3.154 \(\int \frac{\cos ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=248 \[ -\frac{4 (454 A+83 C) \sin ^3(c+d x)}{105 a^4 d}+\frac{4 (454 A+83 C) \sin (c+d x)}{35 a^4 d}-\frac{2 (11 A+2 C) \sin (c+d x) \cos (c+d x)}{a^4 d}-\frac{4 (11 A+2 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^4 d (\sec (c+d x)+1)}-\frac{(178 A+31 C) \sin (c+d x) \cos ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac{2 x (11 A+2 C)}{a^4}-\frac{2 (8 A+C) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

(-2*(11*A + 2*C)*x)/a^4 + (4*(454*A + 83*C)*Sin[c + d*x])/(35*a^4*d) - (2*(11*A + 2*C)*Cos[c + d*x]*Sin[c + d*
x])/(a^4*d) - ((178*A + 31*C)*Cos[c + d*x]^2*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (4*(11*A + 2*C)*
Cos[c + d*x]^2*Sin[c + d*x])/(3*a^4*d*(1 + Sec[c + d*x])) - ((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(7*d*(a + a*
Sec[c + d*x])^4) - (2*(8*A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3) - (4*(454*A + 83*
C)*Sin[c + d*x]^3)/(105*a^4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.687439, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4085, 4020, 3787, 2633, 2635, 8} \[ -\frac{4 (454 A+83 C) \sin ^3(c+d x)}{105 a^4 d}+\frac{4 (454 A+83 C) \sin (c+d x)}{35 a^4 d}-\frac{2 (11 A+2 C) \sin (c+d x) \cos (c+d x)}{a^4 d}-\frac{4 (11 A+2 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^4 d (\sec (c+d x)+1)}-\frac{(178 A+31 C) \sin (c+d x) \cos ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac{2 x (11 A+2 C)}{a^4}-\frac{2 (8 A+C) \sin (c+d x) \cos ^2(c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

(-2*(11*A + 2*C)*x)/a^4 + (4*(454*A + 83*C)*Sin[c + d*x])/(35*a^4*d) - (2*(11*A + 2*C)*Cos[c + d*x]*Sin[c + d*
x])/(a^4*d) - ((178*A + 31*C)*Cos[c + d*x]^2*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (4*(11*A + 2*C)*
Cos[c + d*x]^2*Sin[c + d*x])/(3*a^4*d*(1 + Sec[c + d*x])) - ((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(7*d*(a + a*
Sec[c + d*x])^4) - (2*(8*A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3) - (4*(454*A + 83*
C)*Sin[c + d*x]^3)/(105*a^4*d)

Rule 4085

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*(A + C)*Cot[e + f*x]*(a + b*Csc[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[b*C*n + A*b*(
2*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, C, n}, x]
&& EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4020

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 3787

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 2633

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 2635

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] && Integer
Q[2*n]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{\int \frac{\cos ^3(c+d x) (-a (10 A+3 C)+a (6 A-C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (8 A+C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{\cos ^3(c+d x) \left (-7 a^2 (14 A+3 C)+10 a^2 (8 A+C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{(178 A+31 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (8 A+C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{\cos ^3(c+d x) \left (-12 a^3 (69 A+13 C)+4 a^3 (178 A+31 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac{(178 A+31 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (8 A+C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{4 (11 A+2 C) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{\int \cos ^3(c+d x) \left (-12 a^4 (454 A+83 C)+420 a^4 (11 A+2 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac{(178 A+31 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (8 A+C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{4 (11 A+2 C) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{(4 (11 A+2 C)) \int \cos ^2(c+d x) \, dx}{a^4}+\frac{(4 (454 A+83 C)) \int \cos ^3(c+d x) \, dx}{35 a^4}\\ &=-\frac{2 (11 A+2 C) \cos (c+d x) \sin (c+d x)}{a^4 d}-\frac{(178 A+31 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (8 A+C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{4 (11 A+2 C) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{(2 (11 A+2 C)) \int 1 \, dx}{a^4}-\frac{(4 (454 A+83 C)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{35 a^4 d}\\ &=-\frac{2 (11 A+2 C) x}{a^4}+\frac{4 (454 A+83 C) \sin (c+d x)}{35 a^4 d}-\frac{2 (11 A+2 C) \cos (c+d x) \sin (c+d x)}{a^4 d}-\frac{(178 A+31 C) \cos ^2(c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 (8 A+C) \cos ^2(c+d x) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{4 (11 A+2 C) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{4 (454 A+83 C) \sin ^3(c+d x)}{105 a^4 d}\\ \end{align*}

