3.248 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=267 \[ \frac{\left (60 a^2 A b^2+8 a^4 A+40 a^3 b B+60 a b^3 B+15 A b^4\right ) \tan (c+d x)}{15 d}+\frac{\left (12 a^3 A b+24 a^2 b^2 B+3 a^4 B+16 a A b^3+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac{a \left (60 a^2 A b+15 a^3 B+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{40 d}+\frac{a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d}+\frac{a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d} \]

[Out]

((12*a^3*A*b + 16*a*A*b^3 + 3*a^4*B + 24*a^2*b^2*B + 8*b^4*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((8*a^4*A + 60*a^
2*A*b^2 + 15*A*b^4 + 40*a^3*b*B + 60*a*b^3*B)*Tan[c + d*x])/(15*d) + (a*(60*a^2*A*b + 56*A*b^3 + 15*a^3*B + 11
0*a*b^2*B)*Sec[c + d*x]*Tan[c + d*x])/(40*d) + (a^2*(8*a^2*A + 18*A*b^2 + 25*a*b*B)*Sec[c + d*x]^2*Tan[c + d*x
])/(30*d) + (a*(8*A*b + 5*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (a*A*(a + b*Cos[c
+ d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.722923, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {2989, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac{\left (60 a^2 A b^2+8 a^4 A+40 a^3 b B+60 a b^3 B+15 A b^4\right ) \tan (c+d x)}{15 d}+\frac{\left (12 a^3 A b+24 a^2 b^2 B+3 a^4 B+16 a A b^3+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac{a \left (60 a^2 A b+15 a^3 B+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{40 d}+\frac{a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d}+\frac{a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]

[Out]

((12*a^3*A*b + 16*a*A*b^3 + 3*a^4*B + 24*a^2*b^2*B + 8*b^4*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((8*a^4*A + 60*a^
2*A*b^2 + 15*A*b^4 + 40*a^3*b*B + 60*a*b^3*B)*Tan[c + d*x])/(15*d) + (a*(60*a^2*A*b + 56*A*b^3 + 15*a^3*B + 11
0*a*b^2*B)*Sec[c + d*x]*Tan[c + d*x])/(40*d) + (a^2*(8*a^2*A + 18*A*b^2 + 25*a*b*B)*Sec[c + d*x]^2*Tan[c + d*x
])/(30*d) + (a*(8*A*b + 5*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (a*A*(a + b*Cos[c
+ d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)

Rule 2989

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*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(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 - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && GtQ[m, 1] && 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 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+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx &=\frac{a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cos (c+d x))^2 \left (a (8 A b+5 a B)+\left (4 a^2 A+5 A b^2+10 a b B\right ) \cos (c+d x)+b (a A+5 b B) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{20} \int (a+b \cos (c+d x)) \left (2 a \left (8 a^2 A+18 A b^2+25 a b B\right )+\left (44 a^2 A b+20 A b^3+15 a^3 B+60 a b^2 B\right ) \cos (c+d x)+b \left (12 a A b+5 a^2 B+20 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{60} \int \left (-3 a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right )-4 \left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \cos (c+d x)-3 b^2 \left (12 a A b+5 a^2 B+20 b^2 B\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{120} \int \left (-8 \left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right )-15 \left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{15} \left (-8 a^4 A-60 a^2 A b^2-15 A b^4-40 a^3 b B-60 a b^3 B\right ) \int \sec ^2(c+d x) \, dx-\frac{1}{8} \left (-12 a^3 A b-16 a A b^3-3 a^4 B-24 a^2 b^2 B-8 b^4 B\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{\left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{\left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \tan (c+d x)}{15 d}+\frac{a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac{a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 4.15113, size = 198, normalized size = 0.74 \[ \frac{15 \left (12 a^3 A b+24 a^2 b^2 B+3 a^4 B+16 a A b^3+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (80 a^2 \left (a^2 A+2 a b B+3 A b^2\right ) \tan ^2(c+d x)+15 a \left (12 a^2 A b+3 a^3 B+24 a b^2 B+16 A b^3\right ) \sec (c+d x)+120 \left (6 a^2 A b^2+a^4 A+4 a^3 b B+4 a b^3 B+A b^4\right )+30 a^3 (a B+4 A b) \sec ^3(c+d x)+24 a^4 A \tan ^4(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]

[Out]

