) - 8*a*#1^3 + 3*b*#1^3 - 3*b*#1^5 + b*#1^7) & ])/d

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Maple [A]
time = 0.34, size = 83, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {b \left (\frac {\arctanh \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{d}\)	\(83\)
default	\(\frac {b \left (\frac {\arctanh \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}+b \right ) b}}-\frac {\arctan \left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{2 b \sqrt {\left (\sqrt {a b}-b \right ) b}}\right )}{d}\)	\(83\)
risch	\(\frac {i \left (\munderset {\textit {\_R} =\RootOf \left (-16+\left (a \,b^{3} d^{4}-b^{4} d^{4}\right ) \textit {\_Z}^{4}-8 b^{2} d^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {1}{4} i a \,b^{2} d^{3}+\frac {1}{4} i b^{3} d^{3}\right ) \textit {\_R}^{3}+2 i b d \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}+1\right )\right )}{8}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)

[Out]

1/d*b*(1/2/b/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh(b*cos(d*x+c)/(((a*b)^(1/2)+b)*b)^(1/2))-1/2/b/(((a*b)^(1/2)-b)*
b)^(1/2)*arctan(b*cos(d*x+c)/(((a*b)^(1/2)-b)*b)^(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(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

-integrate(sin(d*x + c)^3/(b*sin(d*x + c)^4 - a), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (79) = 158\).
time = 0.46, size = 703, normalized size = 6.11 \begin {gather*} \frac {1}{4} \, \sqrt {-\frac {{\left (a b - b^{2}\right )} d^{2} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} + 1}{{\left (a b - b^{2}\right )} d^{2}}} \log \left (-{\left ({\left (a b^{2} - b^{3}\right )} d^{3} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} - b d\right )} \sqrt {-\frac {{\left (a b - b^{2}\right )} d^{2} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} + 1}{{\left (a b - b^{2}\right )} d^{2}}} + \cos \left (d x + c\right )\right ) - \frac {1}{4} \, \sqrt {-\frac {{\left (a b - b^{2}\right )} d^{2} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} + 1}{{\left (a b - b^{2}\right )} d^{2}}} \log \left (-{\left ({\left (a b^{2} - b^{3}\right )} d^{3} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} - b d\right )} \sqrt {-\frac {{\left (a b - b^{2}\right )} d^{2} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} + 1}{{\left (a b - b^{2}\right )} d^{2}}} - \cos \left (d x + c\right )\right ) - \frac {1}{4} \, \sqrt {\frac {{\left (a b - b^{2}\right )} d^{2} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} - 1}{{\left (a b - b^{2}\right )} d^{2}}} \log \left (-{\left ({\left (a b^{2} - b^{3}\right )} d^{3} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} + b d\right )} \sqrt {\frac {{\left (a b - b^{2}\right )} d^{2} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} - 1}{{\left (a b - b^{2}\right )} d^{2}}} + \cos \left (d x + c\right )\right ) + \frac {1}{4} \, \sqrt {\frac {{\left (a b - b^{2}\right )} d^{2} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} - 1}{{\left (a b - b^{2}\right )} d^{2}}} \log \left (-{\left ({\left (a b^{2} - b^{3}\right )} d^{3} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} + b d\right )} \sqrt {\frac {{\left (a b - b^{2}\right )} d^{2} \sqrt {\frac {a}{{\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} d^{4}}} - 1}{{\left (a b - b^{2}\right )} d^{2}}} - \cos \left (d x + c\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

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

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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(sin(d*x+c)**3/(a-b*sin(d*x+c)**4),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (79) = 158\).
time = 0.72, size = 166, normalized size = 1.44 \begin {gather*} \frac {\sqrt {-b^{2} - \sqrt {a b} b} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {b d^{2} + \sqrt {{\left (a - b\right )} b d^{4} + b^{2} d^{4}}}{b d^{4}}}}\right )}{2 \, {\left (b + \sqrt {a b}\right )} d {\left | b \right |}} + \frac {\sqrt {-b^{2} + \sqrt {a b} b} \arctan \left (\frac {\cos \left (d x + c\right )}{d \sqrt {-\frac {b d^{2} - \sqrt {{\left (a - b\right )} b d^{4} + b^{2} d^{4}}}{b d^{4}}}}\right )}{2 \, {\left (b - \sqrt {a b}\right )} d {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

