\(\int \frac {\cos ^3(c+d x) (A+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\) [562]

Optimal result

Integrand size = 33, antiderivative size = 233 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x}{8 b^5}-\frac {2 a^3 \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^5 \sqrt {a+b} d}-\frac {a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a C \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3129, 3128, 3102, 2814, 2738, 211} \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}+\frac {x \left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right )}{8 b^5}-\frac {2 a^3 \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^5 d \sqrt {a-b} \sqrt {a+b}}-\frac {a C \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 b d} \]

[In]

Int[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]

[Out]

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

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 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 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]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3129

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

Rubi steps \begin{align*} \text {integral}& = \frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {\cos ^2(c+d x) \left (3 a C+b (4 A+3 C) \cos (c+d x)-4 a C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{4 b} \\ & = -\frac {a C \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {\cos (c+d x) \left (-8 a^2 C+a b C \cos (c+d x)+3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{12 b^2} \\ & = \frac {\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a C \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {3 a \left (4 a^2 C+b^2 (4 A+3 C)\right )+b \left (12 A b^2-4 a^2 C+9 b^2 C\right ) \cos (c+d x)-8 a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^3} \\ & = -\frac {a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a C \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac {\int \frac {3 a b \left (4 a^2 C+b^2 (4 A+3 C)\right )+3 \left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^4} \\ & = \frac {\left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x}{8 b^5}-\frac {a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a C \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d}-\frac {\left (a^3 \left (A b^2+a^2 C\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^5} \\ & = \frac {\left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x}{8 b^5}-\frac {a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a C \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d}-\frac {\left (2 a^3 \left (A b^2+a^2 C\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 )}{b^5 d} \\ & = \frac {\left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x}{8 b^5}-\frac {2 a^3 \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^5 \sqrt {a+b} d}-\frac {a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac {\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a C \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.92 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {12 \left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) (c+d x)+\frac {192 a^3 \left (A b^2+a^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-24 a b \left (4 A b^2+4 a^2 C+3 b^2 C\right ) \sin (c+d x)+24 b^2 \left (A b^2+\left (a^2+b^2\right ) C\right ) \sin (2 (c+d x))-8 a b^3 C \sin (3 (c+d x))+3 b^4 C \sin (4 (c+d x))}{96 b^5 d} \]

[In]

Integrate[(Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]

[Out]

(12*(8*a^4*C + 4*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*(c + d*x) + (192*a^3*(A*b^2 + a^2*C)*ArcTanh[((a - b)*Ta
n[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 24*a*b*(4*A*b^2 + 4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 24*b
^2*(A*b^2 + (a^2 + b^2)*C)*Sin[2*(c + d*x)] - 8*a*b^3*C*Sin[3*(c + d*x)] + 3*b^4*C*Sin[4*(c + d*x)])/(96*b^5*d
)

