\(\int \frac {\cos ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [355]

Optimal result

Integrand size = 40, antiderivative size = 187 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(13 B-6 C) x}{2 a^3}-\frac {8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

1/2*(13*B-6*C)*x/a^3-8/15*(19*B-9*C)*sin(d*x+c)/a^3/d+1/2*(13*B-6*C)*cos(d*x+c)*sin(d*x+c)/a^3/d-1/5*(B-C)*cos
(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(11*B-6*C)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2-4/15*(19*
B-9*C)*cos(d*x+c)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 187, 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, 2715, 8, 2717} \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {4 (19 B-9 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {x (13 B-6 C)}{2 a^3}-\frac {(11 B-6 C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[In]

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

[Out]

((13*B - 6*C)*x)/(2*a^3) - (8*(19*B - 9*C)*Sin[c + d*x])/(15*a^3*d) + ((13*B - 6*C)*Cos[c + d*x]*Sin[c + d*x])
/(2*a^3*d) - ((B - C)*Cos[c + d*x]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((11*B - 6*C)*Cos[c + d*x]*Sin
[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (4*(19*B - 9*C)*Cos[c + d*x]*Sin[c + d*x])/(15*d*(a^3 + a^3*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 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 ^2(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx \\ & = -\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) (a (7 B-2 C)-4 a (B-C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos ^2(c+d x) \left (a^2 (43 B-18 C)-3 a^2 (11 B-6 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \cos ^2(c+d x) \left (15 a^3 (13 B-6 C)-8 a^3 (19 B-9 C) \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(8 (19 B-9 C)) \int \cos (c+d x) \, dx}{15 a^3}+\frac {(13 B-6 C) \int \cos ^2(c+d x) \, dx}{a^3} \\ & = -\frac {8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(13 B-6 C) \int 1 \, dx}{2 a^3} \\ & = \frac {(13 B-6 C) x}{2 a^3}-\frac {8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(435\) vs. \(2(187)=374\).

Time = 1.64 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.33 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (600 (13 B-6 C) d x \cos \left (\frac {d x}{2}\right )+600 (13 B-6 C) d x \cos \left (c+\frac {d x}{2}\right )+3900 B d x \cos \left (c+\frac {3 d x}{2}\right )-1800 C d x \cos \left (c+\frac {3 d x}{2}\right )+3900 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-1800 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+780 B d x \cos \left (2 c+\frac {5 d x}{2}\right )-360 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-360 C d x \cos \left (3 c+\frac {5 d x}{2}\right )-12760 B \sin \left (\frac {d x}{2}\right )+7020 C \sin \left (\frac {d x}{2}\right )+7560 B \sin \left (c+\frac {d x}{2}\right )-4500 C \sin \left (c+\frac {d x}{2}\right )-9230 B \sin \left (c+\frac {3 d x}{2}\right )+4860 C \sin \left (c+\frac {3 d x}{2}\right )+930 B \sin \left (2 c+\frac {3 d x}{2}\right )-900 C \sin \left (2 c+\frac {3 d x}{2}\right )-2782 B \sin \left (2 c+\frac {5 d x}{2}\right )+1452 C \sin \left (2 c+\frac {5 d x}{2}\right )-750 B \sin \left (3 c+\frac {5 d x}{2}\right )+300 C \sin \left (3 c+\frac {5 d x}{2}\right )-105 B \sin \left (3 c+\frac {7 d x}{2}\right )+60 C \sin \left (3 c+\frac {7 d x}{2}\right )-105 B \sin \left (4 c+\frac {7 d x}{2}\right )+60 C \sin \left (4 c+\frac {7 d x}{2}\right )+15 B \sin \left (4 c+\frac {9 d x}{2}\right )+15 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(600*(13*B - 6*C)*d*x*Cos[(d*x)/2] + 600*(13*B - 6*C)*d*x*Cos[c + (d*x)/2] + 3900*B
*d*x*Cos[c + (3*d*x)/2] - 1800*C*d*x*Cos[c + (3*d*x)/2] + 3900*B*d*x*Cos[2*c + (3*d*x)/2] - 1800*C*d*x*Cos[2*c
+ (3*d*x)/2] + 780*B*d*x*Cos[2*c + (5*d*x)/2] - 360*C*d*x*Cos[2*c + (5*d*x)/2] + 780*B*d*x*Cos[3*c + (5*d*x)/
2] - 360*C*d*x*Cos[3*c + (5*d*x)/2] - 12760*B*Sin[(d*x)/2] + 7020*C*Sin[(d*x)/2] + 7560*B*Sin[c + (d*x)/2] - 4
500*C*Sin[c + (d*x)/2] - 9230*B*Sin[c + (3*d*x)/2] + 4860*C*Sin[c + (3*d*x)/2] + 930*B*Sin[2*c + (3*d*x)/2] -
900*C*Sin[2*c + (3*d*x)/2] - 2782*B*Sin[2*c + (5*d*x)/2] + 1452*C*Sin[2*c + (5*d*x)/2] - 750*B*Sin[3*c + (5*d*
x)/2] + 300*C*Sin[3*c + (5*d*x)/2] - 105*B*Sin[3*c + (7*d*x)/2] + 60*C*Sin[3*c + (7*d*x)/2] - 105*B*Sin[4*c +
(7*d*x)/2] + 60*C*Sin[4*c + (7*d*x)/2] + 15*B*Sin[4*c + (9*d*x)/2] + 15*B*Sin[5*c + (9*d*x)/2]))/(480*a^3*d*(1
+ Cos[c + d*x])^3)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.57

