3.556 \(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=250 \[ \frac {a b \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \tan (c+d x) \sec (c+d x)}{30 d}+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{15 d}+\frac {\left (2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)+6 A b^4\right ) \tan (c+d x)}{15 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^4}{5 d}+\frac {A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{5 d}+b^4 C x \]

[Out]

b^4*C*x+1/2*a*b*(4*b^2*(A+2*C)+a^2*(3*A+4*C))*arctanh(sin(d*x+c))/d+1/15*(6*A*b^4+2*a^4*(4*A+5*C)+a^2*b^2*(56*
A+85*C))*tan(d*x+c)/d+1/30*a*b*(6*A*b^2+a^2*(29*A+40*C))*sec(d*x+c)*tan(d*x+c)/d+1/15*(3*A*b^2+a^2*(4*A+5*C))*
(a+b*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d+1/5*A*b*(a+b*cos(d*x+c))^3*sec(d*x+c)^3*tan(d*x+c)/d+1/5*A*(a+b*c
os(d*x+c))^4*sec(d*x+c)^4*tan(d*x+c)/d

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Rubi [A]  time = 0.90, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3048, 3047, 3031, 3021, 2735, 3770} \[ \frac {\left (a^2 b^2 (56 A+85 C)+2 a^4 (4 A+5 C)+6 A b^4\right ) \tan (c+d x)}{15 d}+\frac {a b \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a b \left (a^2 (29 A+40 C)+6 A b^2\right ) \tan (c+d x) \sec (c+d x)}{30 d}+\frac {\left (a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{15 d}+\frac {A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{5 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^4}{5 d}+b^4 C x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

b^4*C*x + (a*b*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(2*d) + ((6*A*b^4 + 2*a^4*(4*A + 5*C
) + a^2*b^2*(56*A + 85*C))*Tan[c + d*x])/(15*d) + (a*b*(6*A*b^2 + a^2*(29*A + 40*C))*Sec[c + d*x]*Tan[c + d*x]
)/(30*d) + ((3*A*b^2 + a^2*(4*A + 5*C))*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (A*b*(a +
b*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(5*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^4*Tan[c + d*x])
/(5*d)

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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 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 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 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 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 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+b \cos (c+d x))^3 \left (4 A b+a (4 A+5 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 \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int (a+b \cos (c+d x))^2 \left (4 \left (3 A b^2+a^2 (4 A+5 C)\right )+4 a b (7 A+10 C) \cos (c+d x)+20 b^2 C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{60} \int (a+b \cos (c+d x)) \left (4 b \left (6 A b^2+a^2 (29 A+40 C)\right )+4 a \left (9 b^2 (3 A+5 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)+60 b^3 C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{120} \int \left (-8 \left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right )-60 a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \cos (c+d x)-120 b^4 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{120} \int \left (-60 a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right )-120 b^4 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^4 C x+\frac {\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} \left (a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=b^4 C x+\frac {a b \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {\left (6 A b^4+2 a^4 (4 A+5 C)+a^2 b^2 (56 A+85 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (6 A b^2+a^2 (29 A+40 C)\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 A b^2+a^2 (4 A+5 C)\right ) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {A b (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{5 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 1.11, size = 169, normalized size = 0.68 \[ \frac {6 a^4 A \tan ^5(c+d x)+10 a^2 \left (a^2 (2 A+C)+6 A b^2\right ) \tan ^3(c+d x)+15 a b \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+15 \tan (c+d x) \left (2 a^3 A b \sec ^3(c+d x)+a b \left (a^2 (3 A+4 C)+4 A b^2\right ) \sec (c+d x)+2 \left (a^4 (A+C)+6 a^2 b^2 (A+C)+A b^4\right )\right )+30 b^4 C d x}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(30*b^4*C*d*x + 15*a*b*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + 15*(2*(A*b^4 + a^4*(A + C)
+ 6*a^2*b^2*(A + C)) + a*b*(4*A*b^2 + a^2*(3*A + 4*C))*Sec[c + d*x] + 2*a^3*A*b*Sec[c + d*x]^3)*Tan[c + d*x] +
10*a^2*(6*A*b^2 + a^2*(2*A + C))*Tan[c + d*x]^3 + 6*a^4*A*Tan[c + d*x]^5)/(30*d)

