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

Optimal. Leaf size=134 \[ \frac {a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac {a^2 (2 A+3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+a^2 C x+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3043, 2975, 2968, 3021, 2735, 3770} \[ \frac {a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac {a^2 (2 A+3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(2 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+a^2 C x+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

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]^4,x]

[Out]

a^2*C*x + (a^2*(2*A + 3*B + 4*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a^2*(2*A + 3*B + 2*C)*Tan[c + d*x])/(2*d) + (
(2*A + 3*B)*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]
^2*Tan[c + d*x])/(3*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 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 3021

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

Rule 3043

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[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 ^4(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^2 (a (2 A+3 B)+3 a C \cos (c+d x)) \sec ^3(c+d x) \, dx}{3 a}\\ &=\frac {(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x)) \left (3 a^2 (2 A+3 B+2 C)+6 a^2 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac {(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (3 a^3 (2 A+3 B+2 C)+\left (6 a^3 C+3 a^3 (2 A+3 B+2 C)\right ) \cos (c+d x)+6 a^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac {a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac {(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (3 a^3 (2 A+3 B+4 C)+6 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=a^2 C x+\frac {a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac {(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} \left (a^2 (2 A+3 B+4 C)\right ) \int \sec (c+d x) \, dx\\ &=a^2 C x+\frac {a^2 (2 A+3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 (2 A+3 B+2 C) \tan (c+d x)}{2 d}+\frac {(2 A+3 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 4.99, size = 315, normalized size = 2.35 \[ \frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-6 (2 A+3 B+4 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \tan (c+d x) \sec ^2(c+d x) ((5 A+6 B+3 C) \cos (2 (c+d x))+A (-\cos (c+d x))+7 A+6 B+3 C)+\frac {7 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {7 A}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+12 A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {3 B}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+18 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+24 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 c C+12 C d x\right )}{48 d} \]

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]^4,x]

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(12*c*C + 12*C*d*x - 6*(2*A + 3*B + 4*C)*Log[Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] + 12*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 18*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]] + 24*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (7*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (3*B)/(Co
s[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - (7*A)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (3*B)/(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^2 + 2*(7*A + 6*B + 3*C - A*Cos[c + d*x] + (5*A + 6*B + 3*C)*Cos[2*(c + d*x)])*Sec[c + d*x
]^2*Tan[c + d*x]))/(48*d)

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fricas [A]  time = 0.55, size = 150, normalized size = 1.12 \[ \frac {12 \, C a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A + 3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + 3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (5 \, A + 6 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

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)^4,x, algorithm="fricas")

[Out]

1/12*(12*C*a^2*d*x*cos(d*x + c)^3 + 3*(2*A + 3*B + 4*C)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*A + 3*
B + 4*C)*a^2*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(5*A + 6*B + 3*C)*a^2*cos(d*x + c)^2 + 3*(2*A + B)*a
^2*cos(d*x + c) + 2*A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.54, size = 250, normalized size = 1.87 \[ \frac {6 \, {\left (d x + c\right )} C a^{2} + 3 \, {\left (2 \, A a^{2} + 3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a^{2} + 3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

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)^4,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*C*a^2 + 3*(2*A*a^2 + 3*B*a^2 + 4*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a^2 + 3*B
*a^2 + 4*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 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 - 16*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
- 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 18*A*a^2*tan(1/2*d*x + 1/2*c) + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*t
an(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 = 0.39, size = 193, normalized size = 1.44 \[ \frac {5 a^{2} A \tan \left (d x +c \right )}{3 d}+\frac {3 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+a^{2} C x +\frac {a^{2} C c}{d}+\frac {a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{2} B \tan \left (d x +c \right )}{d}+\frac {2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} C \tan \left (d x +c \right )}{d} \]

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)^4,x)

[Out]

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

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maxima [A]  time = 0.39, size = 224, normalized size = 1.67 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 12 \, {\left (d x + c\right )} C a^{2} - 6 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{2} \tan \left (d x + c\right ) + 24 \, B a^{2} \tan \left (d x + c\right ) + 12 \, C a^{2} \tan \left (d x + c\right )}{12 \, d} \]

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)^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.20, size = 440, normalized size = 3.28 \[ \frac {\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {5\,A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{4}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{2}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{4}-\frac {A\,a^2\,\cos \left (c+d\,x\right )\,\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 )\,3{}\mathrm {i}}{2}-\frac {B\,a^2\,\cos \left (c+d\,x\right )\,\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 )\,9{}\mathrm {i}}{4}+\frac {3\,C\,a^2\,\cos \left (c+d\,x\right )\,\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 )}{2}-C\,a^2\,\cos \left (c+d\,x\right )\,\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 )\,3{}\mathrm {i}-\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 )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{2}-\frac {B\,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 )\,\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}}{4}+\frac {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 )\,\cos \left (3\,c+3\,d\,x\right )}{2}-C\,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 )\,\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]

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)^4,x)

[Out]

((A*a^2*sin(2*c + 2*d*x))/2 + (5*A*a^2*sin(3*c + 3*d*x))/12 + (B*a^2*sin(2*c + 2*d*x))/4 + (B*a^2*sin(3*c + 3*
d*x))/2 + (C*a^2*sin(3*c + 3*d*x))/4 + (3*A*a^2*sin(c + d*x))/4 + (B*a^2*sin(c + d*x))/2 + (C*a^2*sin(c + d*x)
)/4 - (A*a^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i)/2 - (B*a^2*cos(c + d*x)*atan((s
in(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*9i)/4 + (3*C*a^2*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x
)/2)))/2 - C*a^2*cos(c + d*x)*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*3i - (A*a^2*atan((sin(c/2 + (d*
x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/2 - (B*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))
*cos(3*c + 3*d*x)*3i)/4 + (C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 - C*a^2*atan(
(sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x)*1i)/(d*((3*cos(c + d*x))/4 + cos(3*c + 3*d*x)/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**4,x)

[Out]

Timed out

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