3.3.91 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [291]

Optimal. Leaf size=107 \[ \frac {1}{2} \left (2 a A b+a^2 B+2 b^2 B\right ) x+\frac {\left (2 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{3 d}+\frac {a (2 A b+a B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^2 A \cos ^2(c+d x) \sin (c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4109, 4132, 2717, 4130, 8} \begin {gather*} \frac {\left (2 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x)}{3 d}+\frac {1}{2} x \left (a^2 B+2 a A b+2 b^2 B\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {a (a B+2 A b) \sin (c+d x) \cos (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*A*b + a^2*B + 2*b^2*B)*x)/2 + ((2*a^2*A + 3*A*b^2 + 6*a*b*B)*Sin[c + d*x])/(3*d) + (a*(2*A*b + a*B)*Cos[
c + d*x]*Sin[c + d*x])/(2*d) + (a^2*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 2717

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

Rule 4109

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

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac {a^2 A \cos ^2(c+d x) \sin (c+d x)}{3 d}-\frac {1}{3} \int \cos ^2(c+d x) \left (-3 a (2 A b+a B)+\left (A \left (-2 a^2-3 b^2\right )-6 a b B\right ) \sec (c+d x)-3 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 A \cos ^2(c+d x) \sin (c+d x)}{3 d}-\frac {1}{3} \int \cos ^2(c+d x) \left (-3 a (2 A b+a B)-3 b^2 B \sec ^2(c+d x)\right ) \, dx-\frac {1}{3} \left (-2 a^2 A-3 A b^2-6 a b B\right ) \int \cos (c+d x) \, dx\\ &=\frac {\left (2 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{3 d}+\frac {a (2 A b+a B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^2 A \cos ^2(c+d x) \sin (c+d x)}{3 d}-\frac {1}{2} \left (-2 a A b-a^2 B-2 b^2 B\right ) \int 1 \, dx\\ &=\frac {1}{2} \left (2 a A b+a^2 B+2 b^2 B\right ) x+\frac {\left (2 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{3 d}+\frac {a (2 A b+a B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^2 A \cos ^2(c+d x) \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 90, normalized size = 0.84 \begin {gather*} \frac {6 \left (2 a A b+a^2 B+2 b^2 B\right ) (c+d x)+3 \left (3 a^2 A+4 A b^2+8 a b B\right ) \sin (c+d x)+3 a (2 A b+a B) \sin (2 (c+d x))+a^2 A \sin (3 (c+d x))}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(2*a*A*b + a^2*B + 2*b^2*B)*(c + d*x) + 3*(3*a^2*A + 4*A*b^2 + 8*a*b*B)*Sin[c + d*x] + 3*a*(2*A*b + a*B)*Si
n[2*(c + d*x)] + a^2*A*Sin[3*(c + d*x)])/(12*d)

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Maple [A]
time = 0.32, size = 114, normalized size = 1.07

method	result	size
derivativedivides	\(\frac {\frac {a^{2} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A b a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{2} \sin \left (d x +c \right )+2 B a b \sin \left (d x +c \right )+b^{2} B \left (d x +c \right )}{d}\)	\(114\)
default	\(\frac {\frac {a^{2} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A b a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{2} \sin \left (d x +c \right )+2 B a b \sin \left (d x +c \right )+b^{2} B \left (d x +c \right )}{d}\)	\(114\)
risch	\(A a b x +\frac {B \,a^{2} x}{2}+x \,b^{2} B +\frac {3 a^{2} A \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) A \,b^{2}}{d}+\frac {2 \sin \left (d x +c \right ) B a b}{d}+\frac {a^{2} A \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A b a}{2 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} B}{4 d}\)	\(116\)
norman	\(\frac {\left (A b a +\frac {1}{2} a^{2} B +b^{2} B \right ) x +\left (A b a +\frac {1}{2} a^{2} B +b^{2} B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A b a +\frac {1}{2} a^{2} B +b^{2} B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A b a +\frac {1}{2} a^{2} B +b^{2} B \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A b a -a^{2} B -2 b^{2} B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A b a -a^{2} B -2 b^{2} B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (2 a^{2} A -2 A b a +2 A \,b^{2}-a^{2} B +4 B a b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 a^{2} A +2 A b a +2 A \,b^{2}+a^{2} B +4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 \left (a^{2} A -3 A \,b^{2}-6 B a b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (4 A a -6 A b -3 B a \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (4 A a +6 A b +3 B a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}\)	\(378\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 108, normalized size = 1.01 \begin {gather*} -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 12 \, {\left (d x + c\right )} B b^{2} - 24 \, B a b \sin \left (d x + c\right ) - 12 \, A b^{2} \sin \left (d x + c\right )}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(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^2 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 - 6*(2*d*x + 2*c +
sin(2*d*x + 2*c))*A*a*b - 12*(d*x + c)*B*b^2 - 24*B*a*b*sin(d*x + c) - 12*A*b^2*sin(d*x + c))/d

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Fricas [A]
time = 4.20, size = 85, normalized size = 0.79 \begin {gather*} \frac {3 \, {\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} d x + {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 4 \, A a^{2} + 12 \, B a b + 6 \, A b^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(B*a^2 + 2*A*a*b + 2*B*b^2)*d*x + (2*A*a^2*cos(d*x + c)^2 + 4*A*a^2 + 12*B*a*b + 6*A*b^2 + 3*(B*a^2 + 2
*A*a*b)*cos(d*x + c))*sin(d*x + c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (99) = 198\).
time = 0.46, size = 254, normalized size = 2.37 \begin {gather*} \frac {3 \, {\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{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 )}^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(B*a^2 + 2*A*a*b + 2*B*b^2)*(d*x + c) + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c
)^5 - 6*A*a*b*tan(1/2*d*x + 1/2*c)^5 + 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*
a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1
/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) + 6*A*a*b*tan(1/2*d*x + 1/2*c) + 12*B*a*b*tan(1/2*d*x + 1/2*c)
+ 6*A*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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Mupad [B]
time = 2.10, size = 115, normalized size = 1.07 \begin {gather*} \frac {B\,a^2\,x}{2}+B\,b^2\,x+\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,b^2\,\sin \left (c+d\,x\right )}{d}+A\,a\,b\,x+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,B\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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