3.3.63 \(\int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\) [263]

Optimal. Leaf size=283 \[ -\frac {a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}+\frac {2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^4 (c+d) \sqrt {c^2-d^2} f}-\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))} \]

[Out]

-1/2*a^3*(2*A*(2*c-3*d)*d-B*(6*c^2-12*c*d+7*d^2))*x/d^4-1/2*a^3*(4*A*c*d-B*(6*c^2-3*c*d-5*d^2))*cos(f*x+e)/d^3
/(c+d)/f+1/2*(2*A*d-B*(3*c+d))*cos(f*x+e)*(a^3+a^3*sin(f*x+e))/d^2/(c+d)/f+a*(-A*d+B*c)*cos(f*x+e)*(a+a*sin(f*
x+e))^2/d/(c+d)/f/(c+d*sin(f*x+e))+2*a^3*(c-d)^2*(A*d*(2*c+3*d)-B*(3*c^2+3*c*d-d^2))*arctan((d+c*tan(1/2*f*x+1
/2*e))/(c^2-d^2)^(1/2))/d^4/(c+d)/f/(c^2-d^2)^(1/2)

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Rubi [A]
time = 0.64, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3054, 3055, 3047, 3102, 2814, 2739, 632, 210} \begin {gather*} \frac {2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \text {ArcTan}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{d^4 f (c+d) \sqrt {c^2-d^2}}-\frac {a^3 x \left (2 A d (2 c-3 d)-B \left (6 c^2-12 c d+7 d^2\right )\right )}{2 d^4}-\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 f (c+d)}+\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d^2 f (c+d)}+\frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^2,x]

[Out]

-1/2*(a^3*(2*A*(2*c - 3*d)*d - B*(6*c^2 - 12*c*d + 7*d^2))*x)/d^4 + (2*a^3*(c - d)^2*(A*d*(2*c + 3*d) - B*(3*c
^2 + 3*c*d - d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d^4*(c + d)*Sqrt[c^2 - d^2]*f) - (a^3*(4
*A*c*d - B*(6*c^2 - 3*c*d - 5*d^2))*Cos[e + f*x])/(2*d^3*(c + d)*f) + ((2*A*d - B*(3*c + d))*Cos[e + f*x]*(a^3
+ a^3*Sin[e + f*x]))/(2*d^2*(c + d)*f) + (a*(B*c - A*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(d*(c + d)*f*(c
+ d*Sin[e + f*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 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 3047

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

Rule 3054

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d
*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*
n] || EqQ[c, 0])

Rule 3055

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*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] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx &=\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac {\int \frac {(a+a \sin (e+f x))^2 (-a (B (2 c-d)-3 A d)-a (2 A d-B (3 c+d)) \sin (e+f x))}{c+d \sin (e+f x)} \, dx}{d (c+d)}\\ &=\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac {\int \frac {(a+a \sin (e+f x)) \left (-a^2 \left (2 A (c-3 d) d-B \left (3 c^2-3 c d+2 d^2\right )\right )+a^2 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{2 d^2 (c+d)}\\ &=\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac {\int \frac {-a^3 \left (2 A (c-3 d) d-B \left (3 c^2-3 c d+2 d^2\right )\right )+\left (a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right )-a^3 \left (2 A (c-3 d) d-B \left (3 c^2-3 c d+2 d^2\right )\right )\right ) \sin (e+f x)+a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \sin ^2(e+f x)}{c+d \sin (e+f x)} \, dx}{2 d^2 (c+d)}\\ &=-\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac {\int \frac {-a^3 d \left (2 A (c-3 d) d-B \left (3 c^2-3 c d+2 d^2\right )\right )-a^3 (c+d) \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) \sin (e+f x)}{c+d \sin (e+f x)} \, dx}{2 d^3 (c+d)}\\ &=-\frac {a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}-\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac {\left (a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right )\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{d^4 (c+d)}\\ &=-\frac {a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}-\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}+\frac {\left (2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\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 )}{d^4 (c+d) f}\\ &=-\frac {a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}-\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}-\frac {\left (4 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\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 )}{d^4 (c+d) f}\\ &=-\frac {a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}+\frac {2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^4 (c+d) \sqrt {c^2-d^2} f}-\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 1.03, size = 244, normalized size = 0.86 \begin {gather*} \frac {a^3 (1+\sin (e+f x))^3 \left (2 \left (2 A d (-2 c+3 d)+B \left (6 c^2-12 c d+7 d^2\right )\right ) (e+f x)-\frac {8 (c-d)^2 \left (-A d (2 c+3 d)+B \left (3 c^2+3 c d-d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2}}-4 d (-2 B c+A d+3 B d) \cos (e+f x)+\frac {4 (c-d)^2 d (B c-A d) \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}-B d^2 \sin (2 (e+f x))\right )}{4 d^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^2,x]