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.52

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (-A a \,b^{3}-\frac {1}{2} A \,b^{4}-C \,a^{3} b -\frac {1}{2} C \,a^{2} b^{2}-C a \,b^{3}-\frac {5}{8} C \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 A a \,b^{3}-3 C \,a^{3} b -\frac {5}{3} C a \,b^{3}+\frac {3}{8} C \,b^{4}-\frac {1}{2} A \,b^{4}-\frac {1}{2} C \,a^{2} b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,b^{4}+\frac {1}{2} C \,a^{2} b^{2}-\frac {3}{8} C \,b^{4}-3 A a \,b^{3}-3 C \,a^{3} b -\frac {5}{3} C a \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,b^{4}+\frac {1}{2} C \,a^{2} b^{2}+\frac {5}{8} C \,b^{4}-A a \,b^{3}-C \,a^{3} b -C a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (8 A \,a^{2} b^{2}+4 A \,b^{4}+8 C \,a^{4}+4 C \,a^{2} b^{2}+3 C \,b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{b^{5}}-\frac {2 a^{3} \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(353\)
default	\(\frac {\frac {\frac {2 \left (\left (-A a \,b^{3}-\frac {1}{2} A \,b^{4}-C \,a^{3} b -\frac {1}{2} C \,a^{2} b^{2}-C a \,b^{3}-\frac {5}{8} C \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 A a \,b^{3}-3 C \,a^{3} b -\frac {5}{3} C a \,b^{3}+\frac {3}{8} C \,b^{4}-\frac {1}{2} A \,b^{4}-\frac {1}{2} C \,a^{2} b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,b^{4}+\frac {1}{2} C \,a^{2} b^{2}-\frac {3}{8} C \,b^{4}-3 A a \,b^{3}-3 C \,a^{3} b -\frac {5}{3} C a \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} A \,b^{4}+\frac {1}{2} C \,a^{2} b^{2}+\frac {5}{8} C \,b^{4}-A a \,b^{3}-C \,a^{3} b -C a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (8 A \,a^{2} b^{2}+4 A \,b^{4}+8 C \,a^{4}+4 C \,a^{2} b^{2}+3 C \,b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{b^{5}}-\frac {2 a^{3} \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(353\)
risch	\(\frac {x A \,a^{2}}{b^{3}}+\frac {x A}{2 b}+\frac {x C \,a^{4}}{b^{5}}+\frac {x C \,a^{2}}{2 b^{3}}+\frac {3 C x}{8 b}+\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )} C}{2 b^{4} d}+\frac {3 i a \,{\mathrm e}^{i \left (d x +c \right )} C}{8 b^{2} d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )} A}{2 b^{2} d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} A}{2 b^{2} d}-\frac {3 i a \,{\mathrm e}^{-i \left (d x +c \right )} C}{8 b^{2} d}-\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )} C}{2 b^{4} d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{5}}+\frac {C \sin \left (4 d x +4 c \right )}{32 b d}-\frac {a C \sin \left (3 d x +3 c \right )}{12 b^{2} d}+\frac {\sin \left (2 d x +2 c \right ) A}{4 b d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} C}{4 b^{3} d}+\frac {\sin \left (2 d x +2 c \right ) C}{4 b d}\)	\(564\)

[In]

int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)

[Out]

1/d*(2/b^5*(((-A*a*b^3-1/2*A*b^4-C*a^3*b-1/2*C*a^2*b^2-C*a*b^3-5/8*C*b^4)*tan(1/2*d*x+1/2*c)^7+(-3*A*a*b^3-3*C
*a^3*b-5/3*C*a*b^3+3/8*C*b^4-1/2*A*b^4-1/2*C*a^2*b^2)*tan(1/2*d*x+1/2*c)^5+(1/2*A*b^4+1/2*C*a^2*b^2-3/8*C*b^4-
3*A*a*b^3-3*C*a^3*b-5/3*C*a*b^3)*tan(1/2*d*x+1/2*c)^3+(1/2*A*b^4+1/2*C*a^2*b^2+5/8*C*b^4-A*a*b^3-C*a^3*b-C*a*b
^3)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^4+1/8*(8*A*a^2*b^2+4*A*b^4+8*C*a^4+4*C*a^2*b^2+3*C*b^4)*arcta
n(tan(1/2*d*x+1/2*c)))-2*a^3*(A*b^2+C*a^2)/b^5/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b
))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 609, normalized size of antiderivative = 2.61 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [\frac {3 \, {\left (8 \, C a^{6} + 4 \, {\left (2 \, A - C\right )} a^{4} b^{2} - {\left (4 \, A + C\right )} a^{2} b^{4} - {\left (4 \, A + 3 \, C\right )} b^{6}\right )} d x - 12 \, {\left (C a^{5} + A a^{3} b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (24 \, C a^{5} b + 8 \, {\left (3 \, A - C\right )} a^{3} b^{3} - 8 \, {\left (3 \, A + 2 \, C\right )} a b^{5} - 6 \, {\left (C a^{2} b^{4} - C b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (C a^{3} b^{3} - C a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (4 \, C a^{4} b^{2} + {\left (4 \, A - C\right )} a^{2} b^{4} - {\left (4 \, A + 3 \, C\right )} b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} b^{5} - b^{7}\right )} d}, \frac {3 \, {\left (8 \, C a^{6} + 4 \, {\left (2 \, A - C\right )} a^{4} b^{2} - {\left (4 \, A + C\right )} a^{2} b^{4} - {\left (4 \, A + 3 \, C\right )} b^{6}\right )} d x - 24 \, {\left (C a^{5} + A a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (24 \, C a^{5} b + 8 \, {\left (3 \, A - C\right )} a^{3} b^{3} - 8 \, {\left (3 \, A + 2 \, C\right )} a b^{5} - 6 \, {\left (C a^{2} b^{4} - C b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (C a^{3} b^{3} - C a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (4 \, C a^{4} b^{2} + {\left (4 \, A - C\right )} a^{2} b^{4} - {\left (4 \, A + 3 \, C\right )} b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} b^{5} - b^{7}\right )} d}\right ] \]

