3.99 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=136 \[ \frac {a^2 (7 A+12 C) \sin (c+d x)}{6 d}+\frac {a^2 (7 A+12 C) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} a^2 x (7 A+12 C)+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

[Out]

1/8*a^2*(7*A+12*C)*x+1/6*a^2*(7*A+12*C)*sin(d*x+c)/d+1/24*a^2*(7*A+12*C)*cos(d*x+c)*sin(d*x+c)/d+1/6*A*cos(d*x
+c)^2*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^2*sin(d*x+c)/d

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Rubi [A]  time = 0.31, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4087, 4013, 3788, 2637, 4045, 8} \[ \frac {a^2 (7 A+12 C) \sin (c+d x)}{6 d}+\frac {a^2 (7 A+12 C) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} a^2 x (7 A+12 C)+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(7*A + 12*C)*x)/8 + (a^2*(7*A + 12*C)*Sin[c + d*x])/(6*d) + (a^2*(7*A + 12*C)*Cos[c + d*x]*Sin[c + d*x])/
(24*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]^3*(a + a*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 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 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -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 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^2 (2 a A+a (A+4 C) \sec (c+d x)) \, dx}{4 a}\\ &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {1}{12} (7 A+12 C) \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \, dx\\ &=\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {1}{12} (7 A+12 C) \int \cos ^2(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac {1}{6} \left (a^2 (7 A+12 C)\right ) \int \cos (c+d x) \, dx\\ &=\frac {a^2 (7 A+12 C) \sin (c+d x)}{6 d}+\frac {a^2 (7 A+12 C) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {1}{8} \left (a^2 (7 A+12 C)\right ) \int 1 \, dx\\ &=\frac {1}{8} a^2 (7 A+12 C) x+\frac {a^2 (7 A+12 C) \sin (c+d x)}{6 d}+\frac {a^2 (7 A+12 C) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 73, normalized size = 0.54 \[ \frac {a^2 (48 (3 A+4 C) \sin (c+d x)+24 (2 A+C) \sin (2 (c+d x))+16 A \sin (3 (c+d x))+3 A \sin (4 (c+d x))+84 A d x+144 C d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(84*A*d*x + 144*C*d*x + 48*(3*A + 4*C)*Sin[c + d*x] + 24*(2*A + C)*Sin[2*(c + d*x)] + 16*A*Sin[3*(c + d*x
)] + 3*A*Sin[4*(c + d*x)]))/(96*d)

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fricas [A]  time = 0.44, size = 86, normalized size = 0.63 \[ \frac {3 \, {\left (7 \, A + 12 \, C\right )} a^{2} d x + {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 16 \, A a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) + 16 \, {\left (2 \, A + 3 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*(7*A + 12*C)*a^2*d*x + (6*A*a^2*cos(d*x + c)^3 + 16*A*a^2*cos(d*x + c)^2 + 3*(7*A + 4*C)*a^2*cos(d*x +
c) + 16*(2*A + 3*C)*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.85, size = 176, normalized size = 1.29 \[ \frac {3 \, {\left (7 \, A a^{2} + 12 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (21 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 77 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 132 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 156 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(7*A*a^2 + 12*C*a^2)*(d*x + c) + 2*(21*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 36*C*a^2*tan(1/2*d*x + 1/2*c)^7
+ 77*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 132*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 156*C
*a^2*tan(1/2*d*x + 1/2*c)^3 + 75*A*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*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 = 1.27, size = 142, normalized size = 1.04 \[ \frac {a^{2} A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} C \sin \left (d x +c \right )+a^{2} C \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.34, size = 132, normalized size = 0.97 \[ -\frac {64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 96 \, {\left (d x + c\right )} C a^{2} - 192 \, C a^{2} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/96*(64*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*
A*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 - 96*(d*x + c)*C
*a^2 - 192*C*a^2*sin(d*x + c))/d

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mupad [B]  time = 2.66, size = 117, normalized size = 0.86 \[ \frac {7\,A\,a^2\,x}{8}+\frac {3\,C\,a^2\,x}{2}+\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {A\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2,x)

[Out]

(7*A*a^2*x)/8 + (3*C*a^2*x)/2 + (3*A*a^2*sin(c + d*x))/(2*d) + (2*C*a^2*sin(c + d*x))/d + (A*a^2*sin(2*c + 2*d
*x))/(2*d) + (A*a^2*sin(3*c + 3*d*x))/(6*d) + (A*a^2*sin(4*c + 4*d*x))/(32*d) + (C*a^2*sin(2*c + 2*d*x))/(4*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+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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