3.38 \(\int (a+a \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^8(c+d x) \, dx\)

Optimal. Leaf size=263 \[ \frac {a^4 (454 A+581 C) \tan (c+d x)}{105 d}+\frac {a^4 (11 A+14 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^4 (247 A+308 C) \tan (c+d x) \sec ^2(c+d x)}{210 d}+\frac {a^4 (11 A+14 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {(109 A+126 C) \tan (c+d x) \sec ^3(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{210 d}+\frac {(8 A+7 C) \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{35 d}+\frac {A \tan (c+d x) \sec ^6(c+d x) (a \cos (c+d x)+a)^4}{7 d}+\frac {2 a A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^3}{21 d} \]

[Out]

1/4*a^4*(11*A+14*C)*arctanh(sin(d*x+c))/d+1/105*a^4*(454*A+581*C)*tan(d*x+c)/d+1/4*a^4*(11*A+14*C)*sec(d*x+c)*
tan(d*x+c)/d+1/210*a^4*(247*A+308*C)*sec(d*x+c)^2*tan(d*x+c)/d+1/210*(109*A+126*C)*(a^4+a^4*cos(d*x+c))*sec(d*
x+c)^3*tan(d*x+c)/d+1/35*(8*A+7*C)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d+2/21*a*A*(a+a*cos(d*x+c))^
3*sec(d*x+c)^5*tan(d*x+c)/d+1/7*A*(a+a*cos(d*x+c))^4*sec(d*x+c)^6*tan(d*x+c)/d

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Rubi [A]  time = 0.80, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3044, 2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {a^4 (454 A+581 C) \tan (c+d x)}{105 d}+\frac {a^4 (11 A+14 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^4 (247 A+308 C) \tan (c+d x) \sec ^2(c+d x)}{210 d}+\frac {a^4 (11 A+14 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {(8 A+7 C) \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{35 d}+\frac {(109 A+126 C) \tan (c+d x) \sec ^3(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{210 d}+\frac {A \tan (c+d x) \sec ^6(c+d x) (a \cos (c+d x)+a)^4}{7 d}+\frac {2 a A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^3}{21 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^8,x]

[Out]

(a^4*(11*A + 14*C)*ArcTanh[Sin[c + d*x]])/(4*d) + (a^4*(454*A + 581*C)*Tan[c + d*x])/(105*d) + (a^4*(11*A + 14
*C)*Sec[c + d*x]*Tan[c + d*x])/(4*d) + (a^4*(247*A + 308*C)*Sec[c + d*x]^2*Tan[c + d*x])/(210*d) + ((109*A + 1
26*C)*(a^4 + a^4*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(210*d) + ((8*A + 7*C)*(a^2 + a^2*Cos[c + d*x])^2*
Sec[c + d*x]^4*Tan[c + d*x])/(35*d) + (2*a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^5*Tan[c + d*x])/(21*d) + (A*(
a + a*Cos[c + d*x])^4*Sec[c + d*x]^6*Tan[c + d*x])/(7*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 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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 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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^8(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {\int (a+a \cos (c+d x))^4 (4 a A+a (2 A+7 C) \cos (c+d x)) \sec ^7(c+d x) \, dx}{7 a}\\ &=\frac {2 a A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {\int (a+a \cos (c+d x))^3 \left (6 a^2 (8 A+7 C)+2 a^2 (10 A+21 C) \cos (c+d x)\right ) \sec ^6(c+d x) \, dx}{42 a}\\ &=\frac {(8 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {\int (a+a \cos (c+d x))^2 \left (4 a^3 (109 A+126 C)+98 a^3 (2 A+3 C) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx}{210 a}\\ &=\frac {(109 A+126 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {(8 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {\int (a+a \cos (c+d x)) \left (12 a^4 (247 A+308 C)+24 a^4 (69 A+91 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{840 a}\\ &=\frac {(109 A+126 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {(8 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {\int \left (12 a^5 (247 A+308 C)+\left (24 a^5 (69 A+91 C)+12 a^5 (247 A+308 C)\right ) \cos (c+d x)+24 a^5 (69 A+91 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{840 a}\\ &=\frac {a^4 (247 A+308 C) \sec ^2(c+d x) \tan (c+d x)}{210 d}+\frac {(109 A+126 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {(8 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {\int \left (1260 a^5 (11 A+14 C)+24 a^5 (454 A+581 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{2520 a}\\ &=\frac {a^4 (247 A+308 C) \sec ^2(c+d x) \tan (c+d x)}{210 d}+\frac {(109 A+126 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {(8 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {1}{2} \left (a^4 (11 A+14 C)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{105} \left (a^4 (454 A+581 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^4 (11 A+14 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^4 (247 A+308 C) \sec ^2(c+d x) \tan (c+d x)}{210 d}+\frac {(109 A+126 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {(8 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}+\frac {1}{4} \left (a^4 (11 A+14 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^4 (454 A+581 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 d}\\ &=\frac {a^4 (11 A+14 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^4 (454 A+581 C) \tan (c+d x)}{105 d}+\frac {a^4 (11 A+14 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^4 (247 A+308 C) \sec ^2(c+d x) \tan (c+d x)}{210 d}+\frac {(109 A+126 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{210 d}+\frac {(8 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^4(c+d x) \tan (c+d x)}{35 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{21 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^6(c+d x) \tan (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 3.17, size = 390, normalized size = 1.48 \[ -\frac {a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^7(c+d x) \left (6720 (11 A+14 C) \cos ^7(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-140 (122 A+217 C) \sin (2 c+d x)+16415 A \sin (c+2 d x)+16415 A \sin (3 c+2 d x)+37296 A \sin (2 c+3 d x)-840 A \sin (4 c+3 d x)+7700 A \sin (3 c+4 d x)+7700 A \sin (5 c+4 d x)+12712 A \sin (4 c+5 d x)+1155 A \sin (5 c+6 d x)+1155 A \sin (7 c+6 d x)+1816 A \sin (6 c+7 d x)+560 (83 A+91 C) \sin (d x)+10710 C \sin (c+2 d x)+10710 C \sin (3 c+2 d x)+41244 C \sin (2 c+3 d x)-7560 C \sin (4 c+3 d x)+7560 C \sin (3 c+4 d x)+7560 C \sin (5 c+4 d x)+15848 C \sin (4 c+5 d x)-420 C \sin (6 c+5 d x)+1470 C \sin (5 c+6 d x)+1470 C \sin (7 c+6 d x)+2324 C \sin (6 c+7 d x))\right )}{430080 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^8,x]