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fricas [A]  time = 1.65, size = 251, normalized size = 1.00 \[ \frac {60 \, C b^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, {\left (A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, {\left (A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (30 \, A a^{3} b \cos \left (d x + c\right ) + 6 \, A a^{4} + 2 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{4} + 30 \, {\left (2 \, A + 3 \, C\right )} a^{2} b^{2} + 15 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{3} b + 4 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{4} + 30 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/60*(60*C*b^4*d*x*cos(d*x + c)^5 + 15*((3*A + 4*C)*a^3*b + 4*(A + 2*C)*a*b^3)*cos(d*x + c)^5*log(sin(d*x + c)
+ 1) - 15*((3*A + 4*C)*a^3*b + 4*(A + 2*C)*a*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(30*A*a^3*b*cos(d
*x + c) + 6*A*a^4 + 2*(2*(4*A + 5*C)*a^4 + 30*(2*A + 3*C)*a^2*b^2 + 15*A*b^4)*cos(d*x + c)^4 + 15*((3*A + 4*C)
*a^3*b + 4*A*a*b^3)*cos(d*x + c)^3 + 2*((4*A + 5*C)*a^4 + 30*A*a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d
*x + c)^5)

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giac [B]  time = 0.58, size = 778, normalized size = 3.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/30*(30*(d*x + c)*C*b^4 + 15*(3*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 8*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1
)) - 15*(3*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 8*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(30*A*a^4*tan(1
/2*d*x + 1/2*c)^9 + 30*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 75*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*a^3*b*tan(1/2*d
*x + 1/2*c)^9 + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 180*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*a*b^3*tan(1
/2*d*x + 1/2*c)^9 + 30*A*b^4*tan(1/2*d*x + 1/2*c)^9 - 40*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 80*C*a^4*tan(1/2*d*x +
1/2*c)^7 + 30*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 480*A*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^7 - 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 120*A*b^4*tan(1/2*d*
x + 1/2*c)^7 + 116*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 100*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 600*A*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^5 + 1080*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 180*A*b^4*tan(1/2*d*x + 1/2*c)^5 - 40*A*a^4*tan(1/2*d*x
+ 1/2*c)^3 - 80*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 30*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 120*C*a^3*b*tan(1/2*d*x + 1
/2*c)^3 - 480*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 120*A*a*b^3*tan(1/2*d*
x + 1/2*c)^3 - 120*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 30*A*a^4*tan(1/2*d*x + 1/2*c) + 30*C*a^4*tan(1/2*d*x + 1/2*c
) + 75*A*a^3*b*tan(1/2*d*x + 1/2*c) + 60*C*a^3*b*tan(1/2*d*x + 1/2*c) + 180*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 1
80*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 60*A*a*b^3*tan(1/2*d*x + 1/2*c) + 30*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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maple [A]  time = 0.45, size = 377, normalized size = 1.51 \[ \frac {8 A \,a^{4} \tan \left (d x +c \right )}{15 d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {2 a^{4} C \tan \left (d x +c \right )}{3 d}+\frac {a^{4} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {A \,a^{3} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{d}+\frac {3 A \,a^{3} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a^{3} b C \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {2 a^{3} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 A \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {2 A \,a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {6 C \,a^{2} b^{2} \tan \left (d x +c \right )}{d}+\frac {2 a A \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {2 a A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 C a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{4} \tan \left (d x +c \right )}{d}+b^{4} C x +\frac {C \,b^{4} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*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+2/3/d*a^4*C*t
an(d*x+c)+1/3/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+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))+2/d*a^3*b*C*tan(d*x+c)*sec(d*x+c)+2/d*a^3*b*C*ln(sec(d*x+c)+ta
n(d*x+c))+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+6/d*C*a^2*b^2*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*C*a*b^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*b^4*ta
n(d*x+c)+b^4*C*x+1/d*C*b^4*c

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maxima [A]  time = 0.35, size = 325, normalized size = 1.30 \[ \frac {4 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 20 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 120 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b^{2} + 60 \, {\left (d x + c\right )} C b^{4} - 15 \, A a^{3} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{3} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 360 \, C a^{2} b^{2} \tan \left (d x + c\right ) + 60 \, A b^{4} \tan \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/60*(4*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 20*(tan(d*x + c)^3 + 3*tan(d*x + c))*
C*a^4 + 120*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2*b^2 + 60*(d*x + c)*C*b^4 - 15*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)
) - 60*C*a^3*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*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*C*a*b^3*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 360*C*a^2*b^2*tan(d*x + c) + 60*A*b^4*tan(d*x + c))/d

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mupad [B]  time = 3.96, size = 1738, normalized size = 6.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^4)/cos(c + d*x)^6,x)

[Out]

