3.3.71 \(\int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [271]
Optimal. Leaf size=299 \[ \frac {b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right ) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac {(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]
[Out]
b*(12*A*a^4*b-15*A*a^2*b^3+6*A*b^5-6*B*a^5+5*B*a^3*b^2-2*B*a*b^4)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^
(1/2))/a^4/(a-b)^(5/2)/(a+b)^(5/2)/d-(3*A*b-B*a)*arctanh(sin(d*x+c))/a^4/d+1/2*(2*A*a^4-11*A*a^2*b^2+6*A*b^4+5
*B*a^3*b-2*B*a*b^3)*tan(d*x+c)/a^3/(a^2-b^2)^2/d+1/2*b*(A*b-B*a)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^2+1
/2*b*(6*A*a^2*b-3*A*b^3-4*B*a^3+B*a*b^2)*tan(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
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Rubi [A]
time = 1.14, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3079, 3134,
3080, 3855, 2738, 211} \begin {gather*} -\frac {(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {b (A b-a B) \tan (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {b \left (-4 a^3 B+6 a^2 A b+a b^2 B-3 A b^3\right ) \tan (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {\left (2 a^4 A+5 a^3 b B-11 a^2 A b^2-2 a b^3 B+6 A b^4\right ) \tan (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac {b \left (-6 a^5 B+12 a^4 A b+5 a^3 b^2 B-15 a^2 A b^3-2 a b^4 B+6 A b^5\right ) \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
[Out]
(b*(12*a^4*A*b - 15*a^2*A*b^3 + 6*A*b^5 - 6*a^5*B + 5*a^3*b^2*B - 2*a*b^4*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)
/2])/Sqrt[a + b]])/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d) - ((3*A*b - a*B)*ArcTanh[Sin[c + d*x]])/(a^4*d) + ((2*a
^4*A - 11*a^2*A*b^2 + 6*A*b^4 + 5*a^3*b*B - 2*a*b^3*B)*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + (b*(A*b - a*B)*
Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (b*(6*a^2*A*b - 3*A*b^3 - 4*a^3*B + a*b^2*B)*Tan[c
+ d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))
Rule 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 2738
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*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 3079
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c +
d*Sin[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*
(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n
, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 3080
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3134
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&& !IntegerQ[m]) || EqQ[a, 0])))
Rule 3855
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\int \frac {\left (2 a^2 A-3 A b^2+a b B-2 a (A b-a B) \cos (c+d x)+2 b (A b-a B) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\int \frac {\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B-a \left (4 a^2 A b-A b^3-2 a^3 B-a b^2 B\right ) \cos (c+d x)+b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\int \frac {\left (-2 \left (a^2-b^2\right )^2 (3 A b-a B)+a b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {(3 A b-a B) \int \sec (c+d x) \, dx}{a^4}+\frac {\left (b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac {(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right )^2 d}\\ &=\frac {b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{5/2} (a+b)^{5/2} d}-\frac {(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac {\left (2 a^4 A-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (A b-a B) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {b \left (6 a^2 A b-3 A b^3-4 a^3 B+a b^2 B\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 6.06, size = 352, normalized size = 1.18 \begin {gather*} \frac {-\frac {2 b \left (12 a^4 A b-15 a^2 A b^3+6 A b^5-6 a^5 B+5 a^3 b^2 B-2 a b^4 B\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+2 (3 A b-a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (-3 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 a A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 a A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {a^2 b^2 (-A b+a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac {a b^2 \left (-7 a^2 A b+4 A b^3+5 a^3 B-2 a b^2 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}}{2 a^4 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