[Out]

(a^3*(1 + Sin[e + f*x])^3*(2*(2*A*d*(-2*c + 3*d) + B*(6*c^2 - 12*c*d + 7*d^2))*(e + f*x) - (8*(c - d)^2*(-(A*d
*(2*c + 3*d)) + B*(3*c^2 + 3*c*d - d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)*Sqrt[c^2 -
d^2]) - 4*d*(-2*B*c + A*d + 3*B*d)*Cos[e + f*x] + (4*(c - d)^2*d*(B*c - A*d)*Cos[e + f*x])/((c + d)*(c + d*Si
n[e + f*x])) - B*d^2*Sin[2*(e + f*x)]))/(4*d^4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6)

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Maple [A]
time = 0.60, size = 406, normalized size = 1.43

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {\frac {-\frac {d^{2} \left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {d \left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right )}{c +d}}{c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 A \,c^{3} d -A \,c^{2} d^{2}-4 A c \,d^{3}+3 A \,d^{4}-3 B \,c^{4}+3 B \,c^{3} d +4 B \,c^{2} d^{2}-5 B c \,d^{3}+B \,d^{4}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}}{d^{4}}-\frac {\frac {-\frac {B \,d^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\left (A \,d^{2}-2 B c d +3 B \,d^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A \,d^{2}-2 B c d +3 B \,d^{2}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {\left (4 A c d -6 A \,d^{2}-6 B \,c^{2}+12 B c d -7 B \,d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{4}}\right )}{f}\)	\(406\)
default	\(\frac {2 a^{3} \left (\frac {\frac {-\frac {d^{2} \left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {d \left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right )}{c +d}}{c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 A \,c^{3} d -A \,c^{2} d^{2}-4 A c \,d^{3}+3 A \,d^{4}-3 B \,c^{4}+3 B \,c^{3} d +4 B \,c^{2} d^{2}-5 B c \,d^{3}+B \,d^{4}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}}{d^{4}}-\frac {\frac {-\frac {B \,d^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\left (A \,d^{2}-2 B c d +3 B \,d^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A \,d^{2}-2 B c d +3 B \,d^{2}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {\left (4 A c d -6 A \,d^{2}-6 B \,c^{2}+12 B c d -7 B \,d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{4}}\right )}{f}\)	\(406\)
risch	\(\text {Expression too large to display}\)	\(1083\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/f*a^3*(1/d^4*((-d^2*(A*c^2*d-2*A*c*d^2+A*d^3-B*c^3+2*B*c^2*d-B*c*d^2)/(c+d)/c*tan(1/2*f*x+1/2*e)-d*(A*c^2*d-
2*A*c*d^2+A*d^3-B*c^3+2*B*c^2*d-B*c*d^2)/(c+d))/(c*tan(1/2*f*x+1/2*e)^2+2*d*tan(1/2*f*x+1/2*e)+c)+(2*A*c^3*d-A
*c^2*d^2-4*A*c*d^3+3*A*d^4-3*B*c^4+3*B*c^3*d+4*B*c^2*d^2-5*B*c*d^3+B*d^4)/(c+d)/(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)))-1/d^4*((-1/2*B*d^2*tan(1/2*f*x+1/2*e)^3+(A*d^2-2*B*c*d+3*B*d^2)*ta
n(1/2*f*x+1/2*e)^2+1/2*B*d^2*tan(1/2*f*x+1/2*e)+A*d^2-2*B*c*d+3*B*d^2)/(1+tan(1/2*f*x+1/2*e)^2)^2+1/2*(4*A*c*d
-6*A*d^2-6*B*c^2+12*B*c*d-7*B*d^2)*arctan(tan(1/2*f*x+1/2*e))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^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*d^2-4*c^2>0)', see `assume?`
for more de

