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

Optimal. Leaf size=185 \[ -\frac {(12 A-8 B+5 C) \sin ^3(c+d x)}{3 a^2 d}+\frac {(12 A-8 B+5 C) \sin (c+d x)}{a^2 d}-\frac {(10 A-7 B+4 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(10 A-7 B+4 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {x (10 A-7 B+4 C)}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-1/2*(10*A-7*B+4*C)*x/a^2+(12*A-8*B+5*C)*sin(d*x+c)/a^2/d-1/2*(10*A-7*B+4*C)*cos(d*x+c)*sin(d*x+c)/a^2/d-1/3*(
10*A-7*B+4*C)*cos(d*x+c)^2*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B+C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+
c))^2-1/3*(12*A-8*B+5*C)*sin(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.36, antiderivative size = 185, 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, 2633, 2635, 8} \[ -\frac {(12 A-8 B+5 C) \sin ^3(c+d x)}{3 a^2 d}+\frac {(12 A-8 B+5 C) \sin (c+d x)}{a^2 d}-\frac {(10 A-7 B+4 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(10 A-7 B+4 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {x (10 A-7 B+4 C)}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((10*A - 7*B + 4*C)*x)/(2*a^2) + ((12*A - 8*B + 5*C)*Sin[c + d*x])/(a^2*d) - ((10*A - 7*B + 4*C)*Cos[c + d*x]
*Sin[c + d*x])/(2*a^2*d) - ((10*A - 7*B + 4*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A
- B + C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) - ((12*A - 8*B + 5*C)*Sin[c + d*x]^3)/(3*a
^2*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]

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 ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos ^3(c+d x) (3 a (2 A-B+C)-a (4 A-4 B+C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(10 A-7 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \cos ^3(c+d x) \left (3 a^2 (12 A-8 B+5 C)-3 a^2 (10 A-7 B+4 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(10 A-7 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(10 A-7 B+4 C) \int \cos ^2(c+d x) \, dx}{a^2}+\frac {(12 A-8 B+5 C) \int \cos ^3(c+d x) \, dx}{a^2}\\ &=-\frac {(10 A-7 B+4 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 A-7 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(10 A-7 B+4 C) \int 1 \, dx}{2 a^2}-\frac {(12 A-8 B+5 C) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac {(10 A-7 B+4 C) x}{2 a^2}+\frac {(12 A-8 B+5 C) \sin (c+d x)}{a^2 d}-\frac {(10 A-7 B+4 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 A-7 B+4 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(12 A-8 B+5 C) \sin ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 1.81, size = 473, normalized size = 2.56 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-36 d x (10 A-7 B+4 C) \cos \left (c+\frac {d x}{2}\right )-36 d x (10 A-7 B+4 C) \cos \left (\frac {d x}{2}\right )-156 A \sin \left (c+\frac {d x}{2}\right )+342 A \sin \left (c+\frac {3 d x}{2}\right )+118 A \sin \left (2 c+\frac {3 d x}{2}\right )+30 A \sin \left (2 c+\frac {5 d x}{2}\right )+30 A \sin \left (3 c+\frac {5 d x}{2}\right )-3 A \sin \left (3 c+\frac {7 d x}{2}\right )-3 A \sin \left (4 c+\frac {7 d x}{2}\right )+A \sin \left (4 c+\frac {9 d x}{2}\right )+A \sin \left (5 c+\frac {9 d x}{2}\right )-120 A d x \cos \left (c+\frac {3 d x}{2}\right )-120 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+516 A \sin \left (\frac {d x}{2}\right )+147 B \sin \left (c+\frac {d x}{2}\right )-239 B \sin \left (c+\frac {3 d x}{2}\right )-63 B \sin \left (2 c+\frac {3 d x}{2}\right )-15 B \sin \left (2 c+\frac {5 d x}{2}\right )-15 B \sin \left (3 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {7 d x}{2}\right )+3 B \sin \left (4 c+\frac {7 d x}{2}\right )+84 B d x \cos \left (c+\frac {3 d x}{2}\right )+84 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 B \sin \left (\frac {d x}{2}\right )-120 C \sin \left (c+\frac {d x}{2}\right )+164 C \sin \left (c+\frac {3 d x}{2}\right )+36 C \sin \left (2 c+\frac {3 d x}{2}\right )+12 C \sin \left (2 c+\frac {5 d x}{2}\right )+12 C \sin \left (3 c+\frac {5 d x}{2}\right )-48 C d x \cos \left (c+\frac {3 d x}{2}\right )-48 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+264 C \sin \left (\frac {d x}{2}\right )\right )}{192 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(-36*(10*A - 7*B + 4*C)*d*x*Cos[(d*x)/2] - 36*(10*A - 7*B + 4*C)*d*x*Cos[c + (d*x
)/2] - 120*A*d*x*Cos[c + (3*d*x)/2] + 84*B*d*x*Cos[c + (3*d*x)/2] - 48*C*d*x*Cos[c + (3*d*x)/2] - 120*A*d*x*Co
s[2*c + (3*d*x)/2] + 84*B*d*x*Cos[2*c + (3*d*x)/2] - 48*C*d*x*Cos[2*c + (3*d*x)/2] + 516*A*Sin[(d*x)/2] - 381*
B*Sin[(d*x)/2] + 264*C*Sin[(d*x)/2] - 156*A*Sin[c + (d*x)/2] + 147*B*Sin[c + (d*x)/2] - 120*C*Sin[c + (d*x)/2]
+ 342*A*Sin[c + (3*d*x)/2] - 239*B*Sin[c + (3*d*x)/2] + 164*C*Sin[c + (3*d*x)/2] + 118*A*Sin[2*c + (3*d*x)/2]
- 63*B*Sin[2*c + (3*d*x)/2] + 36*C*Sin[2*c + (3*d*x)/2] + 30*A*Sin[2*c + (5*d*x)/2] - 15*B*Sin[2*c + (5*d*x)/
2] + 12*C*Sin[2*c + (5*d*x)/2] + 30*A*Sin[3*c + (5*d*x)/2] - 15*B*Sin[3*c + (5*d*x)/2] + 12*C*Sin[3*c + (5*d*x
)/2] - 3*A*Sin[3*c + (7*d*x)/2] + 3*B*Sin[3*c + (7*d*x)/2] - 3*A*Sin[4*c + (7*d*x)/2] + 3*B*Sin[4*c + (7*d*x)/
2] + A*Sin[4*c + (9*d*x)/2] + A*Sin[5*c + (9*d*x)/2]))/(192*a^2*d)

