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

Optimal. Leaf size=165 \[ \frac {\tan ^3(c+d x) (4 a A+5 a C+5 b B)}{15 d}+\frac {\tan (c+d x) (4 a A+5 a C+5 b B)}{5 d}+\frac {(3 a B+3 A b+4 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec (c+d x) (3 a B+3 A b+4 b C)}{8 d}+\frac {(a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

[Out]

1/8*(3*A*b+3*B*a+4*C*b)*arctanh(sin(d*x+c))/d+1/5*(4*A*a+5*B*b+5*C*a)*tan(d*x+c)/d+1/8*(3*A*b+3*B*a+4*C*b)*sec
(d*x+c)*tan(d*x+c)/d+1/4*(A*b+B*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a*A*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(4*A*a+5*B
*b+5*C*a)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.26, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3031, 3021, 2748, 3767, 3768, 3770} \[ \frac {\tan ^3(c+d x) (4 a A+5 a C+5 b B)}{15 d}+\frac {\tan (c+d x) (4 a A+5 a C+5 b B)}{5 d}+\frac {(3 a B+3 A b+4 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec (c+d x) (3 a B+3 A b+4 b C)}{8 d}+\frac {(a B+A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A*b + 3*a*B + 4*b*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*a*A + 5*b*B + 5*a*C)*Tan[c + d*x])/(5*d) + ((3*A*b
+ 3*a*B + 4*b*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + ((A*b + a*B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*A*Se
c[c + d*x]^4*Tan[c + d*x])/(5*d) + ((4*a*A + 5*b*B + 5*a*C)*Tan[c + d*x]^3)/(15*d)

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 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 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 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+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{5} \int \left (-5 (A b+a B)-(4 a A+5 b B+5 a C) \cos (c+d x)-5 b C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {(A b+a B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{20} \int (-4 (4 a A+5 b B+5 a C)-5 (3 A b+3 a B+4 b C) \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac {(A b+a B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{5} (-4 a A-5 b B-5 a C) \int \sec ^4(c+d x) \, dx-\frac {1}{4} (-3 A b-3 a B-4 b C) \int \sec ^3(c+d x) \, dx\\ &=\frac {(3 A b+3 a B+4 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(A b+a B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{8} (-3 A b-3 a B-4 b C) \int \sec (c+d x) \, dx-\frac {(4 a A+5 b B+5 a C) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {(3 A b+3 a B+4 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 a A+5 b B+5 a C) \tan (c+d x)}{5 d}+\frac {(3 A b+3 a B+4 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(A b+a B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(4 a A+5 b B+5 a C) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 1.37, size = 123, normalized size = 0.75 \[ \frac {15 (3 a B+3 A b+4 b C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 \tan ^2(c+d x) (a (2 A+C)+b B)+15 (a (A+C)+b B)+3 a A \tan ^4(c+d x)\right )+15 \sec (c+d x) (3 a B+3 A b+4 b C)+30 (a B+A b) \sec ^3(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(3*A*b + 3*a*B + 4*b*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(3*A*b + 3*a*B + 4*b*C)*Sec[c + d*x] + 30
*(A*b + a*B)*Sec[c + d*x]^3 + 8*(15*(b*B + a*(A + C)) + 5*(b*B + a*(2*A + C))*Tan[c + d*x]^2 + 3*a*A*Tan[c + d
*x]^4)))/(120*d)

