\(\int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [98]

Optimal result

Integrand size = 31, antiderivative size = 170 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {(10 A-7 B) x}{2 a^2}+\frac {4 (3 A-2 B) \sin (c+d x)}{a^2 d}-\frac {(10 A-7 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 (3 A-2 B) \sin ^3(c+d x)}{3 a^2 d} \]

[Out]

-1/2*(10*A-7*B)*x/a^2+4*(3*A-2*B)*sin(d*x+c)/a^2/d-1/2*(10*A-7*B)*cos(d*x+c)*sin(d*x+c)/a^2/d-1/3*(10*A-7*B)*c
os(d*x+c)^2*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^2-4/3*(3*A-2*
B)*sin(d*x+c)^3/a^2/d

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4105, 3872, 2713, 2715, 8} \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {4 (3 A-2 B) \sin ^3(c+d x)}{3 a^2 d}+\frac {4 (3 A-2 B) \sin (c+d x)}{a^2 d}-\frac {(10 A-7 B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {x (10 A-7 B)}{2 a^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[In]

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

[Out]

-1/2*((10*A - 7*B)*x)/a^2 + (4*(3*A - 2*B)*Sin[c + d*x])/(a^2*d) - ((10*A - 7*B)*Cos[c + d*x]*Sin[c + d*x])/(2
*a^2*d) - ((10*A - 7*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Cos[c + d*x]^2*Si
n[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) - (4*(3*A - 2*B)*Sin[c + d*x]^3)/(3*a^2*d)

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 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 4105

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos ^3(c+d x) (3 a (2 A-B)-4 a (A-B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2} \\ & = -\frac {(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \cos ^3(c+d x) \left (12 a^2 (3 A-2 B)-3 a^2 (10 A-7 B) \sec (c+d x)\right ) \, dx}{3 a^4} \\ & = -\frac {(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(10 A-7 B) \int \cos ^2(c+d x) \, dx}{a^2}+\frac {(4 (3 A-2 B)) \int \cos ^3(c+d x) \, dx}{a^2} \\ & = -\frac {(10 A-7 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(10 A-7 B) \int 1 \, dx}{2 a^2}-\frac {(4 (3 A-2 B)) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d} \\ & = -\frac {(10 A-7 B) x}{2 a^2}+\frac {4 (3 A-2 B) \sin (c+d x)}{a^2 d}-\frac {(10 A-7 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 (3 A-2 B) \sin ^3(c+d x)}{3 a^2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(369\) vs. \(2(170)=340\).

Time = 2.03 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.17 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-36 (10 A-7 B) d x \cos \left (\frac {d x}{2}\right )-36 (10 A-7 B) d x \cos \left (c+\frac {d x}{2}\right )-120 A d x \cos \left (c+\frac {3 d x}{2}\right )+84 B d x \cos \left (c+\frac {3 d x}{2}\right )-120 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+84 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+516 A \sin \left (\frac {d x}{2}\right )-381 B \sin \left (\frac {d x}{2}\right )-156 A \sin \left (c+\frac {d x}{2}\right )+147 B \sin \left (c+\frac {d x}{2}\right )+342 A \sin \left (c+\frac {3 d x}{2}\right )-239 B \sin \left (c+\frac {3 d x}{2}\right )+118 A \sin \left (2 c+\frac {3 d x}{2}\right )-63 B \sin \left (2 c+\frac {3 d x}{2}\right )+30 A \sin \left (2 c+\frac {5 d x}{2}\right )-15 B \sin \left (2 c+\frac {5 d x}{2}\right )+30 A \sin \left (3 c+\frac {5 d x}{2}\right )-15 B \sin \left (3 c+\frac {5 d x}{2}\right )-3 A \sin \left (3 c+\frac {7 d x}{2}\right )+3 B \sin \left (3 c+\frac {7 d x}{2}\right )-3 A \sin \left (4 c+\frac {7 d x}{2}\right )+3 B \sin \left (4 c+\frac {7 d x}{2}\right )+A \sin \left (4 c+\frac {9 d x}{2}\right )+A \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(10*A - 7*B)*d*x*Cos[(d*x)/2] - 36*(10*A - 7*B)*d*x*Cos[c + (d*x)/2] - 120*A*d
*x*Cos[c + (3*d*x)/2] + 84*B*d*x*Cos[c + (3*d*x)/2] - 120*A*d*x*Cos[2*c + (3*d*x)/2] + 84*B*d*x*Cos[2*c + (3*d
*x)/2] + 516*A*Sin[(d*x)/2] - 381*B*Sin[(d*x)/2] - 156*A*Sin[c + (d*x)/2] + 147*B*Sin[c + (d*x)/2] + 342*A*Sin
[c + (3*d*x)/2] - 239*B*Sin[c + (3*d*x)/2] + 118*A*Sin[2*c + (3*d*x)/2] - 63*B*Sin[2*c + (3*d*x)/2] + 30*A*Sin
[2*c + (5*d*x)/2] - 15*B*Sin[2*c + (5*d*x)/2] + 30*A*Sin[3*c + (5*d*x)/2] - 15*B*Sin[3*c + (5*d*x)/2] - 3*A*Si
n[3*c + (7*d*x)/2] + 3*B*Sin[3*c + (7*d*x)/2] - 3*A*Sin[4*c + (7*d*x)/2] + 3*B*Sin[4*c + (7*d*x)/2] + A*Sin[4*
c + (9*d*x)/2] + A*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (28 A -12 B \right ) \cos \left (2 d x +2 c \right )+\left (-2 A +3 B \right ) \cos \left (3 d x +3 c \right )+A \cos \left (4 d x +4 c \right )+\left (258 A -163 B \right ) \cos \left (d x +c \right )+219 A -140 B \right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-240 d x \left (A -\frac {7 B}{10}\right )}{48 a^{2} d}\)	\(106\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {8 \left (\left (-\frac {5 A}{2}+\frac {5 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {10 A}{3}+2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {3 A}{2}+\frac {3 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-2 \left (10 A -7 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(154\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {8 \left (\left (-\frac {5 A}{2}+\frac {5 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {10 A}{3}+2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {3 A}{2}+\frac {3 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-2 \left (10 A -7 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(154\)
norman	\(\frac {\frac {\left (4 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}+\frac {\left (23 A -15 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {\left (10 A -7 B \right ) x}{2 a}-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {3 \left (10 A -7 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {3 \left (10 A -7 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}-\frac {\left (10 A -7 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}+\frac {5 \left (16 A -11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\left (21 A -13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a}\)	\(230\)
risch	\(-\frac {5 A x}{a^{2}}+\frac {7 x B}{2 a^{2}}+\frac {i A \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B}{8 a^{2} d}-\frac {15 i A \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{a^{2} d}+\frac {15 i A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{a^{2} d}-\frac {i A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{8 a^{2} d}+\frac {2 i \left (15 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B \,{\mathrm e}^{2 i \left (d x +c \right )}+27 \,{\mathrm e}^{i \left (d x +c \right )} A -21 B \,{\mathrm e}^{i \left (d x +c \right )}+14 A -11 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {A \sin \left (3 d x +3 c \right )}{12 a^{2} d}\)	\(263\)