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Fricas [A]
time = 0.46, size = 1048, normalized size = 3.70 \begin {gather*} \left [\frac {{\left (B a^{3} c d^{3} + B a^{3} d^{4}\right )} \cos \left (f x + e\right )^{3} + {\left (6 \, B a^{3} c^{4} - 2 \, {\left (2 \, A + 3 \, B\right )} a^{3} c^{3} d + {\left (2 \, A - 5 \, B\right )} a^{3} c^{2} d^{2} + {\left (6 \, A + 7 \, B\right )} a^{3} c d^{3}\right )} f x + {\left (3 \, B a^{3} c^{4} - 2 \, A a^{3} c^{3} d - {\left (A + 4 \, B\right )} a^{3} c^{2} d^{2} + {\left (3 \, A + B\right )} a^{3} c d^{3} + {\left (3 \, B a^{3} c^{3} d - 2 \, A a^{3} c^{2} d^{2} - {\left (A + 4 \, B\right )} a^{3} c d^{3} + {\left (3 \, A + B\right )} a^{3} d^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + {\left (6 \, B a^{3} c^{3} d - 2 \, {\left (2 \, A + 3 \, B\right )} a^{3} c^{2} d^{2} + {\left (2 \, A - 5 \, B\right )} a^{3} c d^{3} - {\left (2 \, A + B\right )} a^{3} d^{4}\right )} \cos \left (f x + e\right ) + {\left ({\left (6 \, B a^{3} c^{3} d - 2 \, {\left (2 \, A + 3 \, B\right )} a^{3} c^{2} d^{2} + {\left (2 \, A - 5 \, B\right )} a^{3} c d^{3} + {\left (6 \, A + 7 \, B\right )} a^{3} d^{4}\right )} f x + {\left (3 \, B a^{3} c^{2} d^{2} - {\left (2 \, A + 3 \, B\right )} a^{3} c d^{3} - 2 \, {\left (A + 3 \, B\right )} a^{3} d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c d^{5} + d^{6}\right )} f \sin \left (f x + e\right ) + {\left (c^{2} d^{4} + c d^{5}\right )} f\right )}}, \frac {{\left (B a^{3} c d^{3} + B a^{3} d^{4}\right )} \cos \left (f x + e\right )^{3} + {\left (6 \, B a^{3} c^{4} - 2 \, {\left (2 \, A + 3 \, B\right )} a^{3} c^{3} d + {\left (2 \, A - 5 \, B\right )} a^{3} c^{2} d^{2} + {\left (6 \, A + 7 \, B\right )} a^{3} c d^{3}\right )} f x + 2 \, {\left (3 \, B a^{3} c^{4} - 2 \, A a^{3} c^{3} d - {\left (A + 4 \, B\right )} a^{3} c^{2} d^{2} + {\left (3 \, A + B\right )} a^{3} c d^{3} + {\left (3 \, B a^{3} c^{3} d - 2 \, A a^{3} c^{2} d^{2} - {\left (A + 4 \, B\right )} a^{3} c d^{3} + {\left (3 \, A + B\right )} a^{3} d^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) + {\left (6 \, B a^{3} c^{3} d - 2 \, {\left (2 \, A + 3 \, B\right )} a^{3} c^{2} d^{2} + {\left (2 \, A - 5 \, B\right )} a^{3} c d^{3} - {\left (2 \, A + B\right )} a^{3} d^{4}\right )} \cos \left (f x + e\right ) + {\left ({\left (6 \, B a^{3} c^{3} d - 2 \, {\left (2 \, A + 3 \, B\right )} a^{3} c^{2} d^{2} + {\left (2 \, A - 5 \, B\right )} a^{3} c d^{3} + {\left (6 \, A + 7 \, B\right )} a^{3} d^{4}\right )} f x + {\left (3 \, B a^{3} c^{2} d^{2} - {\left (2 \, A + 3 \, B\right )} a^{3} c d^{3} - 2 \, {\left (A + 3 \, B\right )} a^{3} d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c d^{5} + d^{6}\right )} f \sin \left (f x + e\right ) + {\left (c^{2} d^{4} + c d^{5}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/2*((B*a^3*c*d^3 + B*a^3*d^4)*cos(f*x + e)^3 + (6*B*a^3*c^4 - 2*(2*A + 3*B)*a^3*c^3*d + (2*A - 5*B)*a^3*c^2*
d^2 + (6*A + 7*B)*a^3*c*d^3)*f*x + (3*B*a^3*c^4 - 2*A*a^3*c^3*d - (A + 4*B)*a^3*c^2*d^2 + (3*A + B)*a^3*c*d^3
+ (3*B*a^3*c^3*d - 2*A*a^3*c^2*d^2 - (A + 4*B)*a^3*c*d^3 + (3*A + B)*a^3*d^4)*sin(f*x + e))*sqrt(-(c - d)/(c +
d))*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*((c^2 + c*d)*cos(f*x + e)*sin(f*x
+ e) + (c*d + d^2)*cos(f*x + e))*sqrt(-(c - d)/(c + d)))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)
) + (6*B*a^3*c^3*d - 2*(2*A + 3*B)*a^3*c^2*d^2 + (2*A - 5*B)*a^3*c*d^3 - (2*A + B)*a^3*d^4)*cos(f*x + e) + ((6
*B*a^3*c^3*d - 2*(2*A + 3*B)*a^3*c^2*d^2 + (2*A - 5*B)*a^3*c*d^3 + (6*A + 7*B)*a^3*d^4)*f*x + (3*B*a^3*c^2*d^2
- (2*A + 3*B)*a^3*c*d^3 - 2*(A + 3*B)*a^3*d^4)*cos(f*x + e))*sin(f*x + e))/((c*d^5 + d^6)*f*sin(f*x + e) + (c
^2*d^4 + c*d^5)*f), 1/2*((B*a^3*c*d^3 + B*a^3*d^4)*cos(f*x + e)^3 + (6*B*a^3*c^4 - 2*(2*A + 3*B)*a^3*c^3*d + (
2*A - 5*B)*a^3*c^2*d^2 + (6*A + 7*B)*a^3*c*d^3)*f*x + 2*(3*B*a^3*c^4 - 2*A*a^3*c^3*d - (A + 4*B)*a^3*c^2*d^2 +
(3*A + B)*a^3*c*d^3 + (3*B*a^3*c^3*d - 2*A*a^3*c^2*d^2 - (A + 4*B)*a^3*c*d^3 + (3*A + B)*a^3*d^4)*sin(f*x + e
))*sqrt((c - d)/(c + d))*arctan(-(c*sin(f*x + e) + d)*sqrt((c - d)/(c + d))/((c - d)*cos(f*x + e))) + (6*B*a^3
*c^3*d - 2*(2*A + 3*B)*a^3*c^2*d^2 + (2*A - 5*B)*a^3*c*d^3 - (2*A + B)*a^3*d^4)*cos(f*x + e) + ((6*B*a^3*c^3*d
- 2*(2*A + 3*B)*a^3*c^2*d^2 + (2*A - 5*B)*a^3*c*d^3 + (6*A + 7*B)*a^3*d^4)*f*x + (3*B*a^3*c^2*d^2 - (2*A + 3*
B)*a^3*c*d^3 - 2*(A + 3*B)*a^3*d^4)*cos(f*x + e))*sin(f*x + e))/((c*d^5 + d^6)*f*sin(f*x + e) + (c^2*d^4 + c*d
^5)*f)]

