3.17 \(\int (a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=178 \[ \frac {a^2 (18 A+25 C) \tan (c+d x)}{15 d}+\frac {a^2 (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^2 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac {a^2 (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{10 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

[Out]

1/4*a^2*(3*A+4*C)*arctanh(sin(d*x+c))/d+1/15*a^2*(18*A+25*C)*tan(d*x+c)/d+1/4*a^2*(3*A+4*C)*sec(d*x+c)*tan(d*x
+c)/d+1/30*a^2*(9*A+10*C)*sec(d*x+c)^2*tan(d*x+c)/d+1/10*A*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^3*tan(d*x+c)/d+1/5*
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 = 0.47, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {a^2 (18 A+25 C) \tan (c+d x)}{15 d}+\frac {a^2 (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^2 (9 A+10 C) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac {a^2 (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{10 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3*A + 4*C)*ArcTanh[Sin[c + d*x]])/(4*d) + (a^2*(18*A + 25*C)*Tan[c + d*x])/(15*d) + (a^2*(3*A + 4*C)*Sec
[c + d*x]*Tan[c + d*x])/(4*d) + (a^2*(9*A + 10*C)*Sec[c + d*x]^2*Tan[c + d*x])/(30*d) + (A*(a^2 + a^2*Cos[c +
d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(10*d) + (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 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))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^2 (2 a A+a (2 A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x)) \left (2 a^2 (9 A+10 C)+4 a^2 (3 A+5 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (2 a^3 (9 A+10 C)+\left (4 a^3 (3 A+5 C)+2 a^3 (9 A+10 C)\right ) \cos (c+d x)+4 a^3 (3 A+5 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac {a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (30 a^3 (3 A+4 C)+4 a^3 (18 A+25 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac {a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} \left (a^2 (3 A+4 C)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{15} \left (a^2 (18 A+25 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^2 (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (a^2 (3 A+4 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (18 A+25 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {a^2 (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a^2 (18 A+25 C) \tan (c+d x)}{15 d}+\frac {a^2 (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 (9 A+10 C) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {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 = 1.47, size = 292, normalized size = 1.64 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (240 (3 A+4 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (-120 (A+3 C) \sin (2 c+d x)+210 A \sin (c+2 d x)+210 A \sin (3 c+2 d x)+360 A \sin (2 c+3 d x)+45 A \sin (3 c+4 d x)+45 A \sin (5 c+4 d x)+72 A \sin (4 c+5 d x)+40 (15 A+16 C) \sin (d x)+120 C \sin (c+2 d x)+120 C \sin (3 c+2 d x)+440 C \sin (2 c+3 d x)-60 C \sin (4 c+3 d x)+60 C \sin (3 c+4 d x)+60 C \sin (5 c+4 d x)+100 C \sin (4 c+5 d x))\right )}{3840 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3840*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*Sec[c + d*x]^5*(240*(3*A + 4*C)*Cos[c + d*x]^5*(Log[Cos[(
c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(40*(15*A + 16*C)*Sin[d*x
] - 120*(A + 3*C)*Sin[2*c + d*x] + 210*A*Sin[c + 2*d*x] + 120*C*Sin[c + 2*d*x] + 210*A*Sin[3*c + 2*d*x] + 120*
C*Sin[3*c + 2*d*x] + 360*A*Sin[2*c + 3*d*x] + 440*C*Sin[2*c + 3*d*x] - 60*C*Sin[4*c + 3*d*x] + 45*A*Sin[3*c +
4*d*x] + 60*C*Sin[3*c + 4*d*x] + 45*A*Sin[5*c + 4*d*x] + 60*C*Sin[5*c + 4*d*x] + 72*A*Sin[4*c + 5*d*x] + 100*C
*Sin[4*c + 5*d*x])))/d

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fricas [A]  time = 0.60, size = 161, normalized size = 0.90 \[ \frac {15 \, {\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, {\left (18 \, A + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, A a^{2} \cos \left (d x + c\right ) + 12 \, A a^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(3*A + 4*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*A + 4*C)*a^2*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(4*(18*A + 25*C)*a^2*cos(d*x + c)^4 + 15*(3*A + 4*C)*a^2*cos(d*x + c)^3 + 4*(9*A + 5*C)*a^2*co
s(d*x + c)^2 + 30*A*a^2*cos(d*x + c) + 12*A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [A]  time = 0.47, size = 246, normalized size = 1.38 \[ \frac {15 \, {\left (3 \, A a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, A a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 280 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 432 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 560 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 270 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 520 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*(3*A*a^2 + 4*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*A*a^2 + 4*C*a^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(45*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 210*A*a^2*tan(1/2*d*x +
1/2*c)^7 - 280*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 432*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 560*C*a^2*tan(1/2*d*x + 1/2*
c)^5 - 270*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 520*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 195*A*a^2*tan(1/2*d*x + 1/2*c) +
180*C*a^2*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 = 0.49, size = 210, normalized size = 1.18 \[ \frac {6 a^{2} A \tan \left (d x +c \right )}{5 d}+\frac {3 a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {5 a^{2} C \tan \left (d x +c \right )}{3 d}+\frac {a^{2} A \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/5*a^2*A*tan(d*x+c)/d+3/5*a^2*A*sec(d*x+c)^2*tan(d*x+c)/d+5/3/d*a^2*C*tan(d*x+c)+1/2*a^2*A*sec(d*x+c)^3*tan(d
*x+c)/d+3/4*a^2*A*sec(d*x+c)*tan(d*x+c)/d+3/4/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^2*C*sec(d*x+c)*tan(d*x+c
)+1/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+1/5/d*a^2*A*tan(d*x+c)*sec(d*x+c)^4+1/3/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.33, size = 218, normalized size = 1.22 \[ \frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C a^{2} \tan \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))
*A*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 15*A*a^2*(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*C*a^2*(2*sin(d*x + c
)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*C*a^2*tan(d*x + c))/d

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mupad [B]  time = 3.60, size = 222, normalized size = 1.25 \[ \frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,A}{4}+C\right )}{d}-\frac {\left (\frac {3\,A\,a^2}{2}+2\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-7\,A\,a^2-\frac {28\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,A\,a^2}{5}+\frac {56\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-9\,A\,a^2-\frac {52\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a^2}{2}+6\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*atanh(tan(c/2 + (d*x)/2))*((3*A)/4 + C))/d - (tan(c/2 + (d*x)/2)*((13*A*a^2)/2 + 6*C*a^2) + tan(c/2 + (
d*x)/2)^9*((3*A*a^2)/2 + 2*C*a^2) - tan(c/2 + (d*x)/2)^7*(7*A*a^2 + (28*C*a^2)/3) - tan(c/2 + (d*x)/2)^3*(9*A*
a^2 + (52*C*a^2)/3) + tan(c/2 + (d*x)/2)^5*((72*A*a^2)/5 + (56*C*a^2)/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))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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