3.806 \(\int \frac {(B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=189 \[ -\frac {(2 b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {2 b \left (-2 a^3 C+3 a^2 b B+a b^2 C-2 b^3 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

2*b*(3*B*a^2*b-2*B*b^3-2*C*a^3+C*a*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/(a-b)^(3/2)/(a+
b)^(3/2)/d-(2*B*b-C*a)*arctanh(sin(d*x+c))/a^3/d+(B*a^2-2*B*b^2+C*a*b)*tan(d*x+c)/a^2/(a^2-b^2)/d+b*(B*b-C*a)*
tan(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))

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Rubi [A]  time = 0.79, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3029, 3000, 3055, 3001, 3770, 2659, 205} \[ \frac {2 b \left (3 a^2 b B-2 a^3 C+a b^2 C-2 b^3 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {(2 b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(3*a^2*b*B - 2*b^3*B - 2*a^3*C + a*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b
)^(3/2)*(a + b)^(3/2)*d) - ((2*b*B - a*C)*ArcTanh[Sin[c + d*x]])/(a^3*d) + ((a^2*B - 2*b^2*B + a*b*C)*Tan[c +
d*x])/(a^2*(a^2 - b^2)*d) + (b*(b*B - a*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2659

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 3000

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)), Int
[(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 3001

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 3029

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] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3055

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] + Dis
t[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

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 {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\int \frac {(B+C \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\\ &=\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (a^2 B-2 b^2 B+a b C-a (b B-a C) \cos (c+d x)+b (b B-a C) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (a^2 B-2 b^2 B+a b C\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (-\left (a^2-b^2\right ) (2 b B-a C)+a b (b B-a C) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {\left (a^2 B-2 b^2 B+a b C\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {(2 b B-a C) \int \sec (c+d x) \, dx}{a^3}+\frac {\left (b \left (3 a^2 b B-2 b^3 B-2 a^3 C+a b^2 C\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac {(2 b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {\left (a^2 B-2 b^2 B+a b C\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (2 b \left (3 a^2 b B-2 b^3 B-2 a^3 C+a b^2 C\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right ) d}\\ &=\frac {2 b \left (3 a^2 b B-2 b^3 B-2 a^3 C+a b^2 C\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {(2 b B-a C) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac {\left (a^2 B-2 b^2 B+a b C\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.89, size = 240, normalized size = 1.27 \[ \frac {-\frac {2 b \left (2 a^3 C-3 a^2 b B-a b^2 C+2 b^3 B\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac {a b^2 (a C-b B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+a B \tan (c+d x)-a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+a C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 b B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*(-3*a^2*b*B + 2*b^3*B + 2*a^3*C - a*b^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2
+ b^2)^(3/2) + 2*b*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - a*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] -
2*b*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + a*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a*b^2*(-(b*B
) + a*C)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])) + a*B*Tan[c + d*x])/(a^3*d)

