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

Optimal. Leaf size=120 \[ \frac {a^2 (2 A+3 B+2 C) \sin (c+d x)}{2 d}+\frac {1}{2} a^2 x (4 A+3 B+2 C)+\frac {a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(3 B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3045, 2976, 2968, 3023, 2735, 3770} \[ \frac {a^2 (2 A+3 B+2 C) \sin (c+d x)}{2 d}+\frac {1}{2} a^2 x (4 A+3 B+2 C)+\frac {a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(3 B+2 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(4*A + 3*B + 2*C)*x)/2 + (a^2*A*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*A + 3*B + 2*C)*Sin[c + d*x])/(2*d) + (
C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + ((3*B + 2*C)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(6*d)

Rule 2735

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

Rule 2968

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

Rule 2976

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3045

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^2 (3 a A+a (3 B+2 C) \cos (c+d x)) \sec (c+d x) \, dx}{3 a}\\ &=\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 B+2 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x)) \left (6 a^2 A+3 a^2 (2 A+3 B+2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 B+2 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int \left (6 a^3 A+\left (6 a^3 A+3 a^3 (2 A+3 B+2 C)\right ) \cos (c+d x)+3 a^3 (2 A+3 B+2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac {a^2 (2 A+3 B+2 C) \sin (c+d x)}{2 d}+\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 B+2 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int \left (6 a^3 A+3 a^3 (4 A+3 B+2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac {1}{2} a^2 (4 A+3 B+2 C) x+\frac {a^2 (2 A+3 B+2 C) \sin (c+d x)}{2 d}+\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 B+2 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^2 A\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^2 (4 A+3 B+2 C) x+\frac {a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 (2 A+3 B+2 C) \sin (c+d x)}{2 d}+\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 B+2 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 121, normalized size = 1.01 \[ \frac {a^2 \left (3 (4 A+8 B+7 C) \sin (c+d x)-12 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+24 A d x+3 (B+2 C) \sin (2 (c+d x))+18 B d x+C \sin (3 (c+d x))+12 C d x\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(24*A*d*x + 18*B*d*x + 12*C*d*x - 12*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*A*Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]] + 3*(4*A + 8*B + 7*C)*Sin[c + d*x] + 3*(B + 2*C)*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)])
)/(12*d)

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fricas [A]  time = 0.46, size = 108, normalized size = 0.90 \[ \frac {3 \, {\left (4 \, A + 3 \, B + 2 \, C\right )} a^{2} d x + 3 \, A a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, A a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (3 \, A + 6 \, B + 5 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(4*A + 3*B + 2*C)*a^2*d*x + 3*A*a^2*log(sin(d*x + c) + 1) - 3*A*a^2*log(-sin(d*x + c) + 1) + (2*C*a^2*c
os(d*x + c)^2 + 3*(B + 2*C)*a^2*cos(d*x + c) + 2*(3*A + 6*B + 5*C)*a^2)*sin(d*x + c))/d

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giac [B]  time = 0.49, size = 235, normalized size = 1.96 \[ \frac {6 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (4 \, A a^{2} + 3 \, B a^{2} + 2 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(4*A*a^2 + 3*
B*a^2 + 2*C*a^2)*(d*x + c) + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(
1/2*d*x + 1/2*c)^5 + 12*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 16*C*a^2*tan(1/2*d*x
+ 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 18*C*a^2*tan(1/2*d*x + 1/2*c))/(ta
n(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.27, size = 181, normalized size = 1.51 \[ \frac {a^{2} A \sin \left (d x +c \right )}{d}+\frac {B \,a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a^{2} B x}{2}+\frac {3 B \,a^{2} c}{2 d}+\frac {C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{2}}{3 d}+\frac {5 a^{2} C \sin \left (d x +c \right )}{3 d}+2 a^{2} A x +\frac {2 A \,a^{2} c}{d}+\frac {2 B \,a^{2} \sin \left (d x +c \right )}{d}+\frac {a^{2} C \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+a^{2} C x +\frac {a^{2} C c}{d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^2*A*sin(d*x+c)+1/2/d*B*a^2*cos(d*x+c)*sin(d*x+c)+3/2*a^2*B*x+3/2/d*B*a^2*c+1/3/d*C*sin(d*x+c)*cos(d*x+c)
^2*a^2+5/3/d*a^2*C*sin(d*x+c)+2*a^2*A*x+2/d*A*a^2*c+2/d*B*a^2*sin(d*x+c)+1/d*a^2*C*cos(d*x+c)*sin(d*x+c)+a^2*C
*x+1/d*a^2*C*c+1/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.35, size = 153, normalized size = 1.28 \[ \frac {24 \, {\left (d x + c\right )} A a^{2} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 12 \, {\left (d x + c\right )} B a^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 12 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, A a^{2} \sin \left (d x + c\right ) + 24 \, B a^{2} \sin \left (d x + c\right ) + 12 \, C a^{2} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(24*(d*x + c)*A*a^2 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 + 12*(d*x + c)*B*a^2 - 4*(sin(d*x + c)^3 -
3*sin(d*x + c))*C*a^2 + 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 12*A*a^2*log(sec(d*x + c) + tan(d*x + c))
+ 12*A*a^2*sin(d*x + c) + 24*B*a^2*sin(d*x + c) + 12*C*a^2*sin(d*x + c))/d

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mupad [B]  time = 1.63, size = 226, normalized size = 1.88 \[ \frac {A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {7\,C\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {4\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}-\frac {A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a^2*sin(c + d*x))/d + (2*B*a^2*sin(c + d*x))/d + (7*C*a^2*sin(c + d*x))/(4*d) + (4*A*a^2*atan(sin(c/2 + (d*
x)/2)/cos(c/2 + (d*x)/2)))/d - (A*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d + (3*B*a^2*atan(s
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (B*a^2*si
n(2*c + 2*d*x))/(4*d) + (C*a^2*sin(2*c + 2*d*x))/(2*d) + (C*a^2*sin(3*c + 3*d*x))/(12*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*sec(c + d*x), x) + Integral(2*A*cos(c + d*x)*sec(c + d*x), x) + Integral(A*cos(c + d*x)**2*se
c(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(2*B*cos(c + d*x)**2*sec(c + d*x), x) + In
tegral(B*cos(c + d*x)**3*sec(c + d*x), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x), x) + Integral(2*C*cos(c +
d*x)**3*sec(c + d*x), x) + Integral(C*cos(c + d*x)**4*sec(c + d*x), x))

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