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

[Out]

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

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Rubi [A]  time = 0.58, antiderivative size = 180, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 (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*}

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Mathematica [A]  time = 1.79, size = 219, normalized size = 1.22 \[ \frac {2 \cos ^2(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (-\frac {a b \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+\frac {2 \left (a^4 C+3 a^2 A b^2-2 A b^4\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+a A \tan (c+d x)+2 A b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{a^3 d (2 A+C \cos (2 (c+d x))+C)} \]

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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fricas [B]  time = 9.09, size = 842, normalized size = 4.68 \[ \left [\frac {{\left ({\left (C a^{4} b + 3 \, A a^{2} b^{3} - 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{5} + 3 \, A a^{3} b^{2} - 2 \, A a b^{4}\right )} \cos \left (d x + c\right )\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 ) - 2 \, {\left ({\left (A a^{4} b^{2} - 2 \, A a^{2} b^{4} + A b^{6}\right )} \cos \left (d x + c\right )^{2} + {\left (A a^{5} b - 2 \, A a^{3} b^{3} + A a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (A a^{4} b^{2} - 2 \, A a^{2} b^{4} + A b^{6}\right )} \cos \left (d x + c\right )^{2} + {\left (A a^{5} b - 2 \, A a^{3} b^{3} + A a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{6} - 2 \, A a^{4} b^{2} + A a^{2} b^{4} + {\left ({\left (A - C\right )} a^{5} b - {\left (3 \, A - C\right )} a^{3} b^{3} + 2 \, A a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (C a^{4} b + 3 \, A a^{2} b^{3} - 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (C a^{5} + 3 \, A a^{3} b^{2} - 2 \, A a b^{4}\right )} \cos \left (d x + c\right )\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 ({\left (A a^{4} b^{2} - 2 \, A a^{2} b^{4} + A b^{6}\right )} \cos \left (d x + c\right )^{2} + {\left (A a^{5} b - 2 \, A a^{3} b^{3} + A a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A a^{4} b^{2} - 2 \, A a^{2} b^{4} + A b^{6}\right )} \cos \left (d x + c\right )^{2} + {\left (A a^{5} b - 2 \, A a^{3} b^{3} + A a b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{6} - 2 \, A a^{4} b^{2} + A a^{2} b^{4} + {\left ({\left (A - C\right )} a^{5} b - {\left (3 \, A - C\right )} a^{3} b^{3} + 2 \, A a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} d \cos \left (d x + c\right )}\right ] \]

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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giac [B]  time = 0.87, size = 382, normalized size = 2.12 \[ -\frac {2 \, {\left (\frac {{\left (C a^{4} + 3 \, A a^{2} b^{2} - 2 \, A b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {A b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {A b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{4} - a^{2} b^{2}\right )}}\right )}}{d} \]

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

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maple [B]  time = 0.22, size = 394, normalized size = 2.19 \[ -\frac {2 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{d \,a^{2} \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {6 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A \,b^{2}}{d a \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {4 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A \,b^{4}}{d \,a^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 a \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}} \]