Mathematica [B]  time = 2.75575, size = 575, normalized size = 2.32 \[ -\frac{\sec \left (\frac{c}{2}\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right ) \left (58800 d x (11 A+2 C) \cos \left (c+\frac{d x}{2}\right )+687260 A \sin \left (c+\frac{d x}{2}\right )-814107 A \sin \left (c+\frac{3 d x}{2}\right )+204645 A \sin \left (2 c+\frac{3 d x}{2}\right )-357609 A \sin \left (2 c+\frac{5 d x}{2}\right )-18025 A \sin \left (3 c+\frac{5 d x}{2}\right )-72522 A \sin \left (3 c+\frac{7 d x}{2}\right )-24010 A \sin \left (4 c+\frac{7 d x}{2}\right )-2310 A \sin \left (4 c+\frac{9 d x}{2}\right )-2310 A \sin \left (5 c+\frac{9 d x}{2}\right )+175 A \sin \left (5 c+\frac{11 d x}{2}\right )+175 A \sin \left (6 c+\frac{11 d x}{2}\right )-35 A \sin \left (6 c+\frac{13 d x}{2}\right )-35 A \sin \left (7 c+\frac{13 d x}{2}\right )+388080 A d x \cos \left (c+\frac{3 d x}{2}\right )+388080 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+129360 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+129360 A d x \cos \left (3 c+\frac{5 d x}{2}\right )+18480 A d x \cos \left (3 c+\frac{7 d x}{2}\right )+18480 A d x \cos \left (4 c+\frac{7 d x}{2}\right )+58800 d x (11 A+2 C) \cos \left (\frac{d x}{2}\right )-1010660 A \sin \left (\frac{d x}{2}\right )+184520 C \sin \left (c+\frac{d x}{2}\right )-184464 C \sin \left (c+\frac{3 d x}{2}\right )+72240 C \sin \left (2 c+\frac{3 d x}{2}\right )-77168 C \sin \left (2 c+\frac{5 d x}{2}\right )+8400 C \sin \left (3 c+\frac{5 d x}{2}\right )-15164 C \sin \left (3 c+\frac{7 d x}{2}\right )-2940 C \sin \left (4 c+\frac{7 d x}{2}\right )-420 C \sin \left (4 c+\frac{9 d x}{2}\right )-420 C \sin \left (5 c+\frac{9 d x}{2}\right )+70560 C d x \cos \left (c+\frac{3 d x}{2}\right )+70560 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+23520 C d x \cos \left (2 c+\frac{5 d x}{2}\right )+23520 C d x \cos \left (3 c+\frac{5 d x}{2}\right )+3360 C d x \cos \left (3 c+\frac{7 d x}{2}\right )+3360 C d x \cos \left (4 c+\frac{7 d x}{2}\right )-243320 C \sin \left (\frac{d x}{2}\right )\right )}{107520 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^4,x]

[Out]

-(Sec[c/2]*Sec[(c + d*x)/2]^7*(58800*(11*A + 2*C)*d*x*Cos[(d*x)/2] + 58800*(11*A + 2*C)*d*x*Cos[c + (d*x)/2] +
388080*A*d*x*Cos[c + (3*d*x)/2] + 70560*C*d*x*Cos[c + (3*d*x)/2] + 388080*A*d*x*Cos[2*c + (3*d*x)/2] + 70560*
C*d*x*Cos[2*c + (3*d*x)/2] + 129360*A*d*x*Cos[2*c + (5*d*x)/2] + 23520*C*d*x*Cos[2*c + (5*d*x)/2] + 129360*A*d
*x*Cos[3*c + (5*d*x)/2] + 23520*C*d*x*Cos[3*c + (5*d*x)/2] + 18480*A*d*x*Cos[3*c + (7*d*x)/2] + 3360*C*d*x*Cos
[3*c + (7*d*x)/2] + 18480*A*d*x*Cos[4*c + (7*d*x)/2] + 3360*C*d*x*Cos[4*c + (7*d*x)/2] - 1010660*A*Sin[(d*x)/2
] - 243320*C*Sin[(d*x)/2] + 687260*A*Sin[c + (d*x)/2] + 184520*C*Sin[c + (d*x)/2] - 814107*A*Sin[c + (3*d*x)/2
] - 184464*C*Sin[c + (3*d*x)/2] + 204645*A*Sin[2*c + (3*d*x)/2] + 72240*C*Sin[2*c + (3*d*x)/2] - 357609*A*Sin[
2*c + (5*d*x)/2] - 77168*C*Sin[2*c + (5*d*x)/2] - 18025*A*Sin[3*c + (5*d*x)/2] + 8400*C*Sin[3*c + (5*d*x)/2] -
72522*A*Sin[3*c + (7*d*x)/2] - 15164*C*Sin[3*c + (7*d*x)/2] - 24010*A*Sin[4*c + (7*d*x)/2] - 2940*C*Sin[4*c +
(7*d*x)/2] - 2310*A*Sin[4*c + (9*d*x)/2] - 420*C*Sin[4*c + (9*d*x)/2] - 2310*A*Sin[5*c + (9*d*x)/2] - 420*C*S
in[5*c + (9*d*x)/2] + 175*A*Sin[5*c + (11*d*x)/2] + 175*A*Sin[6*c + (11*d*x)/2] - 35*A*Sin[6*c + (13*d*x)/2] -
35*A*Sin[7*c + (13*d*x)/2]))/(107520*a^4*d)