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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((a+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 588 vs. \(2 (278) = 556\).
time = 0.53, size = 588, normalized size = 2.08 \begin {gather*} -\frac {\frac {4 \, {\left (3 \, B a^{3} c^{4} - 2 \, A a^{3} c^{3} d - 3 \, B a^{3} c^{3} d + A a^{3} c^{2} d^{2} - 4 \, B a^{3} c^{2} d^{2} + 4 \, A a^{3} c d^{3} + 5 \, B a^{3} c d^{3} - 3 \, A a^{3} d^{4} - B a^{3} d^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c d^{4} + d^{5}\right )} \sqrt {c^{2} - d^{2}}} - \frac {4 \, {\left (B a^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, B a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B a^{3} c^{4} - A a^{3} c^{3} d - 2 \, B a^{3} c^{3} d + 2 \, A a^{3} c^{2} d^{2} + B a^{3} c^{2} d^{2} - A a^{3} c d^{3}\right )}}{{\left (c^{2} d^{3} + c d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}} - \frac {{\left (6 \, B a^{3} c^{2} - 4 \, A a^{3} c d - 12 \, B a^{3} c d + 6 \, A a^{3} d^{2} + 7 \, B a^{3} d^{2}\right )} {\left (f x + e\right )}}{d^{4}} - \frac {2 \, {\left (B a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 4 \, B a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, A a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, B a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, B a^{3} c - 2 \, A a^{3} d - 6 \, B a^{3} d\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} d^{3}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/2*(4*(3*B*a^3*c^4 - 2*A*a^3*c^3*d - 3*B*a^3*c^3*d + A*a^3*c^2*d^2 - 4*B*a^3*c^2*d^2 + 4*A*a^3*c*d^3 + 5*B*a
^3*c*d^3 - 3*A*a^3*d^4 - B*a^3*d^4)*(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)))/((c*d^4 + d^5)*sqrt(c^2 - d^2)) - 4*(B*a^3*c^3*d*tan(1/2*f*x + 1/2*e) - A*a^3*c^2*d^2*t
an(1/2*f*x + 1/2*e) - 2*B*a^3*c^2*d^2*tan(1/2*f*x + 1/2*e) + 2*A*a^3*c*d^3*tan(1/2*f*x + 1/2*e) + B*a^3*c*d^3*
tan(1/2*f*x + 1/2*e) - A*a^3*d^4*tan(1/2*f*x + 1/2*e) + B*a^3*c^4 - A*a^3*c^3*d - 2*B*a^3*c^3*d + 2*A*a^3*c^2*
d^2 + B*a^3*c^2*d^2 - A*a^3*c*d^3)/((c^2*d^3 + c*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c
)) - (6*B*a^3*c^2 - 4*A*a^3*c*d - 12*B*a^3*c*d + 6*A*a^3*d^2 + 7*B*a^3*d^2)*(f*x + e)/d^4 - 2*(B*a^3*d*tan(1/2
*f*x + 1/2*e)^3 + 4*B*a^3*c*tan(1/2*f*x + 1/2*e)^2 - 2*A*a^3*d*tan(1/2*f*x + 1/2*e)^2 - 6*B*a^3*d*tan(1/2*f*x
+ 1/2*e)^2 - B*a^3*d*tan(1/2*f*x + 1/2*e) + 4*B*a^3*c - 2*A*a^3*d - 6*B*a^3*d)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2
*d^3))/f

