3.79 \(\int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=158 \[ -\frac{4 a^4 (4 A+5 B) \sin ^3(c+d x)}{15 d}+\frac{8 a^4 (4 A+5 B) \sin (c+d x)}{5 d}+\frac{a^4 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{27 a^4 (4 A+5 B) \sin (c+d x) \cos (c+d x)}{40 d}+\frac{7}{8} a^4 x (4 A+5 B)+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

[Out]

(7*a^4*(4*A + 5*B)*x)/8 + (8*a^4*(4*A + 5*B)*Sin[c + d*x])/(5*d) + (27*a^4*(4*A + 5*B)*Cos[c + d*x]*Sin[c + d*
x])/(40*d) + (a^4*(4*A + 5*B)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*S
in[c + d*x])/(5*d) - (4*a^4*(4*A + 5*B)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.201723, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4013, 3791, 2637, 2635, 8, 2633} \[ -\frac{4 a^4 (4 A+5 B) \sin ^3(c+d x)}{15 d}+\frac{8 a^4 (4 A+5 B) \sin (c+d x)}{5 d}+\frac{a^4 (4 A+5 B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{27 a^4 (4 A+5 B) \sin (c+d x) \cos (c+d x)}{40 d}+\frac{7}{8} a^4 x (4 A+5 B)+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(7*a^4*(4*A + 5*B)*x)/8 + (8*a^4*(4*A + 5*B)*Sin[c + d*x])/(5*d) + (27*a^4*(4*A + 5*B)*Cos[c + d*x]*Sin[c + d*
x])/(40*d) + (a^4*(4*A + 5*B)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*S
in[c + d*x])/(5*d) - (4*a^4*(4*A + 5*B)*Sin[c + d*x]^3)/(15*d)

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))^4 (A+B \sec (c+d x)) \, dx &=\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{5} (4 A+5 B) \int \cos ^4(c+d x) (a+a \sec (c+d x))^4 \, dx\\ &=\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{5} (4 A+5 B) \int \left (a^4+4 a^4 \cos (c+d x)+6 a^4 \cos ^2(c+d x)+4 a^4 \cos ^3(c+d x)+a^4 \cos ^4(c+d x)\right ) \, dx\\ &=\frac{1}{5} a^4 (4 A+5 B) x+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{5} \left (a^4 (4 A+5 B)\right ) \int \cos ^4(c+d x) \, dx+\frac{1}{5} \left (4 a^4 (4 A+5 B)\right ) \int \cos (c+d x) \, dx+\frac{1}{5} \left (4 a^4 (4 A+5 B)\right ) \int \cos ^3(c+d x) \, dx+\frac{1}{5} \left (6 a^4 (4 A+5 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{1}{5} a^4 (4 A+5 B) x+\frac{4 a^4 (4 A+5 B) \sin (c+d x)}{5 d}+\frac{3 a^4 (4 A+5 B) \cos (c+d x) \sin (c+d x)}{5 d}+\frac{a^4 (4 A+5 B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{20} \left (3 a^4 (4 A+5 B)\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{5} \left (3 a^4 (4 A+5 B)\right ) \int 1 \, dx-\frac{\left (4 a^4 (4 A+5 B)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{4}{5} a^4 (4 A+5 B) x+\frac{8 a^4 (4 A+5 B) \sin (c+d x)}{5 d}+\frac{27 a^4 (4 A+5 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac{a^4 (4 A+5 B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}-\frac{4 a^4 (4 A+5 B) \sin ^3(c+d x)}{15 d}+\frac{1}{40} \left (3 a^4 (4 A+5 B)\right ) \int 1 \, dx\\ &=\frac{7}{8} a^4 (4 A+5 B) x+\frac{8 a^4 (4 A+5 B) \sin (c+d x)}{5 d}+\frac{27 a^4 (4 A+5 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac{a^4 (4 A+5 B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}-\frac{4 a^4 (4 A+5 B) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.325799, size = 108, normalized size = 0.68 \[ \frac{a^4 (420 (7 A+8 B) \sin (c+d x)+120 (8 A+7 B) \sin (2 (c+d x))+290 A \sin (3 (c+d x))+60 A \sin (4 (c+d x))+6 A \sin (5 (c+d x))+1680 A d x+160 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+2100 B d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(a^4*(1680*A*d*x + 2100*B*d*x + 420*(7*A + 8*B)*Sin[c + d*x] + 120*(8*A + 7*B)*Sin[2*(c + d*x)] + 290*A*Sin[3*
(c + d*x)] + 160*B*Sin[3*(c + d*x)] + 60*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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Maple [A]  time = 0.095, size = 248, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ({\frac{A{a}^{4}\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 ) }+4\,A{a}^{4} \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 ) +B{a}^{4} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +2\,A{a}^{4} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{\frac{4\,B{a}^{4} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+4\,A{a}^{4} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +6\,B{a}^{4} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +A{a}^{4}\sin \left ( dx+c \right ) +4\,B{a}^{4}\sin \left ( dx+c \right ) +B{a}^{4} \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))^4*(A+B*sec(d*x+c)),x)

