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

Optimal. Leaf size=339 \[ -\frac{\left (-a^4 b^2 (12 A+C)+15 a^2 A b^4-5 a^3 b^3 B+6 a^5 b B-2 a^6 C+2 a b^5 B-6 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\tan (c+d x) \left (11 a^2 A b^2+a^4 (-(2 A-3 C))-5 a^3 b B+2 a b^3 B-6 A b^4\right )}{2 a^3 d \left (a^2-b^2\right )^2}-\frac{\tan (c+d x) \left (-a^2 b^2 (6 A+C)+4 a^3 b B-2 a^4 C-a b^3 B+3 A b^4\right )}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]

[Out]

-(((15*a^2*A*b^4 - 6*A*b^6 + 6*a^5*b*B - 5*a^3*b^3*B + 2*a*b^5*B - 2*a^6*C - a^4*b^2*(12*A + C))*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) - ((11*a^2*A*b^2 - 6*A*b^4 - 5*a^3*b*B + 2*a*b^3*B - a^4*(2*A - 3*C))*Tan[c + d*x])/(2*a^3*(a^2
- b^2)^2*d) + ((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((3*A*b^4 +
4*a^3*b*B - a*b^3*B - 2*a^4*C - a^2*b^2*(6*A + C))*Tan[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 3.41858, antiderivative size = 339, 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} \[ -\frac{\left (-a^4 b^2 (12 A+C)+15 a^2 A b^4-5 a^3 b^3 B+6 a^5 b B-2 a^6 C+2 a b^5 B-6 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\tan (c+d x) \left (11 a^2 A b^2+a^4 (-(2 A-3 C))-5 a^3 b B+2 a b^3 B-6 A b^4\right )}{2 a^3 d \left (a^2-b^2\right )^2}-\frac{\tan (c+d x) \left (-a^2 b^2 (6 A+C)+4 a^3 b B-2 a^4 C-a b^3 B+3 A b^4\right )}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((15*a^2*A*b^4 - 6*A*b^6 + 6*a^5*b*B - 5*a^3*b^3*B + 2*a*b^5*B - 2*a^6*C - a^4*b^2*(12*A + C))*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) - ((11*a^2*A*b^2 - 6*A*b^4 - 5*a^3*b*B + 2*a*b^3*B - a^4*(2*A - 3*C))*Tan[c + d*x])/(2*a^3*(a^2
- b^2)^2*d) + ((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((3*A*b^4 +
4*a^3*b*B - a*b^3*B - 2*a^4*C - a^2*b^2*(6*A + C))*Tan[c + d*x])/(2*a^2*(a^2 - b^2)^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 ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (-3 A b^2+a b B+a^2 (2 A-C)-2 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 ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\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 (-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B+a^4 (2 A-3 C)+a \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \cos (c+d x)-\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+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 )^2}\\ &=-\frac{\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\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 \left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+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 )^2}\\ &=-\frac{\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\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 (15 a^2 A b^4-6 A b^6+6 a^5 b B-5 a^3 b^3 B+2 a b^5 B-2 a^6 C-a^4 b^2 (12 A+C)\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 (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (15 a^2 A b^4-6 A b^6+6 a^5 b B-5 a^3 b^3 B+2 a b^5 B-2 a^6 C-a^4 b^2 (12 A+C)\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 )^2 d}\\ &=\frac{\left (12 a^4 A b^2-15 a^2 A b^4+6 A b^6-6 a^5 b B+5 a^3 b^3 B-2 a b^5 B+2 a^6 C+a^4 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)^{5/2} (a+b)^{5/2} d}-\frac{(3 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.40768, size = 444, normalized size = 1.31 \[ \frac{\cos (c+d x) \left (A \sec ^2(c+d x)+B \sec (c+d x)+C\right ) \left (\frac{2 a \sin (c+d x) \left (2 a b \cos (c+d x) \left (a^2 b^2 (C-16 A)+4 a^4 (A-C)+6 a^3 b B-3 a b^3 B+9 A b^4\right )+b^2 \cos (2 (c+d x)) \left (-11 a^2 A b^2+a^4 (2 A-3 C)+5 a^3 b B-2 a b^3 B+6 A b^4\right )-6 a^4 A b^2-7 a^2 A b^4+4 a^6 A+5 a^3 b^3 B-3 a^4 b^2 C-2 a b^5 B+6 A b^6\right )}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac{8 \cos (c+d x) \left (a^4 b^2 (12 A+C)-15 a^2 A b^4+5 a^3 b^3 B-6 a^5 b B+2 a^6 C-2 a b^5 B+6 A b^6\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+8 (3 A b-a B) \cos (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+8 (a B-3 A b) \cos (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 a^4 d (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*((-8*(-15*a^2*A*b^4 + 6*A*b^6 - 6*a^5*b*B + 5*a^3*b^3*B
- 2*a*b^5*B + 2*a^6*C + a^4*b^2*(12*A + C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]*Cos[c + d*x])
/(-a^2 + b^2)^(5/2) + 8*(3*A*b - a*B)*Cos[c + d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*(-3*A*b + a*B)
*Cos[c + d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*a*(4*a^6*A - 6*a^4*A*b^2 - 7*a^2*A*b^4 + 6*A*b^6 +
5*a^3*b^3*B - 2*a*b^5*B - 3*a^4*b^2*C + 2*a*b*(9*A*b^4 + 6*a^3*b*B - 3*a*b^3*B + 4*a^4*(A - C) + a^2*b^2*(-16
*A + C))*Cos[c + d*x] + b^2*(-11*a^2*A*b^2 + 6*A*b^4 + 5*a^3*b*B - 2*a*b^3*B + a^4*(2*A - 3*C))*Cos[2*(c + d*x
)])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2)))/(4*a^4*d*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c +
d*x)]))