((A*a^4*sin(3*c + 3*d*x))/6 + (A*a^4*sin(5*c + 5*d*x))/30 + (3*A*b^4*sin(3*c + 3*d*x))/16 + (A*b^4*sin(5*c + 5
*d*x))/16 + (5*C*a^4*sin(3*c + 3*d*x))/24 + (C*a^4*sin(5*c + 5*d*x))/24 + (A*a^4*sin(c + d*x))/3 + (A*b^4*sin(
c + d*x))/8 + (C*a^4*sin(c + d*x))/6 + (5*C*b^4*cos(c + d*x)*atan((9*A^2*a^6*sin(c/2 + (d*x)/2) + 16*C^2*a^6*s
in(c/2 + (d*x)/2) + 4*C^2*b^6*sin(c/2 + (d*x)/2) + 16*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 24*A^2*a^4*b^2*sin(c/2
+ (d*x)/2) + 64*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 64*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*a^6*sin(c/2 + (d*x
)/2) + 64*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 80*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(9*A^2*a^6 +
16*C^2*a^6 + 4*C^2*b^6 + 16*A^2*a^2*b^4 + 24*A^2*a^4*b^2 + 64*C^2*a^2*b^4 + 64*C^2*a^4*b^2 + 24*A*C*a^6 + 64*
A*C*a^2*b^4 + 80*A*C*a^4*b^2))))/4 + (A*a*b^3*sin(2*c + 2*d*x))/2 + (7*A*a^3*b*sin(2*c + 2*d*x))/8 + (A*a*b^3*
sin(4*c + 4*d*x))/4 + (3*A*a^3*b*sin(4*c + 4*d*x))/16 + A*a^2*b^2*sin(c + d*x) + (C*a^3*b*sin(2*c + 2*d*x))/2
+ (C*a^3*b*sin(4*c + 4*d*x))/4 + (3*C*a^2*b^2*sin(c + d*x))/4 + (5*C*b^4*atan((9*A^2*a^6*sin(c/2 + (d*x)/2) +
16*C^2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*b^6*sin(c/2 + (d*x)/2) + 16*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 24*A^2*a^4*
b^2*sin(c/2 + (d*x)/2) + 64*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 64*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*a^6*si
n(c/2 + (d*x)/2) + 64*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 80*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*
(9*A^2*a^6 + 16*C^2*a^6 + 4*C^2*b^6 + 16*A^2*a^2*b^4 + 24*A^2*a^4*b^2 + 64*C^2*a^2*b^4 + 64*C^2*a^4*b^2 + 24*A
*C*a^6 + 64*A*C*a^2*b^4 + 80*A*C*a^4*b^2)))*cos(3*c + 3*d*x))/8 + (C*b^4*atan((9*A^2*a^6*sin(c/2 + (d*x)/2) +
16*C^2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*b^6*sin(c/2 + (d*x)/2) + 16*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 24*A^2*a^4*
b^2*sin(c/2 + (d*x)/2) + 64*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 64*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*a^6*si
n(c/2 + (d*x)/2) + 64*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 80*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*
(9*A^2*a^6 + 16*C^2*a^6 + 4*C^2*b^6 + 16*A^2*a^2*b^4 + 24*A^2*a^4*b^2 + 64*C^2*a^2*b^4 + 64*C^2*a^4*b^2 + 24*A
*C*a^6 + 64*A*C*a^2*b^4 + 80*A*C*a^4*b^2)))*cos(5*c + 5*d*x))/8 + (5*A*a^2*b^2*sin(3*c + 3*d*x))/4 + (A*a^2*b^
2*sin(5*c + 5*d*x))/4 + (9*C*a^2*b^2*sin(3*c + 3*d*x))/8 + (3*C*a^2*b^2*sin(5*c + 5*d*x))/8 + (5*A*a*b^3*atanh
(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + (15*A*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2))*cos(3*c + 3*d*x))/16 + (A*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/4 + (3
*A*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/16 + (5*C*a*b^3*atanh(sin(c/2 + (d*x)/
2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2 + (5*C*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c +
3*d*x))/4 + (C*a*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/2 + (C*a^3*b*atanh(sin(c/
2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x))/4 + (5*A*a*b^3*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/
2 + (d*x)/2)))/2 + (15*A*a^3*b*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + 5*C*a*b^3*cos(c
+ d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (5*C*a^3*b*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2)))/2)/(d*((5*cos(c + d*x))/8 + (5*cos(3*c + 3*d*x))/16 + cos(5*c + 5*d*x)/16))

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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))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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