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

Optimal. Leaf size=162 \[ \frac {5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{24 d}+\frac {1}{8} a^3 x (28 A+20 B+15 C)+\frac {a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 a d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

[Out]

1/8*a^3*(28*A+20*B+15*C)*x+a^3*A*arctanh(sin(d*x+c))/d+5/8*a^3*(4*A+4*B+3*C)*sin(d*x+c)/d+1/4*C*(a+a*cos(d*x+c
))^3*sin(d*x+c)/d+1/12*(4*B+3*C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d+1/24*(12*A+20*B+15*C)*(a^3+a^3*cos(d*x+
c))*sin(d*x+c)/d

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Rubi [A]  time = 0.48, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac {(12 A+20 B+15 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{24 d}+\frac {1}{8} a^3 x (28 A+20 B+15 C)+\frac {a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(4 B+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{12 a d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(28*A + 20*B + 15*C)*x)/8 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(4*A + 4*B + 3*C)*Sin[c + d*x])/(8*d
) + (C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + ((4*B + 3*C)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(12*
a*d) + ((12*A + 20*B + 15*C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(24*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))^3 \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))^3 \sin (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^3 (4 a A+a (4 B+3 C) \cos (c+d x)) \sec (c+d x) \, dx}{4 a}\\ &=\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {\int (a+a \cos (c+d x))^2 \left (12 a^2 A+a^2 (12 A+20 B+15 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {\int (a+a \cos (c+d x)) \left (24 a^3 A+15 a^3 (4 A+4 B+3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {\int \left (24 a^4 A+\left (24 a^4 A+15 a^4 (4 A+4 B+3 C)\right ) \cos (c+d x)+15 a^4 (4 A+4 B+3 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac {5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {\int \left (24 a^4 A+3 a^4 (28 A+20 B+15 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a}\\ &=\frac {1}{8} a^3 (28 A+20 B+15 C) x+\frac {5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} a^3 (28 A+20 B+15 C) x+\frac {a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 (4 A+4 B+3 C) \sin (c+d x)}{8 d}+\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{24 d}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 147, normalized size = 0.91 \[ \frac {a^3 \left (24 (12 A+15 B+13 C) \sin (c+d x)+24 (A+3 B+4 C) \sin (2 (c+d x))-96 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+96 A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+336 A d x+8 B \sin (3 (c+d x))+240 B d x+24 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))+180 C d x\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(336*A*d*x + 240*B*d*x + 180*C*d*x - 96*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 96*A*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]] + 24*(12*A + 15*B + 13*C)*Sin[c + d*x] + 24*(A + 3*B + 4*C)*Sin[2*(c + d*x)] + 8*B*S
in[3*(c + d*x)] + 24*C*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d)

