\(\int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) [184]

Optimal result

Integrand size = 23, antiderivative size = 259 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {x}{b}-\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b d}-\frac {2 \sqrt [3]{a} \arctan \left (\frac {(-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}} b d}+\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{-1} \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}} b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3299, 3292, 2739, 632, 210} \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}+\frac {2 \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {x}{b} \]

[In]

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

[Out]

x/b - (2*a^(1/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*Sqrt[a^(2/3) - b^(2/
3)]*b*d) - (2*a^(1/3)*ArcTan[((-1)^(2/3)*b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)
]])/(3*Sqrt[a^(2/3) + (-1)^(1/3)*b^(2/3)]*b*d) + (2*a^(1/3)*ArcTan[((-1)^(1/3)*(b^(1/3) + (-1)^(2/3)*a^(1/3)*T
an[(c + d*x)/2]))/Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]])/(3*Sqrt[a^(2/3) - (-1)^(2/3)*b^(2/3)]*b*d)

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3292

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

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

Mathematica [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {\$Aborted} \]

[In]

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

[Out]

$Aborted

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.78 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.40

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}}{d}\)	\(104\)
default	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}+2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}}{d}\)	\(104\)
risch	\(\frac {x}{b}+\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (729 a^{2} b^{6} d^{6}-729 b^{8} d^{6}\right ) \textit {\_Z}^{6}-15552 a^{2} b^{4} d^{4} \textit {\_Z}^{4}+110592 a^{2} b^{2} d^{2} \textit {\_Z}^{2}-262144 a^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {243 i a \,b^{4} d^{5}}{16384}-\frac {243 i b^{6} d^{5}}{16384 a}\right ) \textit {\_R}^{5}+\left (-\frac {81 i a \,b^{3} d^{4}}{4096}+\frac {81 i b^{5} d^{4}}{4096 a}\right ) \textit {\_R}^{4}+\left (-\frac {135 i a \,b^{2} d^{3}}{512}-\frac {27 i b^{4} d^{3}}{512 a}\right ) \textit {\_R}^{3}+\frac {27 i a b \,d^{2} \textit {\_R}^{2}}{64}+\frac {9 i a d \textit {\_R}}{8}-\frac {2 i a}{b}\right )\right )}{8}\)	\(190\)

[In]

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

[Out]

1/d*(2/b*arctan(tan(1/2*d*x+1/2*c))-1/3/b*a*sum((_R^4+2*_R^2+1)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x
+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.25 (sec) , antiderivative size = 29221, normalized size of antiderivative = 112.82 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]

[In]

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

[Out]

(8*a*b*integrate(-(8*a*cos(3*d*x + 3*c)^2 - b*cos(3*d*x + 3*c)*sin(6*d*x + 6*c) + 3*b*cos(3*d*x + 3*c)*sin(4*d
*x + 4*c) + b*cos(6*d*x + 6*c)*sin(3*d*x + 3*c) - 3*b*cos(4*d*x + 4*c)*sin(3*d*x + 3*c) + 8*a*sin(3*d*x + 3*c)
^2 - 3*b*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + (3*b*cos(2*d*x + 2*c) - b)*sin(3*d*x + 3*c))/(b^3*cos(6*d*x + 6*c
)^2 + 9*b^3*cos(4*d*x + 4*c)^2 + 64*a^2*b*cos(3*d*x + 3*c)^2 + 9*b^3*cos(2*d*x + 2*c)^2 + b^3*sin(6*d*x + 6*c)
^2 + 9*b^3*sin(4*d*x + 4*c)^2 + 64*a^2*b*sin(3*d*x + 3*c)^2 - 48*a*b^2*cos(3*d*x + 3*c)*sin(2*d*x + 2*c) + 9*b
^3*sin(2*d*x + 2*c)^2 - 6*b^3*cos(2*d*x + 2*c) + b^3 - 2*(3*b^3*cos(4*d*x + 4*c) - 3*b^3*cos(2*d*x + 2*c) - 8*
a*b^2*sin(3*d*x + 3*c) + b^3)*cos(6*d*x + 6*c) - 6*(3*b^3*cos(2*d*x + 2*c) + 8*a*b^2*sin(3*d*x + 3*c) - b^3)*c
os(4*d*x + 4*c) - 2*(8*a*b^2*cos(3*d*x + 3*c) + 3*b^3*sin(4*d*x + 4*c) - 3*b^3*sin(2*d*x + 2*c))*sin(6*d*x + 6
*c) + 6*(8*a*b^2*cos(3*d*x + 3*c) - 3*b^3*sin(2*d*x + 2*c))*sin(4*d*x + 4*c) + 16*(3*a*b^2*cos(2*d*x + 2*c) -
a*b^2)*sin(3*d*x + 3*c)), x) + x)/b

