\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^3} \, dx\) [285]

Optimal result

Integrand size = 35, antiderivative size = 508 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^3} \, dx=-\frac {d^2 \left (A d \left (20 c^2+30 c d+13 d^2\right )-3 B \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^3 (c-d)^5 (c+d)^2 \sqrt {c^2-d^2} f}-\frac {d \left (3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )+A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac {\left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^2}-\frac {d \left (3 B \left (2 c^4-20 c^3 d-119 c^2 d^2-130 c d^3-48 d^4\right )+A \left (4 c^4-30 c^3 d+142 c^2 d^2+525 c d^3+304 d^4\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^5 (c+d)^2 f (c+d \sin (e+f x))} \]

[Out]

-1/30*d*(3*B*(2*c^3-20*c^2*d-57*c*d^2-30*d^3)+A*(4*c^3-30*c^2*d+146*c*d^2+195*d^3))*cos(f*x+e)/a^3/(c-d)^4/(c+
d)/f/(c+d*sin(f*x+e))^2-1/5*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2-1/15*(2*A*c-11*A*d+
3*B*c+6*B*d)*cos(f*x+e)/a/(c-d)^2/f/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2-1/15*(3*B*(c^2-10*c*d-12*d^2)+A*(2*c
^2-15*c*d+76*d^2))*cos(f*x+e)/(c-d)^3/f/(a^3+a^3*sin(f*x+e))/(c+d*sin(f*x+e))^2-1/30*d*(3*B*(2*c^4-20*c^3*d-11
9*c^2*d^2-130*c*d^3-48*d^4)+A*(4*c^4-30*c^3*d+142*c^2*d^2+525*c*d^3+304*d^4))*cos(f*x+e)/a^3/(c-d)^5/(c+d)^2/f
/(c+d*sin(f*x+e))-d^2*(A*d*(20*c^2+30*c*d+13*d^2)-3*B*(4*c^3+8*c^2*d+7*c*d^2+2*d^3))*arctan((d+c*tan(1/2*f*x+1
/2*e))/(c^2-d^2)^(1/2))/a^3/(c-d)^5/(c+d)^2/f/(c^2-d^2)^(1/2)

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 508, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3057, 2833, 12, 2739, 632, 210} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^3} \, dx=-\frac {d^2 \left (A d \left (20 c^2+30 c d+13 d^2\right )-3 B \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right )\right ) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^3 f (c-d)^5 (c+d)^2 \sqrt {c^2-d^2}}-\frac {\left (A \left (2 c^2-15 c d+76 d^2\right )+3 B \left (c^2-10 c d-12 d^2\right )\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right ) (c+d \sin (e+f x))^2}-\frac {d \left (A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )+3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )\right ) \cos (e+f x)}{30 a^3 f (c-d)^4 (c+d) (c+d \sin (e+f x))^2}-\frac {d \left (A \left (4 c^4-30 c^3 d+142 c^2 d^2+525 c d^3+304 d^4\right )+3 B \left (2 c^4-20 c^3 d-119 c^2 d^2-130 c d^3-48 d^4\right )\right ) \cos (e+f x)}{30 a^3 f (c-d)^5 (c+d)^2 (c+d \sin (e+f x))}-\frac {(2 A c-11 A d+3 B c+6 B d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))^2} \]

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^3),x]

[Out]