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Maple [A]  time = 0.111, size = 402, normalized size = 1.6 \begin{align*} -{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{C}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{11\,A}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{7\,C}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{59\,A}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{23\,C}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{209\,A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{49\,C}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+26\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}A}{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+2\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{124\,A}{3\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}+4\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+18\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-44\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{4}}}-8\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A-1/56/d/a^4*C*tan(1/2*d*x+1/2*c)^7+11/40/d/a^4*tan(1/2*d*x+1/2*c)^5*A+7/40/d
/a^4*C*tan(1/2*d*x+1/2*c)^5-59/24/d/a^4*A*tan(1/2*d*x+1/2*c)^3-23/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3+209/8/d/a^4*
A*tan(1/2*d*x+1/2*c)+49/8/d/a^4*C*tan(1/2*d*x+1/2*c)+26/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*
A+2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)^5+124/3/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x
+1/2*c)^3*A+4/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)^3+18/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*A*ta
n(1/2*d*x+1/2*c)+2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)-44/d/a^4*A*arctan(tan(1/2*d*x+1/2*c))
-8/d/a^4*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [A]  time = 1.45614, size = 547, normalized size = 2.21 \begin{align*} \frac{A{\left (\frac{560 \,{\left (\frac{27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{62 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{39 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{4} + \frac{3 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{21945 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2065 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{231 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{36960 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} + C{\left (\frac{1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac{a^{4} \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{5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{6720 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(A*(560*(27*sin(d*x + c)/(cos(d*x + c) + 1) + 62*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 39*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5)/(a^4 + 3*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^4*sin(d*x + c)^4/(cos(d*x + c) +
1)^4 + a^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (21945*sin(d*x + c)/(cos(d*x + c) + 1) - 2065*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 + 231*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4
- 36960*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) + C*(1680*sin(d*x + c)/((a^4 + a^4*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) - 805*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 6720*arctan(
sin(d*x + c)/(cos(d*x + c) + 1))/a^4))/d

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Fricas [A]  time = 0.529699, size = 676, normalized size = 2.73 \begin{align*} -\frac{210 \,{\left (11 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{4} + 840 \,{\left (11 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{3} + 1260 \,{\left (11 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 840 \,{\left (11 \, A + 2 \, C\right )} d x \cos \left (d x + c\right ) + 210 \,{\left (11 \, A + 2 \, C\right )} d x -{\left (35 \, A \cos \left (d x + c\right )^{6} - 70 \, A \cos \left (d x + c\right )^{5} + 35 \,{\left (14 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} + 8 \,{\left (799 \, A + 148 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (3592 \, A + 659 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (6109 \, A + 1118 \, C\right )} \cos \left (d x + c\right ) + 3632 \, A + 664 \, C\right )} \sin \left (d x + c\right )}{105 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/105*(210*(11*A + 2*C)*d*x*cos(d*x + c)^4 + 840*(11*A + 2*C)*d*x*cos(d*x + c)^3 + 1260*(11*A + 2*C)*d*x*cos(
d*x + c)^2 + 840*(11*A + 2*C)*d*x*cos(d*x + c) + 210*(11*A + 2*C)*d*x - (35*A*cos(d*x + c)^6 - 70*A*cos(d*x +
c)^5 + 35*(14*A + 3*C)*cos(d*x + c)^4 + 8*(799*A + 148*C)*cos(d*x + c)^3 + 4*(3592*A + 659*C)*cos(d*x + c)^2 +
2*(6109*A + 1118*C)*cos(d*x + c) + 3632*A + 664*C)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)
^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.19713, size = 352, normalized size = 1.42 \begin{align*} -\frac{\frac{1680 \,{\left (d x + c\right )}{\left (11 \, A + 2 \, C\right )}}{a^{4}} - \frac{560 \,{\left (39 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 62 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, 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^{4}} + \frac{15 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 231 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 147 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2065 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 805 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 21945 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5145 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(1680*(d*x + c)*(11*A + 2*C)/a^4 - 560*(39*A*tan(1/2*d*x + 1/2*c)^5 + 3*C*tan(1/2*d*x + 1/2*c)^5 + 62*A
*tan(1/2*d*x + 1/2*c)^3 + 6*C*tan(1/2*d*x + 1/2*c)^3 + 27*A*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x + 1/2*c))/(
(tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^4) + (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 - 2
31*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 147*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 2065*A*a^24*tan(1/2*d*x + 1/2*c)^3 + 80
5*C*a^24*tan(1/2*d*x + 1/2*c)^3 - 21945*A*a^24*tan(1/2*d*x + 1/2*c) - 5145*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/
d