3.473 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=123 \[ \frac{b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{4 d}+\frac{3 a \left (a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} a x \left (a^2+4 b^2\right )-\frac{3 a^2 b \sin ^3(c+d x)}{4 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.183282, antiderivative size = 123, 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, 2635, 8, 4044, 3013} \[ \frac{b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{4 d}+\frac{3 a \left (a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} a x \left (a^2+4 b^2\right )-\frac{3 a^2 b \sin ^3(c+d x)}{4 d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3,x]

[Out]

(3*a*(a^2 + 4*b^2)*x)/8 + (b*(11*a^2 + 4*b^2)*Sin[c + d*x])/(4*d) + (3*a*(a^2 + 4*b^2)*Cos[c + d*x]*Sin[c + d*
x])/(8*d) + (a^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])*Sin[c + d*x])/(4*d) - (3*a^2*b*Sin[c + d*x]^3)/(4*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 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 ^4(c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac{a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^3(c+d x) \left (9 a^2 b+3 a \left (a^2+4 b^2\right ) \sec (c+d x)+2 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^3(c+d x) \left (9 a^2 b+2 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac{1}{4} \left (3 a \left (a^2+4 b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{3 a \left (a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos (c+d x) \left (2 b \left (a^2+2 b^2\right )+9 a^2 b \cos ^2(c+d x)\right ) \, dx+\frac{1}{8} \left (3 a \left (a^2+4 b^2\right )\right ) \int 1 \, dx\\ &=\frac{3}{8} a \left (a^2+4 b^2\right ) x+\frac{3 a \left (a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \left (9 a^2 b+2 b \left (a^2+2 b^2\right )-9 a^2 b x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac{3}{8} a \left (a^2+4 b^2\right ) x+\frac{b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{4 d}+\frac{3 a \left (a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}-\frac{3 a^2 b \sin ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.272784, size = 100, normalized size = 0.81 \[ \frac{8 b \left (9 a^2+4 b^2\right ) \sin (c+d x)+a \left (8 \left (a^2+3 b^2\right ) \sin (2 (c+d x))+a^2 \sin (4 (c+d x))+12 a^2 c+12 a^2 d x+8 a b \sin (3 (c+d x))+48 b^2 c+48 b^2 d x\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3,x]

[Out]

(8*b*(9*a^2 + 4*b^2)*Sin[c + d*x] + a*(12*a^2*c + 48*b^2*c + 12*a^2*d*x + 48*b^2*d*x + 8*(a^2 + 3*b^2)*Sin[2*(
c + d*x)] + 8*a*b*Sin[3*(c + d*x)] + a^2*Sin[4*(c + d*x)]))/(32*d)

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Maple [A]  time = 0.058, size = 102, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({a}^{3} \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 ) +{a}^{2}b \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) +3\,a{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{b}^{3}\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.20503, size = 128, normalized size = 1.04 \begin{align*} \frac{{\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} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2} + 32 \, b^{3} \sin \left (d x + c\right )}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/32*((12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2*b
+ 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b^2 + 32*b^3*sin(d*x + c))/d

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Fricas [A]  time = 1.67574, size = 197, normalized size = 1.6 \begin{align*} \frac{3 \,{\left (a^{3} + 4 \, a b^{2}\right )} d x +{\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 8 \, a^{2} b \cos \left (d x + c\right )^{2} + 16 \, a^{2} b + 8 \, b^{3} + 3 \,{\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/8*(3*(a^3 + 4*a*b^2)*d*x + (2*a^3*cos(d*x + c)^3 + 8*a^2*b*cos(d*x + c)^2 + 16*a^2*b + 8*b^3 + 3*(a^3 + 4*a*
b^2)*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)**4*(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.36587, size = 401, normalized size = 3.26 \begin{align*} \frac{3 \,{\left (a^{3} + 4 \, a b^{2}\right )}{\left (d x + c\right )} - \frac{2 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 8 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 40 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 24 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, 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 )}^{4}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(3*(a^3 + 4*a*b^2)*(d*x + c) - 2*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*a*b^
2*tan(1/2*d*x + 1/2*c)^7 - 8*b^3*tan(1/2*d*x + 1/2*c)^7 - 3*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*a^2*b*tan(1/2*d*x
+ 1/2*c)^5 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 24*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^3 -
40*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*b^3*tan(1/2*d*x + 1/2*c)^3 - 5*a^3*tan(
1/2*d*x + 1/2*c) - 24*a^2*b*tan(1/2*d*x + 1/2*c) - 12*a*b^2*tan(1/2*d*x + 1/2*c) - 8*b^3*tan(1/2*d*x + 1/2*c))
/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d