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Mupad [B]
time = 23.88, size = 2500, normalized size = 8.83 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c + d*sin(e + f*x))^2,x)

[Out]

- ((2*(A*a^3*d^3 - 3*B*a^3*c^3 - A*a^3*c*d^2 + 2*A*a^3*c^2*d + 2*B*a^3*c*d^2 + 3*B*a^3*c^2*d))/(d^3*(c + d)) +
(2*tan(e/2 + (f*x)/2)^4*(A*a^3*d^3 - 3*B*a^3*c^3 - B*a^3*d^3 - A*a^3*c*d^2 + 2*A*a^3*c^2*d + B*a^3*c*d^2 + 3*
B*a^3*c^2*d))/(d^3*(c + d)) + (2*tan(e/2 + (f*x)/2)^2*(2*A*a^3*d^3 - 6*B*a^3*c^3 + B*a^3*d^3 - 2*A*a^3*c*d^2 +
4*A*a^3*c^2*d + 5*B*a^3*c*d^2 + 6*B*a^3*c^2*d))/(d^3*(c + d)) + (4*tan(e/2 + (f*x)/2)^3*(A*a^3*d^3 - 3*B*a^3*
c^3 - A*a^3*c*d^2 + 2*A*a^3*c^2*d + 2*B*a^3*c*d^2 + 3*B*a^3*c^2*d))/(c*d^2*(c + d)) + (tan(e/2 + (f*x)/2)^5*(2
*A*a^3*d^3 - 3*B*a^3*c^3 - 4*A*a^3*c*d^2 + 2*A*a^3*c^2*d - 2*B*a^3*c*d^2 + 3*B*a^3*c^2*d))/(c*d^2*(c + d)) + (
tan(e/2 + (f*x)/2)*(2*A*a^3*d^3 - 9*B*a^3*c^3 + 6*A*a^3*c^2*d + 10*B*a^3*c*d^2 + 9*B*a^3*c^2*d))/(c*d^2*(c + d
)))/(f*(c + 2*d*tan(e/2 + (f*x)/2) + 3*c*tan(e/2 + (f*x)/2)^2 + 3*c*tan(e/2 + (f*x)/2)^4 + c*tan(e/2 + (f*x)/2
)^6 + 4*d*tan(e/2 + (f*x)/2)^3 + 2*d*tan(e/2 + (f*x)/2)^5)) - (atan(((((8*(36*A^2*a^6*c^2*d^9 + 24*A^2*a^6*c^3
*d^8 - 44*A^2*a^6*c^4*d^7 - 16*A^2*a^6*c^5*d^6 + 16*A^2*a^6*c^6*d^5 + 49*B^2*a^6*c^2*d^9 - 70*B^2*a^6*c^3*d^8
- 59*B^2*a^6*c^4*d^7 + 144*B^2*a^6*c^5*d^6 - 24*B^2*a^6*c^6*d^5 - 72*B^2*a^6*c^7*d^4 + 36*B^2*a^6*c^8*d^3 + 84
*A*B*a^6*c^2*d^9 - 32*A*B*a^6*c^3*d^8 - 148*A*B*a^6*c^4*d^7 + 88*A*B*a^6*c^5*d^6 + 72*A*B*a^6*c^6*d^5 - 48*A*B
*a^6*c^7*d^4))/(2*c*d^9 + d^10 + c^2*d^8) + (8*tan(e/2 + (f*x)/2)*(144*A^2*a^6*c^2*d^10 - 164*A^2*a^6*c^3*d^9
- 136*A^2*a^6*c^4*d^8 + 136*A^2*a^6*c^5*d^7 + 32*A^2*a^6*c^6*d^6 - 32*A^2*a^6*c^7*d^5 - 100*B^2*a^6*c^2*d^10 -
