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

Optimal. Leaf size=307 \[ -\frac{2 b \left (4 a^2 A b^2-a^2 b^2 C-3 a^3 b B+2 a^4 C+2 a b^3 B-3 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\tan (c+d x) \left (-a^2 b (2 A-C)+a^3 B-2 a b^2 B+3 A b^3\right )}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 (A+2 C)-4 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{\tan (c+d x) \sec (c+d x) \left (a^2 (-(A-2 C))-2 a b B+3 A b^2\right )}{2 a^2 d \left (a^2-b^2\right )}+\frac{\tan (c+d x) \sec (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

[Out]

(-2*b*(4*a^2*A*b^2 - 3*A*b^4 - 3*a^3*b*B + 2*a*b^3*B + 2*a^4*C - a^2*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/
2])/Sqrt[a + b]])/(a^4*(a - b)^(3/2)*(a + b)^(3/2)*d) + ((6*A*b^2 - 4*a*b*B + a^2*(A + 2*C))*ArcTanh[Sin[c + d
*x]])/(2*a^4*d) + ((3*A*b^3 + a^3*B - 2*a*b^2*B - a^2*b*(2*A - C))*Tan[c + d*x])/(a^3*(a^2 - b^2)*d) - ((3*A*b
^2 - 2*a*b*B - a^2*(A - 2*C))*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*(a^2 - b^2)*d) + ((A*b^2 - a*(b*B - a*C))*Sec[
c + d*x]*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.40245, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3055, 3001, 3770, 2659, 205} \[ -\frac{2 b \left (4 a^2 A b^2-a^2 b^2 C-3 a^3 b B+2 a^4 C+2 a b^3 B-3 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\tan (c+d x) \left (-a^2 b (2 A-C)+a^3 B-2 a b^2 B+3 A b^3\right )}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 (A+2 C)-4 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{\tan (c+d x) \sec (c+d x) \left (a^2 (-(A-2 C))-2 a b B+3 A b^2\right )}{2 a^2 d \left (a^2-b^2\right )}+\frac{\tan (c+d x) \sec (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[((A + 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*(4*a^2*A*b^2 - 3*A*b^4 - 3*a^3*b*B + 2*a*b^3*B + 2*a^4*C - a^2*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/
2])/Sqrt[a + b]])/(a^4*(a - b)^(3/2)*(a + b)^(3/2)*d) + ((6*A*b^2 - 4*a*b*B + a^2*(A + 2*C))*ArcTanh[Sin[c + d
*x]])/(2*a^4*d) + ((3*A*b^3 + a^3*B - 2*a*b^2*B - a^2*b*(2*A - C))*Tan[c + d*x])/(a^3*(a^2 - b^2)*d) - ((3*A*b
^2 - 2*a*b*B - a^2*(A - 2*C))*Sec[c + d*x]*Tan[c + d*x])/(2*a^2*(a^2 - b^2)*d) + ((A*b^2 - a*(b*B - a*C))*Sec[
c + d*x]*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

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 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 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]

Rubi steps

\begin{align*} \int \frac{\left (A+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 &=\frac{\left (A b^2-a (b B-a C)\right ) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-3 A b^2+2 a b B+a^2 (A-2 C)-a (A b-a B+b C) \cos (c+d x)+2 \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (2 \left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right )+a \left (A b^2-2 a b B+a^2 (A+2 C)\right ) \cos (c+d x)-b \left (3 A b^2-2 a b B-a^2 (A-2 C)\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 )}\\ &=\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sec (c+d x) \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 ) \left (6 A b^2-4 a b B+a^2 (A+2 C)\right )-a b \left (3 A b^2-2 a b B-a^2 (A-2 C)\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 )}\\ &=\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (b \left (3 A b^4+3 a^3 b B-2 a b^3 B-a^2 b^2 (4 A-C)-2 a^4 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )}+\frac{\left (6 A b^2-4 a b B+a^2 (A+2 C)\right ) \int \sec (c+d x) \, dx}{2 a^4}\\ &=\frac{\left (6 A b^2-4 a b B+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sec (c+d x) \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 b^4+3 a^3 b B-2 a b^3 B-a^2 b^2 (4 A-C)-2 a^4 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^4 \left (a^2-b^2\right ) d}\\ &=-\frac{2 b \left (4 a^2 A b^2-3 A b^4-3 a^3 b B+2 a b^3 B+2 a^4 C-a^2 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^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (6 A b^2-4 a b B+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.23801, size = 389, normalized size = 1.27 \[ \frac{\frac{8 b \left (a^2 b^2 (C-4 A)+3 a^3 b B-2 a^4 C-2 a b^3 B+3 A b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-2 \left (a^2 (A+2 C)-4 a b B+6 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (a^2 (A+2 C)-4 a b B+6 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^2 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2 A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{4 a b^2 \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b) (a+b \cos (c+d x))}+\frac{4 a (a B-2 A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a (a B-2 A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}}{4 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + 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]

