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

Optimal. Leaf size=216 \[ -\frac{4 (34 A+9 C) \sin ^3(c+d x)}{15 a^3 d}+\frac{4 (34 A+9 C) \sin (c+d x)}{5 a^3 d}-\frac{(23 A+6 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{(23 A+6 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{x (23 A+6 C)}{2 a^3}-\frac{(13 A+3 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

-((23*A + 6*C)*x)/(2*a^3) + (4*(34*A + 9*C)*Sin[c + d*x])/(5*a^3*d) - ((23*A + 6*C)*Cos[c + d*x]*Sin[c + d*x])
/(2*a^3*d) - ((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((13*A + 3*C)*Cos[c + d*x]^2
*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((23*A + 6*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*(a^3 + a^3*Se
c[c + d*x])) - (4*(34*A + 9*C)*Sin[c + d*x]^3)/(15*a^3*d)

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Rubi [A]  time = 0.497261, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{4 (34 A+9 C) \sin ^3(c+d x)}{15 a^3 d}+\frac{4 (34 A+9 C) \sin (c+d x)}{5 a^3 d}-\frac{(23 A+6 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{(23 A+6 C) \sin (c+d x) \cos ^2(c+d x)}{3 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{x (23 A+6 C)}{2 a^3}-\frac{(13 A+3 C) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((23*A + 6*C)*x)/(2*a^3) + (4*(34*A + 9*C)*Sin[c + d*x])/(5*a^3*d) - ((23*A + 6*C)*Cos[c + d*x]*Sin[c + d*x])
/(2*a^3*d) - ((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((13*A + 3*C)*Cos[c + d*x]^2
*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((23*A + 6*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*(a^3 + a^3*Se
c[c + d*x])) - (4*(34*A + 9*C)*Sin[c + d*x]^3)/(15*a^3*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))^3} \, dx &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{\int \frac{\cos ^3(c+d x) (-a (8 A+3 C)+5 a A \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(13 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{\int \frac{\cos ^3(c+d x) \left (-9 a^2 (7 A+2 C)+4 a^2 (13 A+3 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(13 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(23 A+6 C) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\int \cos ^3(c+d x) \left (-12 a^3 (34 A+9 C)+15 a^3 (23 A+6 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(13 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(23 A+6 C) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(23 A+6 C) \int \cos ^2(c+d x) \, dx}{a^3}+\frac{(4 (34 A+9 C)) \int \cos ^3(c+d x) \, dx}{5 a^3}\\ &=-\frac{(23 A+6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(13 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(23 A+6 C) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(23 A+6 C) \int 1 \, dx}{2 a^3}-\frac{(4 (34 A+9 C)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 a^3 d}\\ &=-\frac{(23 A+6 C) x}{2 a^3}+\frac{4 (34 A+9 C) \sin (c+d x)}{5 a^3 d}-\frac{(23 A+6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(13 A+3 C) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(23 A+6 C) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{4 (34 A+9 C) \sin ^3(c+d x)}{15 a^3 d}\\ \end{align*}

Mathematica [B]  time = 1.81681, size = 455, normalized size = 2.11 \[ -\frac{\sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right ) \left (600 d x (23 A+6 C) \cos \left (c+\frac{d x}{2}\right )+11110 A \sin \left (c+\frac{d x}{2}\right )-15380 A \sin \left (c+\frac{3 d x}{2}\right )+380 A \sin \left (2 c+\frac{3 d x}{2}\right )-4777 A \sin \left (2 c+\frac{5 d x}{2}\right )-1625 A \sin \left (3 c+\frac{5 d x}{2}\right )-230 A \sin \left (3 c+\frac{7 d x}{2}\right )-230 A \sin \left (4 c+\frac{7 d x}{2}\right )+20 A \sin \left (4 c+\frac{9 d x}{2}\right )+20 A \sin \left (5 c+\frac{9 d x}{2}\right )-5 A \sin \left (5 c+\frac{11 d x}{2}\right )-5 A \sin \left (6 c+\frac{11 d x}{2}\right )+6900 A d x \cos \left (c+\frac{3 d x}{2}\right )+6900 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+1380 A d x \cos \left (2 c+\frac{5 d x}{2}\right )+1380 A d x \cos \left (3 c+\frac{5 d x}{2}\right )+600 d x (23 A+6 C) \cos \left (\frac{d x}{2}\right )-20410 A \sin \left (\frac{d x}{2}\right )+4500 C \sin \left (c+\frac{d x}{2}\right )-4860 C \sin \left (c+\frac{3 d x}{2}\right )+900 C \sin \left (2 c+\frac{3 d x}{2}\right )-1452 C \sin \left (2 c+\frac{5 d x}{2}\right )-300 C \sin \left (3 c+\frac{5 d x}{2}\right )-60 C \sin \left (3 c+\frac{7 d x}{2}\right )-60 C \sin \left (4 c+\frac{7 d x}{2}\right )+1800 C d x \cos \left (c+\frac{3 d x}{2}\right )+1800 C d x \cos \left (2 c+\frac{3 d x}{2}\right )+360 C d x \cos \left (2 c+\frac{5 d x}{2}\right )+360 C d x \cos \left (3 c+\frac{5 d x}{2}\right )-7020 C \sin \left (\frac{d x}{2}\right )\right )}{3840 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c/2]*Sec[(c + d*x)/2]^5*(600*(23*A + 6*C)*d*x*Cos[(d*x)/2] + 600*(23*A + 6*C)*d*x*Cos[c + (d*x)/2] + 690
0*A*d*x*Cos[c + (3*d*x)/2] + 1800*C*d*x*Cos[c + (3*d*x)/2] + 6900*A*d*x*Cos[2*c + (3*d*x)/2] + 1800*C*d*x*Cos[
2*c + (3*d*x)/2] + 1380*A*d*x*Cos[2*c + (5*d*x)/2] + 360*C*d*x*Cos[2*c + (5*d*x)/2] + 1380*A*d*x*Cos[3*c + (5*
d*x)/2] + 360*C*d*x*Cos[3*c + (5*d*x)/2] - 20410*A*Sin[(d*x)/2] - 7020*C*Sin[(d*x)/2] + 11110*A*Sin[c + (d*x)/
2] + 4500*C*Sin[c + (d*x)/2] - 15380*A*Sin[c + (3*d*x)/2] - 4860*C*Sin[c + (3*d*x)/2] + 380*A*Sin[2*c + (3*d*x
)/2] + 900*C*Sin[2*c + (3*d*x)/2] - 4777*A*Sin[2*c + (5*d*x)/2] - 1452*C*Sin[2*c + (5*d*x)/2] - 1625*A*Sin[3*c
+ (5*d*x)/2] - 300*C*Sin[3*c + (5*d*x)/2] - 230*A*Sin[3*c + (7*d*x)/2] - 60*C*Sin[3*c + (7*d*x)/2] - 230*A*Si
n[4*c + (7*d*x)/2] - 60*C*Sin[4*c + (7*d*x)/2] + 20*A*Sin[4*c + (9*d*x)/2] + 20*A*Sin[5*c + (9*d*x)/2] - 5*A*S
in[5*c + (11*d*x)/2] - 5*A*Sin[6*c + (11*d*x)/2]))/(3840*a^3*d)

