3.109 \(\int \cos ^5(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=161 \[ -\frac{a^3 (13 A+20 C) \sin ^3(c+d x)}{60 d}+\frac{a^3 (13 A+20 C) \sin (c+d x)}{5 d}+\frac{3 a^3 (13 A+20 C) \sin (c+d x) \cos (c+d x)}{40 d}+\frac{1}{8} a^3 x (13 A+20 C)+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^3}{5 d}+\frac{3 A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{20 d} \]

[Out]

(a^3*(13*A + 20*C)*x)/8 + (a^3*(13*A + 20*C)*Sin[c + d*x])/(5*d) + (3*a^3*(13*A + 20*C)*Cos[c + d*x]*Sin[c + d
*x])/(40*d) + (3*A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + a*Sec[c
+ d*x])^3*Sin[c + d*x])/(5*d) - (a^3*(13*A + 20*C)*Sin[c + d*x]^3)/(60*d)

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Rubi [A]  time = 0.329812, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4087, 4013, 3791, 2637, 2635, 8, 2633} \[ -\frac{a^3 (13 A+20 C) \sin ^3(c+d x)}{60 d}+\frac{a^3 (13 A+20 C) \sin (c+d x)}{5 d}+\frac{3 a^3 (13 A+20 C) \sin (c+d x) \cos (c+d x)}{40 d}+\frac{1}{8} a^3 x (13 A+20 C)+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^3}{5 d}+\frac{3 A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{20 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(13*A + 20*C)*x)/8 + (a^3*(13*A + 20*C)*Sin[c + d*x])/(5*d) + (3*a^3*(13*A + 20*C)*Cos[c + d*x]*Sin[c + d
*x])/(40*d) + (3*A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + a*Sec[c
+ d*x])^3*Sin[c + d*x])/(5*d) - (a^3*(13*A + 20*C)*Sin[c + d*x]^3)/(60*d)

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])

Rule 4013

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

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 2637

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

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{\int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (3 a A+a (A+5 C) \sec (c+d x)) \, dx}{5 a}\\ &=\frac{3 A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{20} (13 A+20 C) \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac{3 A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{20} (13 A+20 C) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac{1}{20} a^3 (13 A+20 C) x+\frac{3 A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{20} \left (a^3 (13 A+20 C)\right ) \int \cos ^3(c+d x) \, dx+\frac{1}{20} \left (3 a^3 (13 A+20 C)\right ) \int \cos (c+d x) \, dx+\frac{1}{20} \left (3 a^3 (13 A+20 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{1}{20} a^3 (13 A+20 C) x+\frac{3 a^3 (13 A+20 C) \sin (c+d x)}{20 d}+\frac{3 a^3 (13 A+20 C) \cos (c+d x) \sin (c+d x)}{40 d}+\frac{3 A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{40} \left (3 a^3 (13 A+20 C)\right ) \int 1 \, dx-\frac{\left (a^3 (13 A+20 C)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac{1}{8} a^3 (13 A+20 C) x+\frac{a^3 (13 A+20 C) \sin (c+d x)}{5 d}+\frac{3 a^3 (13 A+20 C) \cos (c+d x) \sin (c+d x)}{40 d}+\frac{3 A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac{a^3 (13 A+20 C) \sin ^3(c+d x)}{60 d}\\ \end{align*}

Mathematica [A]  time = 0.3213, size = 97, normalized size = 0.6 \[ \frac{a^3 (60 (23 A+30 C) \sin (c+d x)+120 (4 A+3 C) \sin (2 (c+d x))+170 A \sin (3 (c+d x))+45 A \sin (4 (c+d x))+6 A \sin (5 (c+d x))+780 A d x+40 C \sin (3 (c+d x))+1200 C d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(780*A*d*x + 1200*C*d*x + 60*(23*A + 30*C)*Sin[c + d*x] + 120*(4*A + 3*C)*Sin[2*(c + d*x)] + 170*A*Sin[3*
(c + d*x)] + 40*C*Sin[3*(c + d*x)] + 45*A*Sin[4*(c + d*x)] + 6*A*Sin[5*(c + d*x)]))/(480*d)

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Maple [A]  time = 0.101, size = 197, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{A{a}^{3}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+3\,A{a}^{3} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +A{a}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{\frac{{a}^{3}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+A{a}^{3} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +3\,{a}^{3}C \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{3}C\sin \left ( dx+c \right ) +{a}^{3}C \left ( dx+c \right ) \right ) } \end{align*}

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

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Maxima [A]  time = 0.944548, size = 257, normalized size = 1.6 \begin{align*} \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} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 360 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 480 \,{\left (d x + c\right )} C a^{3} + 1440 \, C a^{3} \sin \left (d x + c\right )}{480 \, d} \end{align*}

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+C*sec(d*x+c)^2),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 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 480*(d
*x + c)*C*a^3 + 1440*C*a^3*sin(d*x + c))/d

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Fricas [A]  time = 0.503565, size = 266, normalized size = 1.65 \begin{align*} \frac{15 \,{\left (13 \, A + 20 \, C\right )} a^{3} d x +{\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 90 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (19 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \,{\left (13 \, A + 12 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \,{\left (38 \, A + 55 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(13*A + 20*C)*a^3*d*x + (24*A*a^3*cos(d*x + c)^4 + 90*A*a^3*cos(d*x + c)^3 + 8*(19*A + 5*C)*a^3*cos(
d*x + c)^2 + 15*(13*A + 12*C)*a^3*cos(d*x + c) + 8*(38*A + 55*C)*a^3)*sin(d*x + c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.2349, size = 284, normalized size = 1.76 \begin{align*} \frac{15 \,{\left (13 \, A a^{3} + 20 \, C 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} + 300 \, C 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} + 1400 \, C 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} + 2560 \, C 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} + 2120 \, C 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 ) + 660 \, C 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} \end{align*}

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+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(13*A*a^3 + 20*C*a^3)*(d*x + c) + 2*(195*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 300*C*a^3*tan(1/2*d*x + 1/2*
c)^9 + 910*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 1400*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*A*a^3*tan(1/2*d*x + 1/2*c)^
5 + 2560*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 1330*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 2120*C*a^3*tan(1/2*d*x + 1/2*c)^3
+ 765*A*a^3*tan(1/2*d*x + 1/2*c) + 660*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d