-((d^2*(A*d*(20*c^2 + 30*c*d + 13*d^2) - 3*B*(4*c^3 + 8*c^2*d + 7*c*d^2 + 2*d^3))*ArcTan[(d + c*Tan[(e + f*x)/
2])/Sqrt[c^2 - d^2]])/(a^3*(c - d)^5*(c + d)^2*Sqrt[c^2 - d^2]*f)) - (d*(3*B*(2*c^3 - 20*c^2*d - 57*c*d^2 - 30
*d^3) + A*(4*c^3 - 30*c^2*d + 146*c*d^2 + 195*d^3))*Cos[e + f*x])/(30*a^3*(c - d)^4*(c + d)*f*(c + d*Sin[e + f
*x])^2) - ((A - B)*Cos[e + f*x])/(5*(c - d)*f*(a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^2) - ((2*A*c + 3*B*c
- 11*A*d + 6*B*d)*Cos[e + f*x])/(15*a*(c - d)^2*f*(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^2) - ((3*B*(c^2
- 10*c*d - 12*d^2) + A*(2*c^2 - 15*c*d + 76*d^2))*Cos[e + f*x])/(15*(c - d)^3*f*(a^3 + a^3*Sin[e + f*x])*(c +
d*Sin[e + f*x])^2) - (d*(3*B*(2*c^4 - 20*c^3*d - 119*c^2*d^2 - 130*c*d^3 - 48*d^4) + A*(4*c^4 - 30*c^3*d + 14
2*c^2*d^2 + 525*c*d^3 + 304*d^4))*Cos[e + f*x])/(30*a^3*(c - d)^5*(c + d)^2*f*(c + d*Sin[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2833

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

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {\int \frac {-a (2 A c+3 B c-7 A d+2 B d)-4 a (A-B) d \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^3} \, dx}{5 a^2 (c-d)} \\ & = -\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}+\frac {\int \frac {a^2 \left (3 B \left (c^2-7 c d-6 d^2\right )+A \left (2 c^2-9 c d+43 d^2\right )\right )+3 a^2 d (A (2 c-11 d)+3 B (c+2 d)) \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx}{15 a^4 (c-d)^2} \\ & = -\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac {\left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^2}-\frac {\int \frac {-3 a^3 d^2 (2 A c+33 B c-65 A d+30 B d)-2 a^3 d \left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx}{15 a^6 (c-d)^3} \\ & = -\frac {d \left (3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )+A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac {\left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^2}+\frac {\int \frac {2 a^3 d^2 \left (2 A c^2+93 B c^2-165 A c d+150 B c d-152 A d^2+72 B d^2\right )+a^3 d \left (3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )+A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{30 a^6 (c-d)^4 (c+d)} \\ & = -\frac {d \left (3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )+A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac {\left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^2}-\frac {d \left (3 B \left (2 c^4-20 c^3 d-119 c^2 d^2-130 c d^3-48 d^4\right )+A \left (4 c^4-30 c^3 d+142 c^2 d^2+525 c d^3+304 d^4\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^5 (c+d)^2 f (c+d \sin (e+f x))}-\frac {\int \frac {15 a^3 d^2 \left (A d \left (20 c^2+30 c d+13 d^2\right )-3 B \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right )\right )}{c+d \sin (e+f x)} \, dx}{30 a^6 (c-d)^5 (c+d)^2} \\ & = -\frac {d \left (3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )+A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac {\left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^2}-\frac {d \left (3 B \left (2 c^4-20 c^3 d-119 c^2 d^2-130 c d^3-48 d^4\right )+A \left (4 c^4-30 c^3 d+142 c^2 d^2+525 c d^3+304 d^4\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^5 (c+d)^2 f (c+d \sin (e+f x))}-\frac {\left (d^2 \left (A d \left (20 c^2+30 c d+13 d^2\right )-3 B \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right )\right )\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 a^3 (c-d)^5 (c+d)^2} \\ & = -\frac {d \left (3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )+A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac {\left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^2}-\frac {d \left (3 B \left (2 c^4-20 c^3 d-119 c^2 d^2-130 c d^3-48 d^4\right )+A \left (4 c^4-30 c^3 d+142 c^2 d^2+525 c d^3+304 d^4\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^5 (c+d)^2 f (c+d \sin (e+f x))}-\frac {\left (d^2 \left (A d \left (20 c^2+30 c d+13 d^2\right )-3 B \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right )\right )\right ) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{a^3 (c-d)^5 (c+d)^2 f} \\ & = -\frac {d \left (3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )+A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac {\left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^2}-\frac {d \left (3 B \left (2 c^4-20 c^3 d-119 c^2 d^2-130 c d^3-48 d^4\right )+A \left (4 c^4-30 c^3 d+142 c^2 d^2+525 c d^3+304 d^4\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^5 (c+d)^2 f (c+d \sin (e+f x))}+\frac {\left (2 d^2 \left (A d \left (20 c^2+30 c d+13 d^2\right )-3 B \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right )\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{a^3 (c-d)^5 (c+d)^2 f} \\ & = -\frac {d^2 \left (A d \left (20 c^2+30 c d+13 d^2\right )-3 B \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^3 (c-d)^5 (c+d)^2 \sqrt {c^2-d^2} f}-\frac {d \left (3 B \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right )+A \left (4 c^3-30 c^2 d+146 c d^2+195 d^3\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2}-\frac {(2 A c+3 B c-11 A d+6 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2}-\frac {\left (3 B \left (c^2-10 c d-12 d^2\right )+A \left (2 c^2-15 c d+76 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))^2}-\frac {d \left (3 B \left (2 c^4-20 c^3 d-119 c^2 d^2-130 c d^3-48 d^4\right )+A \left (4 c^4-30 c^3 d+142 c^2 d^2+525 c d^3+304 d^4\right )\right ) \cos (e+f x)}{30 a^3 (c-d)^5 (c+d)^2 f (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 8.64 (sec) , antiderivative size = 548, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (12 (A-B) (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )+6 (-A+B) (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 (c-d) (A (2 c-17 d)+3 B (c+4 d)) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-2 (c-d) (A (2 c-17 d)+3 B (c+4 d)) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+4 \left (3 B \left (c^2-12 c d-19 d^2\right )+A \left (2 c^2-19 c d+107 d^2\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+\frac {30 d^2 \left (-A d \left (20 c^2+30 c d+13 d^2\right )+3 B \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{(c+d)^2 \sqrt {c^2-d^2}}+\frac {15 (c-d) d^3 (B c-A d) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{(c+d) (c+d \sin (e+f x))^2}+\frac {15 d^3 \left (-3 A d (3 c+2 d)+B \left (7 c^2+6 c d+2 d^2\right )\right ) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{(c+d)^2 (c+d \sin (e+f x))}\right )}{30 a^3 (c-d)^5 f (1+\sin (e+f x))^3} \]