[In]

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[1/24*(3*(8*C*a^6 + 4*(2*A - C)*a^4*b^2 - (4*A + C)*a^2*b^4 - (4*A + 3*C)*b^6)*d*x - 12*(C*a^5 + A*a^3*b^2)*sq
rt(-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)) - (24*C*a^5*b + 8*(3*A - C)*a^3
*b^3 - 8*(3*A + 2*C)*a*b^5 - 6*(C*a^2*b^4 - C*b^6)*cos(d*x + c)^3 + 8*(C*a^3*b^3 - C*a*b^5)*cos(d*x + c)^2 - 3
*(4*C*a^4*b^2 + (4*A - C)*a^2*b^4 - (4*A + 3*C)*b^6)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d), 1/24*(3*
(8*C*a^6 + 4*(2*A - C)*a^4*b^2 - (4*A + C)*a^2*b^4 - (4*A + 3*C)*b^6)*d*x - 24*(C*a^5 + A*a^3*b^2)*sqrt(a^2 -
b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (24*C*a^5*b + 8*(3*A - C)*a^3*b^3 - 8*(3*A
+ 2*C)*a*b^5 - 6*(C*a^2*b^4 - C*b^6)*cos(d*x + c)^3 + 8*(C*a^3*b^3 - C*a*b^5)*cos(d*x + c)^2 - 3*(4*C*a^4*b^2
+ (4*A - C)*a^2*b^4 - (4*A + 3*C)*b^6)*cos(d*x + c))*sin(d*x + c))/((a^2*b^5 - b^7)*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),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. 574 vs. \(2 (214) = 428\).

Time = 0.32 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.46 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {3 \, {\left (8 \, C a^{4} + 8 \, A a^{2} b^{2} + 4 \, C a^{2} b^{2} + 4 \, A b^{4} + 3 \, C b^{4}\right )} {\left (d x + c\right )}}{b^{5}} + \frac {48 \, {\left (C a^{5} + A a^{3} b^{2}\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 )}}{\sqrt {a^{2} - b^{2}} b^{5}} - \frac {2 \, {\left (24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(8*C*a^4 + 8*A*a^2*b^2 + 4*C*a^2*b^2 + 4*A*b^4 + 3*C*b^4)*(d*x + c)/b^5 + 48*(C*a^5 + A*a^3*b^2)*(pi*f
loor(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)))/(sqrt(a^2 - b^2)*b^5) - 2*(24*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 +
24*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 15*C
*b^3*tan(1/2*d*x + 1/2*c)^7 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 72*A*a*b^2
*tan(1/2*d*x + 1/2*c)^5 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 9*C*b^3*tan(1/
2*d*x + 1/2*c)^5 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72*A*a*b^2*tan(1/2*d*
x + 1/2*c)^3 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 9*C*b^3*tan(1/2*d*x + 1/2
*c)^3 + 24*C*a^3*tan(1/2*d*x + 1/2*c) - 12*C*a^2*b*tan(1/2*d*x + 1/2*c) + 24*A*a*b^2*tan(1/2*d*x + 1/2*c) + 24
*C*a*b^2*tan(1/2*d*x + 1/2*c) - 12*A*b^3*tan(1/2*d*x + 1/2*c) - 15*C*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x +
1/2*c)^2 + 1)^4*b^4))/d

