\(\int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [444]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F(-2)]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 187 \[ \int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {5 \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {10 a \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {5 \cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^3 d}-\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))}-\frac {5 \cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^5 d} \]

[Out]

-5/8*(8*a^4-12*a^2*b^2+3*b^4)*x/b^6+10*a*(a^2-b^2)^(3/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^6/
d+5/12*cos(d*x+c)^3*(4*a-3*b*sin(d*x+c))/b^3/d-cos(d*x+c)^5/b/d/(a+b*sin(d*x+c))-5/8*cos(d*x+c)*(8*a*(a^2-b^2)
-b*(4*a^2-3*b^2)*sin(d*x+c))/b^5/d

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2772, 2944, 2814, 2739, 632, 210} \[ \int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {10 a \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^6 d}-\frac {5 \cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^5 d}-\frac {5 x \left (8 a^4-12 a^2 b^2+3 b^4\right )}{8 b^6}+\frac {5 \cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^3 d}-\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))} \]

[In]

Int[Cos[c + d*x]^6/(a + b*Sin[c + d*x])^2,x]

[Out]

(-5*(8*a^4 - 12*a^2*b^2 + 3*b^4)*x)/(8*b^6) + (10*a*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2
 - b^2]])/(b^6*d) + (5*Cos[c + d*x]^3*(4*a - 3*b*Sin[c + d*x]))/(12*b^3*d) - Cos[c + d*x]^5/(b*d*(a + b*Sin[c
+ d*x])) - (5*Cos[c + d*x]*(8*a*(a^2 - b^2) - b*(4*a^2 - 3*b^2)*Sin[c + d*x]))/(8*b^5*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 2772

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[g^2*((p - 1)/(b*(m + 1))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2944

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
 + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) -
 a*d*p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(m + p)*(m + p +
1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))}-\frac {5 \int \frac {\cos ^4(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b} \\ & = \frac {5 \cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^3 d}-\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))}-\frac {5 \int \frac {\cos ^2(c+d x) \left (-a b-\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 b^3} \\ & = \frac {5 \cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^3 d}-\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))}-\frac {5 \cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^5 d}-\frac {5 \int \frac {a b \left (4 a^2-5 b^2\right )+\left (8 a^4-12 a^2 b^2+3 b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{8 b^5} \\ & = -\frac {5 \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {5 \cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^3 d}-\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))}-\frac {5 \cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^5 d}+\frac {\left (5 a \left (a^2-b^2\right )^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^6} \\ & = -\frac {5 \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {5 \cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^3 d}-\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))}-\frac {5 \cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^5 d}+\frac {\left (10 a \left (a^2-b^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^6 d} \\ & = -\frac {5 \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {5 \cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^3 d}-\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))}-\frac {5 \cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^5 d}-\frac {\left (20 a \left (a^2-b^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^6 d} \\ & = -\frac {5 \left (8 a^4-12 a^2 b^2+3 b^4\right ) x}{8 b^6}+\frac {10 a \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^6 d}+\frac {5 \cos ^3(c+d x) (4 a-3 b \sin (c+d x))}{12 b^3 d}-\frac {\cos ^5(c+d x)}{b d (a+b \sin (c+d x))}-\frac {5 \cos (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{8 b^5 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(3679\) vs. \(2(187)=374\).

Time = 7.27 (sec) , antiderivative size = 3679, normalized size of antiderivative = 19.67 \[ \int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Result too large to show} \]

[In]

Integrate[Cos[c + d*x]^6/(a + b*Sin[c + d*x])^2,x]

[Out]

