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

Optimal. Leaf size=84 \[ \frac{(2 a B+3 b C) \sin (c+d x)}{3 d}+\frac{(a C+b B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (a C+b B)+\frac{a B \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.16633, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4072, 3996, 3787, 2635, 8, 2637} \[ \frac{(2 a B+3 b C) \sin (c+d x)}{3 d}+\frac{(a C+b B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (a C+b B)+\frac{a B \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B + a*C)*x)/2 + ((2*a*B + 3*b*C)*Sin[c + d*x])/(3*d) + ((b*B + a*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*
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 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(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) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

Rubi steps

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

Mathematica [A]  time = 0.161308, size = 75, normalized size = 0.89 \[ \frac{3 (3 a B+4 b C) \sin (c+d x)+3 (a C+b B) \sin (2 (c+d x))+a B \sin (3 (c+d x))+6 a c C+6 a C d x+6 b B c+6 b B d x}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.069, size = 85, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{Ba \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+Bb \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +aC \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +C\sin \left ( dx+c \right ) b \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))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.957332, size = 107, normalized size = 1.27 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b - 12 \, C b \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))*(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 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a - 3*(2*d*x + 2*c + sin
(2*d*x + 2*c))*B*b - 12*C*b*sin(d*x + c))/d

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Fricas [A]  time = 0.489494, size = 149, normalized size = 1.77 \begin{align*} \frac{3 \,{\left (C a + B b\right )} d x +{\left (2 \, B a \cos \left (d x + c\right )^{2} + 4 \, B a + 6 \, C b + 3 \,{\left (C a + B 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))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.17514, size = 243, normalized size = 2.89 \begin{align*} \frac{3 \,{\left (C a + B b\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b \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))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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