3.70 \(\int \cos ^5(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=176 \[ \frac {a^3 (38 A+45 B) \sin (c+d x)}{15 d}+\frac {a^3 (43 A+45 B) \sin (c+d x) \cos ^2(c+d x)}{60 d}+\frac {a^3 (13 A+15 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {(7 A+5 B) \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{20 d}+\frac {1}{8} a^3 x (13 A+15 B)+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

[Out]

1/8*a^3*(13*A+15*B)*x+1/15*a^3*(38*A+45*B)*sin(d*x+c)/d+1/8*a^3*(13*A+15*B)*cos(d*x+c)*sin(d*x+c)/d+1/60*a^3*(
43*A+45*B)*cos(d*x+c)^2*sin(d*x+c)/d+1/5*a*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+1/20*(7*A+5*B)*cos(d
*x+c)^3*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.37, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4017, 3996, 3787, 2635, 8, 2637} \[ \frac {a^3 (38 A+45 B) \sin (c+d x)}{15 d}+\frac {a^3 (43 A+45 B) \sin (c+d x) \cos ^2(c+d x)}{60 d}+\frac {a^3 (13 A+15 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {(7 A+5 B) \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{20 d}+\frac {1}{8} a^3 x (13 A+15 B)+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(a^3*(13*A + 15*B)*x)/8 + (a^3*(38*A + 45*B)*Sin[c + d*x])/(15*d) + (a^3*(13*A + 15*B)*Cos[c + d*x]*Sin[c + d*
x])/(8*d) + (a^3*(43*A + 45*B)*Cos[c + d*x]^2*Sin[c + d*x])/(60*d) + (a*A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^
2*Sin[c + d*x])/(5*d) + ((7*A + 5*B)*Cos[c + d*x]^3*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(20*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]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (a (7 A+5 B)+a (2 A+5 B) \sec (c+d x)) \, dx\\ &=\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {(7 A+5 B) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{20 d}+\frac {1}{20} \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (a^2 (43 A+45 B)+2 a^2 (11 A+15 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^3 (43 A+45 B) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {(7 A+5 B) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{20 d}-\frac {1}{60} \int \cos ^2(c+d x) \left (-15 a^3 (13 A+15 B)-4 a^3 (38 A+45 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^3 (43 A+45 B) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {(7 A+5 B) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{20 d}+\frac {1}{4} \left (a^3 (13 A+15 B)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{15} \left (a^3 (38 A+45 B)\right ) \int \cos (c+d x) \, dx\\ &=\frac {a^3 (38 A+45 B) \sin (c+d x)}{15 d}+\frac {a^3 (13 A+15 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 (43 A+45 B) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {(7 A+5 B) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{20 d}+\frac {1}{8} \left (a^3 (13 A+15 B)\right ) \int 1 \, dx\\ &=\frac {1}{8} a^3 (13 A+15 B) x+\frac {a^3 (38 A+45 B) \sin (c+d x)}{15 d}+\frac {a^3 (13 A+15 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 (43 A+45 B) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac {a A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {(7 A+5 B) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{20 d}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 108, normalized size = 0.61 \[ \frac {a^3 (60 (23 A+26 B) \sin (c+d x)+480 (A+B) \sin (2 (c+d x))+170 A \sin (3 (c+d x))+45 A \sin (4 (c+d x))+6 A \sin (5 (c+d x))+780 A c+780 A d x+120 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+900 B d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(a^3*(780*A*c + 780*A*d*x + 900*B*d*x + 60*(23*A + 26*B)*Sin[c + d*x] + 480*(A + B)*Sin[2*(c + d*x)] + 170*A*S
in[3*(c + d*x)] + 120*B*Sin[3*(c + d*x)] + 45*A*Sin[4*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 6*A*Sin[5*(c + d*x)
]))/(480*d)

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fricas [A]  time = 0.45, size = 110, normalized size = 0.62 \[ \frac {15 \, {\left (13 \, A + 15 \, B\right )} a^{3} d x + {\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 30 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (19 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (38 \, A + 45 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(15*(13*A + 15*B)*a^3*d*x + (24*A*a^3*cos(d*x + c)^4 + 30*(3*A + B)*a^3*cos(d*x + c)^3 + 8*(19*A + 15*B)
*a^3*cos(d*x + c)^2 + 15*(13*A + 15*B)*a^3*cos(d*x + c) + 8*(38*A + 45*B)*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.63, size = 210, normalized size = 1.19 \[ \frac {15 \, {\left (13 \, A a^{3} + 15 \, B a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (195 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 225 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 910 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1050 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1920 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1330 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1830 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 735 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(15*(13*A*a^3 + 15*B*a^3)*(d*x + c) + 2*(195*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 225*B*a^3*tan(1/2*d*x + 1/2*
c)^9 + 910*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 1050*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*A*a^3*tan(1/2*d*x + 1/2*c)^
5 + 1920*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 1330*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 1830*B*a^3*tan(1/2*d*x + 1/2*c)^3
+ 765*A*a^3*tan(1/2*d*x + 1/2*c) + 735*B*a^3*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 = 1.80, size = 223, normalized size = 1.27 \[ \frac {\frac {A \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} B \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.36, size = 213, normalized size = 1.21 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 480 \, B a^{3} \sin \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^3 - 480*(sin(d*x + c)^3 - 3*sin(d*x + c
))*A*a^3 + 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3 + 120*(2*d*x + 2*c + sin(2*d*x + 2
*c))*A*a^3 - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x
+ 2*c))*B*a^3 + 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 480*B*a^3*sin(d*x + c))/d

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mupad [B]  time = 4.71, size = 247, normalized size = 1.40 \[ \frac {\left (\frac {13\,A\,a^3}{4}+\frac {15\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {91\,A\,a^3}{6}+\frac {35\,B\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {416\,A\,a^3}{15}+32\,B\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {133\,A\,a^3}{6}+\frac {61\,B\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+\frac {49\,B\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (13\,A+15\,B\right )}{4\,\left (\frac {13\,A\,a^3}{4}+\frac {15\,B\,a^3}{4}\right )}\right )\,\left (13\,A+15\,B\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)*((51*A*a^3)/4 + (49*B*a^3)/4) + tan(c/2 + (d*x)/2)^9*((13*A*a^3)/4 + (15*B*a^3)/4) + tan(c
/2 + (d*x)/2)^7*((91*A*a^3)/6 + (35*B*a^3)/2) + tan(c/2 + (d*x)/2)^3*((133*A*a^3)/6 + (61*B*a^3)/2) + tan(c/2
+ (d*x)/2)^5*((416*A*a^3)/15 + 32*B*a^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^3*atan((a^3*tan(c/2 + (d*x)/2)*(13*A +
15*B))/(4*((13*A*a^3)/4 + (15*B*a^3)/4)))*(13*A + 15*B))/(4*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))**3*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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