(Cos[c + d*x]^5*(-((b*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^(7/2)*(b/(a + b) - (b*Sin[c + d*x])/(a + b))^(
7/2))/(((a*b)/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b))*(a + b*Sin[c + d*x]))) - ((48*Sqrt[2]*(a -
b)*b^3*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^(7/2)*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]*(1 + ((a - b
)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(7/2)*((7*(3/(16*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c +
d*x])/(a - b)))/(2*b))^3) + 1/(2*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^2) + (1 + ((a
 - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(-1)))/12 + (35*b^4*(((a - b)*(-(b/(a - b)) - (b*Sin[c
 + d*x])/(a - b)))/b - ((a - b)^2*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^2)/(3*b^2) + (2*(a - b)^3*(-(b/(a
- b)) - (b*Sin[c + d*x])/(a - b))^3)/(15*b^3) - (Sqrt[2]*Sqrt[a - b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) -
(b*Sin[c + d*x])/(a - b)])/(Sqrt[2]*Sqrt[b])]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[b]*Sqrt[1 +
 ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b)])))/(128*(a - b)^4*(-(b/(a - b)) - (b*Sin[c + d*x])
/(a - b))^4*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^3)))/(7*(a + b)^2*(a^2 - b^2)*Sqrt
[((a + b)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)))/b]) + (5*a*b^2*((8*Sqrt[2]*b*(-(b/(a - b)) - (b*Sin[c + d*x]
)/(a - b))^(5/2)*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a
- b)))/(2*b))^(7/2)*((5/(16*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^3) + 5/(8*(1 + ((a
 - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a
 - b)))/(2*b))^(-1))/2 - (15*b^3*(((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/b - ((a - b)^2*(-(b/(a -
 b)) - (b*Sin[c + d*x])/(a - b))^2)/(3*b^2) - (Sqrt[2]*Sqrt[a - b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b
*Sin[c + d*x])/(a - b)])/(Sqrt[2]*Sqrt[b])]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[b]*Sqrt[1 + (
(a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b)])))/(64*(a - b)^3*(-(b/(a - b)) - (b*Sin[c + d*x])/(a
 - b))^3*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^3)))/(5*(a + b)^2*Sqrt[((a + b)*(b/(a
 + b) - (b*Sin[c + d*x])/(a + b)))/b]) - ((-((a*b)/(a - b)) + b^2/(a - b))*((8*Sqrt[2]*b*(-(b/(a - b)) - (b*Si
n[c + d*x])/(a - b))^(3/2)*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c +
 d*x])/(a - b)))/(2*b))^(7/2)*((3*(5/(8*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^3) + 5
/(6*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[
c + d*x])/(a - b)))/(2*b))^(-1)))/8 + (15*b^2*(((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/b - (Sqrt[2
]*Sqrt[a - b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[2]*Sqrt[b])]*Sqrt[-(b/
(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[b]*Sqrt[1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2
*b)])))/(64*(a - b)^2*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^2*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x
])/(a - b)))/(2*b))^3)))/(3*(a + b)^2*Sqrt[((a + b)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)))/b]) - ((-((a*b)/(a
 - b)) + b^2/(a - b))*((8*Sqrt[2]*b*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)]*Sqrt[b/(a + b) - (b*Sin[c +
d*x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(7/2)*((5*Sqrt[b]*ArcSinh[(Sqrt
[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[2]*Sqrt[b])])/(8*Sqrt[2]*Sqrt[a - b]*Sqrt[-(b/(a
- b)) - (b*Sin[c + d*x])/(a - b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(7/2)) + (15
/(8*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^3) + 5/(4*(1 + ((a - b)*(-(b/(a - b)) - (b
*Sin[c + d*x])/(a - b)))/(2*b))^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(-1))/6))
/((a + b)^2*Sqrt[((a + b)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)))/b]) - ((-((a*b)/(a - b)) + b^2/(a - b))*(-((
(-((a*b)/(a + b)) - b^2/(a + b))*(-(((-((a*b)/(a + b)) - b^2/(a + b))*((2*Sqrt[a - b]*ArcTanh[(Sqrt[a - b]*Sqr
t[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[a + b]*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)])])/(b*Sqrt
[a + b]) - (2*Sqrt[-((a*b)/(a + b)) - b^2/(a + b)]*ArcTanh[(Sqrt[-((a*b)/(a + b)) - b^2/(a + b)]*Sqrt[-(b/(a -
 b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[-((a*b)/(a - b)) + b^2/(a - b)]*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a +
 b)])])/(b*Sqrt[-((a*b)/(a - b)) + b^2/(a - b)])))/b) + (2*Sqrt[2]*(a - b)*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x]
)/(a - b)]*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))
/(2*b))^(3/2)*((Sqrt[b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[2]*Sqrt[b])]
)/(Sqrt[2]*Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c +
d*x])/(a - b)))/(2*b))^(3/2)) + 1/(2*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b)))))/(b*(a
+ b)*Sqrt[((a + b)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)))/b])))/b) + (4*Sqrt[2]*(a - b)*Sqrt[-(b/(a - b)) - (
b*Sin[c + d*x])/(a - b)]*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d
*x])/(a - b)))/(2*b))^(5/2)*((3*Sqrt[b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(S
qrt[2]*Sqrt[b])])/(4*Sqrt[2]*Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)]*(1 + ((a - b)*(-(b/(a -
 b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(5/2)) + (3/(2*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)
))/(2*b))^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(-1))/4))/((a + b)^2*Sqrt[((a +
 b)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)))/b])))/b))/b))/b))/(a^2 - b^2))/(((a*b)/(a - b) - b^2/(a - b))*((a*
b)/(a + b) + b^2/(a + b)))))/(d*(1 - (a + b*Sin[c + d*x])/(a - b))^(5/2)*(1 - (a + b*Sin[c + d*x])/(a + b))^(5
/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(394\) vs. \(2(176)=352\).