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^3),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(12*(A - B)*(c - d)^2*Sin[(e + f*x)/2] + 6*(-A + B)*(c - d)^2*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2]) + 4*(c - d)*(A*(2*c - 17*d) + 3*B*(c + 4*d))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^2 - 2*(c - d)*(A*(2*c - 17*d) + 3*B*(c + 4*d))*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 4*
(3*B*(c^2 - 12*c*d - 19*d^2) + A*(2*c^2 - 19*c*d + 107*d^2))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x
)/2])^4 + (30*d^2*(-(A*d*(20*c^2 + 30*c*d + 13*d^2)) + 3*B*(4*c^3 + 8*c^2*d + 7*c*d^2 + 2*d^3))*ArcTan[(d + c*
Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)/((c + d)^2*Sqrt[c^2 - d^2]) + (15*
(c - d)*d^3*(B*c - A*d)*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)/((c + d)*(c + d*Sin[e + f*x])^2)
+ (15*d^3*(-3*A*d*(3*c + 2*d) + B*(7*c^2 + 6*c*d + 2*d^2))*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])
^5)/((c + d)^2*(c + d*Sin[e + f*x]))))/(30*a^3*(c - d)^5*f*(1 + Sin[e + f*x])^3)

Maple [A] (verified)

Time = 6.12 (sec) , antiderivative size = 639, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {-\frac {-8 A +8 B}{2 \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A -4 B \right )}{5 \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A c +10 d A +2 B c -8 d B}{\left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 A c -14 d A -6 B c +12 d B \right )}{3 \left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A \,c^{2}-5 A c d +10 A \,d^{2}-6 d^{2} B \right )}{\left (c -d \right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 d^{2} \left (\frac {\frac {d^{2} \left (11 c^{2} d A +6 d^{2} c A -2 A \,d^{3}-9 B \,c^{3}-6 c^{2} d B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (10 A \,c^{4} d +6 A \,c^{3} d^{2}+19 A \,c^{2} d^{3}+12 A c \,d^{4}-2 A \,d^{5}-8 B \,c^{5}-6 B \,c^{4} d -17 B \,c^{3} d^{2}-12 B \,c^{2} d^{3}-2 B c \,d^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) c^{2}}+\frac {d^{2} \left (29 c^{2} d A +18 d^{2} c A -2 A \,d^{3}-23 B \,c^{3}-18 c^{2} d B -4 d^{2} c B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (10 c^{2} d A +6 d^{2} c A -A \,d^{3}-8 B \,c^{3}-6 c^{2} d B -d^{2} c B \right )}{2 c^{2}+4 c d +2 d^{2}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {\left (20 c^{2} d A +30 d^{2} c A +13 A \,d^{3}-12 B \,c^{3}-24 c^{2} d B -21 d^{2} c B -6 d^{3} B \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{5}}}{a^{3} f}\)	\(639\)