Mupad [B] (verification not implemented)

Time = 8.39 (sec) , antiderivative size = 5844, normalized size of antiderivative = 25.08 \[ \int \frac {\cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]

[In]

int((cos(c + d*x)^3*(A + C*cos(c + d*x)^2))/(a + b*cos(c + d*x)),x)

[Out]

- ((tan(c/2 + (d*x)/2)^7*(4*A*b^3 + 8*C*a^3 + 5*C*b^3 + 8*A*a*b^2 + 8*C*a*b^2 + 4*C*a^2*b))/(4*b^4) + (tan(c/2
+ (d*x)/2)^3*(72*C*a^3 - 12*A*b^3 + 9*C*b^3 + 72*A*a*b^2 + 40*C*a*b^2 - 12*C*a^2*b))/(12*b^4) + (tan(c/2 + (d
*x)/2)^5*(12*A*b^3 + 72*C*a^3 - 9*C*b^3 + 72*A*a*b^2 + 40*C*a*b^2 + 12*C*a^2*b))/(12*b^4) - (tan(c/2 + (d*x)/2
)*(4*A*b^3 - 8*C*a^3 + 5*C*b^3 - 8*A*a*b^2 - 8*C*a*b^2 + 4*C*a^2*b))/(4*b^4))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*t
an(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (atan(((((tan(c/2 + (d*x)/2)*(16*A
^2*b^11 - 128*C^2*a^11 + 9*C^2*b^11 - 48*A^2*a*b^10 - 27*C^2*a*b^10 + 256*C^2*a^10*b + 112*A^2*a^2*b^9 - 208*A
^2*a^3*b^8 + 256*A^2*a^4*b^7 - 256*A^2*a^5*b^6 + 256*A^2*a^6*b^5 - 128*A^2*a^7*b^4 + 51*C^2*a^2*b^9 - 81*C^2*a
^3*b^8 + 136*C^2*a^4*b^7 - 216*C^2*a^5*b^6 + 256*C^2*a^6*b^5 - 256*C^2*a^7*b^4 + 256*C^2*a^8*b^3 - 256*C^2*a^9
*b^2 + 24*A*C*b^11 - 72*A*C*a*b^10 + 152*A*C*a^2*b^9 - 264*A*C*a^3*b^8 + 368*A*C*a^4*b^7 - 464*A*C*a^5*b^6 + 5
12*A*C*a^6*b^5 - 512*A*C*a^7*b^4 + 512*A*C*a^8*b^3 - 256*A*C*a^9*b^2))/(2*b^8) + (((16*A*b^16 + 12*C*b^16 + 16
*A*a^2*b^14 - 48*A*a^3*b^13 + 32*A*a^4*b^12 + 4*C*a^2*b^14 - 4*C*a^3*b^13 + 16*C*a^4*b^12 - 48*C*a^5*b^11 + 32
*C*a^6*b^10 - 16*A*a*b^15 - 12*C*a*b^15)/b^12 - (tan(c/2 + (d*x)/2)*(128*a*b^12 - 256*a^2*b^11 + 128*a^3*b^10)
*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/(2*b^13))*(b^2*(A*a^2*1i + (C*a^2*1i)
/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/b^5)*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 +
(C*3i)/8))*1i)/b^5 + (((tan(c/2 + (d*x)/2)*(16*A^2*b^11 - 128*C^2*a^11 + 9*C^2*b^11 - 48*A^2*a*b^10 - 27*C^2*a
*b^10 + 256*C^2*a^10*b + 112*A^2*a^2*b^9 - 208*A^2*a^3*b^8 + 256*A^2*a^4*b^7 - 256*A^2*a^5*b^6 + 256*A^2*a^6*b
^5 - 128*A^2*a^7*b^4 + 51*C^2*a^2*b^9 - 81*C^2*a^3*b^8 + 136*C^2*a^4*b^7 - 216*C^2*a^5*b^6 + 256*C^2*a^6*b^5 -