Time = 2.69 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.11

method result size
derivativedivides \(\frac {-\frac {2 \left (\frac {\left (\frac {3}{2} a^{2} b^{2}-\frac {9}{8} b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{3} b -6 a \,b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{2} a^{2} b^{2}-\frac {1}{8} b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 a^{3} b -14 a \,b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {3}{2} a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 a^{3} b -\frac {38}{3} a \,b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {3}{2} a^{2} b^{2}+\frac {9}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b -\frac {14 a \,b^{3}}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {5 \left (8 a^{4}-12 a^{2} b^{2}+3 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{6}}+\frac {\frac {2 \left (-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {10 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{6}}}{d}\) \(395\)
default \(\frac {-\frac {2 \left (\frac {\left (\frac {3}{2} a^{2} b^{2}-\frac {9}{8} b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 a^{3} b -6 a \,b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3}{2} a^{2} b^{2}-\frac {1}{8} b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 a^{3} b -14 a \,b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {3}{2} a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 a^{3} b -\frac {38}{3} a \,b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {3}{2} a^{2} b^{2}+\frac {9}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b -\frac {14 a \,b^{3}}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {5 \left (8 a^{4}-12 a^{2} b^{2}+3 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{6}}+\frac {\frac {2 \left (-\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {10 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{6}}}{d}\) \(395\)
risch \(-\frac {5 x \,a^{4}}{b^{6}}+\frac {15 x \,a^{2}}{2 b^{4}}-\frac {15 x}{8 b^{2}}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{4} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 b^{2} d}-\frac {2 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{b^{5} d}+\frac {9 a \,{\mathrm e}^{i \left (d x +c \right )}}{4 b^{3} d}-\frac {2 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{b^{5} d}+\frac {9 a \,{\mathrm e}^{-i \left (d x +c \right )}}{4 b^{3} d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{4} d}+\frac {2 i \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left (i b +a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{b^{6} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+i b \right )}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{4 b^{2} d}+\frac {5 \sqrt {-a^{2}+b^{2}}\, a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{6}}-\frac {5 \sqrt {-a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}-\frac {5 \sqrt {-a^{2}+b^{2}}\, a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{6}}+\frac {5 \sqrt {-a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}-\frac {\sin \left (4 d x +4 c \right )}{32 b^{2} d}+\frac {a \cos \left (3 d x +3 c \right )}{6 b^{3} d}\) \(499\)