default	\(\frac {-\frac {-8 A +8 B}{2 \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A -4 B \right )}{5 \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A c +10 d A +2 B c -8 d B}{\left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 A c -14 d A -6 B c +12 d B \right )}{3 \left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A \,c^{2}-5 A c d +10 A \,d^{2}-6 d^{2} B \right )}{\left (c -d \right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 d^{2} \left (\frac {\frac {d^{2} \left (11 c^{2} d A +6 d^{2} c A -2 A \,d^{3}-9 B \,c^{3}-6 c^{2} d B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (10 A \,c^{4} d +6 A \,c^{3} d^{2}+19 A \,c^{2} d^{3}+12 A c \,d^{4}-2 A \,d^{5}-8 B \,c^{5}-6 B \,c^{4} d -17 B \,c^{3} d^{2}-12 B \,c^{2} d^{3}-2 B c \,d^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) c^{2}}+\frac {d^{2} \left (29 c^{2} d A +18 d^{2} c A -2 A \,d^{3}-23 B \,c^{3}-18 c^{2} d B -4 d^{2} c B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (10 c^{2} d A +6 d^{2} c A -A \,d^{3}-8 B \,c^{3}-6 c^{2} d B -d^{2} c B \right )}{2 c^{2}+4 c d +2 d^{2}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {\left (20 c^{2} d A +30 d^{2} c A +13 A \,d^{3}-12 B \,c^{3}-24 c^{2} d B -21 d^{2} c B -6 d^{3} B \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{5}}}{a^{3} f}\)	\(639\)
risch	\(\text {Expression too large to display}\)	\(2987\)

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/f/a^3*(-1/4*(-8*A+8*B)/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(4*A-4*B)/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)^5-1/2*(
-4*A*c+10*A*d+2*B*c-8*B*d)/(c-d)^4/(tan(1/2*f*x+1/2*e)+1)^2-1/3*(8*A*c-14*A*d-6*B*c+12*B*d)/(c-d)^4/(tan(1/2*f
*x+1/2*e)+1)^3-(A*c^2-5*A*c*d+10*A*d^2-6*B*d^2)/(c-d)^5/(tan(1/2*f*x+1/2*e)+1)-d^2/(c-d)^5*((1/2*d^2*(11*A*c^2
*d+6*A*c*d^2-2*A*d^3-9*B*c^3-6*B*c^2*d)/c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3+1/2*d*(10*A*c^4*d+6*A*c^3*d^2+1
9*A*c^2*d^3+12*A*c*d^4-2*A*d^5-8*B*c^5-6*B*c^4*d-17*B*c^3*d^2-12*B*c^2*d^3-2*B*c*d^4)/(c^2+2*c*d+d^2)/c^2*tan(
1/2*f*x+1/2*e)^2+1/2*d^2*(29*A*c^2*d+18*A*c*d^2-2*A*d^3-23*B*c^3-18*B*c^2*d-4*B*c*d^2)/c/(c^2+2*c*d+d^2)*tan(1
/2*f*x+1/2*e)+1/2*d*(10*A*c^2*d+6*A*c*d^2-A*d^3-8*B*c^3-6*B*c^2*d-B*c*d^2)/(c^2+2*c*d+d^2))/(tan(1/2*f*x+1/2*e
)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)^2+1/2*(20*A*c^2*d+30*A*c*d^2+13*A*d^3-12*B*c^3-24*B*c^2*d-21*B*c*d^2-6*B*d^3)/
(c^2+2*c*d+d^2)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3599 vs. \(2 (493) = 986\).

