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

Optimal. Leaf size=135 \[ \frac{a^2 (4 A+5 B) \sin (c+d x)}{3 d}+\frac{a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{12 d}+\frac{a^2 (7 A+8 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a^2 x (7 A+8 B)+\frac{A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d} \]

[Out]

(a^2*(7*A + 8*B)*x)/8 + (a^2*(4*A + 5*B)*Sin[c + d*x])/(3*d) + (a^2*(7*A + 8*B)*Cos[c + d*x]*Sin[c + d*x])/(8*
d) + (a^2*(5*A + 4*B)*Cos[c + d*x]^2*Sin[c + d*x])/(12*d) + (A*Cos[c + d*x]^3*(a^2 + a^2*Sec[c + d*x])*Sin[c +
d*x])/(4*d)

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Rubi [A]  time = 0.230517, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4017, 3996, 3787, 2635, 8, 2637} \[ \frac{a^2 (4 A+5 B) \sin (c+d x)}{3 d}+\frac{a^2 (5 A+4 B) \sin (c+d x) \cos ^2(c+d x)}{12 d}+\frac{a^2 (7 A+8 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a^2 x (7 A+8 B)+\frac{A \sin (c+d x) \cos ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*A + 8*B)*x)/8 + (a^2*(4*A + 5*B)*Sin[c + d*x])/(3*d) + (a^2*(7*A + 8*B)*Cos[c + d*x]*Sin[c + d*x])/(8*
d) + (a^2*(5*A + 4*B)*Cos[c + d*x]^2*Sin[c + d*x])/(12*d) + (A*Cos[c + d*x]^3*(a^2 + a^2*Sec[c + d*x])*Sin[c +
d*x])/(4*d)

Rule 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

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+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac{A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^3(c+d x) (a+a \sec (c+d x)) (a (5 A+4 B)+2 a (A+2 B) \sec (c+d x)) \, dx\\ &=\frac{a^2 (5 A+4 B) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac{A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}-\frac{1}{12} \int \cos ^2(c+d x) \left (-3 a^2 (7 A+8 B)-4 a^2 (4 A+5 B) \sec (c+d x)\right ) \, dx\\ &=\frac{a^2 (5 A+4 B) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac{A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}+\frac{1}{3} \left (a^2 (4 A+5 B)\right ) \int \cos (c+d x) \, dx+\frac{1}{4} \left (a^2 (7 A+8 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{a^2 (4 A+5 B) \sin (c+d x)}{3 d}+\frac{a^2 (7 A+8 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (5 A+4 B) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac{A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}+\frac{1}{8} \left (a^2 (7 A+8 B)\right ) \int 1 \, dx\\ &=\frac{1}{8} a^2 (7 A+8 B) x+\frac{a^2 (4 A+5 B) \sin (c+d x)}{3 d}+\frac{a^2 (7 A+8 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (5 A+4 B) \cos ^2(c+d x) \sin (c+d x)}{12 d}+\frac{A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.354537, size = 86, normalized size = 0.64 \[ \frac{a^2 (24 (6 A+7 B) \sin (c+d x)+48 (A+B) \sin (2 (c+d x))+16 A \sin (3 (c+d x))+3 A \sin (4 (c+d x))+84 A c+84 A d x+8 B \sin (3 (c+d x))+96 B d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(84*A*c + 84*A*d*x + 96*B*d*x + 24*(6*A + 7*B)*Sin[c + d*x] + 48*(A + B)*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 = 0.087, size = 154, normalized size = 1.1 \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{B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{2\,{a}^{2}A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+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}A \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 ) \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)),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)+1/3*B*a^2*(2+cos(d*x+c)^2)*sin(d*x+c)+
2/3*a^2*A*(2+cos(d*x+c)^2)*sin(d*x+c)+2*B*a^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^2*A*(1/2*cos(d*x+c)*
sin(d*x+c)+1/2*d*x+1/2*c)+B*a^2*sin(d*x+c))

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Maxima [A]  time = 1.00995, size = 194, normalized size = 1.44 \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} - 96 \, B 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)),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 - 96*B*a^2*sin(d*x + c))/d

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Fricas [A]  time = 0.474397, size = 213, normalized size = 1.58 \begin{align*} \frac{3 \,{\left (7 \, A + 8 \, B\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\right )} a^{2} \cos \left (d x + c\right ) + 8 \,{\left (4 \, A + 5 \, B\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)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.21159, size = 238, normalized size = 1.76 \begin{align*} \frac{3 \,{\left (7 \, A a^{2} + 8 \, B 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} + 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} + 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} + 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 )\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)),x, algorithm="giac")

[Out]

1/24*(3*(7*A*a^2 + 8*B*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 +
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 + 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 + 75*A*a^2*tan(1/2*d*x + 1/2*c) + 72*B*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*
c)^2 + 1)^4)/d