[Out]

1/d*(1/5*A*a^4*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+4*A*a^4*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d
*x+c)+3/8*d*x+3/8*c)+B*a^4*(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*a^4*(2+cos(d*x+c)^
2)*sin(d*x+c)+4/3*B*a^4*(2+cos(d*x+c)^2)*sin(d*x+c)+4*A*a^4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+6*B*a^4*
(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+A*a^4*sin(d*x+c)+4*B*a^4*sin(d*x+c)+B*a^4*(d*x+c))

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Maxima [A]  time = 1.02748, size = 319, normalized size = 2.02 \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^{4} - 960 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} + 60 \,{\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^{4} + 480 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 640 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 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^{4} + 720 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 480 \,{\left (d x + c\right )} B a^{4} + 480 \, A a^{4} \sin \left (d x + c\right ) + 1920 \, B a^{4} \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))^4*(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^4 - 960*(sin(d*x + c)^3 - 3*sin(d*x + c
))*A*a^4 + 60*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 + 480*(2*d*x + 2*c + sin(2*d*x + 2
*c))*A*a^4 - 640*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x
+ 2*c))*B*a^4 + 720*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 + 480*(d*x + c)*B*a^4 + 480*A*a^4*sin(d*x + c) + 19
20*B*a^4*sin(d*x + c))/d

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Fricas [A]  time = 0.485643, size = 279, normalized size = 1.77 \begin{align*} \frac{105 \,{\left (4 \, A + 5 \, B\right )} a^{4} d x +{\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 16 \,{\left (17 \, A + 10 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \,{\left (28 \, A + 27 \, B\right )} a^{4} \cos \left (d x + c\right ) + 8 \,{\left (83 \, A + 100 \, B\right )} a^{4}\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))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(105*(4*A + 5*B)*a^4*d*x + (24*A*a^4*cos(d*x + c)^4 + 30*(4*A + B)*a^4*cos(d*x + c)^3 + 16*(17*A + 10*B)
*a^4*cos(d*x + c)^2 + 15*(28*A + 27*B)*a^4*cos(d*x + c) + 8*(83*A + 100*B)*a^4)*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))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.33091, size = 284, normalized size = 1.8 \begin{align*} \frac{105 \,{\left (4 \, A a^{4} + 5 \, B a^{4}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (420 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 1960 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 2450 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 3584 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3160 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3950 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1500 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1395 \, B a^{4} \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))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(105*(4*A*a^4 + 5*B*a^4)*(d*x + c) + 2*(420*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 525*B*a^4*tan(1/2*d*x + 1/2*c
)^9 + 1960*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 2450*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 3584*A*a^4*tan(1/2*d*x + 1/2*c)^
5 + 4480*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 3160*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 3950*B*a^4*tan(1/2*d*x + 1/2*c)^3
+ 1500*A*a^4*tan(1/2*d*x + 1/2*c) + 1395*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d