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fricas [A]  time = 0.44, size = 182, normalized size = 1.10 \[ \frac {15 \, {\left (3 \, B a + {\left (3 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a + {\left (3 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left ({\left (4 \, A + 5 \, C\right )} a + 5 \, B b\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, B a + {\left (3 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a + 5 \, B b\right )} \cos \left (d x + c\right )^{2} + 24 \, A a + 30 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.37, size = 473, normalized size = 2.87 \[ \frac {15 \, {\left (3 \, B a + 3 \, A b + 4 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, B a + 3 \, A b + 4 \, C b\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 \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 320 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b \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+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(3*B*a + 3*A*b + 4*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B*a + 3*A*b + 4*C*b)*log(abs(tan(
1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a*tan(1/2*d*x + 1/2*c)^9 - 75*B*a*tan(1/2*d*x + 1/2*c)^9 + 120*C*a*tan(1/2*d
*x + 1/2*c)^9 - 75*A*b*tan(1/2*d*x + 1/2*c)^9 + 120*B*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*b*tan(1/2*d*x + 1/2*c)^9
- 160*A*a*tan(1/2*d*x + 1/2*c)^7 + 30*B*a*tan(1/2*d*x + 1/2*c)^7 - 320*C*a*tan(1/2*d*x + 1/2*c)^7 + 30*A*b*ta
n(1/2*d*x + 1/2*c)^7 - 320*B*b*tan(1/2*d*x + 1/2*c)^7 + 120*C*b*tan(1/2*d*x + 1/2*c)^7 + 464*A*a*tan(1/2*d*x +
1/2*c)^5 + 400*C*a*tan(1/2*d*x + 1/2*c)^5 + 400*B*b*tan(1/2*d*x + 1/2*c)^5 - 160*A*a*tan(1/2*d*x + 1/2*c)^3 -
30*B*a*tan(1/2*d*x + 1/2*c)^3 - 320*C*a*tan(1/2*d*x + 1/2*c)^3 - 30*A*b*tan(1/2*d*x + 1/2*c)^3 - 320*B*b*tan(
1/2*d*x + 1/2*c)^3 - 120*C*b*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*tan(1/2*d*x + 1/2*c) + 75*B*a*tan(1/2*d*x + 1/2*
c) + 120*C*a*tan(1/2*d*x + 1/2*c) + 75*A*b*tan(1/2*d*x + 1/2*c) + 120*B*b*tan(1/2*d*x + 1/2*c) + 60*C*b*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 = 0.41, size = 287, normalized size = 1.74 \[ \frac {8 a A \tan \left (d x +c \right )}{15 d}+\frac {a A \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 a A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d}+\frac {a B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a C \tan \left (d x +c \right )}{3 d}+\frac {a C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {A b \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 B b \tan \left (d x +c \right )}{3 d}+\frac {B b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {C b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 266, normalized size = 1.61 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b - 15 \, B a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, A b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*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 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*
C*a + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*b - 15*B*a*(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*A*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)) -
60*C*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 5.30, size = 302, normalized size = 1.83 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,b}{8}+\frac {3\,B\,a}{8}+\frac {C\,b}{2}\right )}{\frac {3\,A\,b}{2}+\frac {3\,B\,a}{2}+2\,C\,b}\right )\,\left (\frac {3\,A\,b}{4}+\frac {3\,B\,a}{4}+C\,b\right )}{d}-\frac {\left (2\,A\,a-\frac {5\,A\,b}{4}-\frac {5\,B\,a}{4}+2\,B\,b+2\,C\,a-C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {A\,b}{2}-\frac {8\,A\,a}{3}+\frac {B\,a}{2}-\frac {16\,B\,b}{3}-\frac {16\,C\,a}{3}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,B\,b}{3}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a}{3}-\frac {A\,b}{2}-\frac {B\,a}{2}-\frac {16\,B\,b}{3}-\frac {16\,C\,a}{3}-2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+\frac {5\,A\,b}{4}+\frac {5\,B\,a}{4}+2\,B\,b+2\,C\,a+C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((3*A*b)/8 + (3*B*a)/8 + (C*b)/2))/((3*A*b)/2 + (3*B*a)/2 + 2*C*b))*((3*A*b)/4 +
(3*B*a)/4 + C*b))/d - (tan(c/2 + (d*x)/2)^9*(2*A*a - (5*A*b)/4 - (5*B*a)/4 + 2*B*b + 2*C*a - C*b) - tan(c/2 +
(d*x)/2)^3*((8*A*a)/3 + (A*b)/2 + (B*a)/2 + (16*B*b)/3 + (16*C*a)/3 + 2*C*b) - tan(c/2 + (d*x)/2)^7*((8*A*a)/3
- (A*b)/2 - (B*a)/2 + (16*B*b)/3 + (16*C*a)/3 - 2*C*b) + tan(c/2 + (d*x)/2)*(2*A*a + (5*A*b)/4 + (5*B*a)/4 +
2*B*b + 2*C*a + C*b) + tan(c/2 + (d*x)/2)^5*((116*A*a)/15 + (20*B*b)/3 + (20*C*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+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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