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

Optimal. Leaf size=149 \[ \frac{a^2 (7 A+8 B+12 C) \sin (c+d x)}{6 d}+\frac{a^2 (7 A+8 B+12 C) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} a^2 x (7 A+8 B+12 C)+\frac{(A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{6 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.327804, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4086, 4013, 3788, 2637, 4045, 8} \[ \frac{a^2 (7 A+8 B+12 C) \sin (c+d x)}{6 d}+\frac{a^2 (7 A+8 B+12 C) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} a^2 x (7 A+8 B+12 C)+\frac{(A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{6 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*A + 8*B + 12*C)*x)/8 + (a^2*(7*A + 8*B + 12*C)*Sin[c + d*x])/(6*d) + (a^2*(7*A + 8*B + 12*C)*Cos[c + d
*x]*Sin[c + d*x])/(24*d) + ((A + 2*B)*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 4086

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_))^(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] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.33116, size = 95, normalized size = 0.64 \[ \frac{a^2 (24 (6 A+7 B+8 C) \sin (c+d x)+24 (2 A+2 B+C) \sin (2 (c+d x))+16 A \sin (3 (c+d x))+3 A \sin (4 (c+d x))+84 A d x+8 B \sin (3 (c+d x))+96 B d x+144 C d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(84*A*d*x + 96*B*d*x + 144*C*d*x + 24*(6*A + 7*B + 8*C)*Sin[c + d*x] + 24*(2*A + 2*B + C)*Sin[2*(c + d*x)
] + 16*A*Sin[3*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 3*A*Sin[4*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.102, size = 203, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \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 ) +{\frac{2\,{a}^{2}A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{2}A \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,B{a}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{2}C \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +B{a}^{2}\sin \left ( dx+c \right ) +2\,{a}^{2}C\sin \left ( dx+c \right ) +{a}^{2}C \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+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+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)+
1/3*B*a^2*(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)+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 = 0.950691, size = 257, normalized size = 1.72 \begin{align*} -\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} + 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 48 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B 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} - 96 \, B a^{2} \sin \left (d x + c\right ) - 192 \, C a^{2} \sin \left (d x + c\right )}{96 \, d} \end{align*}

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+B*sec(d*x+c)+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 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 48*(2*d*x + 2
*c + sin(2*d*x + 2*c))*B*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 - 96*(d*x + c)*C*a^2 - 96*B*a^2*sin(d
*x + c) - 192*C*a^2*sin(d*x + c))/d

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Fricas [A]  time = 0.503528, size = 239, normalized size = 1.6 \begin{align*} \frac{3 \,{\left (7 \, A + 8 \, B + 12 \, C\right )} a^{2} d x +{\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, A + 8 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \,{\left (4 \, A + 5 \, B + 6 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

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

[Out]

1/24*(3*(7*A + 8*B + 12*C)*a^2*d*x + (6*A*a^2*cos(d*x + c)^3 + 8*(2*A + B)*a^2*cos(d*x + c)^2 + 3*(7*A + 8*B +
4*C)*a^2*cos(d*x + c) + 8*(4*A + 5*B + 6*C)*a^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)**4*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.26549, size = 335, normalized size = 2.25 \begin{align*} \frac{3 \,{\left (7 \, A a^{2} + 8 \, B 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} + 24 \, B 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} + 88 \, B 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} + 136 \, B 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 ) + 72 \, B 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} \end{align*}

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

[Out]

1/24*(3*(7*A*a^2 + 8*B*a^2 + 12*C*a^2)*(d*x + c) + 2*(21*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 24*B*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 + 88*B*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 + 136*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 15
6*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 75*A*a^2*tan(1/2*d*x + 1/2*c) + 72*B*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