3.3.48 \(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\) [248]

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

[Out]

1/8*(12*A*a^3*b+16*A*a*b^3+3*B*a^4+24*B*a^2*b^2+8*B*b^4)*arctanh(sin(d*x+c))/d+1/15*(8*A*a^4+60*A*a^2*b^2+15*A
*b^4+40*B*a^3*b+60*B*a*b^3)*tan(d*x+c)/d+1/40*a*(60*A*a^2*b+56*A*b^3+15*B*a^3+110*B*a*b^2)*sec(d*x+c)*tan(d*x+
c)/d+1/30*a^2*(8*A*a^2+18*A*b^2+25*B*a*b)*sec(d*x+c)^2*tan(d*x+c)/d+1/20*a*(8*A*b+5*B*a)*(a+b*cos(d*x+c))^2*se
c(d*x+c)^3*tan(d*x+c)/d+1/5*a*A*(a+b*cos(d*x+c))^3*sec(d*x+c)^4*tan(d*x+c)/d

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Rubi [A]
time = 0.47, antiderivative size = 267, normalized size of antiderivative = 1.00, 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 = {3068, 3126, 3110, 3100, 2827, 3852, 8, 3855} \begin {gather*} \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 (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {\left (8 a^4 A+40 a^3 b B+60 a^2 A b^2+60 a b^3 B+15 A b^4\right ) \tan (c+d x)}{15 d}+\frac {\left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 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} \end {gather*}

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

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 3068

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*Si
n[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 3100

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 3110

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] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3126

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 3852

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 3855

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 ) \text {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*}

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Mathematica [A]
time = 4.38, size = 198, normalized size = 0.74 \begin {gather*} \frac {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 ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (120 \left (a^4 A+6 a^2 A b^2+A b^4+4 a^3 b B+4 a b^3 B\right )+15 a \left (12 a^2 A b+16 A b^3+3 a^3 B+24 a b^2 B\right ) \sec (c+d x)+30 a^3 (4 A b+a B) \sec ^3(c+d x)+80 a^2 \left (a^2 A+3 A b^2+2 a b B\right ) \tan ^2(c+d x)+24 a^4 A \tan ^4(c+d x)\right )}{120 d} \end {gather*}

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 = 0.26, size = 313, normalized size = 1.17 Too large to display

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,method=_RETURNVERBOSE)

[Out]

1/d*(-A*a^4*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+a^4*B*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*t
an(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+4*A*a^3*b*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(
d*x+c)+tan(d*x+c)))-4*B*a^3*b*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)-6*A*a^2*b^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c
)+6*B*a^2*b^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+4*A*a*b^3*(1/2*sec(d*x+c)*tan(d*x+c)+1
/2*ln(sec(d*x+c)+tan(d*x+c)))+4*B*a*b^3*tan(d*x+c)+A*b^4*tan(d*x+c)+B*b^4*ln(sec(d*x+c)+tan(d*x+c)))

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Maxima [A]
time = 0.30, size = 386, normalized size = 1.45 \begin {gather*} \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 {gather*}

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 = 0.37, size = 281, normalized size = 1.05 \begin {gather*} \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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 850 vs. \(2 (255) = 510\).
time = 0.52, size = 850, normalized size = 3.18 \begin {gather*} \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 )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\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 )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 300 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 480 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 720 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 240 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 480 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1280 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1920 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 720 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 480 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1920 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 480 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1600 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2400 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2880 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 720 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1280 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1920 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 720 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 480 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1920 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 480 \, A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 300 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 480 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 720 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 240 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 480 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, A b^{4} \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} \end {gather*}

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

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Mupad [B]
time = 3.88, size = 555, normalized size = 2.08 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^4}{8}+\frac {3\,A\,a^3\,b}{2}+3\,B\,a^2\,b^2+2\,A\,a\,b^3+B\,b^4\right )}{\frac {3\,B\,a^4}{2}+6\,A\,a^3\,b+12\,B\,a^2\,b^2+8\,A\,a\,b^3+4\,B\,b^4}\right )\,\left (\frac {3\,B\,a^4}{4}+3\,A\,a^3\,b+6\,B\,a^2\,b^2+4\,A\,a\,b^3+2\,B\,b^4\right )}{d}-\frac {\left (2\,A\,a^4+2\,A\,b^4-\frac {5\,B\,a^4}{4}+12\,A\,a^2\,b^2-6\,B\,a^2\,b^2-4\,A\,a\,b^3-5\,A\,a^3\,b+8\,B\,a\,b^3+8\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,a^4}{2}-8\,A\,b^4-\frac {8\,A\,a^4}{3}-32\,A\,a^2\,b^2+12\,B\,a^2\,b^2+8\,A\,a\,b^3+2\,A\,a^3\,b-32\,B\,a\,b^3-\frac {64\,B\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^4}{15}+\frac {80\,B\,a^3\,b}{3}+40\,A\,a^2\,b^2+48\,B\,a\,b^3+12\,A\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^4}{3}-8\,A\,b^4-\frac {B\,a^4}{2}-32\,A\,a^2\,b^2-12\,B\,a^2\,b^2-8\,A\,a\,b^3-2\,A\,a^3\,b-32\,B\,a\,b^3-\frac {64\,B\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^4+2\,A\,b^4+\frac {5\,B\,a^4}{4}+12\,A\,a^2\,b^2+6\,B\,a^2\,b^2+4\,A\,a\,b^3+5\,A\,a^3\,b+8\,B\,a\,b^3+8\,B\,a^3\,b\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((3*B*a^4)/8 + B*b^4 + 3*B*a^2*b^2 + 2*A*a*b^3 + (3*A*a^3*b)/2))/((3*B*a^4)/2 + 4
*B*b^4 + 12*B*a^2*b^2 + 8*A*a*b^3 + 6*A*a^3*b))*((3*B*a^4)/4 + 2*B*b^4 + 6*B*a^2*b^2 + 4*A*a*b^3 + 3*A*a^3*b))
/d - (tan(c/2 + (d*x)/2)*(2*A*a^4 + 2*A*b^4 + (5*B*a^4)/4 + 12*A*a^2*b^2 + 6*B*a^2*b^2 + 4*A*a*b^3 + 5*A*a^3*b
+ 8*B*a*b^3 + 8*B*a^3*b) + tan(c/2 + (d*x)/2)^5*((116*A*a^4)/15 + 12*A*b^4 + 40*A*a^2*b^2 + 48*B*a*b^3 + (80*
B*a^3*b)/3) + tan(c/2 + (d*x)/2)^9*(2*A*a^4 + 2*A*b^4 - (5*B*a^4)/4 + 12*A*a^2*b^2 - 6*B*a^2*b^2 - 4*A*a*b^3 -
5*A*a^3*b + 8*B*a*b^3 + 8*B*a^3*b) - tan(c/2 + (d*x)/2)^3*((8*A*a^4)/3 + 8*A*b^4 + (B*a^4)/2 + 32*A*a^2*b^2 +
12*B*a^2*b^2 + 8*A*a*b^3 + 2*A*a^3*b + 32*B*a*b^3 + (64*B*a^3*b)/3) - tan(c/2 + (d*x)/2)^7*((8*A*a^4)/3 + 8*A
*b^4 - (B*a^4)/2 + 32*A*a^2*b^2 - 12*B*a^2*b^2 - 8*A*a*b^3 - 2*A*a^3*b + 32*B*a*b^3 + (64*B*a^3*b)/3))/(d*(5*t
an(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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