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

Optimal. Leaf size=102 \[ \frac {2 a^2 (2 B+3 C) \sin (c+d x)}{3 d}+\frac {a^2 (2 B+3 C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {1}{2} a^2 x (2 B+3 C)+\frac {B \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[Out]

1/2*a^2*(2*B+3*C)*x+2/3*a^2*(2*B+3*C)*sin(d*x+c)/d+1/6*a^2*(2*B+3*C)*cos(d*x+c)*sin(d*x+c)/d+1/3*B*cos(d*x+c)^
2*(a+a*sec(d*x+c))^2*sin(d*x+c)/d

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Rubi [A]  time = 0.23, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4072, 4013, 3788, 2637, 4045, 8} \[ \frac {2 a^2 (2 B+3 C) \sin (c+d x)}{3 d}+\frac {a^2 (2 B+3 C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {1}{2} a^2 x (2 B+3 C)+\frac {B \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(2*B + 3*C)*x)/2 + (2*a^2*(2*B + 3*C)*Sin[c + d*x])/(3*d) + (a^2*(2*B + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(
6*d) + (B*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*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 4072

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

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 61, normalized size = 0.60 \[ \frac {a^2 (3 (7 B+8 C) \sin (c+d x)+3 (2 B+C) \sin (2 (c+d x))+B \sin (3 (c+d x))+12 B d x+18 C d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*B*d*x + 18*C*d*x + 3*(7*B + 8*C)*Sin[c + d*x] + 3*(2*B + C)*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(
12*d)

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fricas [A]  time = 0.46, size = 70, normalized size = 0.69 \[ \frac {3 \, {\left (2 \, B + 3 \, C\right )} a^{2} d x + {\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (5 \, B + 6 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*B + 3*C)*a^2*d*x + (2*B*a^2*cos(d*x + c)^2 + 3*(2*B + C)*a^2*cos(d*x + c) + 2*(5*B + 6*C)*a^2)*sin(d
*x + c))/d

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giac [A]  time = 0.26, size = 142, normalized size = 1.39 \[ \frac {3 \, {\left (2 \, B a^{2} + 3 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*B*a^2 + 3*C*a^2)*(d*x + c) + 2*(6*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 16
*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 18*B*a^2*tan(1/2*d*x + 1/2*c) + 15*C*a^2*tan
(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 1.20, size = 116, normalized size = 1.14 \[ \frac {\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+2 a^{2} C \sin \left (d x +c \right )+a^{2} C \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.39, size = 110, normalized size = 1.08 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 12 \, {\left (d x + c\right )} C a^{2} - 12 \, B a^{2} \sin \left (d x + c\right ) - 24 \, C a^{2} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 - 3*(2*d*x + 2*c +
sin(2*d*x + 2*c))*C*a^2 - 12*(d*x + c)*C*a^2 - 12*B*a^2*sin(d*x + c) - 24*C*a^2*sin(d*x + c))/d

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mupad [B]  time = 2.87, size = 98, normalized size = 0.96 \[ B\,a^2\,x+\frac {3\,C\,a^2\,x}{2}+\frac {7\,B\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*a^2*x + (3*C*a^2*x)/2 + (7*B*a^2*sin(c + d*x))/(4*d) + (2*C*a^2*sin(c + d*x))/d + (B*a^2*sin(2*c + 2*d*x))/(
2*d) + (B*a^2*sin(3*c + 3*d*x))/(12*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*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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