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

Optimal. Leaf size=218 \[ -\frac {4 (34 A-19 B) \sin ^3(c+d x)}{15 a^3 d}+\frac {4 (34 A-19 B) \sin (c+d x)}{5 a^3 d}-\frac {(23 A-13 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {(23 A-13 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {x (23 A-13 B)}{2 a^3}-\frac {(13 A-8 B) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

-1/2*(23*A-13*B)*x/a^3+4/5*(34*A-19*B)*sin(d*x+c)/a^3/d-1/2*(23*A-13*B)*cos(d*x+c)*sin(d*x+c)/a^3/d-1/5*(A-B)*
cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(13*A-8*B)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2-1/
3*(23*A-13*B)*cos(d*x+c)^2*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))-4/15*(34*A-19*B)*sin(d*x+c)^3/a^3/d

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Rubi [A]  time = 0.49, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4020, 3787, 2633, 2635, 8} \[ -\frac {4 (34 A-19 B) \sin ^3(c+d x)}{15 a^3 d}+\frac {4 (34 A-19 B) \sin (c+d x)}{5 a^3 d}-\frac {(23 A-13 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {(23 A-13 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {x (23 A-13 B)}{2 a^3}-\frac {(13 A-8 B) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((23*A - 13*B)*x)/(2*a^3) + (4*(34*A - 19*B)*Sin[c + d*x])/(5*a^3*d) - ((23*A - 13*B)*Cos[c + d*x]*Sin[c + d*
x])/(2*a^3*d) - ((A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((13*A - 8*B)*Cos[c + d*x
]^2*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((23*A - 13*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*(a^3 + a^
3*Sec[c + d*x])) - (4*(34*A - 19*B)*Sin[c + d*x]^3)/(15*a^3*d)

Rule 8

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

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A-B) \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 B)-5 a (A-B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \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 (3 a^2 (21 A-11 B)-4 a^2 (13 A-8 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \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-19 B)-15 a^3 (23 A-13 B) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \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-19 B)) \int \cos ^3(c+d x) \, dx}{5 a^3}-\frac {(23 A-13 B) \int \cos ^2(c+d x) \, dx}{a^3}\\ &=-\frac {(23 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(23 A-13 B) \int 1 \, dx}{2 a^3}-\frac {(4 (34 A-19 B)) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 a^3 d}\\ &=-\frac {(23 A-13 B) x}{2 a^3}+\frac {4 (34 A-19 B) \sin (c+d x)}{5 a^3 d}-\frac {(23 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \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-19 B) \sin ^3(c+d x)}{15 a^3 d}\\ \end {align*}

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Mathematica [B]  time = 1.23, size = 491, normalized size = 2.25 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-600 d x (23 A-13 B) \cos \left (c+\frac {d x}{2}\right )-600 d x (23 A-13 B) \cos \left (\frac {d x}{2}\right )-11110 A \sin \left (c+\frac {d x}{2}\right )+15380 A \sin \left (c+\frac {3 d x}{2}\right )-380 A \sin \left (2 c+\frac {3 d x}{2}\right )+4777 A \sin \left (2 c+\frac {5 d x}{2}\right )+1625 A \sin \left (3 c+\frac {5 d x}{2}\right )+230 A \sin \left (3 c+\frac {7 d x}{2}\right )+230 A \sin \left (4 c+\frac {7 d x}{2}\right )-20 A \sin \left (4 c+\frac {9 d x}{2}\right )-20 A \sin \left (5 c+\frac {9 d x}{2}\right )+5 A \sin \left (5 c+\frac {11 d x}{2}\right )+5 A \sin \left (6 c+\frac {11 d x}{2}\right )-6900 A d x \cos \left (c+\frac {3 d x}{2}\right )-6900 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-1380 A d x \cos \left (2 c+\frac {5 d x}{2}\right )-1380 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+20410 A \sin \left (\frac {d x}{2}\right )+7560 B \sin \left (c+\frac {d x}{2}\right )-9230 B \sin \left (c+\frac {3 d x}{2}\right )+930 B \sin \left (2 c+\frac {3 d x}{2}\right )-2782 B \sin \left (2 c+\frac {5 d x}{2}\right )-750 B \sin \left (3 c+\frac {5 d x}{2}\right )-105 B \sin \left (3 c+\frac {7 d x}{2}\right )-105 B \sin \left (4 c+\frac {7 d x}{2}\right )+15 B \sin \left (4 c+\frac {9 d x}{2}\right )+15 B \sin \left (5 c+\frac {9 d x}{2}\right )+3900 B d x \cos \left (c+\frac {3 d x}{2}\right )+3900 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+780 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-12760 B \sin \left (\frac {d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-600*(23*A - 13*B)*d*x*Cos[(d*x)/2] - 600*(23*A - 13*B)*d*x*Cos[c + (d*x)/2] - 690
0*A*d*x*Cos[c + (3*d*x)/2] + 3900*B*d*x*Cos[c + (3*d*x)/2] - 6900*A*d*x*Cos[2*c + (3*d*x)/2] + 3900*B*d*x*Cos[
2*c + (3*d*x)/2] - 1380*A*d*x*Cos[2*c + (5*d*x)/2] + 780*B*d*x*Cos[2*c + (5*d*x)/2] - 1380*A*d*x*Cos[3*c + (5*
d*x)/2] + 780*B*d*x*Cos[3*c + (5*d*x)/2] + 20410*A*Sin[(d*x)/2] - 12760*B*Sin[(d*x)/2] - 11110*A*Sin[c + (d*x)
/2] + 7560*B*Sin[c + (d*x)/2] + 15380*A*Sin[c + (3*d*x)/2] - 9230*B*Sin[c + (3*d*x)/2] - 380*A*Sin[2*c + (3*d*
x)/2] + 930*B*Sin[2*c + (3*d*x)/2] + 4777*A*Sin[2*c + (5*d*x)/2] - 2782*B*Sin[2*c + (5*d*x)/2] + 1625*A*Sin[3*
c + (5*d*x)/2] - 750*B*Sin[3*c + (5*d*x)/2] + 230*A*Sin[3*c + (7*d*x)/2] - 105*B*Sin[3*c + (7*d*x)/2] + 230*A*
Sin[4*c + (7*d*x)/2] - 105*B*Sin[4*c + (7*d*x)/2] - 20*A*Sin[4*c + (9*d*x)/2] + 15*B*Sin[4*c + (9*d*x)/2] - 20
*A*Sin[5*c + (9*d*x)/2] + 15*B*Sin[5*c + (9*d*x)/2] + 5*A*Sin[5*c + (11*d*x)/2] + 5*A*Sin[6*c + (11*d*x)/2]))/
(480*a^3*d*(1 + Cos[c + d*x])^3)