Time = 0.56 (sec) , antiderivative size = 7283, normalized size of antiderivative = 14.34 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

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

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3/(c+d*sin(f*x+e))**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^3,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*d^2-4*c^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. 1224 vs. \(2 (493) = 986\).

Time = 0.43 (sec) , antiderivative size = 1224, normalized size of antiderivative = 2.41 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(15*(12*B*c^3*d^2 - 20*A*c^2*d^3 + 24*B*c^2*d^3 - 30*A*c*d^4 + 21*B*c*d^4 - 13*A*d^5 + 6*B*d^5)*(pi*floor
(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((a^3*c^7 - 3*a^3*c^6*
d + a^3*c^5*d^2 + 5*a^3*c^4*d^3 - 5*a^3*c^3*d^4 - a^3*c^2*d^5 + 3*a^3*c*d^6 - a^3*d^7)*sqrt(c^2 - d^2)) + 15*(
9*B*c^4*d^4*tan(1/2*f*x + 1/2*e)^3 - 11*A*c^3*d^5*tan(1/2*f*x + 1/2*e)^3 + 6*B*c^3*d^5*tan(1/2*f*x + 1/2*e)^3
- 6*A*c^2*d^6*tan(1/2*f*x + 1/2*e)^3 + 2*A*c*d^7*tan(1/2*f*x + 1/2*e)^3 + 8*B*c^5*d^3*tan(1/2*f*x + 1/2*e)^2 -
10*A*c^4*d^4*tan(1/2*f*x + 1/2*e)^2 + 6*B*c^4*d^4*tan(1/2*f*x + 1/2*e)^2 - 6*A*c^3*d^5*tan(1/2*f*x + 1/2*e)^2
+ 17*B*c^3*d^5*tan(1/2*f*x + 1/2*e)^2 - 19*A*c^2*d^6*tan(1/2*f*x + 1/2*e)^2 + 12*B*c^2*d^6*tan(1/2*f*x + 1/2*
e)^2 - 12*A*c*d^7*tan(1/2*f*x + 1/2*e)^2 + 2*B*c*d^7*tan(1/2*f*x + 1/2*e)^2 + 2*A*d^8*tan(1/2*f*x + 1/2*e)^2 +
23*B*c^4*d^4*tan(1/2*f*x + 1/2*e) - 29*A*c^3*d^5*tan(1/2*f*x + 1/2*e) + 18*B*c^3*d^5*tan(1/2*f*x + 1/2*e) - 1
8*A*c^2*d^6*tan(1/2*f*x + 1/2*e) + 4*B*c^2*d^6*tan(1/2*f*x + 1/2*e) + 2*A*c*d^7*tan(1/2*f*x + 1/2*e) + 8*B*c^5
*d^3 - 10*A*c^4*d^4 + 6*B*c^4*d^4 - 6*A*c^3*d^5 + B*c^3*d^5 + A*c^2*d^6)/((a^3*c^9 - 3*a^3*c^8*d + a^3*c^7*d^2
+ 5*a^3*c^6*d^3 - 5*a^3*c^5*d^4 - a^3*c^4*d^5 + 3*a^3*c^3*d^6 - a^3*c^2*d^7)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*
tan(1/2*f*x + 1/2*e) + c)^2) - 2*(15*A*c^2*tan(1/2*f*x + 1/2*e)^4 - 75*A*c*d*tan(1/2*f*x + 1/2*e)^4 + 150*A*d^
2*tan(1/2*f*x + 1/2*e)^4 - 90*B*d^2*tan(1/2*f*x + 1/2*e)^4 + 30*A*c^2*tan(1/2*f*x + 1/2*e)^3 + 15*B*c^2*tan(1/
2*f*x + 1/2*e)^3 - 195*A*c*d*tan(1/2*f*x + 1/2*e)^3 - 75*B*c*d*tan(1/2*f*x + 1/2*e)^3 + 525*A*d^2*tan(1/2*f*x
+ 1/2*e)^3 - 300*B*d^2*tan(1/2*f*x + 1/2*e)^3 + 40*A*c^2*tan(1/2*f*x + 1/2*e)^2 + 15*B*c^2*tan(1/2*f*x + 1/2*e
)^2 - 245*A*c*d*tan(1/2*f*x + 1/2*e)^2 - 135*B*c*d*tan(1/2*f*x + 1/2*e)^2 + 745*A*d^2*tan(1/2*f*x + 1/2*e)^2 -
420*B*d^2*tan(1/2*f*x + 1/2*e)^2 + 20*A*c^2*tan(1/2*f*x + 1/2*e) + 15*B*c^2*tan(1/2*f*x + 1/2*e) - 145*A*c*d*
tan(1/2*f*x + 1/2*e) - 105*B*c*d*tan(1/2*f*x + 1/2*e) + 485*A*d^2*tan(1/2*f*x + 1/2*e) - 270*B*d^2*tan(1/2*f*x
+ 1/2*e) + 7*A*c^2 + 3*B*c^2 - 44*A*c*d - 21*B*c*d + 127*A*d^2 - 72*B*d^2)/((a^3*c^5 - 5*a^3*c^4*d + 10*a^3*c
^3*d^2 - 10*a^3*c^2*d^3 + 5*a^3*c*d^4 - a^3*d^5)*(tan(1/2*f*x + 1/2*e) + 1)^5))/f