Giac [F]

\[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 14.44 (sec) , antiderivative size = 1672, normalized size of antiderivative = 6.46 \[ \int \frac {\sin ^3(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

symsum(log(134217728*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)*a^7*ta
n(c/2 + (d*x)/2) - 268435456*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k
)^2*a^7*b - 1073741824*a^6*tan(c/2 + (d*x)/2) + 4831838208*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^
4 + 27*a^2*b^2*z^2 + a^2, z, k)^2*a^5*b^3 + 33722204160*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 +
27*a^2*b^2*z^2 + a^2, z, k)^3*a^6*b^3 + 15703474176*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27
*a^2*b^2*z^2 + a^2, z, k)^4*a^5*b^5 - 4831838208*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2
*b^2*z^2 + a^2, z, k)^4*a^7*b^3 - 130459631616*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b
^2*z^2 + a^2, z, k)^5*a^4*b^7 + 154014842880*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2
*z^2 + a^2, z, k)^5*a^6*b^5 + 35332816896*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^
2 + a^2, z, k)^6*a^5*b^7 - 21743271936*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 +
a^2, z, k)^6*a^7*b^5 - 130459631616*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a
^2, z, k)^7*a^4*b^9 + 122305904640*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2
, z, k)^7*a^6*b^7 + 2013265920*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z,
k)*a^6*b - 3221225472*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)*a^5*
b^2*tan(c/2 + (d*x)/2) - 18589155328*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a
^2, z, k)^2*a^6*b^2*tan(c/2 + (d*x)/2) - 17716740096*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27
*a^2*b^2*z^2 + a^2, z, k)^3*a^5*b^4*tan(c/2 + (d*x)/2) + 2818572288*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a
^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^3*a^7*b^2*tan(c/2 + (d*x)/2) + 86973087744*root(729*a^2*b^6*z^6 - 729
*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^4*a^4*b^6*tan(c/2 + (d*x)/2) - 88181047296*root(729*a
^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^4*a^6*b^4*tan(c/2 + (d*x)/2) - 308029
68576*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^5*a^5*b^6*tan(c/2 + (
d*x)/2) + 18119393280*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^5*a^7
*b^4*tan(c/2 + (d*x)/2) + 86973087744*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 +
a^2, z, k)^6*a^4*b^8*tan(c/2 + (d*x)/2) - 70665633792*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243*a^2*b^4*z^4 + 2
7*a^2*b^2*z^2 + a^2, z, k)^6*a^6*b^6*tan(c/2 + (d*x)/2) - 40768634880*root(729*a^2*b^6*z^6 - 729*b^8*z^6 + 243
*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^7*a^5*b^8*tan(c/2 + (d*x)/2) + 32614907904*root(729*a^2*b^6*z^6 - 7
29*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k)^7*a^7*b^6*tan(c/2 + (d*x)/2))*root(729*a^2*b^6*z^6
- 729*b^8*z^6 + 243*a^2*b^4*z^4 + 27*a^2*b^2*z^2 + a^2, z, k), k, 1, 6)/d - (log(tan(c/2 + (d*x)/2) - 1i)*1i)/
(b*d) + (log(tan(c/2 + (d*x)/2) + 1i)*1i)/(b*d)