[Out]
((-2*b*(12*a^4*A*b - 15*a^2*A*b^3 + 6*A*b^5 - 6*a^5*B + 5*a^3*b^2*B - 2*a*b^4*B)*ArcTanh[((a - b)*Tan[(c + d*x
)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 2*(3*A*b - a*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(-3
*A*b + a*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*a*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d
*x)/2]) + (2*a*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (a^2*b^2*(-(A*b) + a*B)*Sin[c + d*x
])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (a*b^2*(-7*a^2*A*b + 4*A*b^3 + 5*a^3*B - 2*a*b^2*B)*Sin[c + d*x]
)/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*a^4*d)
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Maple [A]
time = 1.19, size = 376, normalized size = 1.26
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method |
result |
size |
| | |
derivativedivides |
\(\frac {-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-3 A b +a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}+\frac {2 b \left (\frac {-\frac {\left (8 A \,a^{2} b +A a \,b^{2}-4 A \,b^{3}-6 a^{3} B -a^{2} b B +2 B a \,b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (8 A \,a^{2} b -A a \,b^{2}-4 A \,b^{3}-6 a^{3} B +a^{2} b B +2 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )^{2}}+\frac {\left (12 A \,a^{4} b -15 A \,a^{2} b^{3}+6 A \,b^{5}-6 B \,a^{5}+5 B \,a^{3} b^{2}-2 B a \,b^{4}\right ) \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (3 A b -a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}}{d}\) |
\(376\) |
default |
\(\frac {-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-3 A b +a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4}}+\frac {2 b \left (\frac {-\frac {\left (8 A \,a^{2} b +A a \,b^{2}-4 A \,b^{3}-6 a^{3} B -a^{2} b B +2 B a \,b^{2}\right ) a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (8 A \,a^{2} b -A a \,b^{2}-4 A \,b^{3}-6 a^{3} B +a^{2} b B +2 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )^{2}}+\frac {\left (12 A \,a^{4} b -15 A \,a^{2} b^{3}+6 A \,b^{5}-6 B \,a^{5}+5 B \,a^{3} b^{2}-2 B a \,b^{4}\right ) \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4}}-\frac {A}{a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (3 A b -a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4}}}{d}\) |
\(376\) |
risch |
\(\text {Expression too large to display}\) |
\(1691\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x,method=_RETURNVERBOSE)
[Out]
1/d*(-A/a^3/(tan(1/2*d*x+1/2*c)+1)+1/a^4*(-3*A*b+B*a)*ln(tan(1/2*d*x+1/2*c)+1)+2*b/a^4*((-1/2*(8*A*a^2*b+A*a*b
^2-4*A*b^3-6*B*a^3-B*a^2*b+2*B*a*b^2)*a*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*b*a*(8*A*a^2*b-A*a*b^
2-4*A*b^3-6*B*a^3+B*a^2*b+2*B*a*b^2)/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c))/(a*tan(1/2*d*x+1/2*c)^2-b*tan(1/2*d*x+1
/2*c)^2+a+b)^2+1/2*(12*A*a^4*b-15*A*a^2*b^3+6*A*b^5-6*B*a^5+5*B*a^3*b^2-2*B*a*b^4)/(a^4-2*a^2*b^2+b^4)/((a-b)*
(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2)))-A/a^3/(tan(1/2*d*x+1/2*c)-1)+(3*A*b-B*a)/a^
4*ln(tan(1/2*d*x+1/2*c)-1))
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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+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^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*b^2-4*a^2>0)', see `assume?`
for more de
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1015 vs.
\(2 (284) = 568\).
time = 26.82, size = 2100, normalized size = 7.02 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
