3.484 \(\int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=173 \[ -\frac{a^2 \left (4 a^2+27 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (29 a^2 b^2+4 a^4+5 b^4\right ) \sin (c+d x)}{5 d}+\frac{a b \left (3 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} a b x \left (3 a^2+4 b^2\right )+\frac{3 a^3 b \sin (c+d x) \cos ^3(c+d x)}{5 d}+\frac{a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

[Out]

(a*b*(3*a^2 + 4*b^2)*x)/2 + ((4*a^4 + 29*a^2*b^2 + 5*b^4)*Sin[c + d*x])/(5*d) + (a*b*(3*a^2 + 4*b^2)*Cos[c + d
*x]*Sin[c + d*x])/(2*d) + (3*a^3*b*Cos[c + d*x]^3*Sin[c + d*x])/(5*d) + (a^2*Cos[c + d*x]^4*(a + b*Sec[c + d*x
])^2*Sin[c + d*x])/(5*d) - (a^2*(4*a^2 + 27*b^2)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.351732, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3841, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac{a^2 \left (4 a^2+27 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (29 a^2 b^2+4 a^4+5 b^4\right ) \sin (c+d x)}{5 d}+\frac{a b \left (3 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} a b x \left (3 a^2+4 b^2\right )+\frac{3 a^3 b \sin (c+d x) \cos ^3(c+d x)}{5 d}+\frac{a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*(3*a^2 + 4*b^2)*x)/2 + ((4*a^4 + 29*a^2*b^2 + 5*b^4)*Sin[c + d*x])/(5*d) + (a*b*(3*a^2 + 4*b^2)*Cos[c + d
*x]*Sin[c + d*x])/(2*d) + (3*a^3*b*Cos[c + d*x]^3*Sin[c + d*x])/(5*d) + (a^2*Cos[c + d*x]^4*(a + b*Sec[c + d*x
])^2*Sin[c + d*x])/(5*d) - (a^2*(4*a^2 + 27*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 4074

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_)), 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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[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 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 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (12 a^2 b+a \left (4 a^2+15 b^2\right ) \sec (c+d x)+b \left (2 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{1}{20} \int \cos ^3(c+d x) \left (-4 a^2 \left (4 a^2+27 b^2\right )-20 a b \left (3 a^2+4 b^2\right ) \sec (c+d x)-4 b^2 \left (2 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{1}{20} \int \cos ^3(c+d x) \left (-4 a^2 \left (4 a^2+27 b^2\right )-4 b^2 \left (2 a^2+5 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\left (a b \left (3 a^2+4 b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{a b \left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{1}{20} \int \cos (c+d x) \left (-4 b^2 \left (2 a^2+5 b^2\right )-4 a^2 \left (4 a^2+27 b^2\right ) \cos ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (a b \left (3 a^2+4 b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{2} a b \left (3 a^2+4 b^2\right ) x+\frac{a b \left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{\operatorname{Subst}\left (\int \left (-4 b^2 \left (2 a^2+5 b^2\right )-4 a^2 \left (4 a^2+27 b^2\right )+4 a^2 \left (4 a^2+27 b^2\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac{1}{2} a b \left (3 a^2+4 b^2\right ) x+\frac{\left (4 a^4+29 a^2 b^2+5 b^4\right ) \sin (c+d x)}{5 d}+\frac{a b \left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{3 a^3 b \cos ^3(c+d x) \sin (c+d x)}{5 d}+\frac{a^2 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac{a^2 \left (4 a^2+27 b^2\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.497746, size = 133, normalized size = 0.77 \[ \frac{30 \left (36 a^2 b^2+5 a^4+8 b^4\right ) \sin (c+d x)+a \left (240 b \left (a^2+b^2\right ) \sin (2 (c+d x))+5 \left (5 a^3+24 a b^2\right ) \sin (3 (c+d x))+30 a^2 b \sin (4 (c+d x))+360 a^2 b c+360 a^2 b d x+3 a^3 \sin (5 (c+d x))+480 b^3 c+480 b^3 d x\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*(5*a^4 + 36*a^2*b^2 + 8*b^4)*Sin[c + d*x] + a*(360*a^2*b*c + 480*b^3*c + 360*a^2*b*d*x + 480*b^3*d*x + 240
*b*(a^2 + b^2)*Sin[2*(c + d*x)] + 5*(5*a^3 + 24*a*b^2)*Sin[3*(c + d*x)] + 30*a^2*b*Sin[4*(c + d*x)] + 3*a^3*Si
n[5*(c + d*x)]))/(240*d)