299*B^2*a^6*c^3*d^9 + 494*B^2*a^6*c^4*d^8 + 91*B^2*a^6*c^5*d^7 - 504*B^2*a^6*c^6*d^6 + 156*B^2*a^6*c^7*d^5 +
144*B^2*a^6*c^8*d^4 - 72*B^2*a^6*c^9*d^3 + 36*A^2*a^6*c*d^11 + 94*B^2*a^6*c*d^11 + 88*A*B*a^6*c^2*d^10 - 628*A
*B*a^6*c^3*d^9 + 208*A*B*a^6*c^4*d^8 + 572*A*B*a^6*c^5*d^7 - 320*A*B*a^6*c^6*d^6 - 144*A*B*a^6*c^7*d^5 + 96*A*
B*a^6*c^8*d^4 + 144*A*B*a^6*c*d^11))/(2*c*d^10 + d^11 + c^2*d^9) + (((((8*(4*c^2*d^13 + 8*c^3*d^12 + 4*c^4*d^1
1))/(2*c*d^9 + d^10 + c^2*d^8) + (8*tan(e/2 + (f*x)/2)*(12*c*d^15 + 24*c^2*d^14 + 4*c^3*d^13 - 16*c^4*d^12 - 8
*c^5*d^11))/(2*c*d^10 + d^11 + c^2*d^9))*(B*a^3*c^2*3i + a^3*d^2*(3*A + (7*B)/2)*1i - (a^3*d*(4*A*c + 12*B*c)*
1i)/2))/d^4 - (8*(12*A*a^3*c*d^12 + 14*B*a^3*c*d^12 + 4*A*a^3*c^2*d^11 - 12*A*a^3*c^3*d^10 - 4*A*a^3*c^4*d^9 -
20*B*a^3*c^3*d^10 + 6*B*a^3*c^5*d^8))/(2*c*d^9 + d^10 + c^2*d^8) + (8*tan(e/2 + (f*x)/2)*(24*A*a^3*c*d^13 + 8
*B*a^3*c*d^13 - 8*A*a^3*c^2*d^12 - 40*A*a^3*c^3*d^11 + 8*A*a^3*c^4*d^10 + 16*A*a^3*c^5*d^9 - 32*B*a^3*c^2*d^12
- 8*B*a^3*c^3*d^11 + 56*B*a^3*c^4*d^10 - 24*B*a^3*c^6*d^8))/(2*c*d^10 + d^11 + c^2*d^9))*(B*a^3*c^2*3i + a^3*
d^2*(3*A + (7*B)/2)*1i - (a^3*d*(4*A*c + 12*B*c)*1i)/2))/d^4)*(B*a^3*c^2*3i + a^3*d^2*(3*A + (7*B)/2)*1i - (a^
3*d*(4*A*c + 12*B*c)*1i)/2)*1i)/d^4 + (((8*(36*A^2*a^6*c^2*d^9 + 24*A^2*a^6*c^3*d^8 - 44*A^2*a^6*c^4*d^7 - 16*
A^2*a^6*c^5*d^6 + 16*A^2*a^6*c^6*d^5 + 49*B^2*a^6*c^2*d^9 - 70*B^2*a^6*c^3*d^8 - 59*B^2*a^6*c^4*d^7 + 144*B^2*
a^6*c^5*d^6 - 24*B^2*a^6*c^6*d^5 - 72*B^2*a^6*c^7*d^4 + 36*B^2*a^6*c^8*d^3 + 84*A*B*a^6*c^2*d^9 - 32*A*B*a^6*c
^3*d^8 - 148*A*B*a^6*c^4*d^7 + 88*A*B*a^6*c^5*d^6 + 72*A*B*a^6*c^6*d^5 - 48*A*B*a^6*c^7*d^4))/(2*c*d^9 + d^10
+ c^2*d^8) + (8*tan(e/2 + (f*x)/2)*(144*A^2*a^6*c^2*d^10 - 164*A^2*a^6*c^3*d^9 - 136*A^2*a^6*c^4*d^8 + 136*A^2
