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]

7/8*a^4*(4*A+5*B)*x+8/5*a^4*(4*A+5*B)*sin(d*x+c)/d+27/40*a^4*(4*A+5*B)*cos(d*x+c)*sin(d*x+c)/d+1/20*a^4*(4*A+5
*B)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^4*sin(d*x+c)/d-4/15*a^4*(4*A+5*B)*sin(d*x+c)
^3/d

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Rubi [A]  time = 0.20, antiderivative size = 158, normalized size of antiderivative = 1.00, 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 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 2637

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

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 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]

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*}

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Mathematica [A]  time = 0.36, 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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fricas [A]  time = 0.49, size = 110, normalized size = 0.70 \[ \frac {105 \, {\left (4 \, A + 5 \, B\right )} a^{4} d x + {\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 16 \, {\left (17 \, A + 10 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \, {\left (28 \, A + 27 \, B\right )} a^{4} \cos \left (d x + c\right ) + 8 \, {\left (83 \, A + 100 \, B\right )} a^{4}\right )} \sin \left (d x + c\right )}{120 \, d} \]

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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giac [A]  time = 0.64, size = 210, normalized size = 1.33 \[ \frac {105 \, {\left (4 \, A a^{4} + 5 \, B a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (420 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1960 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2450 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3584 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3160 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3950 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1500 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

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

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maple [A]  time = 1.50, size = 248, normalized size = 1.57 \[ \frac {\frac {A \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+4 A \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{4} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {4 a^{4} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+4 A \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{4} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{4} \sin \left (d x +c \right )+4 a^{4} B \sin \left (d x +c \right )+a^{4} B \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.35, size = 236, normalized size = 1.49 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} + 60 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 480 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 480 \, {\left (d x + c\right )} B a^{4} + 480 \, A a^{4} \sin \left (d x + c\right ) + 1920 \, B a^{4} \sin \left (d x + c\right )}{480 \, d} \]

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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mupad [B]  time = 4.70, size = 248, normalized size = 1.57 \[ \frac {\left (7\,A\,a^4+\frac {35\,B\,a^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {98\,A\,a^4}{3}+\frac {245\,B\,a^4}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {896\,A\,a^4}{15}+\frac {224\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {158\,A\,a^4}{3}+\frac {395\,B\,a^4}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (25\,A\,a^4+\frac {93\,B\,a^4}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {7\,a^4\,\mathrm {atan}\left (\frac {7\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+5\,B\right )}{4\,\left (7\,A\,a^4+\frac {35\,B\,a^4}{4}\right )}\right )\,\left (4\,A+5\,B\right )}{4\,d} \]

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))^4,x)

[Out]

(tan(c/2 + (d*x)/2)*(25*A*a^4 + (93*B*a^4)/4) + tan(c/2 + (d*x)/2)^9*(7*A*a^4 + (35*B*a^4)/4) + tan(c/2 + (d*x
)/2)^7*((98*A*a^4)/3 + (245*B*a^4)/6) + tan(c/2 + (d*x)/2)^3*((158*A*a^4)/3 + (395*B*a^4)/6) + tan(c/2 + (d*x)
/2)^5*((896*A*a^4)/15 + (224*B*a^4)/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)) + (7*a^4*atan((7*a^4*tan(c/2 + (d*x)/2)*(4*A +
5*B))/(4*(7*A*a^4 + (35*B*a^4)/4)))*(4*A + 5*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))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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