[1/4*(((6*B*a^5*b^3 - 12*A*a^4*b^4 - 5*B*a^3*b^5 + 15*A*a^2*b^6 + 2*B*a*b^7 - 6*A*b^8)*cos(d*x + c)^3 + 2*(6*B
*a^6*b^2 - 12*A*a^5*b^3 - 5*B*a^4*b^4 + 15*A*a^3*b^5 + 2*B*a^2*b^6 - 6*A*a*b^7)*cos(d*x + c)^2 + (6*B*a^7*b -
12*A*a^6*b^2 - 5*B*a^5*b^3 + 15*A*a^4*b^4 + 2*B*a^3*b^5 - 6*A*a^2*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a
*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 +
2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 2*((B*a^7*b^2 - 3*A*a^6*b^3 - 3*B*a^5*b^4 + 9*A*a^4*
b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7 - B*a*b^8 + 3*A*b^9)*cos(d*x + c)^3 + 2*(B*a^8*b - 3*A*a^7*b^2 - 3*B*a^6*b^3 +
9*A*a^5*b^4 + 3*B*a^4*b^5 - 9*A*a^3*b^6 - B*a^2*b^7 + 3*A*a*b^8)*cos(d*x + c)^2 + (B*a^9 - 3*A*a^8*b - 3*B*a^
7*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 - B*a^3*b^6 + 3*A*a^2*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1)
- 2*((B*a^7*b^2 - 3*A*a^6*b^3 - 3*B*a^5*b^4 + 9*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7 - B*a*b^8 + 3*A*b^9)*co
s(d*x + c)^3 + 2*(B*a^8*b - 3*A*a^7*b^2 - 3*B*a^6*b^3 + 9*A*a^5*b^4 + 3*B*a^4*b^5 - 9*A*a^3*b^6 - B*a^2*b^7 +
3*A*a*b^8)*cos(d*x + c)^2 + (B*a^9 - 3*A*a^8*b - 3*B*a^7*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 - B*a^3
*b^6 + 3*A*a^2*b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(2*A*a^9 - 6*A*a^7*b^2 + 6*A*a^5*b^4 - 2*A*a^3*b^
6 + (2*A*a^7*b^2 + 5*B*a^6*b^3 - 13*A*a^5*b^4 - 7*B*a^4*b^5 + 17*A*a^3*b^6 + 2*B*a^2*b^7 - 6*A*a*b^8)*cos(d*x
+ c)^2 + (4*A*a^8*b + 6*B*a^7*b^2 - 20*A*a^6*b^3 - 9*B*a^5*b^4 + 25*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7)*cos
(d*x + c))*sin(d*x + c))/((a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d*cos(d*x + c)^3 + 2*(a^11*b - 3*a^9*b^
3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c)^2 + (a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)), -1/2*((
(6*B*a^5*b^3 - 12*A*a^4*b^4 - 5*B*a^3*b^5 + 15*A*a^2*b^6 + 2*B*a*b^7 - 6*A*b^8)*cos(d*x + c)^3 + 2*(6*B*a^6*b^
2 - 12*A*a^5*b^3 - 5*B*a^4*b^4 + 15*A*a^3*b^5 + 2*B*a^2*b^6 - 6*A*a*b^7)*cos(d*x + c)^2 + (6*B*a^7*b - 12*A*a^
6*b^2 - 5*B*a^5*b^3 + 15*A*a^4*b^4 + 2*B*a^3*b^5 - 6*A*a^2*b^6)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d
*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - ((B*a^7*b^2 - 3*A*a^6*b^3 - 3*B*a^5*b^4 + 9*A*a^4*b^5 + 3*B*a^3
*b^6 - 9*A*a^2*b^7 - B*a*b^8 + 3*A*b^9)*cos(d*x + c)^3 + 2*(B*a^8*b - 3*A*a^7*b^2 - 3*B*a^6*b^3 + 9*A*a^5*b^4
+ 3*B*a^4*b^5 - 9*A*a^3*b^6 - B*a^2*b^7 + 3*A*a*b^8)*cos(d*x + c)^2 + (B*a^9 - 3*A*a^8*b - 3*B*a^7*b^2 + 9*A*a
^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 - B*a^3*b^6 + 3*A*a^2*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((B*a^7*b^
2 - 3*A*a^6*b^3 - 3*B*a^5*b^4 + 9*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7 - B*a*b^8 + 3*A*b^9)*cos(d*x + c)^3 +
2*(B*a^8*b - 3*A*a^7*b^2 - 3*B*a^6*b^3 + 9*A*a^5*b^4 + 3*B*a^4*b^5 - 9*A*a^3*b^6 - B*a^2*b^7 + 3*A*a*b^8)*cos(
d*x + c)^2 + (B*a^9 - 3*A*a^8*b - 3*B*a^7*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 - B*a^3*b^6 + 3*A*a^2*
b^7)*cos(d*x + c))*log(-sin(d*x + c) + 1) - (2*A*a^9 - 6*A*a^7*b^2 + 6*A*a^5*b^4 - 2*A*a^3*b^6 + (2*A*a^7*b^2
+ 5*B*a^6*b^3 - 13*A*a^5*b^4 - 7*B*a^4*b^5 + 17*A*a^3*b^6 + 2*B*a^2*b^7 - 6*A*a*b^8)*cos(d*x + c)^2 + (4*A*a^8
*b + 6*B*a^7*b^2 - 20*A*a^6*b^3 - 9*B*a^5*b^4 + 25*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7)*cos(d*x + c))*sin(d*
x + c))/((a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d*cos(d*x + c)^3 + 2*(a^11*b - 3*a^9*b^3 + 3*a^7*b^5 - a
^5*b^7)*d*cos(d*x + c)^2 + (a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c))]
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)**2/(a+b*cos(d*x+c))**3,x)
[Out]
Integral((A + B*cos(c + d*x))*sec(c + d*x)**2/(a + b*cos(c + d*x))**3, x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 574 vs.
