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

Optimal. Leaf size=201 \[ -\frac {2 (76 A-36 B+11 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 A-6 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {(76 A-36 B+11 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {x (13 A-6 B+2 C)}{2 a^3}-\frac {(11 A-6 B+C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

1/2*(13*A-6*B+2*C)*x/a^3-2/15*(76*A-36*B+11*C)*sin(d*x+c)/a^3/d+1/2*(13*A-6*B+2*C)*cos(d*x+c)*sin(d*x+c)/a^3/d
-1/5*(A-B+C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(11*A-6*B+C)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d
*x+c))^2-1/15*(76*A-36*B+11*C)*cos(d*x+c)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]  time = 0.52, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4084, 4020, 3787, 2635, 8, 2637} \[ -\frac {2 (76 A-36 B+11 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 A-6 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {(76 A-36 B+11 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {x (13 A-6 B+2 C)}{2 a^3}-\frac {(11 A-6 B+C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((13*A - 6*B + 2*C)*x)/(2*a^3) - (2*(76*A - 36*B + 11*C)*Sin[c + d*x])/(15*a^3*d) + ((13*A - 6*B + 2*C)*Cos[c
+ d*x]*Sin[c + d*x])/(2*a^3*d) - ((A - B + C)*Cos[c + d*x]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((11*A
- 6*B + C)*Cos[c + d*x]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((76*A - 36*B + 11*C)*Cos[c + d*x]*Si
n[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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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]

Rule 4084

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac {(A-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 A-2 B+2 C)-a (4 A-4 B-C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A-6 B+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 A-18 B+8 C)-3 a^2 (11 A-6 B+C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A-6 B+C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(76 A-36 B+11 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 A-6 B+2 C)-2 a^3 (76 A-36 B+11 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A-6 B+C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(76 A-36 B+11 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(13 A-6 B+2 C) \int \cos ^2(c+d x) \, dx}{a^3}-\frac {(2 (76 A-36 B+11 C)) \int \cos (c+d x) \, dx}{15 a^3}\\ &=-\frac {2 (76 A-36 B+11 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 A-6 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A-6 B+C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(76 A-36 B+11 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(13 A-6 B+2 C) \int 1 \, dx}{2 a^3}\\ &=\frac {(13 A-6 B+2 C) x}{2 a^3}-\frac {2 (76 A-36 B+11 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 A-6 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A-6 B+C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(76 A-36 B+11 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.64, size = 557, normalized size = 2.77 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (600 d x (13 A-6 B+2 C) \cos \left (c+\frac {d x}{2}\right )+600 d x (13 A-6 B+2 C) \cos \left (\frac {d x}{2}\right )+7560 A \sin \left (c+\frac {d x}{2}\right )-9230 A \sin \left (c+\frac {3 d x}{2}\right )+930 A \sin \left (2 c+\frac {3 d x}{2}\right )-2782 A \sin \left (2 c+\frac {5 d x}{2}\right )-750 A \sin \left (3 c+\frac {5 d x}{2}\right )-105 A \sin \left (3 c+\frac {7 d x}{2}\right )-105 A \sin \left (4 c+\frac {7 d x}{2}\right )+15 A \sin \left (4 c+\frac {9 d x}{2}\right )+15 A \sin \left (5 c+\frac {9 d x}{2}\right )+3900 A d x \cos \left (c+\frac {3 d x}{2}\right )+3900 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+780 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-12760 A \sin \left (\frac {d x}{2}\right )-4500 B \sin \left (c+\frac {d x}{2}\right )+4860 B \sin \left (c+\frac {3 d x}{2}\right )-900 B \sin \left (2 c+\frac {3 d x}{2}\right )+1452 B \sin \left (2 c+\frac {5 d x}{2}\right )+300 B \sin \left (3 c+\frac {5 d x}{2}\right )+60 B \sin \left (3 c+\frac {7 d x}{2}\right )+60 B \sin \left (4 c+\frac {7 d x}{2}\right )-1800 B d x \cos \left (c+\frac {3 d x}{2}\right )-1800 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-360 B d x \cos \left (2 c+\frac {5 d x}{2}\right )-360 B d x \cos \left (3 c+\frac {5 d x}{2}\right )+7020 B \sin \left (\frac {d x}{2}\right )+2160 C \sin \left (c+\frac {d x}{2}\right )-1840 C \sin \left (c+\frac {3 d x}{2}\right )+720 C \sin \left (2 c+\frac {3 d x}{2}\right )-512 C \sin \left (2 c+\frac {5 d x}{2}\right )+600 C d x \cos \left (c+\frac {3 d x}{2}\right )+600 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+120 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+120 C d x \cos \left (3 c+\frac {5 d x}{2}\right )-2960 C \sin \left (\frac {d x}{2}\right )\right )}{3840 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(A + B*Sec[c + d*x] + 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*(13*A - 6*B + 2*C)*d*x*Cos[(d*x)/2] + 600*(13*A - 6*B + 2*C)*d*x*Cos[c + (d*
x)/2] + 3900*A*d*x*Cos[c + (3*d*x)/2] - 1800*B*d*x*Cos[c + (3*d*x)/2] + 600*C*d*x*Cos[c + (3*d*x)/2] + 3900*A*
d*x*Cos[2*c + (3*d*x)/2] - 1800*B*d*x*Cos[2*c + (3*d*x)/2] + 600*C*d*x*Cos[2*c + (3*d*x)/2] + 780*A*d*x*Cos[2*
c + (5*d*x)/2] - 360*B*d*x*Cos[2*c + (5*d*x)/2] + 120*C*d*x*Cos[2*c + (5*d*x)/2] + 780*A*d*x*Cos[3*c + (5*d*x)
/2] - 360*B*d*x*Cos[3*c + (5*d*x)/2] + 120*C*d*x*Cos[3*c + (5*d*x)/2] - 12760*A*Sin[(d*x)/2] + 7020*B*Sin[(d*x
)/2] - 2960*C*Sin[(d*x)/2] + 7560*A*Sin[c + (d*x)/2] - 4500*B*Sin[c + (d*x)/2] + 2160*C*Sin[c + (d*x)/2] - 923
0*A*Sin[c + (3*d*x)/2] + 4860*B*Sin[c + (3*d*x)/2] - 1840*C*Sin[c + (3*d*x)/2] + 930*A*Sin[2*c + (3*d*x)/2] -
900*B*Sin[2*c + (3*d*x)/2] + 720*C*Sin[2*c + (3*d*x)/2] - 2782*A*Sin[2*c + (5*d*x)/2] + 1452*B*Sin[2*c + (5*d*
x)/2] - 512*C*Sin[2*c + (5*d*x)/2] - 750*A*Sin[3*c + (5*d*x)/2] + 300*B*Sin[3*c + (5*d*x)/2] - 105*A*Sin[3*c +
(7*d*x)/2] + 60*B*Sin[3*c + (7*d*x)/2] - 105*A*Sin[4*c + (7*d*x)/2] + 60*B*Sin[4*c + (7*d*x)/2] + 15*A*Sin[4*
c + (9*d*x)/2] + 15*A*Sin[5*c + (9*d*x)/2]))/(3840*a^3*d)

