3.251 \(\int (a+a \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=154 \[ \frac {a^3 (9 B+11 C) \tan (c+d x)}{3 d}+\frac {5 a^3 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {(3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

[Out]

5/8*a^3*(3*B+4*C)*arctanh(sin(d*x+c))/d+1/3*a^3*(9*B+11*C)*tan(d*x+c)/d+1/24*a^3*(27*B+28*C)*sec(d*x+c)*tan(d*
x+c)/d+1/6*(3*B+2*C)*(a^3+a^3*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a*B*(a+a*cos(d*x+c))^2*sec(d*x+c)^3*ta
n(d*x+c)/d

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Rubi [A]  time = 0.51, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3029, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac {a^3 (9 B+11 C) \tan (c+d x)}{3 d}+\frac {5 a^3 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {(3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(5*a^3*(3*B + 4*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(9*B + 11*C)*Tan[c + d*x])/(3*d) + (a^3*(27*B + 28*C)*S
ec[c + d*x]*Tan[c + d*x])/(24*d) + ((3*B + 2*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) +
(a*B*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 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]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[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 (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\int (a+a \cos (c+d x))^3 (B+C \cos (c+d x)) \sec ^5(c+d x) \, dx\\ &=\frac {a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+a \cos (c+d x))^2 (2 a (3 B+2 C)+a (B+4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac {(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+a \cos (c+d x)) \left (a^2 (27 B+28 C)+a^2 (9 B+16 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int \left (a^3 (27 B+28 C)+\left (a^3 (9 B+16 C)+a^3 (27 B+28 C)\right ) \cos (c+d x)+a^3 (9 B+16 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{24} \int \left (8 a^3 (9 B+11 C)+15 a^3 (3 B+4 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (5 a^3 (3 B+4 C)\right ) \int \sec (c+d x) \, dx+\frac {1}{3} \left (a^3 (9 B+11 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {5 a^3 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\left (a^3 (9 B+11 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {5 a^3 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (9 B+11 C) \tan (c+d x)}{3 d}+\frac {a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.27, size = 273, normalized size = 1.77 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (120 (3 B+4 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-24 (9 B+11 C) \sin (c)+69 B \sin (2 c+d x)+264 B \sin (c+2 d x)-24 B \sin (3 c+2 d x)+45 B \sin (2 c+3 d x)+45 B \sin (4 c+3 d x)+72 B \sin (3 c+4 d x)+(69 B+36 C) \sin (d x)+36 C \sin (2 c+d x)+280 C \sin (c+2 d x)-72 C \sin (3 c+2 d x)+36 C \sin (2 c+3 d x)+36 C \sin (4 c+3 d x)+88 C \sin (3 c+4 d x))\right )}{1536 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

-1/1536*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^4*(120*(3*B + 4*C)*Cos[c + d*x]^4*(Log[Cos[(
c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-24*(9*B + 11*C)*Sin[c]
+ (69*B + 36*C)*Sin[d*x] + 69*B*Sin[2*c + d*x] + 36*C*Sin[2*c + d*x] + 264*B*Sin[c + 2*d*x] + 280*C*Sin[c + 2*
d*x] - 24*B*Sin[3*c + 2*d*x] - 72*C*Sin[3*c + 2*d*x] + 45*B*Sin[2*c + 3*d*x] + 36*C*Sin[2*c + 3*d*x] + 45*B*Si
n[4*c + 3*d*x] + 36*C*Sin[4*c + 3*d*x] + 72*B*Sin[3*c + 4*d*x] + 88*C*Sin[3*c + 4*d*x])))/d

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fricas [A]  time = 0.47, size = 145, normalized size = 0.94 \[ \frac {15 \, {\left (3 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (9 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \, {\left (5 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, B a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/48*(15*(3*B + 4*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*(3*B + 4*C)*a^3*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(8*(9*B + 11*C)*a^3*cos(d*x + c)^3 + 9*(5*B + 4*C)*a^3*cos(d*x + c)^2 + 8*(3*B + C)*a^3*cos(d*x
+ c) + 6*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.37, size = 212, normalized size = 1.38 \[ \frac {15 \, {\left (3 \, B a^{3} + 4 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, B a^{3} + 4 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 165 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 220 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 219 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 147 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 132 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/24*(15*(3*B*a^3 + 4*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B*a^3 + 4*C*a^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(45*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 165*B*a^3*tan(1/2*d*x +
1/2*c)^5 - 220*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 219*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 292*C*a^3*tan(1/2*d*x + 1/2*
c)^3 - 147*B*a^3*tan(1/2*d*x + 1/2*c) - 132*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 = 0.43, size = 188, normalized size = 1.22 \[ \frac {5 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} B \tan \left (d x +c \right )}{d}+\frac {11 C \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {15 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

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

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maxima [A]  time = 0.70, size = 269, normalized size = 1.75 \[ \frac {48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{3} \tan \left (d x + c\right ) + 144 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/48*(48*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 3*B*a^3*(2*(3*
sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin
(d*x + c) - 1)) - 36*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1
)) - 36*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*C*a^3
*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*B*a^3*tan(d*x + c) + 144*C*a^3*tan(d*x + c))/d

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mupad [B]  time = 3.63, size = 185, normalized size = 1.20 \[ \frac {\left (-\frac {15\,B\,a^3}{4}-5\,C\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {55\,B\,a^3}{4}+\frac {55\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {73\,B\,a^3}{4}-\frac {73\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,B\,a^3}{4}+11\,C\,a^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,B+4\,C\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^3)/cos(c + d*x)^6,x)

[Out]

(tan(c/2 + (d*x)/2)*((49*B*a^3)/4 + 11*C*a^3) - tan(c/2 + (d*x)/2)^7*((15*B*a^3)/4 + 5*C*a^3) + tan(c/2 + (d*x
)/2)^5*((55*B*a^3)/4 + (55*C*a^3)/3) - tan(c/2 + (d*x)/2)^3*((73*B*a^3)/4 + (73*C*a^3)/3))/(d*(6*tan(c/2 + (d*
x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (5*a^3*atanh(tan(c/2
+ (d*x)/2))*(3*B + 4*C))/(4*d)

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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))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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