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Maple [A]  time = 0.062, size = 138, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{{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}^{3}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 ) +2\,{a}^{2}{b}^{2} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) +4\,a{b}^{3} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{b}^{4}\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+b*sec(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 1.17717, size = 180, normalized size = 1.04 \begin{align*} \frac{8 \,{\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^{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 )} a^{3} b - 240 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{3} + 120 \, b^{4} \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))^4,x, algorithm="maxima")

[Out]

1/120*(8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) +
8*sin(2*d*x + 2*c))*a^3*b - 240*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*
c))*a*b^3 + 120*b^4*sin(d*x + c))/d

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Fricas [A]  time = 1.74751, size = 285, normalized size = 1.65 \begin{align*} \frac{15 \,{\left (3 \, a^{3} b + 4 \, a b^{3}\right )} d x +{\left (6 \, a^{4} \cos \left (d x + c\right )^{4} + 30 \, a^{3} b \cos \left (d x + c\right )^{3} + 16 \, a^{4} + 120 \, a^{2} b^{2} + 30 \, b^{4} + 4 \,{\left (2 \, a^{4} + 15 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (3 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, 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))^4,x, algorithm="fricas")

[Out]

1/30*(15*(3*a^3*b + 4*a*b^3)*d*x + (6*a^4*cos(d*x + c)^4 + 30*a^3*b*cos(d*x + c)^3 + 16*a^4 + 120*a^2*b^2 + 30
*b^4 + 4*(2*a^4 + 15*a^2*b^2)*cos(d*x + c)^2 + 15*(3*a^3*b + 4*a*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))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.37699, size = 574, normalized size = 3.32 \begin{align*} \frac{15 \,{\left (3 \, a^{3} b + 4 \, a b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (30 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 75 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 180 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 60 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 30 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 40 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 30 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 480 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 120 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 120 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 116 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 600 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 180 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 480 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 75 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 180 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 60 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 30 \, b^{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}}}{30 \, 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))^4,x, algorithm="giac")

[Out]

1/30*(15*(3*a^3*b + 4*a*b^3)*(d*x + c) + 2*(30*a^4*tan(1/2*d*x + 1/2*c)^9 - 75*a^3*b*tan(1/2*d*x + 1/2*c)^9 +
180*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*b^4*tan(1/2*d*x + 1/2*c)^9 + 40*a^4*
tan(1/2*d*x + 1/2*c)^7 - 30*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 480*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*a*b^3*tan(
1/2*d*x + 1/2*c)^7 + 120*b^4*tan(1/2*d*x + 1/2*c)^7 + 116*a^4*tan(1/2*d*x + 1/2*c)^5 + 600*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^5 + 180*b^4*tan(1/2*d*x + 1/2*c)^5 + 40*a^4*tan(1/2*d*x + 1/2*c)^3 + 30*a^3*b*tan(1/2*d*x + 1/2*c)^3
+ 480*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*b^4*tan(1/2*d*x + 1/2*c)^3 + 30
*a^4*tan(1/2*d*x + 1/2*c) + 75*a^3*b*tan(1/2*d*x + 1/2*c) + 180*a^2*b^2*tan(1/2*d*x + 1/2*c) + 60*a*b^3*tan(1/
2*d*x + 1/2*c) + 30*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d