(15*(12*a^3*A*b + 16*a*A*b^3 + 3*a^4*B + 24*a^2*b^2*B + 8*b^4*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(120*(a^
4*A + 6*a^2*A*b^2 + A*b^4 + 4*a^3*b*B + 4*a*b^3*B) + 15*a*(12*a^2*A*b + 16*A*b^3 + 3*a^3*B + 24*a*b^2*B)*Sec[c
+ d*x] + 30*a^3*(4*A*b + a*B)*Sec[c + d*x]^3 + 80*a^2*(a^2*A + 3*A*b^2 + 2*a*b*B)*Tan[c + d*x]^2 + 24*a^4*A*T
an[c + d*x]^4))/(120*d)

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Maple [A]  time = 0.116, size = 431, normalized size = 1.6 \begin{align*}{\frac{8\,A{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{a}^{4}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{a}^{4}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{4}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{A{a}^{3}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,A{a}^{3}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,A{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{8\,B{a}^{3}b\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,B{a}^{3}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+4\,{\frac{A{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+3\,{\frac{B{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+3\,{\frac{B{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Aa{b}^{3}\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+2\,{\frac{Aa{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Ba{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{A{b}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{B{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^6,x)

[Out]

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

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Maxima [A]  time = 1.05269, size = 521, normalized size = 1.95 \begin{align*} \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^{4} + 320 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} b + 480 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b^{2} - 15 \, B a^{4}{\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} 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 )} - 360 \, B a^{2} 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 )} - 240 \, A a 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 )} + 120 \, B b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 960 \, B a b^{3} \tan \left (d x + c\right ) + 240 \, A b^{4} \tan \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(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^4 + 320*(tan(d*x + c)^3 + 3*tan(d*x + c
))*B*a^3*b + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2*b^2 - 15*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)) - 60*A*a^3*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)) - 360*B*a^2*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x
+ c) - 1)) - 240*A*a*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))
+ 120*B*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 960*B*a*b^3*tan(d*x + c) + 240*A*b^4*tan(d*x +
c))/d

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Fricas [A]  time = 1.57572, size = 687, normalized size = 2.57 \begin{align*} \frac{15 \,{\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 8 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 8 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, A a^{4} + 8 \,{\left (8 \, A a^{4} + 40 \, B a^{3} b + 60 \, A a^{2} b^{2} + 60 \, B a b^{3} + 15 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \,{\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 16 \,{\left (2 \, A a^{4} + 10 \, B a^{3} b + 15 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*cos(d*x + c)^5*log(sin(d*x + c) + 1) -
15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(24*
A*a^4 + 8*(8*A*a^4 + 40*B*a^3*b + 60*A*a^2*b^2 + 60*B*a*b^3 + 15*A*b^4)*cos(d*x + c)^4 + 15*(3*B*a^4 + 12*A*a^
3*b + 24*B*a^2*b^2 + 16*A*a*b^3)*cos(d*x + c)^3 + 16*(2*A*a^4 + 10*B*a^3*b + 15*A*a^2*b^2)*cos(d*x + c)^2 + 30
*(B*a^4 + 4*A*a^3*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)

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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))**4*(A+B*cos(d*x+c))*sec(d*x+c)**6,x)

[Out]

Timed out

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Giac [B]  time = 1.70156, size = 1148, normalized size = 4.3 \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))^4*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15
*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^
4*tan(1/2*d*x + 1/2*c)^9 - 75*B*a^4*tan(1/2*d*x + 1/2*c)^9 - 300*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 480*B*a^3*b*
tan(1/2*d*x + 1/2*c)^9 + 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 360*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 240*A*a
*b^3*tan(1/2*d*x + 1/2*c)^9 + 480*B*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*A*b^4*tan(1/2*d*x + 1/2*c)^9 - 160*A*a^
4*tan(1/2*d*x + 1/2*c)^7 + 30*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 1280*B*a^3*b
*tan(1/2*d*x + 1/2*c)^7 - 1920*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 480*A
*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 1920*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 480*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 464*A
*a^4*tan(1/2*d*x + 1/2*c)^5 + 1600*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 2400*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 28
80*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 720*A*b^4*tan(1/2*d*x + 1/2*c)^5 - 160*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 30*B
*a^4*tan(1/2*d*x + 1/2*c)^3 - 120*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1280*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1920*
A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 720*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 480*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 -
1920*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 480*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^4*tan(1/2*d*x + 1/2*c) + 75*
B*a^4*tan(1/2*d*x + 1/2*c) + 300*A*a^3*b*tan(1/2*d*x + 1/2*c) + 480*B*a^3*b*tan(1/2*d*x + 1/2*c) + 720*A*a^2*b
^2*tan(1/2*d*x + 1/2*c) + 360*B*a^2*b^2*tan(1/2*d*x + 1/2*c) + 240*A*a*b^3*tan(1/2*d*x + 1/2*c) + 480*B*a*b^3*
tan(1/2*d*x + 1/2*c) + 120*A*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d