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

Optimal. Leaf size=107 \[ \frac{\left (2 a^2 B+6 a b C+3 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (a^2 C+2 a b B+2 b^2 C\right )+\frac{a^2 B \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac{a (a C+2 b B) \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.288676, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4072, 4024, 4047, 2637, 4045, 8} \[ \frac{\left (2 a^2 B+6 a b C+3 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (a^2 C+2 a b B+2 b^2 C\right )+\frac{a^2 B \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac{a (a C+2 b B) \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 4024

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2*(csc[(e_.) + (f_.)*(x_)]*(B_
.) + (A_)), x_Symbol] :> Simp[(a^2*A*Cos[e + f*x]*(d*Csc[e + f*x])^(n + 1))/(d*f*n), x] + Dist[1/(d*n), Int[(d
*Csc[e + f*x])^(n + 1)*(a*(2*A*b + a*B)*n + (2*a*b*B*n + A*(b^2*n + a^2*(n + 1)))*Csc[e + f*x] + b^2*B*n*Csc[e
+ f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && 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 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+b \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+b \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac{a^2 B \cos ^2(c+d x) \sin (c+d x)}{3 d}-\frac{1}{3} \int \cos ^2(c+d x) \left (-3 a (2 b B+a C)+\left (\left (-2 a^2-3 b^2\right ) B-6 a b C\right ) \sec (c+d x)-3 b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 B \cos ^2(c+d x) \sin (c+d x)}{3 d}-\frac{1}{3} \int \cos ^2(c+d x) \left (-3 a (2 b B+a C)-3 b^2 C \sec ^2(c+d x)\right ) \, dx-\frac{1}{3} \left (-2 a^2 B-3 b^2 B-6 a b C\right ) \int \cos (c+d x) \, dx\\ &=\frac{\left (2 a^2 B+3 b^2 B+6 a b C\right ) \sin (c+d x)}{3 d}+\frac{a (2 b B+a C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a^2 B \cos ^2(c+d x) \sin (c+d x)}{3 d}-\frac{1}{2} \left (-2 a b B-a^2 C-2 b^2 C\right ) \int 1 \, dx\\ &=\frac{1}{2} \left (2 a b B+a^2 C+2 b^2 C\right ) x+\frac{\left (2 a^2 B+3 b^2 B+6 a b C\right ) \sin (c+d x)}{3 d}+\frac{a (2 b B+a C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a^2 B \cos ^2(c+d x) \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.244285, size = 90, normalized size = 0.84 \[ \frac{6 (c+d x) \left (a^2 C+2 a b B+2 b^2 C\right )+3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \sin (c+d x)+a^2 B \sin (3 (c+d x))+3 a (a C+2 b B) \sin (2 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(2*a*b*B + a^2*C + 2*b^2*C)*(c + d*x) + 3*(3*a^2*B + 4*b^2*B + 8*a*b*C)*Sin[c + d*x] + 3*a*(2*b*B + a*C)*Si
n[2*(c + d*x)] + a^2*B*Sin[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.063, size = 114, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,Bab \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{b}^{2}\sin \left ( dx+c \right ) +2\,abC\sin \left ( dx+c \right ) +{b}^{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+b*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*b*(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*b^2*sin(d*x+c)+2*a*b*C*sin(d*x+c)+b^2*C*(d*x+c))

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Maxima [A]  time = 0.962853, size = 146, normalized size = 1.36 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + 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} - 6 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b - 12 \,{\left (d x + c\right )} C b^{2} - 24 \, C a b \sin \left (d x + c\right ) - 12 \, B b^{2} \sin \left (d x + c\right )}{12 \, 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))^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 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 - 6*(2*d*x + 2*c +
sin(2*d*x + 2*c))*B*a*b - 12*(d*x + c)*C*b^2 - 24*C*a*b*sin(d*x + c) - 12*B*b^2*sin(d*x + c))/d

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Fricas [A]  time = 0.493997, size = 201, normalized size = 1.88 \begin{align*} \frac{3 \,{\left (C a^{2} + 2 \, B a b + 2 \, C b^{2}\right )} d x +{\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + 4 \, B a^{2} + 12 \, C a b + 6 \, B b^{2} + 3 \,{\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, 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))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(3*(C*a^2 + 2*B*a*b + 2*C*b^2)*d*x + (2*B*a^2*cos(d*x + c)^2 + 4*B*a^2 + 12*C*a*b + 6*B*b^2 + 3*(C*a^2 + 2
*B*a*b)*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))**2*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.18597, size = 343, normalized size = 3.21 \begin{align*} \frac{3 \,{\left (C a^{2} + 2 \, B a b + 2 \, C b^{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} - 3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B b^{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} \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))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(C*a^2 + 2*B*a*b + 2*C*b^2)*(d*x + c) + 2*(6*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^2*tan(1/2*d*x + 1/2*c
)^5 - 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 12*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*B*
a^2*tan(1/2*d*x + 1/2*c)^3 + 24*C*a*b*tan(1/2*d*x + 1/2*c)^3 + 12*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^2*tan(1
/2*d*x + 1/2*c) + 3*C*a^2*tan(1/2*d*x + 1/2*c) + 6*B*a*b*tan(1/2*d*x + 1/2*c) + 12*C*a*b*tan(1/2*d*x + 1/2*c)
+ 6*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d