3.548 \(\int (a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=273 \[ \frac{b \left (6 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac{a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a \left (5 a^2 (5 A+6 C)+6 A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{120 d}+\frac{b \left (3 a^2 (4 A+5 C)+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d}+\frac{A b \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d} \]

[Out]

(a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*ArcTanh[Sin[c + d*x]])/(16*d) + (b*(5*b^2*(2*A + 3*C) + 6*a^2*(4*A +
5*C))*Tan[c + d*x])/(15*d) + (a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (b*(
A*b^2 + 3*a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (a*(6*A*b^2 + 5*a^2*(5*A + 6*C))*Sec[c + d*x]
^3*Tan[c + d*x])/(120*d) + (A*b*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(10*d) + (A*(a + b*Cos[c +
d*x])^3*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.786249, antiderivative size = 273, normalized size of antiderivative = 1., 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 = {3048, 3047, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{b \left (6 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac{a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a \left (5 a^2 (5 A+6 C)+6 A b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{120 d}+\frac{b \left (3 a^2 (4 A+5 C)+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{A \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^3}{6 d}+\frac{A b \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{10 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*ArcTanh[Sin[c + d*x]])/(16*d) + (b*(5*b^2*(2*A + 3*C) + 6*a^2*(4*A +
5*C))*Tan[c + d*x])/(15*d) + (a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (b*(
A*b^2 + 3*a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (a*(6*A*b^2 + 5*a^2*(5*A + 6*C))*Sec[c + d*x]
^3*Tan[c + d*x])/(120*d) + (A*b*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(10*d) + (A*(a + b*Cos[c +
d*x])^3*Sec[c + d*x]^5*Tan[c + d*x])/(6*d)

Rule 3048

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/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3047

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/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

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 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]

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]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{6} \int (a+b \cos (c+d x))^2 \left (3 A b+a (5 A+6 C) \cos (c+d x)+2 b (A+3 C) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac{A b (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{30} \int (a+b \cos (c+d x)) \left (6 A b^2+5 a^2 (5 A+6 C)+a b (47 A+60 C) \cos (c+d x)+2 b^2 (8 A+15 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{1}{120} \int \left (-24 b \left (A b^2+3 a^2 (4 A+5 C)\right )-15 a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x)-8 b^3 (8 A+15 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{b \left (A b^2+3 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac{1}{360} \int \left (-45 a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )-24 b \left (5 b^2 (2 A+3 C)+6 a^2 (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{b \left (A b^2+3 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{15} \left (b \left (5 b^2 (2 A+3 C)+6 a^2 (4 A+5 C)\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{b \left (A b^2+3 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac{1}{16} \left (a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int \sec (c+d x) \, dx-\frac{\left (b \left (5 b^2 (2 A+3 C)+6 a^2 (4 A+5 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{b \left (5 b^2 (2 A+3 C)+6 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac{a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{b \left (A b^2+3 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \sec ^3(c+d x) \tan (c+d x)}{120 d}+\frac{A b (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 1.66517, size = 184, normalized size = 0.67 \[ \frac{15 a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (16 b \left (5 \left (3 a^2 (2 A+C)+A b^2\right ) \tan ^2(c+d x)+15 \left (3 a^2+b^2\right ) (A+C)+9 a^2 A \tan ^4(c+d x)\right )+10 a \left (a^2 (5 A+6 C)+18 A b^2\right ) \sec ^3(c+d x)+15 a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \sec (c+d x)+40 a^3 A \sec ^5(c+d x)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^7,x]

[Out]

(15*a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*a*(6*b^2*(3*A + 4*C) + a^
2*(5*A + 6*C))*Sec[c + d*x] + 10*a*(18*A*b^2 + a^2*(5*A + 6*C))*Sec[c + d*x]^3 + 40*a^3*A*Sec[c + d*x]^5 + 16*
b*(15*(3*a^2 + b^2)*(A + C) + 5*(A*b^2 + 3*a^2*(2*A + C))*Tan[c + d*x]^2 + 9*a^2*A*Tan[c + d*x]^4)))/(240*d)

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Maple [A]  time = 0.064, size = 430, normalized size = 1.6 \begin{align*}{\frac{2\,A{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,aA{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,aA{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,aA{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,Ca{b}^{2}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{8\,A{a}^{2}b\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,A{a}^{2}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,A{a}^{2}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+2\,{\frac{{a}^{2}bC\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}bC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x)

[Out]

