3.337 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=253 \[ \frac{a^4 (72 A+83 B+100 C) \tan (c+d x)}{15 d}+\frac{7 a^4 (7 A+8 B+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (417 A+488 B+550 C) \tan (c+d x) \sec (c+d x)}{240 d}+\frac{(37 A+48 B+30 C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{120 d}+\frac{(43 A+52 B+50 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{60 d}+\frac{a (2 A+3 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{15 d}+\frac{A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^4}{6 d} \]

[Out]

(7*a^4*(7*A + 8*B + 10*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^4*(72*A + 83*B + 100*C)*Tan[c + d*x])/(15*d) + (a
^4*(417*A + 488*B + 550*C)*Sec[c + d*x]*Tan[c + d*x])/(240*d) + ((43*A + 52*B + 50*C)*(a^4 + a^4*Cos[c + d*x])
*Sec[c + d*x]^2*Tan[c + d*x])/(60*d) + ((37*A + 48*B + 30*C)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c +
d*x])/(120*d) + (a*(2*A + 3*B)*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(15*d) + (A*(a + a*Cos[c +
d*x])^4*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.833446, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {3043, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac{a^4 (72 A+83 B+100 C) \tan (c+d x)}{15 d}+\frac{7 a^4 (7 A+8 B+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (417 A+488 B+550 C) \tan (c+d x) \sec (c+d x)}{240 d}+\frac{(37 A+48 B+30 C) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{120 d}+\frac{(43 A+52 B+50 C) \tan (c+d x) \sec ^2(c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{60 d}+\frac{a (2 A+3 B) \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^3}{15 d}+\frac{A \tan (c+d x) \sec ^5(c+d x) (a \cos (c+d x)+a)^4}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(7*a^4*(7*A + 8*B + 10*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^4*(72*A + 83*B + 100*C)*Tan[c + d*x])/(15*d) + (a
^4*(417*A + 488*B + 550*C)*Sec[c + d*x]*Tan[c + d*x])/(240*d) + ((43*A + 52*B + 50*C)*(a^4 + a^4*Cos[c + d*x])
*Sec[c + d*x]^2*Tan[c + d*x])/(60*d) + ((37*A + 48*B + 30*C)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c +
d*x])/(120*d) + (a*(2*A + 3*B)*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(15*d) + (A*(a + a*Cos[c +
d*x])^4*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

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 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 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 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 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 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 8

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

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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int (a+a \cos (c+d x))^4 (2 a (2 A+3 B)+a (A+6 C) \cos (c+d x)) \sec ^6(c+d x) \, dx}{6 a}\\ &=\frac{a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int (a+a \cos (c+d x))^3 \left (a^2 (37 A+48 B+30 C)+3 a^2 (3 A+2 B+10 C) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx}{30 a}\\ &=\frac{(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int (a+a \cos (c+d x))^2 \left (6 a^3 (43 A+52 B+50 C)+a^3 (73 A+72 B+150 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{120 a}\\ &=\frac{(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int (a+a \cos (c+d x)) \left (3 a^4 (417 A+488 B+550 C)+3 a^4 (159 A+176 B+250 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{360 a}\\ &=\frac{(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int \left (3 a^5 (417 A+488 B+550 C)+\left (3 a^5 (159 A+176 B+250 C)+3 a^5 (417 A+488 B+550 C)\right ) \cos (c+d x)+3 a^5 (159 A+176 B+250 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{360 a}\\ &=\frac{a^4 (417 A+488 B+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{\int \left (48 a^5 (72 A+83 B+100 C)+315 a^5 (7 A+8 B+10 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{720 a}\\ &=\frac{a^4 (417 A+488 B+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{16} \left (7 a^4 (7 A+8 B+10 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{15} \left (a^4 (72 A+83 B+100 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{7 a^4 (7 A+8 B+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (417 A+488 B+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{\left (a^4 (72 A+83 B+100 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{7 a^4 (7 A+8 B+10 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^4 (72 A+83 B+100 C) \tan (c+d x)}{15 d}+\frac{a^4 (417 A+488 B+550 C) \sec (c+d x) \tan (c+d x)}{240 d}+\frac{(43 A+52 B+50 C) \left (a^4+a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac{(37 A+48 B+30 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{a (2 A+3 B) (a+a \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac{A (a+a \cos (c+d x))^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 2.05746, size = 265, normalized size = 1.05 \[ -\frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (840 (7 A+8 B+10 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sin (c+d x) (16 (672 A+643 B+620 C) \cos (c+d x)+20 (229 A+216 B+174 C) \cos (2 (c+d x))+4032 A \cos (3 (c+d x))+735 A \cos (4 (c+d x))+576 A \cos (5 (c+d x))+4165 A+4408 B \cos (3 (c+d x))+840 B \cos (4 (c+d x))+664 B \cos (5 (c+d x))+3480 B+4640 C \cos (3 (c+d x))+810 C \cos (4 (c+d x))+800 C \cos (5 (c+d x))+2670 C)\right )}{30720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