256*C^2*a^7*b^4 + 256*C^2*a^8*b^3 - 256*C^2*a^9*b^2 + 24*A*C*b^11 - 72*A*C*a*b^10 + 152*A*C*a^2*b^9 - 264*A*C
*a^3*b^8 + 368*A*C*a^4*b^7 - 464*A*C*a^5*b^6 + 512*A*C*a^6*b^5 - 512*A*C*a^7*b^4 + 512*A*C*a^8*b^3 - 256*A*C*a
^9*b^2))/(2*b^8) - (((16*A*b^16 + 12*C*b^16 + 16*A*a^2*b^14 - 48*A*a^3*b^13 + 32*A*a^4*b^12 + 4*C*a^2*b^14 - 4
*C*a^3*b^13 + 16*C*a^4*b^12 - 48*C*a^5*b^11 + 32*C*a^6*b^10 - 16*A*a*b^15 - 12*C*a*b^15)/b^12 + (tan(c/2 + (d*
x)/2)*(128*a*b^12 - 256*a^2*b^11 + 128*a^3*b^10)*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (
C*3i)/8)))/(2*b^13))*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/b^5)*(b^2*(A*a^2*
1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8))*1i)/b^5)/((64*C^3*a^14 - 96*C^3*a^13*b - 16*A^3*a^3
*b^11 + 32*A^3*a^4*b^10 - 80*A^3*a^5*b^9 + 96*A^3*a^6*b^8 - 96*A^3*a^7*b^7 + 64*A^3*a^8*b^6 - 9*C^3*a^5*b^9 +
18*C^3*a^6*b^8 - 33*C^3*a^7*b^7 + 48*C^3*a^8*b^6 - 88*C^3*a^9*b^5 + 104*C^3*a^10*b^4 - 104*C^3*a^11*b^3 + 96*C
^3*a^12*b^2 - 9*A*C^2*a^3*b^11 + 18*A*C^2*a^4*b^10 - 57*A*C^2*a^5*b^9 + 96*A*C^2*a^6*b^8 - 192*A*C^2*a^7*b^7 +
240*A*C^2*a^8*b^6 - 288*A*C^2*a^9*b^5 + 288*A*C^2*a^10*b^4 - 288*A*C^2*a^11*b^3 + 192*A*C^2*a^12*b^2 - 24*A^2
*C*a^3*b^11 + 48*A^2*C*a^4*b^10 - 120*A^2*C*a^5*b^9 + 168*A^2*C*a^6*b^8 - 264*A^2*C*a^7*b^7 + 288*A^2*C*a^8*b^
6 - 288*A^2*C*a^9*b^5 + 192*A^2*C*a^10*b^4)/b^12 + (((tan(c/2 + (d*x)/2)*(16*A^2*b^11 - 128*C^2*a^11 + 9*C^2*b
^11 - 48*A^2*a*b^10 - 27*C^2*a*b^10 + 256*C^2*a^10*b + 112*A^2*a^2*b^9 - 208*A^2*a^3*b^8 + 256*A^2*a^4*b^7 - 2
56*A^2*a^5*b^6 + 256*A^2*a^6*b^5 - 128*A^2*a^7*b^4 + 51*C^2*a^2*b^9 - 81*C^2*a^3*b^8 + 136*C^2*a^4*b^7 - 216*C
^2*a^5*b^6 + 256*C^2*a^6*b^5 - 256*C^2*a^7*b^4 + 256*C^2*a^8*b^3 - 256*C^2*a^9*b^2 + 24*A*C*b^11 - 72*A*C*a*b^
10 + 152*A*C*a^2*b^9 - 264*A*C*a^3*b^8 + 368*A*C*a^4*b^7 - 464*A*C*a^5*b^6 + 512*A*C*a^6*b^5 - 512*A*C*a^7*b^4
+ 512*A*C*a^8*b^3 - 256*A*C*a^9*b^2))/(2*b^8) + (((16*A*b^16 + 12*C*b^16 + 16*A*a^2*b^14 - 48*A*a^3*b^13 + 32
*A*a^4*b^12 + 4*C*a^2*b^14 - 4*C*a^3*b^13 + 16*C*a^4*b^12 - 48*C*a^5*b^11 + 32*C*a^6*b^10 - 16*A*a*b^15 - 12*C