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fricas [A]  time = 0.48, size = 131, normalized size = 0.81 \[ \frac {3 \, {\left (28 \, A + 20 \, B + 15 \, C\right )} a^{3} d x + 12 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, C a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 12 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (9 \, A + 11 \, B + 9 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(28*A + 20*B + 15*C)*a^3*d*x + 12*A*a^3*log(sin(d*x + c) + 1) - 12*A*a^3*log(-sin(d*x + c) + 1) + (6*C
*a^3*cos(d*x + c)^3 + 8*(B + 3*C)*a^3*cos(d*x + c)^2 + 3*(4*A + 12*B + 15*C)*a^3*cos(d*x + c) + 8*(9*A + 11*B
+ 9*C)*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.60, size = 286, normalized size = 1.77 \[ \frac {24 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (28 \, A a^{3} + 20 \, B a^{3} + 15 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (60 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 204 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 228 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 84 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 147 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(28*A*a^3
+ 20*B*a^3 + 15*C*a^3)*(d*x + c) + 2*(60*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 45*C
*a^3*tan(1/2*d*x + 1/2*c)^7 + 204*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 220*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 165*C*a^3*
tan(1/2*d*x + 1/2*c)^5 + 228*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 292*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 219*C*a^3*tan(1
/2*d*x + 1/2*c)^3 + 84*A*a^3*tan(1/2*d*x + 1/2*c) + 132*B*a^3*tan(1/2*d*x + 1/2*c) + 147*C*a^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 = 0.34, size = 251, normalized size = 1.55 \[ \frac {A \,a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {7 A x \,a^{3}}{2}+\frac {7 A \,a^{3} c}{2 d}+\frac {B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{3 d}+\frac {11 a^{3} B \sin \left (d x +c \right )}{3 d}+\frac {C \,a^{3} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {15 C \,a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {15 a^{3} C x}{8}+\frac {15 C \,a^{3} c}{8 d}+\frac {3 a^{3} A \sin \left (d x +c \right )}{d}+\frac {3 a^{3} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a^{3} B x}{2}+\frac {5 a^{3} B c}{2 d}+\frac {C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{d}+\frac {3 a^{3} C \sin \left (d x +c \right )}{d}+\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*A*a^3*cos(d*x+c)*sin(d*x+c)+7/2*A*x*a^3+7/2/d*A*a^3*c+1/3/d*B*cos(d*x+c)^2*sin(d*x+c)*a^3+11/3*a^3*B*sin
(d*x+c)/d+1/4/d*C*a^3*sin(d*x+c)*cos(d*x+c)^3+15/8/d*C*a^3*cos(d*x+c)*sin(d*x+c)+15/8*a^3*C*x+15/8/d*C*a^3*c+3
*a^3*A*sin(d*x+c)/d+3/2/d*a^3*B*cos(d*x+c)*sin(d*x+c)+5/2*a^3*B*x+5/2/d*a^3*B*c+1/d*C*cos(d*x+c)^2*sin(d*x+c)*
a^3+3*a^3*C*sin(d*x+c)/d+1/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.49, size = 233, normalized size = 1.44 \[ \frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 288 \, {\left (d x + c\right )} A a^{3} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 96 \, {\left (d x + c\right )} B a^{3} - 96 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 96 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 288 \, A a^{3} \sin \left (d x + c\right ) + 288 \, B a^{3} \sin \left (d x + c\right ) + 96 \, C a^{3} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 288*(d*x + c)*A*a^3 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B
*a^3 + 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 96*(d*x + c)*B*a^3 - 96*(sin(d*x + c)^3 - 3*sin(d*x + c))*C
*a^3 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^3 + 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*C
*a^3 + 96*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 288*A*a^3*sin(d*x + c) + 288*B*a^3*sin(d*x + c) + 96*C*a^3*
sin(d*x + c))/d

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mupad [B]  time = 1.81, size = 242, normalized size = 1.49 \[ \frac {7\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+5\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {15\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12}+C\,a^3\,\sin \left (2\,c+2\,d\,x\right )+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32}+3\,A\,a^3\,\sin \left (c+d\,x\right )+\frac {15\,B\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {13\,C\,a^3\,\sin \left (c+d\,x\right )}{4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) +
5*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (15*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4
+ (A*a^3*sin(2*c + 2*d*x))/4 + (3*B*a^3*sin(2*c + 2*d*x))/4 + (B*a^3*sin(3*c + 3*d*x))/12 + C*a^3*sin(2*c + 2
*d*x) + (C*a^3*sin(3*c + 3*d*x))/4 + (C*a^3*sin(4*c + 4*d*x))/32 + 3*A*a^3*sin(c + d*x) + (15*B*a^3*sin(c + d*
x))/4 + (13*C*a^3*sin(c + d*x))/4)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 3 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 C \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(A*sec(c + d*x), x) + Integral(3*A*cos(c + d*x)*sec(c + d*x), x) + Integral(3*A*cos(c + d*x)**2*
sec(c + d*x), x) + Integral(A*cos(c + d*x)**3*sec(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + In
tegral(3*B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(3*B*cos(c + d*x)**3*sec(c + d*x), x) + Integral(B*cos(c
+ d*x)**4*sec(c + d*x), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x), x) + Integral(3*C*cos(c + d*x)**3*sec(c
+ d*x), x) + Integral(3*C*cos(c + d*x)**4*sec(c + d*x), x) + Integral(C*cos(c + d*x)**5*sec(c + d*x), x))

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