\(2 (284) = 568\).
time = 0.52, size = 574, normalized size = 1.92 \begin {gather*} \frac {\frac {{\left (6 \, B a^{5} b - 12 \, A a^{4} b^{2} - 5 \, B a^{3} b^{3} + 15 \, A a^{2} b^{4} + 2 \, B a b^{5} - 6 \, A b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, B a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, A a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, B a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, A a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, A a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, B a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, A a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, A a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} + \frac {{\left (B a - 3 \, A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {{\left (B a - 3 \, A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}}}{d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="giac")
[Out]
((6*B*a^5*b - 12*A*a^4*b^2 - 5*B*a^3*b^3 + 15*A*a^2*b^4 + 2*B*a*b^5 - 6*A*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/
2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^8 - 2*a^6
*b^2 + a^4*b^4)*sqrt(a^2 - b^2)) + (6*B*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 -
5*B*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 7*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 3*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 +
5*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*B*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*A*b^6*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^
4*b^2*tan(1/2*d*x + 1/2*c) - 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c) + 5*B*a^3*b^3*tan(1/2*d*x + 1/2*c) - 7*A*a^2*b^4
*tan(1/2*d*x + 1/2*c) - 3*B*a^2*b^4*tan(1/2*d*x + 1/2*c) + 5*A*a*b^5*tan(1/2*d*x + 1/2*c) - 2*B*a*b^5*tan(1/2*
d*x + 1/2*c) + 4*A*b^6*tan(1/2*d*x + 1/2*c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/
2*d*x + 1/2*c)^2 + a + b)^2) + (B*a - 3*A*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - (B*a - 3*A*b)*log(abs(ta
n(1/2*d*x + 1/2*c) - 1))/a^4 - 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3))/d
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Mupad [B]
time = 12.91, size = 2500, normalized size = 8.36 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B*cos(c + d*x))/(cos(c + d*x)^2*(a + b*cos(c + d*x))^3),x)
[Out]
((tan(c/2 + (d*x)/2)^5*(6*A*b^5 - 2*A*a^5 - 12*A*a^2*b^3 + 4*A*a^3*b^2 + B*a^2*b^3 + 6*B*a^3*b^2 - 3*A*a*b^4 +