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fricas [A]  time = 0.50, size = 205, normalized size = 0.94 \[ -\frac {15 \, {\left (23 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (23 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (23 \, A - 13 \, B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (23 \, A - 13 \, B\right )} d x - {\left (10 \, A \cos \left (d x + c\right )^{5} - 15 \, {\left (A - B\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (19 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (869 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (429 \, A - 239 \, B\right )} \cos \left (d x + c\right ) + 544 \, A - 304 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*(23*A - 13*B)*d*x*cos(d*x + c)^3 + 45*(23*A - 13*B)*d*x*cos(d*x + c)^2 + 45*(23*A - 13*B)*d*x*cos(d*
x + c) + 15*(23*A - 13*B)*d*x - (10*A*cos(d*x + c)^5 - 15*(A - B)*cos(d*x + c)^4 + 5*(19*A - 9*B)*cos(d*x + c)
^3 + (869*A - 479*B)*cos(d*x + c)^2 + 3*(429*A - 239*B)*cos(d*x + c) + 544*A - 304*B)*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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giac [A]  time = 0.31, size = 228, normalized size = 1.05 \[ -\frac {\frac {30 \, {\left (d x + c\right )} {\left (23 \, A - 13 \, B\right )}}{a^{3}} - \frac {20 \, {\left (51 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 76 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 50 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 735 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(30*(d*x + c)*(23*A - 13*B)/a^3 - 20*(51*A*tan(1/2*d*x + 1/2*c)^5 - 21*B*tan(1/2*d*x + 1/2*c)^5 + 76*A*t
an(1/2*d*x + 1/2*c)^3 - 36*B*tan(1/2*d*x + 1/2*c)^3 + 33*A*tan(1/2*d*x + 1/2*c) - 15*B*tan(1/2*d*x + 1/2*c))/(
(tan(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*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 50*
A*a^12*tan(1/2*d*x + 1/2*c)^3 + 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 735*A*a^12*tan(1/2*d*x + 1/2*c) - 465*B*a^1
2*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [A]  time = 1.28, size = 362, normalized size = 1.66 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {5 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{3}}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {31 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {76 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {12 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {11 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {23 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3}}+\frac {13 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5-1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5-5/6/d/a^3*tan(1/2*d*x+1/2*c)^3*A+2/3/d/a^3
*B*tan(1/2*d*x+1/2*c)^3+49/4/d/a^3*A*tan(1/2*d*x+1/2*c)-31/4/d/a^3*B*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-7/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*B+76/3/d/a^3/(1+t
an(1/2*d*x+1/2*c)^2)^3*A*tan(1/2*d*x+1/2*c)^3-12/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*B*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)-5/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)-23
/d/a^3*arctan(tan(1/2*d*x+1/2*c))*A+13/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B

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maxima [B]  time = 0.45, size = 412, normalized size = 1.89 \[ \frac {A {\left (\frac {20 \, {\left (\frac {33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {76 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {51 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {735 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {50 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {1380 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(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) - 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))/d

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mupad [B]  time = 2.05, size = 237, normalized size = 1.09 \[ \frac {\left (17\,A-7\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {76\,A}{3}-12\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (11\,A-5\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A-B\right )}{2\,a^3}+\frac {6\,A-4\,B}{a^3}+\frac {15\,A-5\,B}{4\,a^3}\right )}{d}-\frac {x\,\left (23\,A-13\,B\right )}{2\,a^3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{3\,a^3}+\frac {6\,A-4\,B}{12\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^5*(17*A - 7*B) + tan(c/2 + (d*x)/2)^3*((76*A)/3 - 12*B) + tan(c/2 + (d*x)/2)*(11*A - 5*B))
/(d*(3*a^3*tan(c/2 + (d*x)/2)^2 + 3*a^3*tan(c/2 + (d*x)/2)^4 + a^3*tan(c/2 + (d*x)/2)^6 + a^3)) + (tan(c/2 + (
d*x)/2)*((5*(A - B))/(2*a^3) + (6*A - 4*B)/a^3 + (15*A - 5*B)/(4*a^3)))/d - (x*(23*A - 13*B))/(2*a^3) - (tan(c
/2 + (d*x)/2)^3*((A - B)/(3*a^3) + (6*A - 4*B)/(12*a^3)))/d + (tan(c/2 + (d*x)/2)^5*(A - B))/(20*a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \cos ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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