3.27 \(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=185 \[ \frac {a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+15 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (43 A+45 B) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac {a^3 (13 A+15 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {(7 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{20 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

[Out]

1/8*a^3*(13*A+15*B)*arctanh(sin(d*x+c))/d+1/15*a^3*(38*A+45*B)*tan(d*x+c)/d+1/8*a^3*(13*A+15*B)*sec(d*x+c)*tan
(d*x+c)/d+1/60*a^3*(43*A+45*B)*sec(d*x+c)^2*tan(d*x+c)/d+1/20*(7*A+5*B)*(a^3+a^3*cos(d*x+c))*sec(d*x+c)^3*tan(
d*x+c)/d+1/5*a*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d

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Rubi [A]  time = 0.45, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+15 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (43 A+45 B) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac {a^3 (13 A+15 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {(7 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{20 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(13*A + 15*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(38*A + 45*B)*Tan[c + d*x])/(15*d) + (a^3*(13*A + 15*B)
*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^3*(43*A + 45*B)*Sec[c + d*x]^2*Tan[c + d*x])/(60*d) + ((7*A + 5*B)*(a^3
+ a^3*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (a*A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c +
d*x])/(5*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 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))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx &=\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+a \cos (c+d x))^2 (a (7 A+5 B)+a (2 A+5 B) \cos (c+d x)) \sec ^5(c+d x) \, dx\\ &=\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int (a+a \cos (c+d x)) \left (a^2 (43 A+45 B)+2 a^2 (11 A+15 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int \left (a^3 (43 A+45 B)+\left (2 a^3 (11 A+15 B)+a^3 (43 A+45 B)\right ) \cos (c+d x)+2 a^3 (11 A+15 B) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{60} \int \left (15 a^3 (13 A+15 B)+4 a^3 (38 A+45 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (a^3 (13 A+15 B)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{15} \left (a^3 (38 A+45 B)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^3 (13 A+15 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^3 (13 A+15 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (38 A+45 B)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {a^3 (13 A+15 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+15 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 1.60, size = 294, normalized size = 1.59 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (13 A+15 B) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (3 A+5 B) \sin (2 c+d x)+80 (29 A+30 B) \sin (d x)+750 A \sin (c+2 d x)+750 A \sin (3 c+2 d x)+1520 A \sin (2 c+3 d x)+195 A \sin (3 c+4 d x)+195 A \sin (5 c+4 d x)+304 A \sin (4 c+5 d x)+570 B \sin (c+2 d x)+570 B \sin (3 c+2 d x)+1680 B \sin (2 c+3 d x)-120 B \sin (4 c+3 d x)+225 B \sin (3 c+4 d x)+225 B \sin (5 c+4 d x)+360 B \sin (4 c+5 d x))\right )}{15360 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15360*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^5*(240*(13*A + 15*B)*Cos[c + d*x]^5*(Log[Co
s[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(80*(29*A + 30*B)*Sin[
d*x] - 240*(3*A + 5*B)*Sin[2*c + d*x] + 750*A*Sin[c + 2*d*x] + 570*B*Sin[c + 2*d*x] + 750*A*Sin[3*c + 2*d*x] +
570*B*Sin[3*c + 2*d*x] + 1520*A*Sin[2*c + 3*d*x] + 1680*B*Sin[2*c + 3*d*x] - 120*B*Sin[4*c + 3*d*x] + 195*A*S
in[3*c + 4*d*x] + 225*B*Sin[3*c + 4*d*x] + 195*A*Sin[5*c + 4*d*x] + 225*B*Sin[5*c + 4*d*x] + 304*A*Sin[4*c + 5
*d*x] + 360*B*Sin[4*c + 5*d*x])))/d

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fricas [A]  time = 0.73, size = 165, normalized size = 0.89 \[ \frac {15 \, {\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (38 \, A + 45 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (19 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 24 \, A a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(13*A + 15*B)*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(13*A + 15*B)*a^3*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 2*(8*(38*A + 45*B)*a^3*cos(d*x + c)^4 + 15*(13*A + 15*B)*a^3*cos(d*x + c)^3 + 8*(19*A + 15*B
)*a^3*cos(d*x + c)^2 + 30*(3*A + B)*a^3*cos(d*x + c) + 24*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [A]  time = 1.72, size = 246, normalized size = 1.33 \[ \frac {15 \, {\left (13 \, A a^{3} + 15 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (13 \, A a^{3} + 15 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (195 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 225 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 910 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1050 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1920 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1330 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1830 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 735 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

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))*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(13*A*a^3 + 15*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(13*A*a^3 + 15*B*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(195*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 225*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 910*A*a^3*tan(1/
2*d*x + 1/2*c)^7 - 1050*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1920*B*a^3*tan(1/2*
d*x + 1/2*c)^5 - 1330*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 1830*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 765*A*a^3*tan(1/2*d*x
+ 1/2*c) + 735*B*a^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 = 0.17, size = 234, normalized size = 1.26 \[ \frac {13 A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {13 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a^{3} B \tan \left (d x +c \right )}{d}+\frac {38 A \,a^{3} \tan \left (d x +c \right )}{15 d}+\frac {19 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {15 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {15 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/8/d*A*a^3*sec(d*x+c)*tan(d*x+c)+13/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^3*B*tan(d*x+c)+38/15/d*A*a^3*t
an(d*x+c)+19/15/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+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))+3/4/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+1/d*a^3*B*tan(d*x+c)*sec(d*x+c)^2+1/5/d*A*a^3*tan(d*x+c)*sec(d
*x+c)^4+1/4/d*a^3*B*tan(d*x+c)*sec(d*x+c)^3

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maxima [A]  time = 0.68, size = 337, normalized size = 1.82 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 45 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B a^{3} \tan \left (d x + c\right )}{240 \, d} \]

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))*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*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c
))*A*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 - 45*A*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)) - 15*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)) - 60*A*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)) -
180*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)) + 240*B*a^3*ta
n(d*x + c))/d

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mupad [B]  time = 2.82, size = 224, normalized size = 1.21 \[ \frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (13\,A+15\,B\right )}{4\,d}-\frac {\left (\frac {13\,A\,a^3}{4}+\frac {15\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {91\,A\,a^3}{6}-\frac {35\,B\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {416\,A\,a^3}{15}+32\,B\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {133\,A\,a^3}{6}-\frac {61\,B\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+\frac {49\,B\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*atanh(tan(c/2 + (d*x)/2))*(13*A + 15*B))/(4*d) - (tan(c/2 + (d*x)/2)*((51*A*a^3)/4 + (49*B*a^3)/4) + tan(
c/2 + (d*x)/2)^9*((13*A*a^3)/4 + (15*B*a^3)/4) - tan(c/2 + (d*x)/2)^7*((91*A*a^3)/6 + (35*B*a^3)/2) - tan(c/2
+ (d*x)/2)^3*((133*A*a^3)/6 + (61*B*a^3)/2) + tan(c/2 + (d*x)/2)^5*((416*A*a^3)/15 + 32*B*a^3))/(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+a*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c)**6,x)

[Out]

Timed out

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