[In]

int(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/48*(tan(1/2*d*x+1/2*c)*((28*A-12*B)*cos(2*d*x+2*c)+(-2*A+3*B)*cos(3*d*x+3*c)+A*cos(4*d*x+4*c)+(258*A-163*B)*
cos(d*x+c)+219*A-140*B)*sec(1/2*d*x+1/2*c)^2-240*d*x*(A-7/10*B))/a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {3 \, {\left (10 \, A - 7 \, B\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (10 \, A - 7 \, B\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (10 \, A - 7 \, B\right )} d x - {\left (2 \, A \cos \left (d x + c\right )^{4} - {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (66 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 48 \, A - 32 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(3*(10*A - 7*B)*d*x*cos(d*x + c)^2 + 6*(10*A - 7*B)*d*x*cos(d*x + c) + 3*(10*A - 7*B)*d*x - (2*A*cos(d*x
+ c)^4 - (2*A - 3*B)*cos(d*x + c)^3 + 6*(2*A - B)*cos(d*x + c)^2 + (66*A - 43*B)*cos(d*x + c) + 48*A - 32*B)*s
in(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [F]

\[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \cos ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(cos(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(A*cos(c + d*x)**3/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*cos(c + d*x)**3*sec(c + d*
x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (160) = 320\).

Time = 0.30 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.19 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {A {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(c
os(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arct
an(sin(d*x + c)/(cos(d*x + c) + 1))/a^2))/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} {\left (10 \, A - 7 \, B\right )}}{a^{2}} - \frac {2 \, {\left (30 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(3*(d*x + c)*(10*A - 7*B)/a^2 - 2*(30*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 + 40*A*tan(1
/2*d*x + 1/2*c)^3 - 24*B*tan(1/2*d*x + 1/2*c)^3 + 18*A*tan(1/2*d*x + 1/2*c) - 9*B*tan(1/2*d*x + 1/2*c))/((tan(
1/2*d*x + 1/2*c)^2 + 1)^3*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 - 27*A*a^4*tan(1
/2*d*x + 1/2*c) + 21*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 13.86 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {\left (10\,A-5\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {40\,A}{3}-8\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A-3\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {x\,\left (10\,A-7\,B\right )}{2\,a^2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A-B\right )}{a^2}+\frac {5\,A-3\,B}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d} \]

[In]

int((cos(c + d*x)^3*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)^5*(10*A - 5*B) + tan(c/2 + (d*x)/2)^3*((40*A)/3 - 8*B) + tan(c/2 + (d*x)/2)*(6*A - 3*B))/(
d*(3*a^2*tan(c/2 + (d*x)/2)^2 + 3*a^2*tan(c/2 + (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 + a^2)) - (x*(10*A - 7*B
))/(2*a^2) + (tan(c/2 + (d*x)/2)*((2*(A - B))/a^2 + (5*A - 3*B)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^3*(A - B))/(
6*a^2*d)