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fricas [B]  time = 12.36, size = 1088, normalized size = 5.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((2*C*a^3*b^2 - 3*B*a^2*b^3 - C*a*b^4 + 2*B*b^5)*cos(d*x + c)^2 + (2*C*a^4*b - 3*B*a^3*b^2 - C*a^2*b^3
+ 2*B*a*b^4)*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)) -
((C*a^5*b - 2*B*a^4*b^2 - 2*C*a^3*b^3 + 4*B*a^2*b^4 + C*a*b^5 - 2*B*b^6)*cos(d*x + c)^2 + (C*a^6 - 2*B*a^5*b -
2*C*a^4*b^2 + 4*B*a^3*b^3 + C*a^2*b^4 - 2*B*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((C*a^5*b - 2*B*a^4*
b^2 - 2*C*a^3*b^3 + 4*B*a^2*b^4 + C*a*b^5 - 2*B*b^6)*cos(d*x + c)^2 + (C*a^6 - 2*B*a^5*b - 2*C*a^4*b^2 + 4*B*a
^3*b^3 + C*a^2*b^4 - 2*B*a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(B*a^6 - 2*B*a^4*b^2 + B*a^2*b^4 + (B
*a^5*b + C*a^4*b^2 - 3*B*a^3*b^3 - C*a^2*b^4 + 2*B*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b - 2*a^5*b^3 + a^
3*b^5)*d*cos(d*x + c)^2 + (a^8 - 2*a^6*b^2 + a^4*b^4)*d*cos(d*x + c)), -1/2*(2*((2*C*a^3*b^2 - 3*B*a^2*b^3 - C
*a*b^4 + 2*B*b^5)*cos(d*x + c)^2 + (2*C*a^4*b - 3*B*a^3*b^2 - C*a^2*b^3 + 2*B*a*b^4)*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))) - ((C*a^5*b - 2*B*a^4*b^2 - 2*C*a^3*b^3 + 4*
B*a^2*b^4 + C*a*b^5 - 2*B*b^6)*cos(d*x + c)^2 + (C*a^6 - 2*B*a^5*b - 2*C*a^4*b^2 + 4*B*a^3*b^3 + C*a^2*b^4 - 2
*B*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((C*a^5*b - 2*B*a^4*b^2 - 2*C*a^3*b^3 + 4*B*a^2*b^4 + C*a*b^5
- 2*B*b^6)*cos(d*x + c)^2 + (C*a^6 - 2*B*a^5*b - 2*C*a^4*b^2 + 4*B*a^3*b^3 + C*a^2*b^4 - 2*B*a*b^5)*cos(d*x +
c))*log(-sin(d*x + c) + 1) - 2*(B*a^6 - 2*B*a^4*b^2 + B*a^2*b^4 + (B*a^5*b + C*a^4*b^2 - 3*B*a^3*b^3 - C*a^2*b
^4 + 2*B*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b - 2*a^5*b^3 + a^3*b^5)*d*cos(d*x + c)^2 + (a^8 - 2*a^6*b^2
+ a^4*b^4)*d*cos(d*x + c))]

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giac [B]  time = 0.58, size = 404, normalized size = 2.14 \[ \frac {\frac {2 \, {\left (2 \, C a^{3} b - 3 \, B a^{2} b^{2} - C a b^{3} + 2 \, B b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{4} - a^{2} b^{2}\right )}} + \frac {{\left (C a - 2 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (C a - 2 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(2*C*a^3*b - 3*B*a^2*b^2 - C*a*b^3 + 2*B*b^4)*(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^5 - a^3*b^2)*sqrt(a^2 - b^2)) - 2*(B*a^
3*tan(1/2*d*x + 1/2*c)^3 - B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - C*a*b^2*tan(1/2*d
*x + 1/2*c)^3 + 2*B*b^3*tan(1/2*d*x + 1/2*c)^3 + B*a^3*tan(1/2*d*x + 1/2*c) + B*a^2*b*tan(1/2*d*x + 1/2*c) - B
*a*b^2*tan(1/2*d*x + 1/2*c) + C*a*b^2*tan(1/2*d*x + 1/2*c) - 2*B*b^3*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1
/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^2 - a - b)*(a^4 - a^2*b^2)) + (C*a - 2*B*b)*log(
abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (C*a - 2*B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3)/d

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maple [B]  time = 0.23, size = 502, normalized size = 2.66 \[ -\frac {2 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d \,a^{2} \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {6 b^{2} \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 ) B}{d a \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {4 b^{4} \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 ) B}{d \,a^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {4 \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 ) C b}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 b^{3} \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 ) C}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {B}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B b}{d \,a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,a^{2}}-\frac {B}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B b}{d \,a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*b^3/a^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*B+2/d*b^2/a/(a^2
-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*C+6/d*b^2/a/(a-b)/(a+b)/((a-b)*(a
+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B-4/d*b^4/a^3/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*
arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B-4/d/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+
1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*C*b+2/d*b^3/a^2/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a
-b)/((a-b)*(a+b))^(1/2))*C-1/d/a^2/(tan(1/2*d*x+1/2*c)-1)*B+2/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)*B*b-1/d/a^2*ln(ta
n(1/2*d*x+1/2*c)-1)*C-1/d/a^2/(tan(1/2*d*x+1/2*c)+1)*B-2/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*B*b+1/d/a^2*ln(tan(1/2
*d*x+1/2*c)+1)*C