1/2*sqrt(-b^2 - sqrt(a*b)*b)*arctan(cos(d*x + c)/(d*sqrt(-(b*d^2 + sqrt((a - b)*b*d^4 + b^2*d^4))/(b*d^4))))/(
(b + sqrt(a*b))*d*abs(b)) + 1/2*sqrt(-b^2 + sqrt(a*b)*b)*arctan(cos(d*x + c)/(d*sqrt(-(b*d^2 - sqrt((a - b)*b*
d^4 + b^2*d^4))/(b*d^4))))/((b - sqrt(a*b))*d*abs(b))

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Mupad [B]
time = 0.51, size = 976, normalized size = 8.49 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {8\,a\,b^2\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b+\frac {2\,a\,b^5}{a\,b^3-b^4}-\frac {2\,a\,b^3\,\sqrt {a\,b^3}}{a\,b^3-b^4}}-\frac {8\,a\,b^6\,\cos \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b^5-2\,a^2\,b^4-\frac {2\,a^2\,b^8}{a\,b^3-b^4}+\frac {2\,a\,b^9}{a\,b^3-b^4}+\frac {2\,a^2\,b^6\,\sqrt {a\,b^3}}{a\,b^3-b^4}-\frac {2\,a\,b^7\,\sqrt {a\,b^3}}{a\,b^3-b^4}}+\frac {8\,a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {a\,b^3}\,\sqrt {\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b^5-2\,a^2\,b^4-\frac {2\,a^2\,b^8}{a\,b^3-b^4}+\frac {2\,a\,b^9}{a\,b^3-b^4}+\frac {2\,a^2\,b^6\,\sqrt {a\,b^3}}{a\,b^3-b^4}-\frac {2\,a\,b^7\,\sqrt {a\,b^3}}{a\,b^3-b^4}}\right )\,\sqrt {-\frac {b^2-\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{d}-\frac {2\,\mathrm {atanh}\left (\frac {8\,a\,b^6\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}-\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b^5-2\,a^2\,b^4-\frac {2\,a^2\,b^8}{a\,b^3-b^4}+\frac {2\,a\,b^9}{a\,b^3-b^4}-\frac {2\,a^2\,b^6\,\sqrt {a\,b^3}}{a\,b^3-b^4}+\frac {2\,a\,b^7\,\sqrt {a\,b^3}}{a\,b^3-b^4}}-\frac {8\,a\,b^2\,\cos \left (c+d\,x\right )\,\sqrt {-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}-\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b+\frac {2\,a\,b^5}{a\,b^3-b^4}+\frac {2\,a\,b^3\,\sqrt {a\,b^3}}{a\,b^3-b^4}}+\frac {8\,a\,b^4\,\cos \left (c+d\,x\right )\,\sqrt {a\,b^3}\,\sqrt {-\frac {b^2}{16\,\left (a\,b^3-b^4\right )}-\frac {\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{2\,a\,b^5-2\,a^2\,b^4-\frac {2\,a^2\,b^8}{a\,b^3-b^4}+\frac {2\,a\,b^9}{a\,b^3-b^4}-\frac {2\,a^2\,b^6\,\sqrt {a\,b^3}}{a\,b^3-b^4}+\frac {2\,a\,b^7\,\sqrt {a\,b^3}}{a\,b^3-b^4}}\right )\,\sqrt {-\frac {b^2+\sqrt {a\,b^3}}{16\,\left (a\,b^3-b^4\right )}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^3/(a - b*sin(c + d*x)^4),x)

[Out]

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

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