[In]

int(cos(d*x+c)^6/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/b^6*(((3/2*a^2*b^2-9/8*b^4)*tan(1/2*d*x+1/2*c)^7+(4*a^3*b-6*a*b^3)*tan(1/2*d*x+1/2*c)^6+(3/2*a^2*b^2-1
/8*b^4)*tan(1/2*d*x+1/2*c)^5+(12*a^3*b-14*a*b^3)*tan(1/2*d*x+1/2*c)^4+(-3/2*a^2*b^2+1/8*b^4)*tan(1/2*d*x+1/2*c
)^3+(12*a^3*b-38/3*a*b^3)*tan(1/2*d*x+1/2*c)^2+(-3/2*a^2*b^2+9/8*b^4)*tan(1/2*d*x+1/2*c)+4*a^3*b-14/3*a*b^3)/(
1+tan(1/2*d*x+1/2*c)^2)^4+5/8*(8*a^4-12*a^2*b^2+3*b^4)*arctan(tan(1/2*d*x+1/2*c)))+2/b^6*((-(a^4-2*a^2*b^2+b^4
)*b^2/a*tan(1/2*d*x+1/2*c)-b*(a^4-2*a^2*b^2+b^4))/(a*tan(1/2*d*x+1/2*c)^2+2*b*tan(1/2*d*x+1/2*c)+a)+5*a*(a^4-2
*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 599, normalized size of antiderivative = 3.20 \[ \int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [\frac {6 \, b^{5} \cos \left (d x + c\right )^{5} - 5 \, {\left (4 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x - 60 \, {\left (a^{4} - a^{2} b^{2} + {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 15 \, {\left (8 \, a^{4} b - 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right ) + 5 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (8 \, a^{4} b - 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} d x - 3 \, {\left (4 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}}, \frac {6 \, b^{5} \cos \left (d x + c\right )^{5} - 5 \, {\left (4 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (8 \, a^{5} - 12 \, a^{3} b^{2} + 3 \, a b^{4}\right )} d x - 120 \, {\left (a^{4} - a^{2} b^{2} + {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 15 \, {\left (8 \, a^{4} b - 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right ) + 5 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - 3 \, {\left (8 \, a^{4} b - 12 \, a^{2} b^{3} + 3 \, b^{5}\right )} d x - 3 \, {\left (4 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}}\right ] \]

[In]

integrate(cos(d*x+c)^6/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/24*(6*b^5*cos(d*x + c)^5 - 5*(4*a^2*b^3 - 3*b^5)*cos(d*x + c)^3 - 15*(8*a^5 - 12*a^3*b^2 + 3*a*b^4)*d*x - 6
0*(a^4 - a^2*b^2 + (a^3*b - a*b^3)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*si
n(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^
2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 15*(8*a^4*b - 12*a^2*b^3 + 3*b^5)*cos(d*x + c) + 5*(2*a*b^4*cos(d*x + c
)^3 - 3*(8*a^4*b - 12*a^2*b^3 + 3*b^5)*d*x - 3*(4*a^3*b^2 - 5*a*b^4)*cos(d*x + c))*sin(d*x + c))/(b^7*d*sin(d*
x + c) + a*b^6*d), 1/24*(6*b^5*cos(d*x + c)^5 - 5*(4*a^2*b^3 - 3*b^5)*cos(d*x + c)^3 - 15*(8*a^5 - 12*a^3*b^2
+ 3*a*b^4)*d*x - 120*(a^4 - a^2*b^2 + (a^3*b - a*b^3)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) +
b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 15*(8*a^4*b - 12*a^2*b^3 + 3*b^5)*cos(d*x + c) + 5*(2*a*b^4*cos(d*x + c)^
3 - 3*(8*a^4*b - 12*a^2*b^3 + 3*b^5)*d*x - 3*(4*a^3*b^2 - 5*a*b^4)*cos(d*x + c))*sin(d*x + c))/(b^7*d*sin(d*x
+ c) + a*b^6*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cos(d*x+c)^6/(a+b*sin(d*x+c))^2,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(4*b^2-4*a^2>0)', see `assume?`
 for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (175) = 350\).