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fricas [A]  time = 0.44, size = 173, normalized size = 0.94 \[ -\frac {3 \, {\left (10 \, A - 7 \, B + 4 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (10 \, A - 7 \, B + 4 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (10 \, A - 7 \, B + 4 \, C\right )} d x - {\left (2 \, A \cos \left (d x + c\right )^{4} - {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, A - B + C\right )} \cos \left (d x + c\right )^{2} + {\left (66 \, A - 43 \, B + 28 \, C\right )} \cos \left (d x + c\right ) + 48 \, A - 32 \, B + 20 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(10*A - 7*B + 4*C)*d*x*cos(d*x + c)^2 + 6*(10*A - 7*B + 4*C)*d*x*cos(d*x + c) + 3*(10*A - 7*B + 4*C)*d
*x - (2*A*cos(d*x + c)^4 - (2*A - 3*B)*cos(d*x + c)^3 + 6*(2*A - B + C)*cos(d*x + c)^2 + (66*A - 43*B + 28*C)*
cos(d*x + c) + 48*A - 32*B + 20*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [A]  time = 0.25, size = 266, normalized size = 1.44 \[ -\frac {\frac {3 \, {\left (d x + c\right )} {\left (10 \, A - 7 \, B + 4 \, C\right )}}{a^{2}} - \frac {2 \, {\left (30 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(d*x + c)*(10*A - 7*B + 4*C)/a^2 - 2*(30*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*
tan(1/2*d*x + 1/2*c)^5 + 40*A*tan(1/2*d*x + 1/2*c)^3 - 24*B*tan(1/2*d*x + 1/2*c)^3 + 12*C*tan(1/2*d*x + 1/2*c)
^3 + 18*A*tan(1/2*d*x + 1/2*c) - 9*B*tan(1/2*d*x + 1/2*c) + 6*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2
+ 1)^3*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 2
7*A*a^4*tan(1/2*d*x + 1/2*c) + 21*B*a^4*tan(1/2*d*x + 1/2*c) - 15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 1.29, size = 482, normalized size = 2.61 \[ -\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}+\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {9 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {40 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {3 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {10 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2}}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A+1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3-1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*
A*tan(1/2*d*x+1/2*c)-7/2/d/a^2*B*tan(1/2*d*x+1/2*c)+5/2/d/a^2*C*tan(1/2*d*x+1/2*c)+10/d/a^2/(1+tan(1/2*d*x+1/2
*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A-5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*B+2/d/a^2/(1+tan(1/2*d
*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*C+40/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A-8/d/a^2/(1+
tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*B+4/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*C+6/d/a
^2/(1+tan(1/2*d*x+1/2*c)^2)^3*A*tan(1/2*d*x+1/2*c)-3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)+2/d
/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)-10/d/a^2*arctan(tan(1/2*d*x+1/2*c))*A+7/d/a^2*arctan(tan(
1/2*d*x+1/2*c))*B-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

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maxima [B]  time = 0.58, size = 487, normalized size = 2.63 \[ \frac {A {\left (\frac {4 \, {\left (\frac {9 \, \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 {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} + C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \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 )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(c
os(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arct
an(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) + C*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(
cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 3.34, size = 202, normalized size = 1.09 \[ \frac {\left (10\,A-5\,B+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {40\,A}{3}-8\,B+4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A-3\,B+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {x\,\left (10\,A-7\,B+4\,C\right )}{2\,a^2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,A-3\,B+C}{2\,a^2}+\frac {2\,\left (A-B+C\right )}{a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(6*A - 3*B + 2*C) + tan(c/2 + (d*x)/2)^5*(10*A - 5*B + 2*C) + tan(c/2 + (d*x)/2)^3*((40*A)
/3 - 8*B + 4*C))/(d*(3*a^2*tan(c/2 + (d*x)/2)^2 + 3*a^2*tan(c/2 + (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 + a^2)
) - (x*(10*A - 7*B + 4*C))/(2*a^2) + (tan(c/2 + (d*x)/2)*((5*A - 3*B + C)/(2*a^2) + (2*(A - B + C))/a^2))/d -
(tan(c/2 + (d*x)/2)^3*(A - B + C))/(6*a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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