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

Optimal. Leaf size=180 \[ \frac{2 \left (3 a^2 A b^2+a^4 C-2 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{2 A b \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.581978, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3056, 3055, 3001, 3770, 2659, 205} \[ \frac{2 \left (3 a^2 A b^2+a^4 C-2 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{2 A b \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + 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] + 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*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 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+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 A b^2+a^2 (A-C)-a b (A+C) \cos (c+d x)+\left (A b^2+a^2 C\right ) \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 (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 A b \left (a^2-b^2\right )+a \left (A b^2+a^2 C\right ) \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 (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{(2 A b) \int \sec (c+d x) \, dx}{a^3}+\frac{\left (3 a^2 A b^2-2 A b^4+a^4 C\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac{2 A b \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (2 \left (3 a^2 A b^2-2 A b^4+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^3 \left (a^2-b^2\right ) d}\\ &=\frac{2 \left (3 a^2 A b^2-2 A b^4+a^4 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 A b \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{\left (2 A b^2-a^2 (A-C)\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.74338, size = 219, normalized size = 1.22 \[ \frac{2 \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (-\frac{a b \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+\frac{2 \left (3 a^2 A b^2+a^4 C-2 A b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+a A \tan (c+d x)+2 A b \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{a^3 d (2 A+C \cos (2 (c+d x))+C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.076, size = 394, normalized size = 2.2 \begin{align*} -2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) A}{d{a}^{2} \left ({a}^{2}-{b}^{2} \right ) \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}-2\,{\frac{b\tan \left ( 1/2\,dx+c/2 \right ) C}{d \left ({a}^{2}-{b}^{2} \right ) \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+6\,{\frac{A{b}^{2}}{da \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-4\,{\frac{A{b}^{4}}{d{a}^{3} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{aC}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{Ab\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{3}}}-{\frac{A}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{A}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{Ab\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 20.9006, size = 1887, normalized size = 10.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((C*a^4*b + 3*A*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^2 + (C*a^5 + 3*A*a^3*b^2 - 2*A*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)) - 2*((A*a^4*b^2 - 2*A*a^2*b^4 + A
*b^6)*cos(d*x + c)^2 + (A*a^5*b - 2*A*a^3*b^3 + A*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*((A*a^4*b^2 -
2*A*a^2*b^4 + A*b^6)*cos(d*x + c)^2 + (A*a^5*b - 2*A*a^3*b^3 + A*a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1)
+ 2*(A*a^6 - 2*A*a^4*b^2 + A*a^2*b^4 + ((A - C)*a^5*b - (3*A - C)*a^3*b^3 + 2*A*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)), (((C*a^4*b
+ 3*A*a^2*b^3 - 2*A*b^5)*cos(d*x + c)^2 + (C*a^5 + 3*A*a^3*b^2 - 2*A*a*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arc
tan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - ((A*a^4*b^2 - 2*A*a^2*b^4 + A*b^6)*cos(d*x + c)^2
+ (A*a^5*b - 2*A*a^3*b^3 + A*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((A*a^4*b^2 - 2*A*a^2*b^4 + A*b^6)*c
os(d*x + c)^2 + (A*a^5*b - 2*A*a^3*b^3 + A*a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) + (A*a^6 - 2*A*a^4*b^2
+ A*a^2*b^4 + ((A - C)*a^5*b - (3*A - C)*a^3*b^3 + 2*A*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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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+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.30631, size = 516, normalized size = 2.87 \begin{align*} -\frac{2 \,{\left (\frac{{\left (C a^{4} + 3 \, A a^{2} b^{2} - 2 \, A 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{A b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{A b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\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 )}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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