2*A*a^4*b - 2*B*a*b^4))/((a^3*b - a^4)*(a + b)^2) + (tan(c/2 + (d*x)/2)*(2*A*a^5 + 6*A*b^5 - 12*A*a^2*b^3 - 4
*A*a^3*b^2 - B*a^2*b^3 + 6*B*a^3*b^2 + 3*A*a*b^4 + 2*A*a^4*b - 2*B*a*b^4))/((a + b)*(a^5 - 2*a^4*b + a^3*b^2))
- (2*tan(c/2 + (d*x)/2)^3*(2*A*a^6 - 6*A*b^6 + 13*A*a^2*b^4 - 6*A*a^4*b^2 - 5*B*a^3*b^3 + 2*B*a*b^5))/(a*(a^2
*b - a^3)*(a + b)^2*(a - b)))/(d*(2*a*b - tan(c/2 + (d*x)/2)^2*(2*a*b - a^2 + 3*b^2) - tan(c/2 + (d*x)/2)^6*(a
^2 - 2*a*b + b^2) + a^2 + b^2 - tan(c/2 + (d*x)/2)^4*(2*a*b + a^2 - 3*b^2))) + (atan(((((8*tan(c/2 + (d*x)/2)*
(72*A^2*b^12 + 4*B^2*a^12 - 72*A^2*a*b^11 - 8*B^2*a^11*b - 288*A^2*a^2*b^10 + 288*A^2*a^3*b^9 + 441*A^2*a^4*b^
8 - 432*A^2*a^5*b^7 - 288*A^2*a^6*b^6 + 288*A^2*a^7*b^5 + 36*A^2*a^8*b^4 - 72*A^2*a^9*b^3 + 36*A^2*a^10*b^2 +
8*B^2*a^2*b^10 - 8*B^2*a^3*b^9 - 32*B^2*a^4*b^8 + 32*B^2*a^5*b^7 + 57*B^2*a^6*b^6 - 48*B^2*a^7*b^5 - 52*B^2*a^
8*b^4 + 32*B^2*a^9*b^3 + 24*B^2*a^10*b^2 - 48*A*B*a*b^11 - 24*A*B*a^11*b + 48*A*B*a^2*b^10 + 192*A*B*a^3*b^9 -
192*A*B*a^4*b^8 - 318*A*B*a^5*b^7 + 288*A*B*a^6*b^6 + 252*A*B*a^7*b^5 - 192*A*B*a^8*b^4 - 72*A*B*a^9*b^3 + 48
*A*B*a^10*b^2))/(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) + (((8*(
4*B*a^18 + 12*A*a^8*b^10 - 6*A*a^9*b^9 - 54*A*a^10*b^8 + 24*A*a^11*b^7 + 96*A*a^12*b^6 - 42*A*a^13*b^5 - 78*A*
a^14*b^4 + 36*A*a^15*b^3 + 24*A*a^16*b^2 - 4*B*a^9*b^9 + 2*B*a^10*b^8 + 18*B*a^11*b^7 - 4*B*a^12*b^6 - 36*B*a^
13*b^5 + 6*B*a^14*b^4 + 34*B*a^15*b^3 - 8*B*a^16*b^2 - 12*A*a^17*b - 12*B*a^17*b))/(a^15*b + a^16 - a^9*b^7 -
a^10*b^6 + 3*a^11*b^5 + 3*a^12*b^4 - 3*a^13*b^3 - 3*a^14*b^2) + (8*tan(c/2 + (d*x)/2)*(3*A*b - B*a)*(8*a^17*b
- 8*a^8*b^10 + 8*a^9*b^9 + 32*a^10*b^8 - 32*a^11*b^7 - 48*a^12*b^6 + 48*a^13*b^5 + 32*a^14*b^4 - 32*a^15*b^3 -
8*a^16*b^2))/(a^4*(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2)))*(3*
A*b - B*a))/a^4)*(3*A*b - B*a)*1i)/a^4 + (((8*tan(c/2 + (d*x)/2)*(72*A^2*b^12 + 4*B^2*a^12 - 72*A^2*a*b^11 - 8
*B^2*a^11*b - 288*A^2*a^2*b^10 + 288*A^2*a^3*b^9 + 441*A^2*a^4*b^8 - 432*A^2*a^5*b^7 - 288*A^2*a^6*b^6 + 288*A
^2*a^7*b^5 + 36*A^2*a^8*b^4 - 72*A^2*a^9*b^3 + 36*A^2*a^10*b^2 + 8*B^2*a^2*b^10 - 8*B^2*a^3*b^9 - 32*B^2*a^4*b
^8 + 32*B^2*a^5*b^7 + 57*B^2*a^6*b^6 - 48*B^2*a^7*b^5 - 52*B^2*a^8*b^4 + 32*B^2*a^9*b^3 + 24*B^2*a^10*b^2 - 48
*A*B*a*b^11 - 24*A*B*a^11*b + 48*A*B*a^2*b^10 + 192*A*B*a^3*b^9 - 192*A*B*a^4*b^8 - 318*A*B*a^5*b^7 + 288*A*B*
a^6*b^6 + 252*A*B*a^7*b^5 - 192*A*B*a^8*b^4 - 72*A*B*a^9*b^3 + 48*A*B*a^10*b^2))/(a^12*b + a^13 - a^6*b^7 - a^
