3.69 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=124 \[ -\frac{a^3 (3 A+4 B) \sin ^3(c+d x)}{12 d}+\frac{a^3 (3 A+4 B) \sin (c+d x)}{d}+\frac{3 a^3 (3 A+4 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 x (3 A+4 B)+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

[Out]

(5*a^3*(3*A + 4*B)*x)/8 + (a^3*(3*A + 4*B)*Sin[c + d*x])/d + (3*a^3*(3*A + 4*B)*Cos[c + d*x]*Sin[c + d*x])/(8*
d) + (A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) - (a^3*(3*A + 4*B)*Sin[c + d*x]^3)/(12*d)

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Rubi [A]  time = 0.169572, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4013, 3791, 2637, 2635, 8, 2633} \[ -\frac{a^3 (3 A+4 B) \sin ^3(c+d x)}{12 d}+\frac{a^3 (3 A+4 B) \sin (c+d x)}{d}+\frac{3 a^3 (3 A+4 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 x (3 A+4 B)+\frac{A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(5*a^3*(3*A + 4*B)*x)/8 + (a^3*(3*A + 4*B)*Sin[c + d*x])/d + (3*a^3*(3*A + 4*B)*Cos[c + d*x]*Sin[c + d*x])/(8*
d) + (A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) - (a^3*(3*A + 4*B)*Sin[c + d*x]^3)/(12*d)

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 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, 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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} (3 A+4 B) \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} (3 A+4 B) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac{1}{4} a^3 (3 A+4 B) x+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \left (a^3 (3 A+4 B)\right ) \int \cos ^3(c+d x) \, dx+\frac{1}{4} \left (3 a^3 (3 A+4 B)\right ) \int \cos (c+d x) \, dx+\frac{1}{4} \left (3 a^3 (3 A+4 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{1}{4} a^3 (3 A+4 B) x+\frac{3 a^3 (3 A+4 B) \sin (c+d x)}{4 d}+\frac{3 a^3 (3 A+4 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{8} \left (3 a^3 (3 A+4 B)\right ) \int 1 \, dx-\frac{\left (a^3 (3 A+4 B)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac{5}{8} a^3 (3 A+4 B) x+\frac{a^3 (3 A+4 B) \sin (c+d x)}{d}+\frac{3 a^3 (3 A+4 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac{a^3 (3 A+4 B) \sin ^3(c+d x)}{12 d}\\ \end{align*}

Mathematica [A]  time = 0.270153, size = 86, normalized size = 0.69 \[ \frac{a^3 (24 (13 A+15 B) \sin (c+d x)+24 (4 A+3 B) \sin (2 (c+d x))+24 A \sin (3 (c+d x))+3 A \sin (4 (c+d x))+180 A d x+8 B \sin (3 (c+d x))+240 B d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(a^3*(180*A*d*x + 240*B*d*x + 24*(13*A + 15*B)*Sin[c + d*x] + 24*(4*A + 3*B)*Sin[2*(c + d*x)] + 24*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 = 0.086, size = 176, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( A{a}^{3} \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 ) +A{a}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{\frac{B{a}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+3\,A{a}^{3} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,B{a}^{3} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +A{a}^{3}\sin \left ( dx+c \right ) +3\,B{a}^{3}\sin \left ( dx+c \right ) +B{a}^{3} \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))^3*(A+B*sec(d*x+c)),x)

[Out]

1/d*(A*a^3*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+A*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+1/3*
B*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+3*A*a^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+3*B*a^3*(1/2*cos(d*x+c)*si
n(d*x+c)+1/2*d*x+1/2*c)+A*a^3*sin(d*x+c)+3*B*a^3*sin(d*x+c)+B*a^3*(d*x+c))

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Maxima [A]  time = 1.00001, size = 225, normalized size = 1.81 \begin{align*} -\frac{96 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 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^{3} - 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 96 \,{\left (d x + c\right )} B a^{3} - 96 \, A a^{3} \sin \left (d x + c\right ) - 288 \, B a^{3} \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))^3*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/96*(96*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*
A*a^3 - 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 72*(2*d*x + 2
*c + sin(2*d*x + 2*c))*B*a^3 - 96*(d*x + c)*B*a^3 - 96*A*a^3*sin(d*x + c) - 288*B*a^3*sin(d*x + c))/d

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Fricas [A]  time = 0.481609, size = 216, normalized size = 1.74 \begin{align*} \frac{15 \,{\left (3 \, A + 4 \, B\right )} a^{3} d x +{\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 9 \,{\left (5 \, A + 4 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \,{\left (9 \, A + 11 \, B\right )} a^{3}\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))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(15*(3*A + 4*B)*a^3*d*x + (6*A*a^3*cos(d*x + c)^3 + 8*(3*A + B)*a^3*cos(d*x + c)^2 + 9*(5*A + 4*B)*a^3*co
s(d*x + c) + 8*(9*A + 11*B)*a^3)*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))**3*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.34596, size = 238, normalized size = 1.92 \begin{align*} \frac{15 \,{\left (3 \, A a^{3} + 4 \, B a^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (45 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 165 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 220 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 219 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 147 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 132 \, B a^{3} \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))^3*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(15*(3*A*a^3 + 4*B*a^3)*(d*x + c) + 2*(45*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*B*a^3*tan(1/2*d*x + 1/2*c)^7
+ 165*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 220*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 219*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 292
*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 147*A*a^3*tan(1/2*d*x + 1/2*c) + 132*B*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 + 1)^4)/d