*a^6*c^5*d^7 + 32*A^2*a^6*c^6*d^6 - 32*A^2*a^6*c^7*d^5 - 100*B^2*a^6*c^2*d^10 - 299*B^2*a^6*c^3*d^9 + 494*B^2*
a^6*c^4*d^8 + 91*B^2*a^6*c^5*d^7 - 504*B^2*a^6*c^6*d^6 + 156*B^2*a^6*c^7*d^5 + 144*B^2*a^6*c^8*d^4 - 72*B^2*a^
6*c^9*d^3 + 36*A^2*a^6*c*d^11 + 94*B^2*a^6*c*d^11 + 88*A*B*a^6*c^2*d^10 - 628*A*B*a^6*c^3*d^9 + 208*A*B*a^6*c^
4*d^8 + 572*A*B*a^6*c^5*d^7 - 320*A*B*a^6*c^6*d^6 - 144*A*B*a^6*c^7*d^5 + 96*A*B*a^6*c^8*d^4 + 144*A*B*a^6*c*d
^11))/(2*c*d^10 + d^11 + c^2*d^9) + (((8*(12*A*a^3*c*d^12 + 14*B*a^3*c*d^12 + 4*A*a^3*c^2*d^11 - 12*A*a^3*c^3*
d^10 - 4*A*a^3*c^4*d^9 - 20*B*a^3*c^3*d^10 + 6*B*a^3*c^5*d^8))/(2*c*d^9 + d^10 + c^2*d^8) + (((8*(4*c^2*d^13 +
8*c^3*d^12 + 4*c^4*d^11))/(2*c*d^9 + d^10 + c^2*d^8) + (8*tan(e/2 + (f*x)/2)*(12*c*d^15 + 24*c^2*d^14 + 4*c^3
*d^13 - 16*c^4*d^12 - 8*c^5*d^11))/(2*c*d^10 + d^11 + c^2*d^9))*(B*a^3*c^2*3i + a^3*d^2*(3*A + (7*B)/2)*1i - (
a^3*d*(4*A*c + 12*B*c)*1i)/2))/d^4 - (8*tan(e/2 + (f*x)/2)*(24*A*a^3*c*d^13 + 8*B*a^3*c*d^13 - 8*A*a^3*c^2*d^1
2 - 40*A*a^3*c^3*d^11 + 8*A*a^3*c^4*d^10 + 16*A*a^3*c^5*d^9 - 32*B*a^3*c^2*d^12 - 8*B*a^3*c^3*d^11 + 56*B*a^3*
c^4*d^10 - 24*B*a^3*c^6*d^8))/(2*c*d^10 + d^11 + c^2*d^9))*(B*a^3*c^2*3i + a^3*d^2*(3*A + (7*B)/2)*1i - (a^3*d
*(4*A*c + 12*B*c)*1i)/2))/d^4)*(B*a^3*c^2*3i + a^3*d^2*(3*A + (7*B)/2)*1i - (a^3*d*(4*A*c + 12*B*c)*1i)/2)*1i)
/d^4)/((16*(132*A^3*a^9*c^3*d^6 - 252*A^3*a^9*c^2*d^7 - 54*B^3*a^9*c^9 + 76*A^3*a^9*c^4*d^5 - 80*A^3*a^9*c^5*d
^4 + 16*A^3*a^9*c^6*d^3 - 115*B^3*a^9*c^2*d^7 + 350*B^3*a^9*c^3*d^6 - 537*B^3*a^9*c^4*d^5 + 387*B^3*a^9*c^5*d^
4 + 36*B^3*a^9*c^6*d^3 - 297*B^3*a^9*c^7*d^2 + 108*A^3*a^9*c*d^8 + 14*B^3*a^9*c*d^8 + 216*B^3*a^9*c^8*d + 96*A
*B^2*a^9*c*d^8 + 108*A*B^2*a^9*c^8*d + 198*A^2*...

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