3.771 \(\int \cos ^3(c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=52 \[ \frac{(a C+b B) \sin (c+d x)}{d}+\frac{1}{2} x (a B+2 b C)+\frac{a B \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.141558, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {4072, 3996, 3787, 2637, 8} \[ \frac{(a C+b B) \sin (c+d x)}{d}+\frac{1}{2} x (a B+2 b C)+\frac{a B \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((a*B + 2*b*C)*x)/2 + ((b*B + a*C)*Sin[c + d*x])/d + (a*B*Cos[c + d*x]*Sin[c + d*x])/(2*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 2637

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

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0866127, size = 51, normalized size = 0.98 \[ \frac{4 (a C+b B) \sin (c+d x)+a B \sin (2 (c+d x))+2 a B c+2 a B d x+4 b C d x}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

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Maple [A]  time = 0.053, size = 57, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( Ba \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +Bb\sin \left ( dx+c \right ) +aC\sin \left ( dx+c \right ) +Cb \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.954808, size = 74, normalized size = 1.42 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 4 \,{\left (d x + c\right )} C b + 4 \, C a \sin \left (d x + c\right ) + 4 \, B b \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*B*a + 4*(d*x + c)*C*b + 4*C*a*sin(d*x + c) + 4*B*b*sin(d*x + c))/d

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Fricas [A]  time = 0.482725, size = 104, normalized size = 2. \begin{align*} \frac{{\left (B a + 2 \, C b\right )} d x +{\left (B a \cos \left (d x + c\right ) + 2 \, C a + 2 \, B b\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/2*((B*a + 2*C*b)*d*x + (B*a*cos(d*x + c) + 2*C*a + 2*B*b)*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)**3*(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 = 1.17867, size = 163, normalized size = 3.13 \begin{align*} \frac{{\left (B a + 2 \, C b\right )}{\left (d x + c\right )} - \frac{2 \,{\left (B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B 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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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