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

Optimal. Leaf size=77 \[ \frac {a (2 A+3 B) \sin (c+d x)}{3 d}+\frac {a (A+B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} a x (A+B)+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3996, 3787, 2635, 8, 2637} \[ \frac {a (2 A+3 B) \sin (c+d x)}{3 d}+\frac {a (A+B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} a x (A+B)+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2637

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 65, normalized size = 0.84 \[ \frac {a (3 (3 A+4 B) \sin (c+d x)+3 (A+B) \sin (2 (c+d x))+A \sin (3 (c+d x))+6 A c+6 A d x+6 B c+6 B d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 56, normalized size = 0.73 \[ \frac {3 \, {\left (A + B\right )} a d x + {\left (2 \, A a \cos \left (d x + c\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, {\left (2 \, A + 3 \, B\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(A + B)*a*d*x + (2*A*a*cos(d*x + c)^2 + 3*(A + B)*a*cos(d*x + c) + 2*(2*A + 3*B)*a)*sin(d*x + c))/d

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giac [A]  time = 0.31, size = 124, normalized size = 1.61 \[ \frac {3 \, {\left (A a + B a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(A*a + B*a)*(d*x + c) + 2*(3*A*a*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*tan(1/2*d*x + 1/2*c)^5 + 4*A*a*tan(1/2*
d*x + 1/2*c)^3 + 12*B*a*tan(1/2*d*x + 1/2*c)^3 + 9*A*a*tan(1/2*d*x + 1/2*c) + 9*B*a*tan(1/2*d*x + 1/2*c))/(tan
(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 1.13, size = 85, normalized size = 1.10 \[ \frac {\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a B \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.32, size = 79, normalized size = 1.03 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 12 \, B a \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 3*(2*d*x + 2*c + sin
(2*d*x + 2*c))*B*a - 12*B*a*sin(d*x + c))/d

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mupad [B]  time = 2.10, size = 84, normalized size = 1.09 \[ \frac {A\,a\,x}{2}+\frac {B\,a\,x}{2}+\frac {3\,A\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + a/cos(c + d*x)),x)

[Out]

(A*a*x)/2 + (B*a*x)/2 + (3*A*a*sin(c + d*x))/(4*d) + (B*a*sin(c + d*x))/d + (A*a*sin(2*c + 2*d*x))/(4*d) + (A*
a*sin(3*c + 3*d*x))/(12*d) + (B*a*sin(2*c + 2*d*x))/(4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int A \cos ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

a*(Integral(A*cos(c + d*x)**3, x) + Integral(A*cos(c + d*x)**3*sec(c + d*x), x) + Integral(B*cos(c + d*x)**3*s
ec(c + d*x), x) + Integral(B*cos(c + d*x)**3*sec(c + d*x)**2, x))

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