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

Optimal. Leaf size=145 \[ \frac {5 a^3 b \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {4 a b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac {1}{8} x \left (3 a^4+24 a^2 b^2+8 b^4\right ) \]

[Out]

1/8*(3*a^4+24*a^2*b^2+8*b^4)*x+4/3*a*b*(2*a^2+3*b^2)*sin(d*x+c)/d+1/8*a^2*(3*a^2+22*b^2)*cos(d*x+c)*sin(d*x+c)
/d+5/6*a^3*b*cos(d*x+c)^2*sin(d*x+c)/d+1/4*a^2*cos(d*x+c)^3*(a+b*sec(d*x+c))^2*sin(d*x+c)/d

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Rubi [A]  time = 0.31, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3841, 4074, 4047, 2637, 4045, 8} \[ \frac {4 a b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (24 a^2 b^2+3 a^4+8 b^4\right )+\frac {5 a^3 b \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^4 + 24*a^2*b^2 + 8*b^4)*x)/8 + (4*a*b*(2*a^2 + 3*b^2)*Sin[c + d*x])/(3*d) + (a^2*(3*a^2 + 22*b^2)*Cos[c
+ d*x]*Sin[c + d*x])/(8*d) + (5*a^3*b*Cos[c + d*x]^2*Sin[c + d*x])/(6*d) + (a^2*Cos[c + d*x]^3*(a + b*Sec[c +
d*x])^2*Sin[c + d*x])/(4*d)

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]

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 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 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 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]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (10 a^2 b+3 a \left (a^2+4 b^2\right ) \sec (c+d x)+b \left (a^2+4 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {5 a^3 b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{12} \int \cos ^2(c+d x) \left (-3 a^2 \left (3 a^2+22 b^2\right )-16 a b \left (2 a^2+3 b^2\right ) \sec (c+d x)-3 b^2 \left (a^2+4 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {5 a^3 b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{12} \int \cos ^2(c+d x) \left (-3 a^2 \left (3 a^2+22 b^2\right )-3 b^2 \left (a^2+4 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{3} \left (4 a b \left (2 a^2+3 b^2\right )\right ) \int \cos (c+d x) \, dx\\ &=\frac {4 a b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {5 a^3 b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{8} \left (-3 a^4-24 a^2 b^2-8 b^4\right ) \int 1 \, dx\\ &=\frac {1}{8} \left (3 a^4+24 a^2 b^2+8 b^4\right ) x+\frac {4 a b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {5 a^3 b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 104, normalized size = 0.72 \[ \frac {3 a^4 \sin (4 (c+d x))+32 a^3 b \sin (3 (c+d x))+24 a^2 \left (a^2+6 b^2\right ) \sin (2 (c+d x))+96 a b \left (3 a^2+4 b^2\right ) \sin (c+d x)+12 \left (3 a^4+24 a^2 b^2+8 b^4\right ) (c+d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(3*a^4 + 24*a^2*b^2 + 8*b^4)*(c + d*x) + 96*a*b*(3*a^2 + 4*b^2)*Sin[c + d*x] + 24*a^2*(a^2 + 6*b^2)*Sin[2*
(c + d*x)] + 32*a^3*b*Sin[3*(c + d*x)] + 3*a^4*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.47, size = 96, normalized size = 0.66 \[ \frac {3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} d x + {\left (6 \, a^{4} \cos \left (d x + c\right )^{3} + 32 \, a^{3} b \cos \left (d x + c\right )^{2} + 64 \, a^{3} b + 96 \, a b^{3} + 9 \, {\left (a^{4} + 8 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*d*x + (6*a^4*cos(d*x + c)^3 + 32*a^3*b*cos(d*x + c)^2 + 64*a^3*b + 96*a*b
^3 + 9*(a^4 + 8*a^2*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.24, size = 318, normalized size = 2.19 \[ \frac {3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 288 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*(d*x + c) - 2*(15*a^4*tan(1/2*d*x + 1/2*c)^7 - 96*a^3*b*tan(1/2*d*x + 1/2
*c)^7 + 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 96*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 9*a^4*tan(1/2*d*x + 1/2*c)^5 - 1
60*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 288*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 9*a^4
*tan(1/2*d*x + 1/2*c)^3 - 160*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 288*a*b^3*tan
(1/2*d*x + 1/2*c)^3 - 15*a^4*tan(1/2*d*x + 1/2*c) - 96*a^3*b*tan(1/2*d*x + 1/2*c) - 72*a^2*b^2*tan(1/2*d*x + 1
/2*c) - 96*a*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.95, size = 116, normalized size = 0.80 \[ \frac {a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a^{3} b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+6 a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a \,b^{3} \sin \left (d x +c \right )+b^{4} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^4*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+4/3*a^3*b*(2+cos(d*x+c)^2)*sin(d*x+c)+6*
a^2*b^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*a*b^3*sin(d*x+c)+b^4*(d*x+c))

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maxima [A]  time = 0.46, size = 109, normalized size = 0.75 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} b + 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b^{2} + 96 \, {\left (d x + c\right )} b^{4} + 384 \, a b^{3} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^4 - 128*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^
3*b + 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2*b^2 + 96*(d*x + c)*b^4 + 384*a*b^3*sin(d*x + c))/d

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mupad [B]  time = 0.88, size = 123, normalized size = 0.85 \[ \frac {3\,a^4\,x}{8}+b^4\,x+3\,a^2\,b^2\,x+\frac {a^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {3\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {4\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))**4,x)

[Out]

Timed out

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