Time = 0.34 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {15 \, {\left (8 \, a^{4} - 12 \, a^{2} b^{2} + 3 \, b^{4}\right )} {\left (d x + c\right )}}{b^{6}} - \frac {240 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{6}} + \frac {48 \, {\left (a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a b^{5}} + \frac {2 \, {\left (36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 27 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 96 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 144 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 336 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 288 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 304 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a^{3} - 112 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{5}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^6/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/24*(15*(8*a^4 - 12*a^2*b^2 + 3*b^4)*(d*x + c)/b^6 - 240*(a^5 - 2*a^3*b^2 + a*b^4)*(pi*floor(1/2*(d*x + c)/p
i + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^6) + 48*(a^4*b*tan(
1/2*d*x + 1/2*c) - 2*a^2*b^3*tan(1/2*d*x + 1/2*c) + b^5*tan(1/2*d*x + 1/2*c) + a^5 - 2*a^3*b^2 + a*b^4)/((a*ta
n(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a*b^5) + 2*(36*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 27*b^3*tan(
1/2*d*x + 1/2*c)^7 + 96*a^3*tan(1/2*d*x + 1/2*c)^6 - 144*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 36*a^2*b*tan(1/2*d*x +
 1/2*c)^5 - 3*b^3*tan(1/2*d*x + 1/2*c)^5 + 288*a^3*tan(1/2*d*x + 1/2*c)^4 - 336*a*b^2*tan(1/2*d*x + 1/2*c)^4 -
 36*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 3*b^3*tan(1/2*d*x + 1/2*c)^3 + 288*a^3*tan(1/2*d*x + 1/2*c)^2 - 304*a*b^2*t
an(1/2*d*x + 1/2*c)^2 - 36*a^2*b*tan(1/2*d*x + 1/2*c) + 27*b^3*tan(1/2*d*x + 1/2*c) + 96*a^3 - 112*a*b^2)/((ta
n(1/2*d*x + 1/2*c)^2 + 1)^4*b^5))/d

Mupad [B] (verification not implemented)

Time = 7.30 (sec) , antiderivative size = 2530, normalized size of antiderivative = 13.53 \[ \int \frac {\cos ^6(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int(cos(c + d*x)^6/(a + b*sin(c + d*x))^2,x)

[Out]