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(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A*b^2-4/d/a^3/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan(t
an(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A*b^4+2/d*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))*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 = 8.81, size = 4118, normalized size = 22.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^3*(A*a^3 + 2*A*b^3 - A*a*b^2 - A*a^2*b + C*a^2*b))/(a^2*(a + b)*(a - b)) - (2*tan(c/2 +
(d*x)/2)*(2*A*b^3 - A*a^3 + A*a*b^2 - A*a^2*b + C*a^2*b))/(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)) + (A*b*atan(((A*b*((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A
^2*a*b^7 - 16*A^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3 + 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6
*A*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (2*A*b*((32*(C*a^12 + 2*A*a^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4
+ 3*A*a^9*b^3 + 3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*
b^2) - (64*A*b*tan(c/2 + (d*x)/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^
3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2))))/a^3)*2i)/a^3 + (A*b*((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^
2*a*b^7 - 16*A^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3 + 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6*
A*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (2*A*b*((32*(C*a^12 + 2*A*a^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4
+ 3*A*a^9*b^3 + 3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b
^2) + (64*A*b*tan(c/2 + (d*x)/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^3
*(a^6*b + a^7 - a^4*b^3 - a^5*b^2))))/a^3)*2i)/a^3)/((64*(8*A^3*b^8 - 4*A^3*a*b^7 - 20*A^3*a^2*b^6 + 6*A^3*a^3
*b^5 + 12*A^3*a^4*b^4 + 2*A*C^2*a^7*b - 4*A^2*C*a^3*b^5 - 4*A^2*C*a^4*b^4 + 8*A^2*C*a^5*b^3 + 4*A^2*C*a^6*b^2)
)/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (2*A*b*((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^2*a*b^7 - 16*A
^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3 + 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C*a^6*b^2))/
(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (2*A*b*((32*(C*a^12 + 2*A*a^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4 + 3*A*a^9*b^3
+ 3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (64*A*b*
tan(c/2 + (d*x)/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^3*(a^6*b + a^7
- a^4*b^3 - a^5*b^2))))/a^3))/a^3 - (2*A*b*((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^2*a*b^7 - 16*A^2
*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3 + 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C*a^6*b^2))/(a
^6*b + a^7 - a^4*b^3 - a^5*b^2) + (2*A*b*((32*(C*a^12 + 2*A*a^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4 + 3*A*a^9*b^3 +
3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (64*A*b*ta
n(c/2 + (d*x)/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^3*(a^6*b + a^7 -
a^4*b^3 - a^5*b^2))))/a^3))/a^3))*4i)/(a^3*d) + (atan(((((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^2*a
*b^7 - 16*A^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3 + 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C
*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (((32*(C*a^12 + 2*A*a^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4 + 3*A*a^9
*b^3 + 3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32
*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(C*a^4 - 2*A*b^4 + 3*A*a^2*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)*(C*a^4 - 2*A*b^4 + 3*A*a^2*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)*(C*a^4 - 2*A*b^4 + 3*A*a^2*b^2)*1i)/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2) +
(((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^2*a*b^7 - 16*A^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4
- 8*A^2*a^5*b^3 + 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (((32*(
C*a^12 + 2*A*a^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4 + 3*A*a^9*b^3 + 3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^1
1*b - C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (32*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(C*a^
4 - 2*A*b^4 + 3*A*a^2*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)*(C*a^4 - 2*A*b
^4 + 3*A*a^2*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)*(C*a^4 - 2*A*b^4 + 3*
A*a^2*b^2)*1i)/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))/((64*(8*A^3*b^8 - 4*A^3*a*b^7 - 20*A^3*a^2*b^6 + 6*A^3
*a^3*b^5 + 12*A^3*a^4*b^4 + 2*A*C^2*a^7*b - 4*A^2*C*a^3*b^5 - 4*A^2*C*a^4*b^4 + 8*A^2*C*a^5*b^3 + 4*A^2*C*a^6*
b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (((32*tan(c/2 + (d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^2*a*b^7 - 16*A^2
*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3 + 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C*a^6*b^2))/(a
^6*b + a^7 - a^4*b^3 - a^5*b^2) + (((32*(C*a^12 + 2*A*a^6*b^6 - A*a^7*b^5 - 5*A*a^8*b^4 + 3*A*a^9*b^3 + 3*A*a^
10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32*tan(c/2 + (d
*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(C*a^4 - 2*A*b^4 + 3*A*a^2*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)*(C*a^4 - 2*A*b^4 + 3*A*a^2*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)*(C*a^4 - 2*A*b^4 + 3*A*a^2*b^2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2) + (((32*tan(c/2 +
(d*x)/2)*(8*A^2*b^8 + C^2*a^8 - 8*A^2*a*b^7 - 16*A^2*a^2*b^6 + 16*A^2*a^3*b^5 + 5*A^2*a^4*b^4 - 8*A^2*a^5*b^3
+ 4*A^2*a^6*b^2 - 4*A*C*a^4*b^4 + 6*A*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (((32*(C*a^12 + 2*A*a^6
*b^6 - A*a^7*b^5 - 5*A*a^8*b^4 + 3*A*a^9*b^3 + 3*A*a^10*b^2 + C*a^9*b^3 - C*a^10*b^2 - 2*A*a^11*b - C*a^11*b))
/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (32*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(C*a^4 - 2*A*b^4 + 3*
A*a^2*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)*(C*a^4 - 2*A*b^4 + 3*A*a^2*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)*(C*a^4 - 2*A*b^4 + 3*A*a^2*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)*(C*a^4 - 2*A*b^4 + 3*A*a^2*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 \[ \int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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