((8*b*(3*A*b^4 + 3*a^3*b*B - 2*a*b^3*B - 2*a^4*C + a^2*b^2*(-4*A + C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt
[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) - 2*(6*A*b^2 - 4*a*b*B + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)
/2]] + 2*(6*A*b^2 - 4*a*b*B + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*A)/(Cos[(c + d*x)
/2] - Sin[(c + d*x)/2])^2 + (4*a*(-2*A*b + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (a^2
*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a*(-2*A*b + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]) + (4*a*b^2*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])))/(4*a^4
*d)

________________________________________________________________________________________

Maple [B]  time = 0.101, size = 914, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+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]

-1/d/a^2*B/(tan(1/2*d*x+1/2*c)+1)-1/d/a^2*B/(tan(1/2*d*x+1/2*c)-1)-1/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)*C+1/2/d/a^
2*A/(tan(1/2*d*x+1/2*c)-1)^2+1/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)*C-1/2/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)^2+1/2/d/a^2
*A/(tan(1/2*d*x+1/2*c)-1)+1/2/d/a^2*A/(tan(1/2*d*x+1/2*c)+1)+1/2/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)-1/2/d/a^2*A*
ln(tan(1/2*d*x+1/2*c)-1)-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-8/d/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*b^3+2
/d*b^3/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-4/d/(a+b)/(a
-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C*b+6/d*b^2/a/(a+b)/(a-b)/((a+b)*
(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-4/d*b^4/a^3/(a+b)/(a-b)/((a+b)*(a-b))^(1/2
)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+6/d/a^4*b^5/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a
-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+3/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*A*b^2+2/d/a^3/(tan(1/2*d*x+1/2*
c)-1)*A*b-3/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*A*b^2+2/d/a^3/(tan(1/2*d*x+1/2*c)+1)*A*b+2/d/a^3*ln(tan(1/2*d*x+1/2
*c)-1)*b*B-2/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*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+2/d/a^3*b^4/(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)*A

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.27294, size = 571, normalized size = 1.86 \begin{align*} \frac{\frac{4 \,{\left (2 \, C a^{4} b - 3 \, B a^{3} b^{2} + 4 \, A a^{2} b^{3} - C a^{2} b^{3} + 2 \, B a b^{4} - 3 \, A b^{5}\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^{6} - a^{4} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{4 \,{\left (C a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\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 )}} + \frac{{\left (A a^{2} + 2 \, C a^{2} - 4 \, B a b + 6 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{{\left (A a^{2} + 2 \, C a^{2} - 4 \, B a b + 6 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{2 \,{\left (A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, A b \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 )}^{2} a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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]

1/2*(4*(2*C*a^4*b - 3*B*a^3*b^2 + 4*A*a^2*b^3 - C*a^2*b^3 + 2*B*a*b^4 - 3*A*b^5)*(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^6 - a^4
*b^2)*sqrt(a^2 - b^2)) + 4*(C*a^2*b^2*tan(1/2*d*x + 1/2*c) - B*a*b^3*tan(1/2*d*x + 1/2*c) + A*b^4*tan(1/2*d*x
+ 1/2*c))/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)) + (A*a^2 + 2*C*a^2 -
4*B*a*b + 6*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - (A*a^2 + 2*C*a^2 - 4*B*a*b + 6*A*b^2)*log(abs(tan
(1/2*d*x + 1/2*c) - 1))/a^4 + 2*(A*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*a*tan(1/2*d*x + 1/2*c)^3 + 4*A*b*tan(1/2*d*x
+ 1/2*c)^3 + A*a*tan(1/2*d*x + 1/2*c) + 2*B*a*tan(1/2*d*x + 1/2*c) - 4*A*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*
x + 1/2*c)^2 - 1)^2*a^3))/d