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Maple [A]  time = 0.105, size = 362, normalized size = 1.7 \begin{align*}{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{5\,A}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{49\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+17\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}A}{d{a}^{3} \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}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{76\,A}{3\,d{a}^{3}} \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}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+11\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \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}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-23\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{3}}} \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))^3,x)

[Out]

1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*A+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5-5/6/d/a^3*tan(1/2*d*x+1/2*c)^3*A-1/2/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+49/4/d/a^3*A*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)+17/d/a^3/(1+tan(1/2*d*
x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)^5+76/3/d/a^3/(1+t
an(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A+4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)^3+11/d/a
^3/(1+tan(1/2*d*x+1/2*c)^2)^3*A*tan(1/2*d*x+1/2*c)+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)-23/
d/a^3*A*arctan(tan(1/2*d*x+1/2*c))-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [A]  time = 1.44132, size = 493, normalized size = 2.28 \begin{align*} \frac{A{\left (\frac{20 \,{\left (\frac{33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{76 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{51 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{735 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{50 \, \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{1380 \, \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} \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))^3,x, algorithm="maxima")

[Out]

1/60*(A*(20*(33*sin(d*x + c)/(cos(d*x + c) + 1) + 76*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 51*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5)/(a^3 + 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 + a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (735*sin(d*x + c)/(cos(d*x + c) + 1) - 50*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 - 1380*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

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Fricas [A]  time = 0.510645, size = 525, normalized size = 2.43 \begin{align*} -\frac{15 \,{\left (23 \, A + 6 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (23 \, A + 6 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (23 \, A + 6 \, C\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (23 \, A + 6 \, C\right )} d x -{\left (10 \, A \cos \left (d x + c\right )^{5} - 15 \, A \cos \left (d x + c\right )^{4} + 5 \,{\left (19 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (869 \, A + 234 \, C\right )} \cos \left (d x + c\right )^{2} + 9 \,{\left (143 \, A + 38 \, C\right )} \cos \left (d x + c\right ) + 544 \, A + 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 )}} \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))^3,x, algorithm="fricas")

[Out]

-1/30*(15*(23*A + 6*C)*d*x*cos(d*x + c)^3 + 45*(23*A + 6*C)*d*x*cos(d*x + c)^2 + 45*(23*A + 6*C)*d*x*cos(d*x +
c) + 15*(23*A + 6*C)*d*x - (10*A*cos(d*x + c)^5 - 15*A*cos(d*x + c)^4 + 5*(19*A + 6*C)*cos(d*x + c)^3 + (869*
A + 234*C)*cos(d*x + c)^2 + 9*(143*A + 38*C)*cos(d*x + c) + 544*A + 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)

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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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20886, size = 308, normalized size = 1.43 \begin{align*} -\frac{\frac{30 \,{\left (d x + c\right )}{\left (23 \, A + 6 \, C\right )}}{a^{3}} - \frac{20 \,{\left (51 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 76 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 33 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, 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^{3}} - \frac{3 \, A 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} - 50 \, A 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} + 735 \, A 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} \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))^3,x, algorithm="giac")

[Out]

-1/60*(30*(d*x + c)*(23*A + 6*C)/a^3 - 20*(51*A*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5 + 76*A*tan
(1/2*d*x + 1/2*c)^3 + 12*C*tan(1/2*d*x + 1/2*c)^3 + 33*A*tan(1/2*d*x + 1/2*c) + 6*C*tan(1/2*d*x + 1/2*c))/((ta
n(1/2*d*x + 1/2*c)^2 + 1)^3*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 50*A*a
^12*tan(1/2*d*x + 1/2*c)^3 - 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 735*A*a^12*tan(1/2*d*x + 1/2*c) + 255*C*a^12*t
an(1/2*d*x + 1/2*c))/a^15)/d