Mupad [B] (verification not implemented)

Time = 19.58 (sec) , antiderivative size = 2387, normalized size of antiderivative = 4.70 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c + d*sin(e + f*x))^3),x)

[Out]

((15*A*d^6 - 14*A*c^6 - 6*B*c^6 - 404*A*c^2*d^4 - 420*A*c^3*d^3 - 92*A*c^4*d^2 + 234*B*c^2*d^4 + 450*B*c^3*d^3
+ 222*B*c^4*d^2 - 90*A*c*d^5 + 60*A*c^5*d + 15*B*c*d^5 + 30*B*c^5*d)/(15*(c + d)^2*(c - d)*(c^4 - 4*c^3*d - 4
*c*d^3 + d^4 + 6*c^2*d^2)) + (tan(e/2 + (f*x)/2)^7*(2*A*d^8 - 4*A*c^8 - 2*B*c^8 - 49*A*c^2*d^6 - 141*A*c^3*d^5
- 200*A*c^4*d^4 - 122*A*c^5*d^3 + 2*A*c^6*d^2 + 12*B*c^2*d^6 + 95*B*c^3*d^5 + 187*B*c^4*d^4 + 146*B*c^5*d^3 +
58*B*c^6*d^2 - 2*A*c*d^7 + 10*A*c^7*d + 2*B*c*d^7 + 6*B*c^7*d))/(c^2*(c - d)*(2*c*d + c^2 + d^2)*(c^4 - 4*c^3
*d - 4*c*d^3 + d^4 + 6*c^2*d^2)) + (tan(e/2 + (f*x)/2)^6*(30*A*d^8 - 28*A*c^8 - 6*B*c^8 - 759*A*c^2*d^6 - 1707
*A*c^3*d^5 - 1960*A*c^4*d^4 - 870*A*c^5*d^3 + 62*A*c^6*d^2 + 336*B*c^2*d^6 + 1257*B*c^3*d^5 + 1893*B*c^4*d^4 +
1350*B*c^5*d^3 + 414*B*c^6*d^2 - 114*A*c*d^7 + 54*A*c^7*d + 30*B*c*d^7 + 18*B*c^7*d))/(3*c^2*(c - d)*(2*c*d +
c^2 + d^2)*(c^4 - 4*c^3*d - 4*c*d^3 + d^4 + 6*c^2*d^2)) + (tan(e/2 + (f*x)/2)^5*(60*A*d^8 - 32*A*c^8 - 18*B*c
^8 - 1857*A*c^2*d^6 - 3763*A*c^3*d^5 - 3560*A*c^4*d^4 - 1294*A*c^5*d^3 + 70*A*c^6*d^2 + 900*B*c^2*d^6 + 2859*B
*c^3*d^5 + 3705*B*c^4*d^4 + 2358*B*c^5*d^3 + 678*B*c^6*d^2 - 270*A*c*d^7 + 62*A*c^7*d + 60*B*c*d^7 + 42*B*c^7*
d))/(3*c^2*(c - d)*(2*c*d + c^2 + d^2)*(c^4 - 4*c^3*d - 4*c*d^3 + d^4 + 6*c^2*d^2)) + (tan(e/2 + (f*x)/2)^2*(3
0*A*d^8 - 108*A*c^8 - 42*B*c^8 - 2501*A*c^2*d^6 - 8725*A*c^3*d^5 - 10616*A*c^4*d^4 - 4810*A*c^5*d^3 + 10*A*c^6
*d^2 + 1056*B*c^2*d^6 + 5235*B*c^3*d^5 + 9891*B*c^4*d^4 + 7770*B*c^5*d^3 + 2370*B*c^6*d^2 - 30*A*c*d^7 + 290*A
*c^7*d + 30*B*c*d^7 + 150*B*c^7*d))/(15*c^2*(c + d)^2*(c - d)*(c^4 - 4*c^3*d - 4*c*d^3 + d^4 + 6*c^2*d^2)) + (
tan(e/2 + (f*x)/2)^3*(150*A*d^8 - 140*A*c^8 - 90*B*c^8 - 7945*A*c^2*d^6 - 19441*A*c^3*d^5 - 18600*A*c^4*d^4 -
6898*A*c^5*d^3 + 210*A*c^6*d^2 + 3660*B*c^2*d^6 + 13311*B*c^3*d^5 + 19455*B*c^4*d^4 + 12618*B*c^5*d^3 + 3570*B
*c^6*d^2 - 570*A*c*d^7 + 314*A*c^7*d + 150*B*c*d^7 + 246*B*c^7*d))/(15*c^2*(c + d)^2*(c - d)*(c^4 - 4*c^3*d -
