\(\int \frac {\cos ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [347]

Optimal result

Integrand size = 40, antiderivative size = 170 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {(10 B-7 C) x}{2 a^2}+\frac {4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac {(10 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 (3 B-2 C) \sin ^3(c+d x)}{3 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4157, 4105, 3872, 2713, 2715, 8} \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {4 (3 B-2 C) \sin ^3(c+d x)}{3 a^2 d}+\frac {4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac {(10 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(10 B-7 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {x (10 B-7 C)}{2 a^2}-\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[In]

Int[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]

[Out]

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

Rule 4157

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^3(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx \\ & = -\frac {(B-C) \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 B-C)-4 a (B-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2} \\ & = -\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \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 B-2 C)-3 a^2 (10 B-7 C) \sec (c+d x)\right ) \, dx}{3 a^4} \\ & = -\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(10 B-7 C) \int \cos ^2(c+d x) \, dx}{a^2}+\frac {(4 (3 B-2 C)) \int \cos ^3(c+d x) \, dx}{a^2} \\ & = -\frac {(10 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(10 B-7 C) \int 1 \, dx}{2 a^2}-\frac {(4 (3 B-2 C)) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d} \\ & = -\frac {(10 B-7 C) x}{2 a^2}+\frac {4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac {(10 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 (3 B-2 C) \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 = 1.33 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.17 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-36 (10 B-7 C) d x \cos \left (\frac {d x}{2}\right )-36 (10 B-7 C) d x \cos \left (c+\frac {d x}{2}\right )-120 B d x \cos \left (c+\frac {3 d x}{2}\right )+84 C d x \cos \left (c+\frac {3 d x}{2}\right )-120 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+516 B \sin \left (\frac {d x}{2}\right )-381 C \sin \left (\frac {d x}{2}\right )-156 B \sin \left (c+\frac {d x}{2}\right )+147 C \sin \left (c+\frac {d x}{2}\right )+342 B \sin \left (c+\frac {3 d x}{2}\right )-239 C \sin \left (c+\frac {3 d x}{2}\right )+118 B \sin \left (2 c+\frac {3 d x}{2}\right )-63 C \sin \left (2 c+\frac {3 d x}{2}\right )+30 B \sin \left (2 c+\frac {5 d x}{2}\right )-15 C \sin \left (2 c+\frac {5 d x}{2}\right )+30 B \sin \left (3 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {5 d x}{2}\right )-3 B \sin \left (3 c+\frac {7 d x}{2}\right )+3 C \sin \left (3 c+\frac {7 d x}{2}\right )-3 B \sin \left (4 c+\frac {7 d x}{2}\right )+3 C \sin \left (4 c+\frac {7 d x}{2}\right )+B \sin \left (4 c+\frac {9 d x}{2}\right )+B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \]

[In]

Integrate[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (28 B -12 C \right ) \cos \left (2 d x +2 c \right )+\left (-2 B +3 C \right ) \cos \left (3 d x +3 c \right )+B \cos \left (4 d x +4 c \right )+\left (258 B -163 C \right ) \cos \left (d x +c \right )+219 B -140 C \right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-240 \left (B -\frac {7 C}{10}\right ) x d}{48 a^{2} d}\)	\(106\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {8 \left (\left (-\frac {5 B}{2}+\frac {5 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {10 B}{3}+2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {3 B}{2}+\frac {3 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-2 \left (10 B -7 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(154\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {8 \left (\left (-\frac {5 B}{2}+\frac {5 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {10 B}{3}+2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {3 B}{2}+\frac {3 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-2 \left (10 B -7 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(154\)
risch	\(-\frac {5 B x}{a^{2}}+\frac {7 x C}{2 a^{2}}+\frac {i B \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} C}{8 a^{2} d}-\frac {15 i B \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{a^{2} d}+\frac {15 i B \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{a^{2} d}-\frac {i B \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} C}{8 a^{2} d}+\frac {2 i \left (15 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}+27 B \,{\mathrm e}^{i \left (d x +c \right )}-21 C \,{\mathrm e}^{i \left (d x +c \right )}+14 B -11 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {B \sin \left (3 d x +3 c \right )}{12 a^{2} d}\)	\(263\)
norman	\(\frac {\frac {\left (4 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{a d}+\frac {\left (10 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}+\frac {\left (10 B -7 C \right ) x}{2 a}-\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{6 a d}+\frac {3 \left (10 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {\left (10 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {3 \left (10 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}-\frac {\left (10 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 a}-\frac {5 \left (16 B -11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}-\frac {\left (21 B -13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (25 B -17 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 a d}+\frac {2 \left (34 B -23 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}+\frac {\left (139 B -91 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a}\)	\(345\)

[In]

int(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {3 \, {\left (10 \, B - 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (10 \, B - 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (10 \, B - 7 \, C\right )} d x - {\left (2 \, B \cos \left (d x + c\right )^{4} - {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, B - C\right )} \cos \left (d x + c\right )^{2} + {\left (66 \, B - 43 \, C\right )} \cos \left (d x + c\right ) + 48 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(3*(10*B - 7*C)*d*x*cos(d*x + c)^2 + 6*(10*B - 7*C)*d*x*cos(d*x + c) + 3*(10*B - 7*C)*d*x - (2*B*cos(d*x
+ c)^4 - (2*B - 3*C)*cos(d*x + c)^3 + 6*(2*B - C)*cos(d*x + c)^2 + (66*B - 43*C)*cos(d*x + c) + 48*B - 32*C)*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]

\[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {B \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(cos(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

(Integral(B*cos(c + d*x)**4*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*cos(c + d*x)*
*4*sec(c + d*x)**2/(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 = 0.31 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.19 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {B {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(B*(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) - C*(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 = 0.30 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} {\left (10 \, B - 7 \, C\right )}}{a^{2}} - \frac {2 \, {\left (30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.90 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\left (10\,B-5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {40\,B}{3}-8\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,B-3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {x\,\left (10\,B-7\,C\right )}{2\,a^2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (B-C\right )}{a^2}+\frac {5\,B-3\,C}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \]

[In]

int((cos(c + d*x)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^2,x)

[Out]

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