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 9.79, size = 5464, normalized size = 28.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((((2*B*b - C*a)*((32*tan(c/2 + (d*x)/2)*(8*B^2*b^8 + C^2*a^8 - 8*B^2*a*b^7 - 2*C^2*a^7*b - 16*B^2*a^2*b^
6 + 16*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 5*C^2*a^4
*b^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^2*b^6 + 18*B*C*a^3*b^5 - 16*B*C*a^4
*b^4 - 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (((32*(B*a^7*b^5 - 2*B*a^6*b^6 - C*
a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^2 + C*a^7*b^5 - 3*C*a^9*b^3 + C*a^10*b^2 + 2*B*a^11*b + 2*C*a^11
*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32*tan(c/2 + (d*x)/2)*(2*B*b - C*a)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^
5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))*(2*B*b - C*a))/a^3)*1i)/a^3
+ ((2*B*b - C*a)*((32*tan(c/2 + (d*x)/2)*(8*B^2*b^8 + C^2*a^8 - 8*B^2*a*b^7 - 2*C^2*a^7*b - 16*B^2*a^2*b^6 + 1
6*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 5*C^2*a^4*b^4
+ 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^2*b^6 + 18*B*C*a^3*b^5 - 16*B*C*a^4*b^4
- 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (((32*(B*a^7*b^5 - 2*B*a^6*b^6 - C*a^12
+ 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^2 + C*a^7*b^5 - 3*C*a^9*b^3 + C*a^10*b^2 + 2*B*a^11*b + 2*C*a^11*b))/
(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (32*tan(c/2 + (d*x)/2)*(2*B*b - C*a)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4
*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))*(2*B*b - C*a))/a^3)*1i)/a^3)/((64
*(8*B^3*b^8 - 4*B^3*a*b^7 - 2*C^3*a^7*b - 20*B^3*a^2*b^6 + 6*B^3*a^3*b^5 + 12*B^3*a^4*b^4 - C^3*a^3*b^5 + C^3*
a^4*b^4 + 3*C^3*a^5*b^3 - 2*C^3*a^6*b^2 - 12*B^2*C*a*b^7 + 6*B*C^2*a^2*b^6 - 5*B*C^2*a^3*b^5 - 17*B*C^2*a^4*b^
4 + 9*B*C^2*a^5*b^3 + 11*B*C^2*a^6*b^2 + 8*B^2*C*a^2*b^6 + 32*B^2*C*a^3*b^5 - 13*B^2*C*a^4*b^4 - 20*B^2*C*a^5*
b^3))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + ((2*B*b - C*a)*((32*tan(c/2 + (d*x)/2)*(8*B^2*b^8 + C^2*a^8 - 8*B^2*
a*b^7 - 2*C^2*a^7*b - 16*B^2*a^2*b^6 + 16*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*
a^2*b^6 - 2*C^2*a^3*b^5 - 5*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^
2*b^6 + 18*B*C*a^3*b^5 - 16*B*C*a^4*b^4 - 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) +
(((32*(B*a^7*b^5 - 2*B*a^6*b^6 - C*a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^2 + C*a^7*b^5 - 3*C*a^9*b^3 +
C*a^10*b^2 + 2*B*a^11*b + 2*C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32*tan(c/2 + (d*x)/2)*(2*B*b - C*
a)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*
b^2)))*(2*B*b - C*a))/a^3))/a^3 - ((2*B*b - C*a)*((32*tan(c/2 + (d*x)/2)*(8*B^2*b^8 + C^2*a^8 - 8*B^2*a*b^7 -
2*C^2*a^7*b - 16*B^2*a^2*b^6 + 16*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6
- 2*C^2*a^3*b^5 - 5*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^2*b^6 +
18*B*C*a^3*b^5 - 16*B*C*a^4*b^4 - 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (((32*(B
*a^7*b^5 - 2*B*a^6*b^6 - C*a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^2 + C*a^7*b^5 - 3*C*a^9*b^3 + C*a^10*
b^2 + 2*B*a^11*b + 2*C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (32*tan(c/2 + (d*x)/2)*(2*B*b - C*a)*(2*a^
11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))*(
2*B*b - C*a))/a^3))/a^3))*(2*B*b - C*a)*2i)/(a^3*d) - ((2*tan(c/2 + (d*x)/2)^3*(B*a*b^2 - 2*B*b^3 - B*a^3 + B*
a^2*b + C*a*b^2))/(a^2*(a + b)*(a - b)) - (2*tan(c/2 + (d*x)/2)*(B*a^3 - 2*B*b^3 - B*a*b^2 + B*a^2*b + C*a*b^2
))/(a^2*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/2)^4*(a - b) - 2*b*tan(c/2 + (d*x)/2)^2)) + (b*atan(((b*
(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(8*B^2*b^8 + C^2*a^8 - 8*B^2*a*b^7 - 2*C^2*a^7*b - 16*B^2
*a^2*b^6 + 16*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 5*
C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^2*b^6 + 18*B*C*a^3*b^5 - 16*
B*C*a^4*b^4 - 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (b*((32*(B*a^7*b^5 - 2*B*a^6
*b^6 - C*a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^2 + C*a^7*b^5 - 3*C*a^9*b^3 + C*a^10*b^2 + 2*B*a^11*b +