- ((2*(15*a^4 + 3*b^4 - 20*a^2*b^2))/(3*b^5) - (5*tan(c/2 + (d*x)/2)^8*(b^4 - 4*a^4 + 4*a^2*b^2))/(2*b^5) + (5
*tan(c/2 + (d*x)/2)^6*(16*a^4 + 3*b^4 - 20*a^2*b^2))/(2*b^5) + (5*tan(c/2 + (d*x)/2)^2*(48*a^4 + 15*b^4 - 68*a
^2*b^2))/(6*b^5) + (5*tan(c/2 + (d*x)/2)^4*(72*a^4 + 15*b^4 - 100*a^2*b^2))/(6*b^5) + (tan(c/2 + (d*x)/2)*(180
*a^4 + 24*b^4 - 245*a^2*b^2))/(12*a*b^4) + (4*tan(c/2 + (d*x)/2)^5*(15*a^4 + 3*b^4 - 20*a^2*b^2))/(a*b^4) + (t
an(c/2 + (d*x)/2)^9*(20*a^4 + 8*b^4 - 25*a^2*b^2))/(4*a*b^4) + (tan(c/2 + (d*x)/2)^7*(60*a^4 + 16*b^4 - 85*a^2
*b^2))/(2*a*b^4) + (tan(c/2 + (d*x)/2)^3*(300*a^4 + 48*b^4 - 385*a^2*b^2))/(6*a*b^4))/(d*(a + 2*b*tan(c/2 + (d
*x)/2) + 5*a*tan(c/2 + (d*x)/2)^2 + 10*a*tan(c/2 + (d*x)/2)^4 + 10*a*tan(c/2 + (d*x)/2)^6 + 5*a*tan(c/2 + (d*x
)/2)^8 + a*tan(c/2 + (d*x)/2)^10 + 8*b*tan(c/2 + (d*x)/2)^3 + 12*b*tan(c/2 + (d*x)/2)^5 + 8*b*tan(c/2 + (d*x)/
2)^7 + 2*b*tan(c/2 + (d*x)/2)^9)) - (atan((((a^4*8i + b^4*3i - a^2*b^2*12i)*(((225*a^2*b^13)/2 - 900*a^4*b^11
+ 2400*a^6*b^9 - 2400*a^8*b^7 + 800*a^10*b^5)/b^14 + (tan(c/2 + (d*x)/2)*(450*a*b^15 - 5425*a^3*b^13 + 17800*a
^5*b^11 - 24000*a^7*b^9 + 14400*a^9*b^7 - 3200*a^11*b^5))/(2*b^15) - (5*(a^4*8i + b^4*3i - a^2*b^2*12i)*((60*a
*b^16 - 140*a^3*b^14 + 80*a^5*b^12)/b^14 - (5*(32*a^2*b^3 + (tan(c/2 + (d*x)/2)*(192*a*b^19 - 128*a^3*b^17))/(
2*b^15))*(a^4*8i + b^4*3i - a^2*b^2*12i))/(8*b^6) + (tan(c/2 + (d*x)/2)*(640*a^2*b^16 - 1280*a^4*b^14 + 640*a^
6*b^12))/(2*b^15)))/(8*b^6))*5i)/(8*b^6) + ((a^4*8i + b^4*3i - a^2*b^2*12i)*(((225*a^2*b^13)/2 - 900*a^4*b^11
+ 2400*a^6*b^9 - 2400*a^8*b^7 + 800*a^10*b^5)/b^14 + (tan(c/2 + (d*x)/2)*(450*a*b^15 - 5425*a^3*b^13 + 17800*a
^5*b^11 - 24000*a^7*b^9 + 14400*a^9*b^7 - 3200*a^11*b^5))/(2*b^15) + (5*(a^4*8i + b^4*3i - a^2*b^2*12i)*((60*a
*b^16 - 140*a^3*b^14 + 80*a^5*b^12)/b^14 + (5*(32*a^2*b^3 + (tan(c/2 + (d*x)/2)*(192*a*b^19 - 128*a^3*b^17))/(
2*b^15))*(a^4*8i + b^4*3i - a^2*b^2*12i))/(8*b^6) + (tan(c/2 + (d*x)/2)*(640*a^2*b^16 - 1280*a^4*b^14 + 640*a^
6*b^12))/(2*b^15)))/(8*b^6))*5i)/(8*b^6))/((4000*a^13 - 1875*a^3*b^10 + 12750*a^5*b^8 - 30875*a^7*b^6 + 35000*
a^9*b^4 - 19000*a^11*b^2)/b^14 - (5*(a^4*8i + b^4*3i - a^2*b^2*12i)*(((225*a^2*b^13)/2 - 900*a^4*b^11 + 2400*a
^6*b^9 - 2400*a^8*b^7 + 800*a^10*b^5)/b^14 + (tan(c/2 + (d*x)/2)*(450*a*b^15 - 5425*a^3*b^13 + 17800*a^5*b^11
- 24000*a^7*b^9 + 14400*a^9*b^7 - 3200*a^11*b^5))/(2*b^15) - (5*(a^4*8i + b^4*3i - a^2*b^2*12i)*((60*a*b^16 -