2/3/d*A*b^3*tan(d*x+c)+1/3/d*A*b^3*tan(d*x+c)*sec(d*x+c)^2+1/d*C*b^3*tan(d*x+c)+3/4/d*a*A*b^2*tan(d*x+c)*sec(d
*x+c)^3+9/8/d*a*A*b^2*sec(d*x+c)*tan(d*x+c)+9/8/d*a*A*b^2*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*C*a*b^2*tan(d*x+c)*s
ec(d*x+c)+3/2/d*C*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+8/5/d*A*a^2*b*tan(d*x+c)+3/5/d*A*a^2*b*tan(d*x+c)*sec(d*x+c)
^4+4/5/d*A*a^2*b*tan(d*x+c)*sec(d*x+c)^2+2/d*a^2*b*C*tan(d*x+c)+1/d*a^2*b*C*tan(d*x+c)*sec(d*x+c)^2+1/6/d*A*a^
3*tan(d*x+c)*sec(d*x+c)^5+5/24/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+5/16/d*A*a^3*sec(d*x+c)*tan(d*x+c)+5/16/d*A*a^3
*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*a^3*C*tan(d*x+c)*sec(d*x+c)^3+3/8/d*a^3*C*sec(d*x+c)*tan(d*x+c)+3/8/d*a^3*C*l
n(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.05218, size = 521, normalized size = 1.91 \begin{align*} \frac{96 \,{\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} b + 480 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{3} - 5 \, A a^{3}{\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 )} - 30 \, C 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 )} - 90 \, A a b^{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 )} - 360 \, C a b^{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 )} + 480 \, C b^{3} \tan \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")

[Out]

1/480*(96*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2*b + 480*(tan(d*x + c)^3 + 3*tan(d*x +
c))*C*a^2*b + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^3 - 5*A*a^3*(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)) - 30*C*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)) - 90*A*a*b^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)) - 360*C*a*b^2*(2*s
in(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 480*C*b^3*tan(d*x + c))/d

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Fricas [A]  time = 1.55409, size = 636, normalized size = 2.33 \begin{align*} \frac{15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 6 \,{\left (3 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 6 \,{\left (3 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (6 \,{\left (4 \, A + 5 \, C\right )} a^{2} b + 5 \,{\left (2 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} + 144 \, A a^{2} b \cos \left (d x + c\right ) + 15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 6 \,{\left (3 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{4} + 40 \, A a^{3} + 16 \,{\left (3 \,{\left (4 \, A + 5 \, C\right )} a^{2} b + 5 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\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+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/480*(15*((5*A + 6*C)*a^3 + 6*(3*A + 4*C)*a*b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*((5*A + 6*C)*a^3 +
6*(3*A + 4*C)*a*b^2)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(16*(6*(4*A + 5*C)*a^2*b + 5*(2*A + 3*C)*b^3)*
cos(d*x + c)^5 + 144*A*a^2*b*cos(d*x + c) + 15*((5*A + 6*C)*a^3 + 6*(3*A + 4*C)*a*b^2)*cos(d*x + c)^4 + 40*A*a
^3 + 16*(3*(4*A + 5*C)*a^2*b + 5*A*b^3)*cos(d*x + c)^3 + 10*((5*A + 6*C)*a^3 + 18*A*a*b^2)*cos(d*x + c)^2)*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+b*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**7,x)

[Out]

Timed out

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Giac [B]  time = 1.64299, size = 1258, normalized size = 4.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="giac")

[Out]

1/240*(15*(5*A*a^3 + 6*C*a^3 + 18*A*a*b^2 + 24*C*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(5*A*a^3 + 6*C
*a^3 + 18*A*a*b^2 + 24*C*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(165*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 15
0*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 720*A*a^2*b*tan(1/2*d*x + 1/2*c)^11 - 720*C*a^2*b*tan(1/2*d*x + 1/2*c)^11 +
450*A*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 360*C*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 240*A*b^3*tan(1/2*d*x + 1/2*c)^11
- 240*C*b^3*tan(1/2*d*x + 1/2*c)^11 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 210*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 168
0*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 2640*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 630*A*a*b^2*tan(1/2*d*x + 1/2*c)^9 -
1080*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 880*A*b^3*tan(1/2*d*x + 1/2*c)^9 + 1200*C*b^3*tan(1/2*d*x + 1/2*c)^9 + 4
50*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 4320
*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 180*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 14
40*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 2400*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 450*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 60*C*
a^3*tan(1/2*d*x + 1/2*c)^5 + 3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 4320*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 180*A
*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 1440*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 2400*
C*b^3*tan(1/2*d*x + 1/2*c)^5 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 210*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1680*A*a^2
*b*tan(1/2*d*x + 1/2*c)^3 - 2640*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 630*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 1080*C*
a*b^2*tan(1/2*d*x + 1/2*c)^3 - 880*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 1200*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 165*A*a^
3*tan(1/2*d*x + 1/2*c) + 150*C*a^3*tan(1/2*d*x + 1/2*c) + 720*A*a^2*b*tan(1/2*d*x + 1/2*c) + 720*C*a^2*b*tan(1
/2*d*x + 1/2*c) + 450*A*a*b^2*tan(1/2*d*x + 1/2*c) + 360*C*a*b^2*tan(1/2*d*x + 1/2*c) + 240*A*b^3*tan(1/2*d*x
+ 1/2*c) + 240*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d