3.437 \(\int (a+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=169 \[ \frac {a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \left (9 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {11 a^2 b \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))}{5 d} \]

[Out]

1/8*b*(9*a^2+4*b^2)*arctanh(sin(d*x+c))/d+1/5*a*(4*a^2+15*b^2)*tan(d*x+c)/d+1/8*b*(9*a^2+4*b^2)*sec(d*x+c)*tan
(d*x+c)/d+11/20*a^2*b*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a^2*(a+b*cos(d*x+c))*sec(d*x+c)^4*tan(d*x+c)/d+1/15*a*(4*a
^2+15*b^2)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.22, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2792, 3021, 2748, 3767, 3768, 3770} \[ \frac {a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \left (9 a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {11 a^2 b \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(9*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(4*a^2 + 15*b^2)*Tan[c + d*x])/(5*d) + (b*(9*a^2 + 4*b^2)
*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (11*a^2*b*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (a^2*(a + b*Cos[c + d*x])*
Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (a*(4*a^2 + 15*b^2)*Tan[c + d*x]^3)/(15*d)

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 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

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 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+b \cos (c+d x))^3 \sec ^6(c+d x) \, dx &=\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \left (11 a^2 b+a \left (4 a^2+15 b^2\right ) \cos (c+d x)+b \left (3 a^2+5 b^2\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int \left (4 a \left (4 a^2+15 b^2\right )+5 b \left (9 a^2+4 b^2\right ) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (b \left (9 a^2+4 b^2\right )\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (a \left (4 a^2+15 b^2\right )\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {b \left (9 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (b \left (9 a^2+4 b^2\right )\right ) \int \sec (c+d x) \, dx-\frac {\left (a \left (4 a^2+15 b^2\right )\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {b \left (9 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {11 a^2 b \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a \left (4 a^2+15 b^2\right ) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 120, normalized size = 0.71 \[ \frac {15 b \left (9 a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 a \left (5 \left (2 a^2+3 b^2\right ) \tan ^2(c+d x)+15 \left (a^2+3 b^2\right )+3 a^2 \tan ^4(c+d x)\right )+15 b \left (9 a^2+4 b^2\right ) \sec (c+d x)+90 a^2 b \sec ^3(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*b*(9*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*b*(9*a^2 + 4*b^2)*Sec[c + d*x] + 90*a^2*b*Sec[c
+ d*x]^3 + 8*a*(15*(a^2 + 3*b^2) + 5*(2*a^2 + 3*b^2)*Tan[c + d*x]^2 + 3*a^2*Tan[c + d*x]^4)))/(120*d)

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fricas [A]  time = 0.89, size = 170, normalized size = 1.01 \[ \frac {15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 90 \, a^{2} b \cos \left (d x + c\right ) + 15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 24 \, a^{3} + 8 \, {\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(9*a^2*b + 4*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(9*a^2*b + 4*b^3)*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 2*(16*(4*a^3 + 15*a*b^2)*cos(d*x + c)^4 + 90*a^2*b*cos(d*x + c) + 15*(9*a^2*b + 4*b^3)*cos(d
*x + c)^3 + 24*a^3 + 8*(4*a^3 + 15*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [B]  time = 0.66, size = 367, normalized size = 2.17 \[ \frac {15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 960 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 960 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(9*a^2*b + 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(9*a^2*b + 4*b^3)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) - 2*(120*a^3*tan(1/2*d*x + 1/2*c)^9 - 225*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*a*b^2*tan(1/2*d*x + 1/
2*c)^9 - 60*b^3*tan(1/2*d*x + 1/2*c)^9 - 160*a^3*tan(1/2*d*x + 1/2*c)^7 + 90*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 96
0*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*a*b^2*
tan(1/2*d*x + 1/2*c)^5 - 160*a^3*tan(1/2*d*x + 1/2*c)^3 - 90*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*a*b^2*tan(1/2*
d*x + 1/2*c)^3 - 120*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*a^3*tan(1/2*d*x + 1/2*c) + 225*a^2*b*tan(1/2*d*x + 1/2*c
) + 360*a*b^2*tan(1/2*d*x + 1/2*c) + 60*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 0.12, size = 206, normalized size = 1.22 \[ \frac {8 a^{3} \tan \left (d x +c \right )}{15 d}+\frac {a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {3 a^{2} b \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {9 a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {9 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 b^{2} a \tan \left (d x +c \right )}{d}+\frac {b^{2} a \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*sec(d*x+c)^6,x)

[Out]

8/15*a^3*tan(d*x+c)/d+1/5/d*a^3*tan(d*x+c)*sec(d*x+c)^4+4/15/d*a^3*tan(d*x+c)*sec(d*x+c)^2+3/4*a^2*b*sec(d*x+c
)^3*tan(d*x+c)/d+9/8*a^2*b*sec(d*x+c)*tan(d*x+c)/d+9/8/d*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+2/d*b^2*a*tan(d*x+c)+
1/d*b^2*a*tan(d*x+c)*sec(d*x+c)^2+1/2/d*b^3*tan(d*x+c)*sec(d*x+c)+1/2/d*b^3*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.65, size = 181, normalized size = 1.07 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{2} - 45 \, a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))
*a*b^2 - 45*a^2*b*(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)) - 60*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1)
+ log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 4.22, size = 260, normalized size = 1.54 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {9\,a^2\,b}{4}+b^3\right )}{d}-\frac {\left (2\,a^3-\frac {15\,a^2\,b}{4}+6\,a\,b^2-b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {8\,a^3}{3}+\frac {3\,a^2\,b}{2}-16\,a\,b^2+2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^3}{15}+20\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,a^3}{3}-\frac {3\,a^2\,b}{2}-16\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^3+\frac {15\,a^2\,b}{4}+6\,a\,b^2+b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^3/cos(c + d*x)^6,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((9*a^2*b)/4 + b^3))/d - (tan(c/2 + (d*x)/2)^9*(6*a*b^2 - (15*a^2*b)/4 + 2*a^3 - b^
3) - tan(c/2 + (d*x)/2)^3*(16*a*b^2 + (3*a^2*b)/2 + (8*a^3)/3 + 2*b^3) - tan(c/2 + (d*x)/2)^7*(16*a*b^2 - (3*a
^2*b)/2 + (8*a^3)/3 - 2*b^3) + tan(c/2 + (d*x)/2)*(6*a*b^2 + (15*a^2*b)/4 + 2*a^3 + b^3) + tan(c/2 + (d*x)/2)^
5*(20*a*b^2 + (116*a^3)/15))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 -
5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 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+b*cos(d*x+c))**3*sec(d*x+c)**6,x)

[Out]

Timed out

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