*a*b^15)/b^12 - (tan(c/2 + (d*x)/2)*(128*a*b^12 - 256*a^2*b^11 + 128*a^3*b^10)*(b^2*(A*a^2*1i + (C*a^2*1i)/2)
+ C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/(2*b^13))*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 +
(C*3i)/8)))/b^5)*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/b^5 - (((tan(c/2 + (
d*x)/2)*(16*A^2*b^11 - 128*C^2*a^11 + 9*C^2*b^11 - 48*A^2*a*b^10 - 27*C^2*a*b^10 + 256*C^2*a^10*b + 112*A^2*a^
2*b^9 - 208*A^2*a^3*b^8 + 256*A^2*a^4*b^7 - 256*A^2*a^5*b^6 + 256*A^2*a^6*b^5 - 128*A^2*a^7*b^4 + 51*C^2*a^2*b
^9 - 81*C^2*a^3*b^8 + 136*C^2*a^4*b^7 - 216*C^2*a^5*b^6 + 256*C^2*a^6*b^5 - 256*C^2*a^7*b^4 + 256*C^2*a^8*b^3
- 256*C^2*a^9*b^2 + 24*A*C*b^11 - 72*A*C*a*b^10 + 152*A*C*a^2*b^9 - 264*A*C*a^3*b^8 + 368*A*C*a^4*b^7 - 464*A*
C*a^5*b^6 + 512*A*C*a^6*b^5 - 512*A*C*a^7*b^4 + 512*A*C*a^8*b^3 - 256*A*C*a^9*b^2))/(2*b^8) - (((16*A*b^16 + 1
2*C*b^16 + 16*A*a^2*b^14 - 48*A*a^3*b^13 + 32*A*a^4*b^12 + 4*C*a^2*b^14 - 4*C*a^3*b^13 + 16*C*a^4*b^12 - 48*C*
a^5*b^11 + 32*C*a^6*b^10 - 16*A*a*b^15 - 12*C*a*b^15)/b^12 + (tan(c/2 + (d*x)/2)*(128*a*b^12 - 256*a^2*b^11 +
128*a^3*b^10)*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/(2*b^13))*(b^2*(A*a^2*1i
+ (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8)))/b^5)*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4
*((A*1i)/2 + (C*3i)/8)))/b^5))*(b^2*(A*a^2*1i + (C*a^2*1i)/2) + C*a^4*1i + b^4*((A*1i)/2 + (C*3i)/8))*2i)/(b^5
*d) - (a^3*atan(((a^3*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((tan(c/2 + (d*x)/2)*(16*A^2*b^11 - 128*C^2*a^1
1 + 9*C^2*b^11 - 48*A^2*a*b^10 - 27*C^2*a*b^10 + 256*C^2*a^10*b + 112*A^2*a^2*b^9 - 208*A^2*a^3*b^8 + 256*A^2*
a^4*b^7 - 256*A^2*a^5*b^6 + 256*A^2*a^6*b^5 - 128*A^2*a^7*b^4 + 51*C^2*a^2*b^9 - 81*C^2*a^3*b^8 + 136*C^2*a^4*
b^7 - 216*C^2*a^5*b^6 + 256*C^2*a^6*b^5 - 256*C^2*a^7*b^4 + 256*C^2*a^8*b^3 - 256*C^2*a^9*b^2 + 24*A*C*b^11 -
72*A*C*a*b^10 + 152*A*C*a^2*b^9 - 264*A*C*a^3*b^8 + 368*A*C*a^4*b^7 - 464*A*C*a^5*b^6 + 512*A*C*a^6*b^5 - 512*
A*C*a^7*b^4 + 512*A*C*a^8*b^3 - 256*A*C*a^9*b^2))/(2*b^8) + (a^3*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((16
