3.474 \(\int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=160 \[ -\frac{a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac{b \left (9 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} b x \left (9 a^2+4 b^2\right )+\frac{11 a^2 b \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d} \]

[Out]

(b*(9*a^2 + 4*b^2)*x)/8 + (a*(4*a^2 + 15*b^2)*Sin[c + d*x])/(5*d) + (b*(9*a^2 + 4*b^2)*Cos[c + d*x]*Sin[c + d*
x])/(8*d) + (11*a^2*b*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (a^2*Cos[c + d*x]^4*(a + b*Sec[c + d*x])*Sin[c + d
*x])/(5*d) - (a*(4*a^2 + 15*b^2)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.191936, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3841, 4047, 2633, 4045, 2635, 8} \[ -\frac{a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac{b \left (9 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} b x \left (9 a^2+4 b^2\right )+\frac{11 a^2 b \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(9*a^2 + 4*b^2)*x)/8 + (a*(4*a^2 + 15*b^2)*Sin[c + d*x])/(5*d) + (b*(9*a^2 + 4*b^2)*Cos[c + d*x]*Sin[c + d*
x])/(8*d) + (11*a^2*b*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (a^2*Cos[c + d*x]^4*(a + b*Sec[c + d*x])*Sin[c + d
*x])/(5*d) - (a*(4*a^2 + 15*b^2)*Sin[c + d*x]^3)/(15*d)

Rule 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^4(c+d x) \left (11 a^2 b+a \left (4 a^2+15 b^2\right ) \sec (c+d x)+b \left (3 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^4(c+d x) \left (11 a^2 b+b \left (3 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac{1}{5} \left (a \left (4 a^2+15 b^2\right )\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac{11 a^2 b \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{4} \left (b \left (9 a^2+4 b^2\right )\right ) \int \cos ^2(c+d x) \, dx-\frac{\left (a \left (4 a^2+15 b^2\right )\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac{b \left (9 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{11 a^2 b \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}-\frac{a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{1}{8} \left (b \left (9 a^2+4 b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{8} b \left (9 a^2+4 b^2\right ) x+\frac{a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac{b \left (9 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{11 a^2 b \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}-\frac{a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.29678, size = 130, normalized size = 0.81 \[ \frac{60 a \left (5 a^2+18 b^2\right ) \sin (c+d x)+120 \left (3 a^2 b+b^3\right ) \sin (2 (c+d x))+45 a^2 b \sin (4 (c+d x))+540 a^2 b c+540 a^2 b d x+50 a^3 \sin (3 (c+d x))+6 a^3 \sin (5 (c+d x))+120 a b^2 \sin (3 (c+d x))+240 b^3 c+240 b^3 d x}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(540*a^2*b*c + 240*b^3*c + 540*a^2*b*d*x + 240*b^3*d*x + 60*a*(5*a^2 + 18*b^2)*Sin[c + d*x] + 120*(3*a^2*b + b
^3)*Sin[2*(c + d*x)] + 50*a^3*Sin[3*(c + d*x)] + 120*a*b^2*Sin[3*(c + d*x)] + 45*a^2*b*Sin[4*(c + d*x)] + 6*a^
3*Sin[5*(c + d*x)])/(480*d)

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Maple [A]  time = 0.057, size = 123, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{{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}^{2}b \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{b}^{2} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) +{b}^{3} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20398, size = 161, normalized size = 1.01 \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^{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^{2} b - 480 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3,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^3 + 45*(12*d*x + 12*c + sin(4*d*x + 4*c)
+ 8*sin(2*d*x + 2*c))*a^2*b - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*a*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c
))*b^3)/d

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Fricas [A]  time = 1.69645, size = 265, normalized size = 1.66 \begin{align*} \frac{15 \,{\left (9 \, a^{2} b + 4 \, b^{3}\right )} d x +{\left (24 \, a^{3} \cos \left (d x + c\right )^{4} + 90 \, a^{2} b \cos \left (d x + c\right )^{3} + 64 \, a^{3} + 240 \, a b^{2} + 8 \,{\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )\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+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/120*(15*(9*a^2*b + 4*b^3)*d*x + (24*a^3*cos(d*x + c)^4 + 90*a^2*b*cos(d*x + c)^3 + 64*a^3 + 240*a*b^2 + 8*(4
*a^3 + 15*a*b^2)*cos(d*x + c)^2 + 15*(9*a^2*b + 4*b^3)*cos(d*x + c))*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+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.29605, size = 448, normalized size = 2.8 \begin{align*} \frac{15 \,{\left (9 \, a^{2} b + 4 \, b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (120 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 225 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 360 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 60 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 160 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 90 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 960 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 120 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 464 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1200 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 160 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 90 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 960 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 225 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 360 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 60 \, b^{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+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/120*(15*(9*a^2*b + 4*b^3)*(d*x + c) + 2*(120*a^3*tan(1/2*d*x + 1/2*c)^9 - 225*a^2*b*tan(1/2*d*x + 1/2*c)^9 +
360*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*b^3*tan(1/2*d*x + 1/2*c)^9 + 160*a^3*tan(1/2*d*x + 1/2*c)^7 - 90*a^2*b*
tan(1/2*d*x + 1/2*c)^7 + 960*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*a^3*tan(1/2*d
*x + 1/2*c)^5 + 1200*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 160*a^3*tan(1/2*d*x + 1/2*c)^3 + 90*a^2*b*tan(1/2*d*x + 1/
2*c)^3 + 960*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*a^3*tan(1/2*d*x + 1/2*c) + 22
5*a^2*b*tan(1/2*d*x + 1/2*c) + 360*a*b^2*tan(1/2*d*x + 1/2*c) + 60*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/
2*c)^2 + 1)^5)/d