\(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [455]

Optimal result

Integrand size = 41, antiderivative size = 108 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {(3 A-2 B+2 C) x}{2 a}-\frac {(2 A-2 B+C) \sin (c+d x)}{a d}+\frac {(3 A-2 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

1/2*(3*A-2*B+2*C)*x/a-(2*A-2*B+C)*sin(d*x+c)/a/d+1/2*(3*A-2*B+2*C)*cos(d*x+c)*sin(d*x+c)/a/d-(A-B+C)*cos(d*x+c
)*sin(d*x+c)/d/(a+a*sec(d*x+c))

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4169, 3872, 2715, 8, 2717} \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {(2 A-2 B+C) \sin (c+d x)}{a d}+\frac {(3 A-2 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}+\frac {x (3 A-2 B+2 C)}{2 a} \]

[In]

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

[Out]

((3*A - 2*B + 2*C)*x)/(2*a) - ((2*A - 2*B + C)*Sin[c + d*x])/(a*d) + ((3*A - 2*B + 2*C)*Cos[c + d*x]*Sin[c + d
*x])/(2*a*d) - ((A - B + C)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4169

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*C
sc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m -
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {\int \cos ^2(c+d x) (a (3 A-2 B+2 C)-a (2 A-2 B+C) \sec (c+d x)) \, dx}{a^2} \\ & = -\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(2 A-2 B+C) \int \cos (c+d x) \, dx}{a}+\frac {(3 A-2 B+2 C) \int \cos ^2(c+d x) \, dx}{a} \\ & = -\frac {(2 A-2 B+C) \sin (c+d x)}{a d}+\frac {(3 A-2 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {(3 A-2 B+2 C) \int 1 \, dx}{2 a} \\ & = \frac {(3 A-2 B+2 C) x}{2 a}-\frac {(2 A-2 B+C) \sin (c+d x)}{a d}+\frac {(3 A-2 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.97 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (4 (3 A-2 B+2 C) d x \cos \left (\frac {d x}{2}\right )+4 (3 A-2 B+2 C) d x \cos \left (c+\frac {d x}{2}\right )-20 A \sin \left (\frac {d x}{2}\right )+20 B \sin \left (\frac {d x}{2}\right )-16 C \sin \left (\frac {d x}{2}\right )-4 A \sin \left (c+\frac {d x}{2}\right )+4 B \sin \left (c+\frac {d x}{2}\right )-3 A \sin \left (c+\frac {3 d x}{2}\right )+4 B \sin \left (c+\frac {3 d x}{2}\right )-3 A \sin \left (2 c+\frac {3 d x}{2}\right )+4 B \sin \left (2 c+\frac {3 d x}{2}\right )+A \sin \left (2 c+\frac {5 d x}{2}\right )+A \sin \left (3 c+\frac {5 d x}{2}\right )\right )}{8 a d (1+\cos (c+d x))} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(4*(3*A - 2*B + 2*C)*d*x*Cos[(d*x)/2] + 4*(3*A - 2*B + 2*C)*d*x*Cos[c + (d*x)/2] -
20*A*Sin[(d*x)/2] + 20*B*Sin[(d*x)/2] - 16*C*Sin[(d*x)/2] - 4*A*Sin[c + (d*x)/2] + 4*B*Sin[c + (d*x)/2] - 3*A*
Sin[c + (3*d*x)/2] + 4*B*Sin[c + (3*d*x)/2] - 3*A*Sin[2*c + (3*d*x)/2] + 4*B*Sin[2*c + (3*d*x)/2] + A*Sin[2*c
+ (5*d*x)/2] + A*Sin[3*c + (5*d*x)/2]))/(8*a*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\(\frac {\left (A \cos \left (2 d x +2 c \right )+\left (-2 A +4 B \right ) \cos \left (d x +c \right )-7 A +8 B -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+6 x d \left (A -\frac {2 B}{3}+\frac {2 C}{3}\right )}{4 d a}\)	\(67\)
derivativedivides	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {2 \left (-\frac {3 A}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-\frac {A}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (3 A -2 B +2 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(115\)
default	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {2 \left (-\frac {3 A}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+2 \left (-\frac {A}{2}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (3 A -2 B +2 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(115\)
risch	\(\frac {3 A x}{2 a}-\frac {B x}{a}+\frac {x C}{a}+\frac {i A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {i B \,{\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {i A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a d}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {A \sin \left (2 d x +2 c \right )}{4 a d}\)	\(185\)
norman	\(\frac {\frac {\left (2 A -3 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\left (3 A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}-\frac {\left (3 A -2 B +2 C \right ) x}{2 a}-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (3 A -2 B +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}+\frac {\left (3 A -2 B +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}+\frac {\left (3 A -2 B +2 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}-\frac {\left (4 A -3 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(232\)

[In]

int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/4*((A*cos(2*d*x+2*c)+(-2*A+4*B)*cos(d*x+c)-7*A+8*B-4*C)*tan(1/2*d*x+1/2*c)+6*x*d*(A-2/3*B+2/3*C))/d/a

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {{\left (3 \, A - 2 \, B + 2 \, C\right )} d x \cos \left (d x + c\right ) + {\left (3 \, A - 2 \, B + 2 \, C\right )} d x + {\left (A \cos \left (d x + c\right )^{2} - {\left (A - 2 \, B\right )} \cos \left (d x + c\right ) - 4 \, A + 4 \, B - 2 \, C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

1/2*((3*A - 2*B + 2*C)*d*x*cos(d*x + c) + (3*A - 2*B + 2*C)*d*x + (A*cos(d*x + c)^2 - (A - 2*B)*cos(d*x + c) -
4*A + 4*B - 2*C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \cos ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

(Integral(A*cos(c + d*x)**2/(sec(c + d*x) + 1), x) + Integral(B*cos(c + d*x)**2*sec(c + d*x)/(sec(c + d*x) + 1
), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2/(sec(c + d*x) + 1), x))/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (104) = 208\).

Time = 0.32 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.53 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {A {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + B {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]

[In]

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

[Out]

-(A*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a + 2*a*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + sin(d*x
+ c)/(a*(cos(d*x + c) + 1))) + B*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 2*sin(d*x + c)/((a + a*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))) - C*(2*arctan(sin(d*
x + c)/(cos(d*x + c) + 1))/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} {\left (3 \, A - 2 \, B + 2 \, C\right )}}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 16.80 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {x\,\left (3\,A-2\,B+2\,C\right )}{2\,a}-\frac {\left (3\,A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A-2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d} \]

[In]

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

[Out]

(x*(3*A - 2*B + 2*C))/(2*a) - (tan(c/2 + (d*x)/2)^3*(3*A - 2*B) + tan(c/2 + (d*x)/2)*(A - 2*B))/(d*(a + 2*a*ta
n(c/2 + (d*x)/2)^2 + a*tan(c/2 + (d*x)/2)^4)) - (tan(c/2 + (d*x)/2)*(A - B + C))/(a*d)