2*C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32*b*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*B*b
^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(
(a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*B*
b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))*(2*B*b^3 + 2*C*a^3 - 3*B*a^2*b
- C*a*b^2)*1i)/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2) + (b*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/
2)*(8*B^2*b^8 + C^2*a^8 - 8*B^2*a*b^7 - 2*C^2*a^7*b - 16*B^2*a^2*b^6 + 16*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*
a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 5*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*
C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^2*b^6 + 18*B*C*a^3*b^5 - 16*B*C*a^4*b^4 - 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a^6
*b + a^7 - a^4*b^3 - a^5*b^2) - (b*((32*(B*a^7*b^5 - 2*B*a^6*b^6 - C*a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^
10*b^2 + C*a^7*b^5 - 3*C*a^9*b^3 + C*a^10*b^2 + 2*B*a^11*b + 2*C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) -
(32*b*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*B*b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2)*(2*a^11*b - 2*
a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 +
3*a^5*b^4 - 3*a^7*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*B*b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2))/(a^9 - a^3*b
^6 + 3*a^5*b^4 - 3*a^7*b^2))*(2*B*b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2)*1i)/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*
b^2))/((64*(8*B^3*b^8 - 4*B^3*a*b^7 - 2*C^3*a^7*b - 20*B^3*a^2*b^6 + 6*B^3*a^3*b^5 + 12*B^3*a^4*b^4 - C^3*a^3*
b^5 + C^3*a^4*b^4 + 3*C^3*a^5*b^3 - 2*C^3*a^6*b^2 - 12*B^2*C*a*b^7 + 6*B*C^2*a^2*b^6 - 5*B*C^2*a^3*b^5 - 17*B*
C^2*a^4*b^4 + 9*B*C^2*a^5*b^3 + 11*B*C^2*a^6*b^2 + 8*B^2*C*a^2*b^6 + 32*B^2*C*a^3*b^5 - 13*B^2*C*a^4*b^4 - 20*
B^2*C*a^5*b^3))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (b*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(8
*B^2*b^8 + C^2*a^8 - 8*B^2*a*b^7 - 2*C^2*a^7*b - 16*B^2*a^2*b^6 + 16*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*a^5*b
^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 5*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b
^7 - 4*B*C*a^7*b + 8*B*C*a^2*b^6 + 18*B*C*a^3*b^5 - 16*B*C*a^4*b^4 - 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a^6*b +
a^7 - a^4*b^3 - a^5*b^2) + (b*((32*(B*a^7*b^5 - 2*B*a^6*b^6 - C*a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^
2 + C*a^7*b^5 - 3*C*a^9*b^3 + C*a^10*b^2 + 2*B*a^11*b + 2*C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32*b
*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*B*b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2)*(2*a^11*b - 2*a^6*b
^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^
5*b^4 - 3*a^7*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*B*b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2))/(a^9 - a^3*b^6 +
3*a^5*b^4 - 3*a^7*b^2))*(2*B*b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2) - (
b*(-(a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(8*B^2*b^8 + C^2*a^8 - 8*B^2*a*b^7 - 2*C^2*a^7*b - 16*B
^2*a^2*b^6 + 16*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 -
5*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^2*b^6 + 18*B*C*a^3*b^5 - 1
6*B*C*a^4*b^4 - 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (b*((32*(B*a^7*b^5 - 2*B*a
^6*b^6 - C*a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^2 + C*a^7*b^5 - 3*C*a^9*b^3 + C*a^10*b^2 + 2*B*a^11*b
+ 2*C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (32*b*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*B
*b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))
/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*
B*b^3 + 2*C*a^3 - 3*B*a^2*b - C*a*b^2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))*(2*B*b^3 + 2*C*a^3 - 3*B*a^2*
b - C*a*b^2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*B*b^3 + 2*C*a^3 - 3*B*
a^2*b - C*a*b^2)*2i)/(d*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((B + C*cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**3/(a + b*cos(c + d*x))**2, x)

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