7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2) - (((8*(4*B*a^18 + 12*A*a^8*b^10 - 6*A*a^9*b^9 - 54*A
*a^10*b^8 + 24*A*a^11*b^7 + 96*A*a^12*b^6 - 42*A*a^13*b^5 - 78*A*a^14*b^4 + 36*A*a^15*b^3 + 24*A*a^16*b^2 - 4*
B*a^9*b^9 + 2*B*a^10*b^8 + 18*B*a^11*b^7 - 4*B*a^12*b^6 - 36*B*a^13*b^5 + 6*B*a^14*b^4 + 34*B*a^15*b^3 - 8*B*a
^16*b^2 - 12*A*a^17*b - 12*B*a^17*b))/(a^15*b + a^16 - a^9*b^7 - a^10*b^6 + 3*a^11*b^5 + 3*a^12*b^4 - 3*a^13*b
^3 - 3*a^14*b^2) - (8*tan(c/2 + (d*x)/2)*(3*A*b - B*a)*(8*a^17*b - 8*a^8*b^10 + 8*a^9*b^9 + 32*a^10*b^8 - 32*a
^11*b^7 - 48*a^12*b^6 + 48*a^13*b^5 + 32*a^14*b^4 - 32*a^15*b^3 - 8*a^16*b^2))/(a^4*(a^12*b + a^13 - a^6*b^7 -
a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*b^2)))*(3*A*b - B*a))/a^4)*(3*A*b - B*a)*1i)/a^4)/((16*
(108*A^3*b^12 - 54*A^3*a*b^11 - 12*B^3*a^11*b - 486*A^3*a^2*b^10 + 243*A^3*a^3*b^9 + 864*A^3*a^4*b^8 - 378*A^3
*a^5*b^7 - 702*A^3*a^6*b^6 + 216*A^3*a^7*b^5 + 216*A^3*a^8*b^4 - 4*B^3*a^3*b^9 + 2*B^3*a^4*b^8 + 18*B^3*a^5*b^
7 - 13*B^3*a^6*b^6 - 36*B^3*a^7*b^5 + 26*B^3*a^8*b^4 + 34*B^3*a^9*b^3 - 24*B^3*a^10*b^2 - 108*A^2*B*a*b^11 + 3
6*A*B^2*a^2*b^10 - 18*A*B^2*a^3*b^9 - 162*A*B^2*a^4*b^8 + 105*A*B^2*a^5*b^7 + 312*A*B^2*a^6*b^6 - 198*A*B^2*a^
7*b^5 - 282*A*B^2*a^8*b^4 + 156*A*B^2*a^9*b^3 + 96*A*B^2*a^10*b^2 + 54*A^2*B*a^2*b^10 + 486*A^2*B*a^3*b^9 - 27
9*A^2*B*a^4*b^8 - 900*A^2*B*a^5*b^7 + 486*A^2*B*a^6*b^6 + 774*A^2*B*a^7*b^5 - 324*A^2*B*a^8*b^4 - 252*A^2*B*a^
9*b^3))/(a^15*b + a^16 - a^9*b^7 - a^10*b^6 + 3*a^11*b^5 + 3*a^12*b^4 - 3*a^13*b^3 - 3*a^14*b^2) - (((8*tan(c/
2 + (d*x)/2)*(72*A^2*b^12 + 4*B^2*a^12 - 72*A^2*a*b^11 - 8*B^2*a^11*b - 288*A^2*a^2*b^10 + 288*A^2*a^3*b^9 + 4
41*A^2*a^4*b^8 - 432*A^2*a^5*b^7 - 288*A^2*a^6*b^6 + 288*A^2*a^7*b^5 + 36*A^2*a^8*b^4 - 72*A^2*a^9*b^3 + 36*A^
2*a^10*b^2 + 8*B^2*a^2*b^10 - 8*B^2*a^3*b^9 - 32*B^2*a^4*b^8 + 32*B^2*a^5*b^7 + 57*B^2*a^6*b^6 - 48*B^2*a^7*b^
5 - 52*B^2*a^8*b^4 + 32*B^2*a^9*b^3 + 24*B^2*a^10*b^2 - 48*A*B*a*b^11 - 24*A*B*a^11*b + 48*A*B*a^2*b^10 + 192*
A*B*a^3*b^9 - 192*A*B*a^4*b^8 - 318*A*B*a^5*b^7 + 288*A*B*a^6*b^6 + 252*A*B*a^7*b^5 - 192*A*B*a^8*b^4 - 72*A*B
*a^9*b^3 + 48*A*B*a^10*b^2))/(a^12*b + a^13 - a^6*b^7 - a^7*b^6 + 3*a^8*b^5 + 3*a^9*b^4 - 3*a^10*b^3 - 3*a^11*
b^2) + (((8*(4*B*a^18 + 12*A*a^8*b^10 - 6*A*a^9*b^9 - 54*A*a^10*b^8 + 24*A*a^11*b^7 + 96*A*a^12*b^6 - 42*A*a^1
3*b^5 - 78*A*a^14*b^4 + 36*A*a^15*b^3 + 24*A*a^...
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