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

Optimal. Leaf size=187 \[ \frac{a^2 (18 A+20 B+25 C) \sin (c+d x)}{15 d}+\frac{a^2 (18 A+25 B+20 C) \sin (c+d x) \cos ^2(c+d x)}{60 d}+\frac{a^2 (6 A+7 B+8 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{(2 A+5 B) \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{20 d}+\frac{1}{8} a^2 x (6 A+7 B+8 C)+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

[Out]

(a^2*(6*A + 7*B + 8*C)*x)/8 + (a^2*(18*A + 20*B + 25*C)*Sin[c + d*x])/(15*d) + (a^2*(6*A + 7*B + 8*C)*Cos[c +
d*x]*Sin[c + d*x])/(8*d) + (a^2*(18*A + 25*B + 20*C)*Cos[c + d*x]^2*Sin[c + d*x])/(60*d) + (A*Cos[c + d*x]^4*(
a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((2*A + 5*B)*Cos[c + d*x]^3*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])
/(20*d)

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Rubi [A]  time = 0.414596, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4086, 4017, 3996, 3787, 2635, 8, 2637} \[ \frac{a^2 (18 A+20 B+25 C) \sin (c+d x)}{15 d}+\frac{a^2 (18 A+25 B+20 C) \sin (c+d x) \cos ^2(c+d x)}{60 d}+\frac{a^2 (6 A+7 B+8 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{(2 A+5 B) \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{20 d}+\frac{1}{8} a^2 x (6 A+7 B+8 C)+\frac{A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(6*A + 7*B + 8*C)*x)/8 + (a^2*(18*A + 20*B + 25*C)*Sin[c + d*x])/(15*d) + (a^2*(6*A + 7*B + 8*C)*Cos[c +
d*x]*Sin[c + d*x])/(8*d) + (a^2*(18*A + 25*B + 20*C)*Cos[c + d*x]^2*Sin[c + d*x])/(60*d) + (A*Cos[c + d*x]^4*(
a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((2*A + 5*B)*Cos[c + d*x]^3*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])
/(20*d)

Rule 4086

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

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 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 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 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 2637

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

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+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 (a (2 A+5 B)+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{(2 A+5 B) \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{20 d}+\frac{\int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (a^2 (18 A+25 B+20 C)+2 a^2 (6 A+5 B+10 C) \sec (c+d x)\right ) \, dx}{20 a}\\ &=\frac{a^2 (18 A+25 B+20 C) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{(2 A+5 B) \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{20 d}-\frac{\int \cos ^2(c+d x) \left (-15 a^3 (6 A+7 B+8 C)-4 a^3 (18 A+20 B+25 C) \sec (c+d x)\right ) \, dx}{60 a}\\ &=\frac{a^2 (18 A+25 B+20 C) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{(2 A+5 B) \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{20 d}+\frac{1}{4} \left (a^2 (6 A+7 B+8 C)\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{15} \left (a^2 (18 A+20 B+25 C)\right ) \int \cos (c+d x) \, dx\\ &=\frac{a^2 (18 A+20 B+25 C) \sin (c+d x)}{15 d}+\frac{a^2 (6 A+7 B+8 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (18 A+25 B+20 C) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{(2 A+5 B) \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{20 d}+\frac{1}{8} \left (a^2 (6 A+7 B+8 C)\right ) \int 1 \, dx\\ &=\frac{1}{8} a^2 (6 A+7 B+8 C) x+\frac{a^2 (18 A+20 B+25 C) \sin (c+d x)}{15 d}+\frac{a^2 (6 A+7 B+8 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (18 A+25 B+20 C) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac{A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{(2 A+5 B) \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{20 d}\\ \end{align*}

Mathematica [A]  time = 0.603488, size = 132, normalized size = 0.71 \[ \frac{a^2 (60 (11 A+12 B+14 C) \sin (c+d x)+240 (A+B+C) \sin (2 (c+d x))+90 A \sin (3 (c+d x))+30 A \sin (4 (c+d x))+6 A \sin (5 (c+d x))+240 A c+360 A d x+80 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+420 B c+420 B d x+40 C \sin (3 (c+d x))+480 C d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(240*A*c + 420*B*c + 360*A*d*x + 420*B*d*x + 480*C*d*x + 60*(11*A + 12*B + 14*C)*Sin[c + d*x] + 240*(A +
B + C)*Sin[2*(c + d*x)] + 90*A*Sin[3*(c + d*x)] + 80*B*Sin[3*(c + d*x)] + 40*C*Sin[3*(c + d*x)] + 30*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 = 0.11, size = 247, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}A\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 ) }+B{a}^{2} \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 ) +{\frac{{a}^{2}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,{a}^{2}A \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 ) +{\frac{2\,B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,{a}^{2}C \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{\frac{{a}^{2}A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+B{a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +{a}^{2}C\sin \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))^2*(A+B*sec(d*x+c)+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)+B*a^2*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x
+c)+3/8*d*x+3/8*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)+2/3*B*a^2*(2+cos(d*x+c)^2)*sin(d*x+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)+B*a^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^2*C*sin(d*x+c))

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Maxima [A]  time = 0.957742, size = 319, normalized size = 1.71 \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^{2} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} + 30 \,{\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} - 320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B 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 )} B a^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 480 \, C a^{2} \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))^2*(A+B*sec(d*x+c)+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^2 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c
))*A*a^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2 - 320*(sin(d*x + c)^3 - 3*sin(d*x
+ c))*B*a^2 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 + 120*(2*d*x + 2*c + sin(2*d*x
+ 2*c))*B*a^2 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 480
*C*a^2*sin(d*x + c))/d

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Fricas [A]  time = 0.509778, size = 305, normalized size = 1.63 \begin{align*} \frac{15 \,{\left (6 \, A + 7 \, B + 8 \, C\right )} a^{2} d x +{\left (24 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \,{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (9 \, A + 10 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \,{\left (6 \, A + 7 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \,{\left (18 \, A + 20 \, B + 25 \, C\right )} a^{2}\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))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(6*A + 7*B + 8*C)*a^2*d*x + (24*A*a^2*cos(d*x + c)^4 + 30*(2*A + B)*a^2*cos(d*x + c)^3 + 8*(9*A + 10
*B + 5*C)*a^2*cos(d*x + c)^2 + 15*(6*A + 7*B + 8*C)*a^2*cos(d*x + c) + 8*(18*A + 20*B + 25*C)*a^2)*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))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.24251, size = 404, normalized size = 2.16 \begin{align*} \frac{15 \,{\left (6 \, A a^{2} + 7 \, B a^{2} + 8 \, C a^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (90 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 105 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 120 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 420 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 490 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 560 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 864 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 800 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1120 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 540 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 790 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1040 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 390 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 375 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 360 \, 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}}}{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))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(6*A*a^2 + 7*B*a^2 + 8*C*a^2)*(d*x + c) + 2*(90*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 105*B*a^2*tan(1/2*d*x
+ 1/2*c)^9 + 120*C*a^2*tan(1/2*d*x + 1/2*c)^9 + 420*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 490*B*a^2*tan(1/2*d*x + 1/
2*c)^7 + 560*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 864*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*B*a^2*tan(1/2*d*x + 1/2*c)^
5 + 1120*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 540*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 790*B*a^2*tan(1/2*d*x + 1/2*c)^3 +
1040*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 390*A*a^2*tan(1/2*d*x + 1/2*c) + 375*B*a^2*tan(1/2*d*x + 1/2*c) + 360*C*a^
2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d