[Out]

-1/430080*(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*Sec[c + d*x]^7*(6720*(11*A + 14*C)*Cos[c + d*x]^7*(Log[
Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(560*(83*A + 91*C)*S
in[d*x] - 140*(122*A + 217*C)*Sin[2*c + d*x] + 16415*A*Sin[c + 2*d*x] + 10710*C*Sin[c + 2*d*x] + 16415*A*Sin[3
*c + 2*d*x] + 10710*C*Sin[3*c + 2*d*x] + 37296*A*Sin[2*c + 3*d*x] + 41244*C*Sin[2*c + 3*d*x] - 840*A*Sin[4*c +
3*d*x] - 7560*C*Sin[4*c + 3*d*x] + 7700*A*Sin[3*c + 4*d*x] + 7560*C*Sin[3*c + 4*d*x] + 7700*A*Sin[5*c + 4*d*x
] + 7560*C*Sin[5*c + 4*d*x] + 12712*A*Sin[4*c + 5*d*x] + 15848*C*Sin[4*c + 5*d*x] - 420*C*Sin[6*c + 5*d*x] + 1
155*A*Sin[5*c + 6*d*x] + 1470*C*Sin[5*c + 6*d*x] + 1155*A*Sin[7*c + 6*d*x] + 1470*C*Sin[7*c + 6*d*x] + 1816*A*
Sin[6*c + 7*d*x] + 2324*C*Sin[6*c + 7*d*x])))/d

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fricas [A]  time = 0.73, size = 201, normalized size = 0.76 \[ \frac {105 \, {\left (11 \, A + 14 \, C\right )} a^{4} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (11 \, A + 14 \, C\right )} a^{4} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, {\left (454 \, A + 581 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} + 105 \, {\left (11 \, A + 14 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} + 4 \, {\left (227 \, A + 238 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 70 \, {\left (11 \, A + 6 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 12 \, {\left (48 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 280 \, A a^{4} \cos \left (d x + c\right ) + 60 \, A a^{4}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="fricas")

[Out]

1/840*(105*(11*A + 14*C)*a^4*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*(11*A + 14*C)*a^4*cos(d*x + c)^7*log(-
sin(d*x + c) + 1) + 2*(4*(454*A + 581*C)*a^4*cos(d*x + c)^6 + 105*(11*A + 14*C)*a^4*cos(d*x + c)^5 + 4*(227*A
+ 238*C)*a^4*cos(d*x + c)^4 + 70*(11*A + 6*C)*a^4*cos(d*x + c)^3 + 12*(48*A + 7*C)*a^4*cos(d*x + c)^2 + 280*A*
a^4*cos(d*x + c) + 60*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^7)

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giac [A]  time = 0.79, size = 314, normalized size = 1.19 \[ \frac {105 \, {\left (11 \, A a^{4} + 14 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 105 \, {\left (11 \, A a^{4} + 14 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (1155 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 1470 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 7700 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 9800 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 21791 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 27734 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 33792 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 43008 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 31521 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 39914 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 14700 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21560 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5565 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5250 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{7}}}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="giac")

[Out]