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fricas [A]  time = 0.48, size = 211, normalized size = 1.05 \[ \frac {15 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (13 \, A - 6 \, B + 2 \, C\right )} d x + {\left (15 \, A \cos \left (d x + c\right )^{4} - 15 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} - {\left (479 \, A - 234 \, B + 64 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (239 \, A - 114 \, B + 34 \, C\right )} \cos \left (d x + c\right ) - 304 \, A + 144 \, B - 44 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+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*A - 6*B + 2*C)*d*x*cos(d*x + c)^3 + 45*(13*A - 6*B + 2*C)*d*x*cos(d*x + c)^2 + 45*(13*A - 6*B + 2
*C)*d*x*cos(d*x + c) + 15*(13*A - 6*B + 2*C)*d*x + (15*A*cos(d*x + c)^4 - 15*(3*A - 2*B)*cos(d*x + c)^3 - (479
*A - 234*B + 64*C)*cos(d*x + c)^2 - 3*(239*A - 114*B + 34*C)*cos(d*x + c) - 304*A + 144*B - 44*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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giac [A]  time = 0.29, size = 252, normalized size = 1.25 \[ \frac {\frac {30 \, {\left (d x + c\right )} {\left (13 \, A - 6 \, B + 2 \, C\right )}}{a^{3}} - \frac {60 \, {\left (7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+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*A - 6*B + 2*C)/a^3 - 60*(7*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + 5*A*
tan(1/2*d*x + 1/2*c) - 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*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 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 40*A*a^12*tan(1/2*d*x + 1/2*c
)^3 + 30*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) -
255*B*a^12*tan(1/2*d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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maple [A]  time = 1.78, size = 369, normalized size = 1.84 \[ -\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{20 d \,a^{3}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3 d \,a^{3}}-\frac {31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {17 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {13 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3}}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(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)^5-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*C+2/3/d/a
^3*tan(1/2*d*x+1/2*c)^3*A-1/2/d/a^3*B*tan(1/2*d*x+1/2*c)^3+1/3/d/a^3*tan(1/2*d*x+1/2*c)^3*C-31/4/d/a^3*A*tan(1
/2*d*x+1/2*c)+17/4/d/a^3*B*tan(1/2*d*x+1/2*c)-7/4/d/a^3*C*tan(1/2*d*x+1/2*c)-7/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^
2*tan(1/2*d*x+1/2*c)^3*A+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)^3-5/d/a^3/(1+tan(1/2*d*x+1/2*
c)^2)^2*A*tan(1/2*d*x+1/2*c)+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)+13/d/a^3*arctan(tan(1/2*d
*x+1/2*c))*A-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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maxima [B]  time = 0.46, size = 411, normalized size = 2.04 \[ -\frac {A {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - 3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(A*(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*B*(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) + C*((105*sin(d*x + c)/(cos(d*x + c) +
1) - 20*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 - 120*arctan(sin(d*x
+ c)/(cos(d*x + c) + 1))/a^3))/d

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mupad [B]  time = 3.36, size = 215, normalized size = 1.07 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,A-3\,B+C}{12\,a^3}+\frac {A-B+C}{4\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (5\,A-3\,B+C\right )}{4\,a^3}-\frac {2\,B-10\,A+2\,C}{4\,a^3}+\frac {3\,\left (A-B+C\right )}{2\,a^3}\right )}{d}-\frac {\left (7\,A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (5\,A-2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {x\,\left (13\,A-6\,B+2\,C\right )}{2\,a^3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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