3.100 \(\int \cos ^5(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=169 \[ \frac {a^2 (18 A+25 C) \sin (c+d x)}{15 d}+\frac {a^2 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{30 d}+\frac {a^2 (3 A+4 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{10 d}+\frac {1}{4} a^2 x (3 A+4 C)+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

[Out]

1/4*a^2*(3*A+4*C)*x+1/15*a^2*(18*A+25*C)*sin(d*x+c)/d+1/4*a^2*(3*A+4*C)*cos(d*x+c)*sin(d*x+c)/d+1/30*a^2*(9*A+
10*C)*cos(d*x+c)^2*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+1/10*A*cos(d*x+c)^3*(a^2+a^
2*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.39, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4087, 4017, 3996, 3787, 2635, 8, 2637} \[ \frac {a^2 (18 A+25 C) \sin (c+d x)}{15 d}+\frac {a^2 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{30 d}+\frac {a^2 (3 A+4 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{10 d}+\frac {1}{4} a^2 x (3 A+4 C)+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(3*A + 4*C)*x)/4 + (a^2*(18*A + 25*C)*Sin[c + d*x])/(15*d) + (a^2*(3*A + 4*C)*Cos[c + d*x]*Sin[c + d*x])/
(4*d) + (a^2*(9*A + 10*C)*Cos[c + d*x]^2*Sin[c + d*x])/(30*d) + (A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2*Sin[c
+ d*x])/(5*d) + (A*Cos[c + d*x]^3*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(10*d)

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 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

Rule 4087

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (2 a A+a (2 A+5 C) \sec (c+d x)) \, dx}{5 a}\\ &=\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{10 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (2 a^2 (9 A+10 C)+4 a^2 (3 A+5 C) \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac {a^2 (9 A+10 C) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{10 d}-\frac {\int \cos ^2(c+d x) \left (-30 a^3 (3 A+4 C)-4 a^3 (18 A+25 C) \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac {a^2 (9 A+10 C) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{10 d}+\frac {1}{2} \left (a^2 (3 A+4 C)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{15} \left (a^2 (18 A+25 C)\right ) \int \cos (c+d x) \, dx\\ &=\frac {a^2 (18 A+25 C) \sin (c+d x)}{15 d}+\frac {a^2 (3 A+4 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 (9 A+10 C) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{10 d}+\frac {1}{4} \left (a^2 (3 A+4 C)\right ) \int 1 \, dx\\ &=\frac {1}{4} a^2 (3 A+4 C) x+\frac {a^2 (18 A+25 C) \sin (c+d x)}{15 d}+\frac {a^2 (3 A+4 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 (9 A+10 C) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{10 d}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 97, normalized size = 0.57 \[ \frac {a^2 (30 (11 A+14 C) \sin (c+d x)+120 (A+C) \sin (2 (c+d x))+45 A \sin (3 (c+d x))+15 A \sin (4 (c+d x))+3 A \sin (5 (c+d x))+120 A c+180 A d x+20 C \sin (3 (c+d x))+240 C d x)}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(120*A*c + 180*A*d*x + 240*C*d*x + 30*(11*A + 14*C)*Sin[c + d*x] + 120*(A + C)*Sin[2*(c + d*x)] + 45*A*Si
n[3*(c + d*x)] + 20*C*Sin[3*(c + d*x)] + 15*A*Sin[4*(c + d*x)] + 3*A*Sin[5*(c + d*x)]))/(240*d)

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fricas [A]  time = 0.43, size = 106, normalized size = 0.63 \[ \frac {15 \, {\left (3 \, A + 4 \, C\right )} a^{2} d x + {\left (12 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \, A a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) + 4 \, {\left (18 \, A + 25 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/60*(15*(3*A + 4*C)*a^2*d*x + (12*A*a^2*cos(d*x + c)^4 + 30*A*a^2*cos(d*x + c)^3 + 4*(9*A + 5*C)*a^2*cos(d*x
+ c)^2 + 15*(3*A + 4*C)*a^2*cos(d*x + c) + 4*(18*A + 25*C)*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.26, size = 210, normalized size = 1.24 \[ \frac {15 \, {\left (3 \, A a^{2} + 4 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 280 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 432 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 560 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 270 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/60*(15*(3*A*a^2 + 4*C*a^2)*(d*x + c) + 2*(45*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9
+ 210*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 280*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 432*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 560
*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 270*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 520*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 195*A*a^
2*tan(1/2*d*x + 1/2*c) + 180*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 1.49, size = 160, normalized size = 0.95 \[ \frac {\frac {a^{2} A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 a^{2} A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{2} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} C \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*(1/5*a^2*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+1/3*a^2*C*(2+cos(d*x+c)^2)*sin(d*x+c)+2*a^2*A*(1
/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+2*a^2*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1
/3*a^2*A*(2+cos(d*x+c)^2)*sin(d*x+c)+a^2*C*sin(d*x+c))

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maxima [A]  time = 0.36, size = 156, normalized size = 0.92 \[ \frac {16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} - 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 240 \, C a^{2} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/240*(16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^2 - 80*(sin(d*x + c)^3 - 3*sin(d*x + c)
)*A*a^2 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2 - 80*(sin(d*x + c)^3 - 3*sin(d*x +
c))*C*a^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 240*C*a^2*sin(d*x + c))/d

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mupad [B]  time = 5.34, size = 247, normalized size = 1.46 \[ \frac {\left (\frac {3\,A\,a^2}{2}+2\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (7\,A\,a^2+\frac {28\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,A\,a^2}{5}+\frac {56\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (9\,A\,a^2+\frac {52\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a^2}{2}+6\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,C\right )}{2\,\left (\frac {3\,A\,a^2}{2}+2\,C\,a^2\right )}\right )\,\left (3\,A+4\,C\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)*((13*A*a^2)/2 + 6*C*a^2) + tan(c/2 + (d*x)/2)^9*((3*A*a^2)/2 + 2*C*a^2) + tan(c/2 + (d*x)/
2)^7*(7*A*a^2 + (28*C*a^2)/3) + tan(c/2 + (d*x)/2)^3*(9*A*a^2 + (52*C*a^2)/3) + tan(c/2 + (d*x)/2)^5*((72*A*a^
2)/5 + (56*C*a^2)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c
/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (a^2*atan((a^2*tan(c/2 + (d*x)/2)*(3*A + 4*C))/(2*((3*A*a^2)/2
+ 2*C*a^2)))*(3*A + 4*C))/(2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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