4*c*d^3 + d^4 + 6*c^2*d^2)) + (tan(e/2 + (f*x)/2)*(30*A*d^7 - 40*A*c^7 - 30*B*c^7 - 1901*A*c^2*d^5 - 3400*A*c^
3*d^4 - 2018*A*c^4*d^3 - 190*A*c^5*d^2 + 921*B*c^2*d^5 + 2655*B*c^3*d^4 + 2778*B*c^4*d^3 + 1050*B*c^5*d^2 - 19
5*A*c*d^6 + 154*A*c^6*d + 60*B*c*d^6 + 126*B*c^6*d))/(15*c*(c + d)^2*(c - d)*(c^4 - 4*c^3*d - 4*c*d^3 + d^4 +
6*c^2*d^2)) - (tan(e/2 + (f*x)/2)^8*(2*A*c^7 - 2*A*d^7 + 11*A*c^2*d^5 + 20*A*c^3*d^4 + 30*A*c^4*d^3 + 2*A*c^5*
d^2 - 6*B*c^2*d^5 - 21*B*c^3*d^4 - 24*B*c^4*d^3 - 12*B*c^5*d^2 + 6*A*c*d^6 - 6*A*c^6*d))/(c*(c - d)*(2*c*d + c
^2 + d^2)*(c^4 - 4*c^3*d - 4*c*d^3 + d^4 + 6*c^2*d^2)) + (tan(e/2 + (f*x)/2)^4*(300*A*d^7 - 204*A*c^7 - 66*B*c
^7 - 10235*A*c^2*d^5 - 14330*A*c^3*d^4 - 7254*A*c^4*d^3 - 316*A*c^5*d^2 + 5460*B*c^2*d^5 + 12675*B*c^3*d^4 + 1
0764*B*c^4*d^3 + 3666*B*c^5*d^2 - 1650*A*c*d^6 + 614*A*c^6*d + 300*B*c*d^6 + 276*B*c^6*d))/(15*c^2*(c + d)*(c
- d)*(c^4 - 4*c^3*d - 4*c*d^3 + d^4 + 6*c^2*d^2)))/(f*(tan(e/2 + (f*x)/2)*(5*a^3*c^2 + 4*a^3*c*d) + tan(e/2 +
(f*x)/2)^2*(12*a^3*c^2 + 4*a^3*d^2 + 20*a^3*c*d) + tan(e/2 + (f*x)/2)^7*(12*a^3*c^2 + 4*a^3*d^2 + 20*a^3*c*d)
+ tan(e/2 + (f*x)/2)^3*(20*a^3*c^2 + 20*a^3*d^2 + 44*a^3*c*d) + tan(e/2 + (f*x)/2)^6*(20*a^3*c^2 + 20*a^3*d^2
+ 44*a^3*c*d) + tan(e/2 + (f*x)/2)^4*(26*a^3*c^2 + 40*a^3*d^2 + 60*a^3*c*d) + tan(e/2 + (f*x)/2)^5*(26*a^3*c^2
+ 40*a^3*d^2 + 60*a^3*c*d) + tan(e/2 + (f*x)/2)^8*(5*a^3*c^2 + 4*a^3*c*d) + a^3*c^2 + a^3*c^2*tan(e/2 + (f*x)
/2)^9)) - (d^2*atan(((d^2*(12*B*c^3 - 13*A*d^3 + 6*B*d^3 - 30*A*c*d^2 - 20*A*c^2*d + 21*B*c*d^2 + 24*B*c^2*d)*
(2*a^3*d^8 - 6*a^3*c*d^7 - 2*a^3*c^7*d + 2*a^3*c^2*d^6 + 10*a^3*c^3*d^5 - 10*a^3*c^4*d^4 - 2*a^3*c^5*d^3 + 6*a
^3*c^6*d^2))/(2*a^3*(c + d)^(5/2)*(c - d)^(11/2)) - (c*d^2*tan(e/2 + (f*x)/2)*(12*B*c^3 - 13*A*d^3 + 6*B*d^3 -
30*A*c*d^2 - 20*A*c^2*d + 21*B*c*d^2 + 24*B*c^2*d)*(a^3*c^7 - a^3*d^7 + 3*a^3*c*d^6 - 3*a^3*c^6*d - a^3*c^2*d
^5 - 5*a^3*c^3*d^4 + 5*a^3*c^4*d^3 + a^3*c^5*d^2))/(a^3*(c + d)^(5/2)*(c - d)^(11/2)))/(6*B*d^5 - 13*A*d^5 - 2
0*A*c^2*d^3 + 24*B*c^2*d^3 + 12*B*c^3*d^2 - 30*A*c*d^4 + 21*B*c*d^4))*(12*B*c^3 - 13*A*d^3 + 6*B*d^3 - 30*A*c*
d^2 - 20*A*c^2*d + 21*B*c*d^2 + 24*B*c^2*d))/(a^3*f*(c + d)^(5/2)*(c - d)^(11/2))