-(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*Sec[c + d*x]^6*(840*(7*A + 8*B + 10*C)*Cos[c + d*x]^6*(Log[Cos[(
c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - (4165*A + 3480*B + 2670*C + 16*(
672*A + 643*B + 620*C)*Cos[c + d*x] + 20*(229*A + 216*B + 174*C)*Cos[2*(c + d*x)] + 4032*A*Cos[3*(c + d*x)] +
4408*B*Cos[3*(c + d*x)] + 4640*C*Cos[3*(c + d*x)] + 735*A*Cos[4*(c + d*x)] + 840*B*Cos[4*(c + d*x)] + 810*C*Co
s[4*(c + d*x)] + 576*A*Cos[5*(c + d*x)] + 664*B*Cos[5*(c + d*x)] + 800*C*Cos[5*(c + d*x)])*Sin[c + d*x]))/(307
20*d)

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Maple [A]  time = 0.106, size = 385, normalized size = 1.5 \begin{align*}{\frac{20\,{a}^{4}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{83\,{a}^{4}B\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{34\,{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{41\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{49\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{27\,{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{7\,{a}^{4}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{24\,A{a}^{4}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{4\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{12\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{4\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{7\,{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{35\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{49\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)

[Out]

20/3/d*a^4*C*tan(d*x+c)+83/15/d*a^4*B*tan(d*x+c)+1/5/d*a^4*B*tan(d*x+c)*sec(d*x+c)^4+34/15/d*a^4*B*tan(d*x+c)*
sec(d*x+c)^2+1/6/d*A*a^4*tan(d*x+c)*sec(d*x+c)^5+41/24/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+49/16/d*A*a^4*sec(d*x+c
)*tan(d*x+c)+1/4/d*a^4*C*tan(d*x+c)*sec(d*x+c)^3+27/8/d*a^4*C*sec(d*x+c)*tan(d*x+c)+1/d*a^4*B*tan(d*x+c)*sec(d
*x+c)^3+7/2/d*a^4*B*sec(d*x+c)*tan(d*x+c)+24/5/d*A*a^4*tan(d*x+c)+4/5/d*A*a^4*tan(d*x+c)*sec(d*x+c)^4+12/5/d*A
*a^4*tan(d*x+c)*sec(d*x+c)^2+4/3/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+7/2/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+35/8/d*
a^4*C*ln(sec(d*x+c)+tan(d*x+c))+49/16/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [B]  time = 1.06076, size = 871, normalized size = 3.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")

[Out]

1/480*(128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 640*(tan(d*x + c)^3 + 3*tan(d*x +
c))*A*a^4 + 32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^4 + 960*(tan(d*x + c)^3 + 3*tan(d*
x + c))*B*a^4 + 640*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 - 5*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) + 1
5*log(sin(d*x + c) - 1)) - 180*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)) - 120*B*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)) - 30*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)) - 120*A*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)) - 480*B*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)) - 720*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)) + 240*C*a^4*(log(sin
(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 480*B*a^4*tan(d*x + c) + 1920*C*a^4*tan(d*x + c))/d

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Fricas [A]  time = 2.09994, size = 539, normalized size = 2.13 \begin{align*} \frac{105 \,{\left (7 \, A + 8 \, B + 10 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (7 \, A + 8 \, B + 10 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (72 \, A + 83 \, B + 100 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} + 15 \,{\left (49 \, A + 56 \, B + 54 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 32 \,{\left (18 \, A + 17 \, B + 10 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 10 \,{\left (41 \, A + 24 \, B + 6 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 48 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 40 \, A a^{4}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/480*(105*(7*A + 8*B + 10*C)*a^4*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 105*(7*A + 8*B + 10*C)*a^4*cos(d*x +
c)^6*log(-sin(d*x + c) + 1) + 2*(16*(72*A + 83*B + 100*C)*a^4*cos(d*x + c)^5 + 15*(49*A + 56*B + 54*C)*a^4*cos
(d*x + c)^4 + 32*(18*A + 17*B + 10*C)*a^4*cos(d*x + c)^3 + 10*(41*A + 24*B + 6*C)*a^4*cos(d*x + c)^2 + 48*(4*A
+ B)*a^4*cos(d*x + c) + 40*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)

[Out]

Timed out

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Giac [A]  time = 1.33507, size = 529, normalized size = 2.09 \begin{align*} \frac{105 \,{\left (7 \, A a^{4} + 8 \, B a^{4} + 10 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 105 \,{\left (7 \, A a^{4} + 8 \, B a^{4} + 10 \, 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 (735 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 840 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 1050 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 4165 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 4760 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 5950 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 9702 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 11088 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 13860 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 11802 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 13488 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 16860 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 7355 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9320 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 10690 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3105 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3000 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2790 \, 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 )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="giac")

[Out]

1/240*(105*(7*A*a^4 + 8*B*a^4 + 10*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(7*A*a^4 + 8*B*a^4 + 10*C*a
^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(735*A*a^4*tan(1/2*d*x + 1/2*c)^11 + 840*B*a^4*tan(1/2*d*x + 1/2*c)
^11 + 1050*C*a^4*tan(1/2*d*x + 1/2*c)^11 - 4165*A*a^4*tan(1/2*d*x + 1/2*c)^9 - 4760*B*a^4*tan(1/2*d*x + 1/2*c)
^9 - 5950*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 9702*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 11088*B*a^4*tan(1/2*d*x + 1/2*c)^
7 + 13860*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 11802*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 13488*B*a^4*tan(1/2*d*x + 1/2*c)
^5 - 16860*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 7355*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 9320*B*a^4*tan(1/2*d*x + 1/2*c)^
3 + 10690*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 3105*A*a^4*tan(1/2*d*x + 1/2*c) - 3000*B*a^4*tan(1/2*d*x + 1/2*c) - 2
790*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d