method	result	size
parallelrisch	\(\frac {15 \left (\left (-\frac {1856 B}{15}+\frac {312 C}{5}\right ) \cos \left (2 d x +2 c \right )+\left (-6 B +4 C \right ) \cos \left (3 d x +3 c \right )+B \cos \left (4 d x +4 c \right )+\left (-\frac {2002 B}{5}+\frac {972 C}{5}\right ) \cos \left (d x +c \right )-\frac {4303 B}{15}+\frac {696 C}{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6240 \left (B -\frac {6 C}{13}\right ) x d}{960 a^{3} d}\)	\(107\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {16 \left (-\frac {7 B}{4}+\frac {C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+16 \left (-\frac {5 B}{4}+\frac {C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+4 \left (13 B -6 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(163\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {16 \left (-\frac {7 B}{4}+\frac {C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+16 \left (-\frac {5 B}{4}+\frac {C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+4 \left (13 B -6 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(163\)
risch	\(\frac {13 B x}{2 a^{3}}-\frac {3 x C}{a^{3}}-\frac {i B \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{3} d}+\frac {3 i B \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a^{3} d}-\frac {3 i B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a^{3} d}+\frac {i B \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{3} d}-\frac {2 i \left (150 B \,{\mathrm e}^{4 i \left (d x +c \right )}-90 C \,{\mathrm e}^{4 i \left (d x +c \right )}+525 B \,{\mathrm e}^{3 i \left (d x +c \right )}-300 C \,{\mathrm e}^{3 i \left (d x +c \right )}+745 B \,{\mathrm e}^{2 i \left (d x +c \right )}-420 C \,{\mathrm e}^{2 i \left (d x +c \right )}+485 B \,{\mathrm e}^{i \left (d x +c \right )}-270 C \,{\mathrm e}^{i \left (d x +c \right )}+127 B -72 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(255\)
norman	\(\frac {\frac {\left (13 B -6 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {\left (13 B -6 C \right ) x}{2 a}-\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}-\frac {\left (13 B -6 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (13 B -6 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}+\frac {\left (17 B -12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{30 a d}-\frac {4 \left (28 B -13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 a d}+\frac {\left (51 B -25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (77 B -39 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}+\frac {\left (131 B -60 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {\left (377 B -177 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{2}}\)	\(298\)

[In]

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

[Out]

1/960*(15*((-1856/15*B+312/5*C)*cos(2*d*x+2*c)+(-6*B+4*C)*cos(3*d*x+3*c)+B*cos(4*d*x+4*c)+(-2002/5*B+972/5*C)*
cos(d*x+c)-4303/15*B+696/5*C)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4+6240*(B-6/13*C)*x*d)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (13 \, B - 6 \, C\right )} d x + {\left (15 \, B \cos \left (d x + c\right )^{4} - 15 \, {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (479 \, B - 234 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (239 \, B - 114 \, C\right )} \cos \left (d x + c\right ) - 304 \, B + 144 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/30*(15*(13*B - 6*C)*d*x*cos(d*x + c)^3 + 45*(13*B - 6*C)*d*x*cos(d*x + c)^2 + 45*(13*B - 6*C)*d*x*cos(d*x +
c) + 15*(13*B - 6*C)*d*x + (15*B*cos(d*x + c)^4 - 15*(3*B - 2*C)*cos(d*x + c)^3 - (479*B - 234*C)*cos(d*x + c)
^2 - 3*(239*B - 114*C)*cos(d*x + c) - 304*B + 144*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c
)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.72 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - 3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

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

[Out]

-1/60*(B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
- 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 780*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^3) - 3*C*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x +
c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(
d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (d x + c\right )} {\left (13 \, B - 6 \, C\right )}}{a^{3}} - \frac {60 \, {\left (7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

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

[Out]

1/60*(30*(d*x + c)*(13*B - 6*C)/a^3 - 60*(7*B*tan(1/2*d*x + 1/2*c)^3 - 2*C*tan(1/2*d*x + 1/2*c)^3 + 5*B*tan(1/
2*d*x + 1/2*c) - 2*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3) - (3*B*a^12*tan(1/2*d*x + 1/2*
c)^5 - 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 +
465*B*a^12*tan(1/2*d*x + 1/2*c) - 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 15.92 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {x\,\left (13\,B-6\,C\right )}{2\,a^3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^3}+\frac {3\,\left (5\,B-3\,C\right )}{4\,a^3}+\frac {10\,B-2\,C}{4\,a^3}\right )}{d}-\frac {\left (7\,B-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (5\,B-2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{4\,a^3}+\frac {5\,B-3\,C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d} \]

[In]

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

[Out]

(x*(13*B - 6*C))/(2*a^3) - (tan(c/2 + (d*x)/2)*((3*(B - C))/(2*a^3) + (3*(5*B - 3*C))/(4*a^3) + (10*B - 2*C)/(
4*a^3)))/d - (tan(c/2 + (d*x)/2)^3*(7*B - 2*C) + tan(c/2 + (d*x)/2)*(5*B - 2*C))/(d*(2*a^3*tan(c/2 + (d*x)/2)^
2 + a^3*tan(c/2 + (d*x)/2)^4 + a^3)) + (tan(c/2 + (d*x)/2)^3*((B - C)/(4*a^3) + (5*B - 3*C)/(12*a^3)))/d - (ta
n(c/2 + (d*x)/2)^5*(B - C))/(20*a^3*d)