*A*b^16 + 12*C*b^16 + 16*A*a^2*b^14 - 48*A*a^3*b^13 + 32*A*a^4*b^12 + 4*C*a^2*b^14 - 4*C*a^3*b^13 + 16*C*a^4*b
^12 - 48*C*a^5*b^11 + 32*C*a^6*b^10 - 16*A*a*b^15 - 12*C*a*b^15)/b^12 - (a^3*tan(c/2 + (d*x)/2)*(-(a + b)*(a -
b))^(1/2)*(A*b^2 + C*a^2)*(128*a*b^12 - 256*a^2*b^11 + 128*a^3*b^10))/(2*b^8*(b^7 - a^2*b^5))))/(b^7 - a^2*b^
5))*1i)/(b^7 - a^2*b^5) + (a^3*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((tan(c/2 + (d*x)/2)*(16*A^2*b^11 - 12
8*C^2*a^11 + 9*C^2*b^11 - 48*A^2*a*b^10 - 27*C^2*a*b^10 + 256*C^2*a^10*b + 112*A^2*a^2*b^9 - 208*A^2*a^3*b^8 +
256*A^2*a^4*b^7 - 256*A^2*a^5*b^6 + 256*A^2*a^6*b^5 - 128*A^2*a^7*b^4 + 51*C^2*a^2*b^9 - 81*C^2*a^3*b^8 + 136
*C^2*a^4*b^7 - 216*C^2*a^5*b^6 + 256*C^2*a^6*b^5 - 256*C^2*a^7*b^4 + 256*C^2*a^8*b^3 - 256*C^2*a^9*b^2 + 24*A*
C*b^11 - 72*A*C*a*b^10 + 152*A*C*a^2*b^9 - 264*A*C*a^3*b^8 + 368*A*C*a^4*b^7 - 464*A*C*a^5*b^6 + 512*A*C*a^6*b
^5 - 512*A*C*a^7*b^4 + 512*A*C*a^8*b^3 - 256*A*C*a^9*b^2))/(2*b^8) - (a^3*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*
a^2)*((16*A*b^16 + 12*C*b^16 + 16*A*a^2*b^14 - 48*A*a^3*b^13 + 32*A*a^4*b^12 + 4*C*a^2*b^14 - 4*C*a^3*b^13 + 1
6*C*a^4*b^12 - 48*C*a^5*b^11 + 32*C*a^6*b^10 - 16*A*a*b^15 - 12*C*a*b^15)/b^12 + (a^3*tan(c/2 + (d*x)/2)*(-(a
+ b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(128*a*b^12 - 256*a^2*b^11 + 128*a^3*b^10))/(2*b^8*(b^7 - a^2*b^5))))/(b^7
- a^2*b^5))*1i)/(b^7 - a^2*b^5))/((64*C^3*a^14 - 96*C^3*a^13*b - 16*A^3*a^3*b^11 + 32*A^3*a^4*b^10 - 80*A^3*a
^5*b^9 + 96*A^3*a^6*b^8 - 96*A^3*a^7*b^7 + 64*A^3*a^8*b^6 - 9*C^3*a^5*b^9 + 18*C^3*a^6*b^8 - 33*C^3*a^7*b^7 +
48*C^3*a^8*b^6 - 88*C^3*a^9*b^5 + 104*C^3*a^10*b^4 - 104*C^3*a^11*b^3 + 96*C^3*a^12*b^2 - 9*A*C^2*a^3*b^11 + 1
8*A*C^2*a^4*b^10 - 57*A*C^2*a^5*b^9 + 96*A*C^2*a^6*b^8 - 192*A*C^2*a^7*b^7 + 240*A*C^2*a^8*b^6 - 288*A*C^2*a^9
*b^5 + 288*A*C^2*a^10*b^4 - 288*A*C^2*a^11*b^3 + 192*A*C^2*a^12*b^2 - 24*A^2*C*a^3*b^11 + 48*A^2*C*a^4*b^10 -
120*A^2*C*a^5*b^9 + 168*A^2*C*a^6*b^8 - 264*A^2*C*a^7*b^7 + 288*A^2*C*a^8*b^6 - 288*A^2*C*a^9*b^5 + 192*A^2*C*