1/420*(105*(11*A*a^4 + 14*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(11*A*a^4 + 14*C*a^4)*log(abs(tan(1/
2*d*x + 1/2*c) - 1)) - 2*(1155*A*a^4*tan(1/2*d*x + 1/2*c)^13 + 1470*C*a^4*tan(1/2*d*x + 1/2*c)^13 - 7700*A*a^4
*tan(1/2*d*x + 1/2*c)^11 - 9800*C*a^4*tan(1/2*d*x + 1/2*c)^11 + 21791*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 27734*C*a
^4*tan(1/2*d*x + 1/2*c)^9 - 33792*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 43008*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 31521*A*
a^4*tan(1/2*d*x + 1/2*c)^5 + 39914*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 14700*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 21560*C
*a^4*tan(1/2*d*x + 1/2*c)^3 + 5565*A*a^4*tan(1/2*d*x + 1/2*c) + 5250*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^7)/d

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maple [A]  time = 0.62, size = 303, normalized size = 1.15 \[ \frac {454 A \,a^{4} \tan \left (d x +c \right )}{105 d}+\frac {227 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{105 d}+\frac {83 a^{4} C \tan \left (d x +c \right )}{15 d}+\frac {11 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{6 d}+\frac {11 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {11 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {7 a^{4} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {7 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {48 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{35 d}+\frac {34 a^{4} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {2 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{6}\left (d x +c \right )\right )}{7 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x)

[Out]

454/105/d*A*a^4*tan(d*x+c)+227/105/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+83/15/d*a^4*C*tan(d*x+c)+11/6/d*A*a^4*tan(d
*x+c)*sec(d*x+c)^3+11/4/d*A*a^4*sec(d*x+c)*tan(d*x+c)+11/4/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+7/2/d*a^4*C*sec(d
*x+c)*tan(d*x+c)+7/2/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+48/35/d*A*a^4*tan(d*x+c)*sec(d*x+c)^4+34/15/d*a^4*C*tan
(d*x+c)*sec(d*x+c)^2+2/3/d*A*a^4*tan(d*x+c)*sec(d*x+c)^5+1/d*a^4*C*tan(d*x+c)*sec(d*x+c)^3+1/7/d*A*a^4*tan(d*x
+c)*sec(d*x+c)^6+1/5/d*a^4*C*tan(d*x+c)*sec(d*x+c)^4

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maxima [A]  time = 0.40, size = 462, normalized size = 1.76 \[ \frac {24 \, {\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} A a^{4} + 336 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 280 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 56 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{4} + 1680 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 35 \, A a^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 210 \, A a^{4} {\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 )} - 210 \, C a^{4} {\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 )} - 840 \, C a^{4} {\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 )} + 840 \, C a^{4} \tan \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^8,x, algorithm="maxima")

[Out]

1/840*(24*(5*tan(d*x + c)^7 + 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*A*a^4 + 336*(3*tan(d*x
+ c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 280*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 56*(3*tan(
d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^4 + 1680*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 - 35*A*
a^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*
x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 210*A*a^4*(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)) - 210*C
*a^4*(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)) - 840*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(
d*x + c) - 1)) + 840*C*a^4*tan(d*x + c))/d

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mupad [B]  time = 3.70, size = 301, normalized size = 1.14 \[ \frac {a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (11\,A+14\,C\right )}{2\,d}-\frac {\left (\frac {11\,A\,a^4}{2}+7\,C\,a^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (-\frac {110\,A\,a^4}{3}-\frac {140\,C\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {3113\,A\,a^4}{30}+\frac {1981\,C\,a^4}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {5632\,A\,a^4}{35}-\frac {1024\,C\,a^4}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {1501\,A\,a^4}{10}+\frac {2851\,C\,a^4}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-70\,A\,a^4-\frac {308\,C\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {53\,A\,a^4}{2}+25\,C\,a^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^4)/cos(c + d*x)^8,x)

[Out]

(a^4*atanh(tan(c/2 + (d*x)/2))*(11*A + 14*C))/(2*d) - (tan(c/2 + (d*x)/2)*((53*A*a^4)/2 + 25*C*a^4) + tan(c/2
+ (d*x)/2)^13*((11*A*a^4)/2 + 7*C*a^4) - tan(c/2 + (d*x)/2)^11*((110*A*a^4)/3 + (140*C*a^4)/3) - tan(c/2 + (d*
x)/2)^3*(70*A*a^4 + (308*C*a^4)/3) + tan(c/2 + (d*x)/2)^5*((1501*A*a^4)/10 + (2851*C*a^4)/15) + tan(c/2 + (d*x
)/2)^9*((3113*A*a^4)/30 + (1981*C*a^4)/15) - tan(c/2 + (d*x)/2)^7*((5632*A*a^4)/35 + (1024*C*a^4)/5))/(d*(7*ta
n(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 - 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2
+ (d*x)/2)^10 - 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 - 1))

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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))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**8,x)

[Out]

Timed out

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