140*a^3*b^14 + 80*a^5*b^12)/b^14 - (5*(32*a^2*b^3 + (tan(c/2 + (d*x)/2)*(192*a*b^19 - 128*a^3*b^17))/(2*b^15))
*(a^4*8i + b^4*3i - a^2*b^2*12i))/(8*b^6) + (tan(c/2 + (d*x)/2)*(640*a^2*b^16 - 1280*a^4*b^14 + 640*a^6*b^12))
/(2*b^15)))/(8*b^6)))/(8*b^6) + (5*(a^4*8i + b^4*3i - a^2*b^2*12i)*(((225*a^2*b^13)/2 - 900*a^4*b^11 + 2400*a^
6*b^9 - 2400*a^8*b^7 + 800*a^10*b^5)/b^14 + (tan(c/2 + (d*x)/2)*(450*a*b^15 - 5425*a^3*b^13 + 17800*a^5*b^11 -
 24000*a^7*b^9 + 14400*a^9*b^7 - 3200*a^11*b^5))/(2*b^15) + (5*(a^4*8i + b^4*3i - a^2*b^2*12i)*((60*a*b^16 - 1
40*a^3*b^14 + 80*a^5*b^12)/b^14 + (5*(32*a^2*b^3 + (tan(c/2 + (d*x)/2)*(192*a*b^19 - 128*a^3*b^17))/(2*b^15))*
(a^4*8i + b^4*3i - a^2*b^2*12i))/(8*b^6) + (tan(c/2 + (d*x)/2)*(640*a^2*b^16 - 1280*a^4*b^14 + 640*a^6*b^12))/
(2*b^15)))/(8*b^6)))/(8*b^6) + (tan(c/2 + (d*x)/2)*(16000*a^14 + 2250*a^2*b^12 - 22500*a^4*b^10 + 86250*a^6*b^
8 - 162000*a^8*b^6 + 160000*a^10*b^4 - 80000*a^12*b^2))/b^15))*(a^4*8i + b^4*3i - a^2*b^2*12i)*5i)/(4*b^6*d) -
 (10*a*atanh((1125*a^3*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(3250*a^5*b - 1125*a^3*b^3 - (3125*a^7)/b +
(1000*a^9)/b^3 - 6250*a^6*tan(c/2 + (d*x)/2) - 2250*a^2*b^4*tan(c/2 + (d*x)/2) + 6500*a^4*b^2*tan(c/2 + (d*x)/
2) + (2000*a^8*tan(c/2 + (d*x)/2))/b^2) + (1000*a^5*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(3125*a^7*b + 1
125*a^3*b^5 - 3250*a^5*b^3 - (1000*a^9)/b - 2000*a^8*tan(c/2 + (d*x)/2) + 2250*a^2*b^6*tan(c/2 + (d*x)/2) - 65
00*a^4*b^4*tan(c/2 + (d*x)/2) + 6250*a^6*b^2*tan(c/2 + (d*x)/2)) + (2250*a^2*tan(c/2 + (d*x)/2)*(b^6 - a^6 - 3
*a^2*b^4 + 3*a^4*b^2)^(1/2))/(3250*a^5 - 1125*a^3*b^2 - (3125*a^7)/b^2 + (1000*a^9)/b^4 + 6500*a^4*b*tan(c/2 +
 (d*x)/2) - 2250*a^2*b^3*tan(c/2 + (d*x)/2) - (6250*a^6*tan(c/2 + (d*x)/2))/b + (2000*a^8*tan(c/2 + (d*x)/2))/
b^3) + (3125*a^4*tan(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(3125*a^7 + 1125*a^3*b^4 - 3250
*a^5*b^2 - (1000*a^9)/b^2 + 6250*a^6*b*tan(c/2 + (d*x)/2) + 2250*a^2*b^5*tan(c/2 + (d*x)/2) - 6500*a^4*b^3*tan
(c/2 + (d*x)/2) - (2000*a^8*tan(c/2 + (d*x)/2))/b) + (1000*a^6*tan(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a
^4*b^2)^(1/2))/(1000*a^9 - 1125*a^3*b^6 + 3250*a^5*b^4 - 3125*a^7*b^2 + 2000*a^8*b*tan(c/2 + (d*x)/2) - 2250*a
^2*b^7*tan(c/2 + (d*x)/2) + 6500*a^4*b^5*tan(c/2 + (d*x)/2) - 6250*a^6*b^3*tan(c/2 + (d*x)/2)))*(-(a + b)^3*(a
 - b)^3)^(1/2))/(b^6*d)