a^10*b^4)/b^12 + (a^3*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((tan(c/2 + (d*x)/2)*(16*A^2*b^11 - 128*C^2*a^1
1 + 9*C^2*b^11 - 48*A^2*a*b^10 - 27*C^2*a*b^10 + 256*C^2*a^10*b + 112*A^2*a^2*b^9 - 208*A^2*a^3*b^8 + 256*A^2*
a^4*b^7 - 256*A^2*a^5*b^6 + 256*A^2*a^6*b^5 - 128*A^2*a^7*b^4 + 51*C^2*a^2*b^9 - 81*C^2*a^3*b^8 + 136*C^2*a^4*
b^7 - 216*C^2*a^5*b^6 + 256*C^2*a^6*b^5 - 256*C^2*a^7*b^4 + 256*C^2*a^8*b^3 - 256*C^2*a^9*b^2 + 24*A*C*b^11 -
72*A*C*a*b^10 + 152*A*C*a^2*b^9 - 264*A*C*a^3*b^8 + 368*A*C*a^4*b^7 - 464*A*C*a^5*b^6 + 512*A*C*a^6*b^5 - 512*
A*C*a^7*b^4 + 512*A*C*a^8*b^3 - 256*A*C*a^9*b^2))/(2*b^8) + (a^3*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((16
*A*b^16 + 12*C*b^16 + 16*A*a^2*b^14 - 48*A*a^3*b^13 + 32*A*a^4*b^12 + 4*C*a^2*b^14 - 4*C*a^3*b^13 + 16*C*a^4*b
^12 - 48*C*a^5*b^11 + 32*C*a^6*b^10 - 16*A*a*b^15 - 12*C*a*b^15)/b^12 - (a^3*tan(c/2 + (d*x)/2)*(-(a + b)*(a -
b))^(1/2)*(A*b^2 + C*a^2)*(128*a*b^12 - 256*a^2*b^11 + 128*a^3*b^10))/(2*b^8*(b^7 - a^2*b^5))))/(b^7 - a^2*b^
5)))/(b^7 - a^2*b^5) - (a^3*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((tan(c/2 + (d*x)/2)*(16*A^2*b^11 - 128*C
^2*a^11 + 9*C^2*b^11 - 48*A^2*a*b^10 - 27*C^2*a*b^10 + 256*C^2*a^10*b + 112*A^2*a^2*b^9 - 208*A^2*a^3*b^8 + 25
6*A^2*a^4*b^7 - 256*A^2*a^5*b^6 + 256*A^2*a^6*b^5 - 128*A^2*a^7*b^4 + 51*C^2*a^2*b^9 - 81*C^2*a^3*b^8 + 136*C^
2*a^4*b^7 - 216*C^2*a^5*b^6 + 256*C^2*a^6*b^5 - 256*C^2*a^7*b^4 + 256*C^2*a^8*b^3 - 256*C^2*a^9*b^2 + 24*A*C*b
^11 - 72*A*C*a*b^10 + 152*A*C*a^2*b^9 - 264*A*C*a^3*b^8 + 368*A*C*a^4*b^7 - 464*A*C*a^5*b^6 + 512*A*C*a^6*b^5
- 512*A*C*a^7*b^4 + 512*A*C*a^8*b^3 - 256*A*C*a^9*b^2))/(2*b^8) - (a^3*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2
)*((16*A*b^16 + 12*C*b^16 + 16*A*a^2*b^14 - 48*A*a^3*b^13 + 32*A*a^4*b^12 + 4*C*a^2*b^14 - 4*C*a^3*b^13 + 16*C
*a^4*b^12 - 48*C*a^5*b^11 + 32*C*a^6*b^10 - 16*A*a*b^15 - 12*C*a*b^15)/b^12 + (a^3*tan(c/2 + (d*x)/2)*(-(a + b
)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(128*a*b^12 - 256*a^2*b^11 + 128*a^3*b^10))/(2*b^8*(b^7 - a^2*b^5))))/(b^7 -
a^2*b^5)))/(b^7 - a^2*b^5)))*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*2i)/(d*(b^7 - a^2*b^5))