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Maple [B]  time = 0.104, size = 1750, normalized size = 5.2 \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)^2/(a+b*cos(d*x+c))^3,x)

[Out]

-4/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*a*b*C+1/
d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B-2/d
/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^4/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B-1/
d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^3/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-2/d/a^2/(a*ta
n(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^4/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B+12/d*b^2/(a^4-2*a^2*b^
2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+4/d/a^3/(a*tan(1/2*d*x+1/2*c
)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*b^5/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+1/d/(a*tan(1/2*d*x+1/2*c)^2
-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*b^2*C+2/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1
/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C*a^2+1/d/a^3*B*ln(tan(1/2*d*x+1/2*c)+1)-1/d/a^3*B*ln
(tan(1/2*d*x+1/2*c)-1)+1/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b
))^(1/2))*b^2*C-1/d/a^3*A/(tan(1/2*d*x+1/2*c)-1)-1/d/a^3*A/(tan(1/2*d*x+1/2*c)+1)-6/d*a*b/(a^4-2*a^2*b^2+b^4)/
((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-1/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2
*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b^2*C+6/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*
x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b^2*B-8/d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x
+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A*b^3+4/d*b^5/a^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*
c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A-8/d/a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(
a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^3-1/d/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/
(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^4+1/d/a^2/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2
/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A*b^4-4/d*a/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b
)^2*tan(1/2*d*x+1/2*c)*b*C-2/d/a^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a
+b)*(a-b))^(1/2))*b^5*B+3/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)*A*b+5/d/a/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arc
tan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*b^3*B+6/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b
)^2*b^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-15/d/a^2/(a^4-2*a^2*b^2+b^4)/((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^4+6/d/a^4/(a^4-2*a^2*b^2+b^4)/((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^6-3/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)*A*b

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

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [B]  time = 1.34014, size = 944, normalized size = 2.78 \begin{align*} \text{result too large to display} \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)^2/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-((2*C*a^6 - 6*B*a^5*b + 12*A*a^4*b^2 + C*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*floo
r(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)) + (4*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 6*B*a^4*b^2*tan(
1/2*d*x + 1/2*c)^3 - 3*C*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 - C*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 + 4*C*a^5*b*tan(1/2*d*x + 1/2*c) - 6*B*a^4*b^2*tan(1/2*d*x + 1/2*c) + 3*C*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) - C*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(tan(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