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

Optimal. Leaf size=171 \[ \frac {b \left (6 a^2 C+20 a b B+9 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {\left (3 a^3 C+16 a^2 b B+12 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{6 d}+\frac {1}{8} x \left (8 a^3 B+12 a^2 b C+12 a b^2 B+3 b^3 C\right )+\frac {(3 a C+4 b B) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

[Out]

1/8*(8*B*a^3+12*B*a*b^2+12*C*a^2*b+3*C*b^3)*x+1/6*(16*B*a^2*b+4*B*b^3+3*C*a^3+12*C*a*b^2)*sin(d*x+c)/d+1/24*b*
(20*B*a*b+6*C*a^2+9*C*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/12*(4*B*b+3*C*a)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/4*C*(a
+b*cos(d*x+c))^3*sin(d*x+c)/d

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Rubi [A]  time = 0.26, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {3029, 2753, 2734} \[ \frac {\left (16 a^2 b B+3 a^3 C+12 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{6 d}+\frac {b \left (6 a^2 C+20 a b B+9 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (12 a^2 b C+8 a^3 B+12 a b^2 B+3 b^3 C\right )+\frac {(3 a C+4 b B) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^3*B + 12*a*b^2*B + 12*a^2*b*C + 3*b^3*C)*x)/8 + ((16*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*Sin[c + d
*x])/(6*d) + (b*(20*a*b*B + 6*a^2*C + 9*b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*b*B + 3*a*C)*(a + b*Cos
[c + d*x])^2*Sin[c + d*x])/(12*d) + (C*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d)

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=\int (a+b \cos (c+d x))^3 (B+C \cos (c+d x)) \, dx\\ &=\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \int (a+b \cos (c+d x))^2 (4 a B+3 b C+(4 b B+3 a C) \cos (c+d x)) \, dx\\ &=\frac {(4 b B+3 a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{12} \int (a+b \cos (c+d x)) \left (12 a^2 B+8 b^2 B+15 a b C+\left (20 a b B+6 a^2 C+9 b^2 C\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {1}{8} \left (8 a^3 B+12 a b^2 B+12 a^2 b C+3 b^3 C\right ) x+\frac {\left (16 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \sin (c+d x)}{6 d}+\frac {b \left (20 a b B+6 a^2 C+9 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {(4 b B+3 a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 140, normalized size = 0.82 \[ \frac {24 b \left (3 a^2 C+3 a b B+b^2 C\right ) \sin (2 (c+d x))+12 (c+d x) \left (8 a^3 B+12 a^2 b C+12 a b^2 B+3 b^3 C\right )+24 \left (4 a^3 C+12 a^2 b B+9 a b^2 C+3 b^3 B\right ) \sin (c+d x)+8 b^2 (3 a C+b B) \sin (3 (c+d x))+3 b^3 C \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*a^3*B + 12*a*b^2*B + 12*a^2*b*C + 3*b^3*C)*(c + d*x) + 24*(12*a^2*b*B + 3*b^3*B + 4*a^3*C + 9*a*b^2*C)*
Sin[c + d*x] + 24*b*(3*a*b*B + 3*a^2*C + b^2*C)*Sin[2*(c + d*x)] + 8*b^2*(b*B + 3*a*C)*Sin[3*(c + d*x)] + 3*b^
3*C*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.48, size = 136, normalized size = 0.80 \[ \frac {3 \, {\left (8 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 3 \, C b^{3}\right )} d x + {\left (6 \, C b^{3} \cos \left (d x + c\right )^{3} + 24 \, C a^{3} + 72 \, B a^{2} b + 48 \, C a b^{2} + 16 \, B b^{3} + 8 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 9 \, {\left (4 \, C a^{2} b + 4 \, B a b^{2} + C b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(8*B*a^3 + 12*C*a^2*b + 12*B*a*b^2 + 3*C*b^3)*d*x + (6*C*b^3*cos(d*x + c)^3 + 24*C*a^3 + 72*B*a^2*b +
48*C*a*b^2 + 16*B*b^3 + 8*(3*C*a*b^2 + B*b^3)*cos(d*x + c)^2 + 9*(4*C*a^2*b + 4*B*a*b^2 + C*b^3)*cos(d*x + c))
*sin(d*x + c))/d

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giac [B]  time = 0.28, size = 536, normalized size = 3.13 \[ \frac {3 \, {\left (8 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 3 \, C b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 216 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 216 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C b^{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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(8*B*a^3 + 12*C*a^2*b + 12*B*a*b^2 + 3*C*b^3)*(d*x + c) + 2*(24*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 72*B*a^
2*b*tan(1/2*d*x + 1/2*c)^7 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 72*C*a*b^
2*tan(1/2*d*x + 1/2*c)^7 + 24*B*b^3*tan(1/2*d*x + 1/2*c)^7 - 15*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 72*C*a^3*tan(1/
2*d*x + 1/2*c)^5 + 216*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 36*B*a*b^2*tan(1/2
*d*x + 1/2*c)^5 + 120*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 40*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 9*C*b^3*tan(1/2*d*x +
1/2*c)^5 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 216*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 36*C*a^2*b*tan(1/2*d*x + 1/
2*c)^3 + 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 40*B*b^3*tan(1/2*d*x + 1/2*c
)^3 - 9*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 24*C*a^3*tan(1/2*d*x + 1/2*c) + 72*B*a^2*b*tan(1/2*d*x + 1/2*c) + 36*C*
a^2*b*tan(1/2*d*x + 1/2*c) + 36*B*a*b^2*tan(1/2*d*x + 1/2*c) + 72*C*a*b^2*tan(1/2*d*x + 1/2*c) + 24*B*b^3*tan(
1/2*d*x + 1/2*c) + 15*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.26, size = 180, normalized size = 1.05 \[ \frac {b^{3} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b B \sin \left (d x +c \right )+C \,a^{3} \sin \left (d x +c \right )+B \left (d x +c \right ) a^{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 171, normalized size = 1.00 \[ \frac {96 \, {\left (d x + c\right )} B a^{3} + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} - 96 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b^{2} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{3} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} + 96 \, C a^{3} \sin \left (d x + c\right ) + 288 \, B a^{2} b \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.84, size = 202, normalized size = 1.18 \[ B\,a^3\,x+\frac {3\,C\,b^3\,x}{8}+\frac {3\,B\,a\,b^2\,x}{2}+\frac {3\,C\,a^2\,b\,x}{2}+\frac {3\,B\,b^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {B\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,C\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{d}+\frac {9\,C\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*a^3*x + (3*C*b^3*x)/8 + (3*B*a*b^2*x)/2 + (3*C*a^2*b*x)/2 + (3*B*b^3*sin(c + d*x))/(4*d) + (C*a^3*sin(c + d*
x))/d + (B*b^3*sin(3*c + 3*d*x))/(12*d) + (C*b^3*sin(2*c + 2*d*x))/(4*d) + (C*b^3*sin(4*c + 4*d*x))/(32*d) + (
3*B*a*b^2*sin(2*c + 2*d*x))/(4*d) + (3*C*a^2*b*sin(2*c + 2*d*x))/(4*d) + (C*a*b^2*sin(3*c + 3*d*x))/(4*d) + (3
*B*a^2*b*sin(c + d*x))/d + (9*C*a*b^2*sin(c + d*x))/(4*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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