3.465 \(\int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=111 \[ -\frac{\left (2 a^2+b^2\right ) \sin ^3(c+d x)}{3 d}+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac{a^2 \sin ^5(c+d x)}{5 d}+\frac{a b \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a b x}{4} \]

[Out]

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

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Rubi [A]  time = 0.122443, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3788, 2635, 8, 4044, 3013, 373} \[ -\frac{\left (2 a^2+b^2\right ) \sin ^3(c+d x)}{3 d}+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac{a^2 \sin ^5(c+d x)}{5 d}+\frac{a b \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a b x}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), 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 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]

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.163394, size = 85, normalized size = 0.77 \[ \frac{-80 \left (2 a^2+b^2\right ) \sin ^3(c+d x)+240 \left (a^2+b^2\right ) \sin (c+d x)+48 a^2 \sin ^5(c+d x)+15 a b (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(240*(a^2 + b^2)*Sin[c + d*x] - 80*(2*a^2 + b^2)*Sin[c + d*x]^3 + 48*a^2*Sin[c + d*x]^5 + 15*a*b*(12*(c + d*x)
+ 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(240*d)

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Maple [A]  time = 0.054, size = 95, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}\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 ) }+2\,ab \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{{b}^{2} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) }{3}} \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.03851, size = 127, normalized size = 1.14 \begin{align*} \frac{16 \,{\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^{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 )} a b - 80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{2}}{240 \, 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))^2,x, algorithm="maxima")

[Out]

1/240*(16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^2 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c)
+ 8*sin(2*d*x + 2*c))*a*b - 80*(sin(d*x + c)^3 - 3*sin(d*x + c))*b^2)/d

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Fricas [A]  time = 1.70132, size = 215, normalized size = 1.94 \begin{align*} \frac{45 \, a b d x +{\left (12 \, a^{2} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right )^{3} + 45 \, a b \cos \left (d x + c\right ) + 4 \,{\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 32 \, a^{2} + 40 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \, 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))^2,x, algorithm="fricas")

[Out]

1/60*(45*a*b*d*x + (12*a^2*cos(d*x + c)^4 + 30*a*b*cos(d*x + c)^3 + 45*a*b*cos(d*x + c) + 4*(4*a^2 + 5*b^2)*co
s(d*x + c)^2 + 32*a^2 + 40*b^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+b*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.30324, size = 333, normalized size = 3. \begin{align*} \frac{45 \,{\left (d x + c\right )} a b + \frac{2 \,{\left (60 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 75 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 60 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 80 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 30 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 160 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 232 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 200 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 80 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 160 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 60 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 75 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 60 \, b^{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}}}{60 \, 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))^2,x, algorithm="giac")

[Out]

1/60*(45*(d*x + c)*a*b + 2*(60*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*b^2*tan(1/2*d*x
+ 1/2*c)^9 + 80*a^2*tan(1/2*d*x + 1/2*c)^7 - 30*a*b*tan(1/2*d*x + 1/2*c)^7 + 160*b^2*tan(1/2*d*x + 1/2*c)^7 +
232*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*b^2*tan(1/2*d*x + 1/2*c)^5 + 80*a^2*tan(1/2*d*x + 1/2*c)^3 + 30*a*b*tan(
1/2*d*x + 1/2*c)^3 + 160*b^2*tan(1/2*d*x + 1/2*c)^3 + 60*a^2*tan(1/2*d*x + 1/2*c) + 75*a*b*tan(1/2*d*x + 1/2*c
) + 60*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d