\(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [554]

Optimal result

Integrand size = 33, antiderivative size = 251 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {1}{2} b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) x+\frac {2 a b \left (2 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{3 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

1/2*b^2*(2*A*b^2+(12*a^2+b^2)*C)*x+2*a*b*(2*A*b^2+a^2*(A+2*C))*arctanh(sin(d*x+c))/d-2/3*a*b*(b^2*(11*A-6*C)+a
^2*(2*A+3*C))*sin(d*x+c)/d-1/6*b^2*(3*b^2*(6*A-C)+a^2*(4*A+6*C))*cos(d*x+c)*sin(d*x+c)/d+1/3*(6*A*b^2+a^2*(2*A
+3*C))*(a+b*cos(d*x+c))^2*tan(d*x+c)/d+2/3*A*b*(a+b*cos(d*x+c))^3*sec(d*x+c)*tan(d*x+c)/d+1/3*A*(a+b*cos(d*x+c
))^4*sec(d*x+c)^2*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3127, 3126, 3112, 3102, 2814, 3855} \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2 a b \left (a^2 (A+2 C)+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \left (a^2 (2 A+3 C)+b^2 (11 A-6 C)\right ) \sin (c+d x)}{3 d}-\frac {b^2 \left (a^2 (4 A+6 C)+3 b^2 (6 A-C)\right ) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {\left (a^2 (2 A+3 C)+6 A b^2\right ) \tan (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {1}{2} b^2 x \left (C \left (12 a^2+b^2\right )+2 A b^2\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}+\frac {2 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{3 d} \]

[In]

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

[Out]

(b^2*(2*A*b^2 + (12*a^2 + b^2)*C)*x)/2 + (2*a*b*(2*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/d - (2*a*b*(b
^2*(11*A - 6*C) + a^2*(2*A + 3*C))*Sin[c + d*x])/(3*d) - (b^2*(3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Cos[c + d*x]
*Sin[c + d*x])/(6*d) + ((6*A*b^2 + a^2*(2*A + 3*C))*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/(3*d) + (2*A*b*(a + b
*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(3*d) + (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^2*Tan[c + d*x])/(3*
d)

Rule 2814

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 3102

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

Rule 3112

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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a +
b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*
c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 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 3127

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*Si
n[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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+b \cos (c+d x))^3 \left (4 A b+a (2 A+3 C) \cos (c+d x)-b (2 A-3 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {2 A b (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int (a+b \cos (c+d x))^2 \left (2 \left (6 A b^2+\frac {1}{2} a^2 (4 A+6 C)\right )+4 a b (A+3 C) \cos (c+d x)-6 b^2 (2 A-C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{3 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int (a+b \cos (c+d x)) \left (12 b \left (2 A b^2+a^2 (A+2 C)\right )-2 a b^2 (4 A-9 C) \cos (c+d x)-2 b \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{3 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{12} \int \left (24 a b \left (2 A b^2+a^2 (A+2 C)\right )+6 b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \cos (c+d x)-8 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{3 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{12} \int \left (24 a b \left (2 A b^2+a^2 (A+2 C)\right )+6 b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {1}{2} b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) x-\frac {2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{3 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\left (2 a b \left (2 A b^2+a^2 (A+2 C)\right )\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) x+\frac {2 a b \left (2 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \cos (c+d x))^2 \tan (c+d x)}{3 d}+\frac {2 A b (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.29 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.64 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {6 b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) (c+d x)-24 a b \left (2 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+24 a b \left (2 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^3 A (a+12 b)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 a^4 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 a^2 \left (18 A b^2+a^2 (2 A+3 C)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 a^4 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {a^3 A (a+12 b)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a^2 \left (18 A b^2+a^2 (2 A+3 C)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+48 a b^3 C \sin (c+d x)+3 b^4 C \sin (2 (c+d x))}{12 d} \]

[In]

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

[Out]

(6*b^2*(2*A*b^2 + (12*a^2 + b^2)*C)*(c + d*x) - 24*a*b*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] + 24*a*b*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^3*A*(a + 12*b))/(
Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*a^4*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 +
(4*a^2*(18*A*b^2 + a^2*(2*A + 3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*a^4*A*Sin[(c
+ d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (a^3*A*(a + 12*b))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^
2 + (4*a^2*(18*A*b^2 + a^2*(2*A + 3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 48*a*b^3*C*S
in[c + d*x] + 3*b^4*C*Sin[2*(c + d*x)])/(12*d)

Maple [A] (verified)

Time = 9.20 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.79

method	result	size
parts	\(-\frac {a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{4}+6 C \,a^{2} b^{2}\right ) \left (d x +c \right )}{d}+\frac {\left (4 A a \,b^{3}+4 C \,a^{3} b \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (6 A \,a^{2} b^{2}+C \,a^{4}\right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 A \,a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {4 \sin \left (d x +c \right ) C a \,b^{3}}{d}\)	\(198\)
derivativedivides	\(\frac {-a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \,a^{4} \tan \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 C \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 A \,a^{2} b^{2} \tan \left (d x +c \right )+6 C \,a^{2} b^{2} \left (d x +c \right )+4 A a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 C \sin \left (d x +c \right ) a \,b^{3}+A \,b^{4} \left (d x +c \right )+C \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(200\)
default	\(\frac {-a^{4} A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \,a^{4} \tan \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 C \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 A \,a^{2} b^{2} \tan \left (d x +c \right )+6 C \,a^{2} b^{2} \left (d x +c \right )+4 A a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 C \sin \left (d x +c \right ) a \,b^{3}+A \,b^{4} \left (d x +c \right )+C \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(200\)
parallelrisch	\(\frac {-144 b \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) a \left (2 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+144 b \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) a \left (2 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 x \,b^{2} \left (\left (A +\frac {C}{2}\right ) b^{2}+6 a^{2} C \right ) d \cos \left (3 d x +3 c \right )+\left (9 C \,b^{4}+144 A \,a^{2} b^{2}+16 \left (A +\frac {3 C}{2}\right ) a^{4}\right ) \sin \left (3 d x +3 c \right )+96 \left (A \,a^{3} b +C a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+48 C \sin \left (4 d x +4 c \right ) a \,b^{3}+3 C \sin \left (5 d x +5 c \right ) b^{4}+72 x \,b^{2} \left (\left (A +\frac {C}{2}\right ) b^{2}+6 a^{2} C \right ) d \cos \left (d x +c \right )+48 \left (\frac {C \,b^{4}}{8}+3 A \,a^{2} b^{2}+a^{4} \left (A +\frac {C}{2}\right )\right ) \sin \left (d x +c \right )}{24 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(312\)
risch	\(x A \,b^{4}+6 x C \,a^{2} b^{2}+\frac {b^{4} C x}{2}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} C \,b^{4}}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} C a \,b^{3}}{d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} C a \,b^{3}}{d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} C \,b^{4}}{8 d}-\frac {2 i a^{2} \left (6 A a b \,{\mathrm e}^{5 i \left (d x +c \right )}-18 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-36 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 A a b \,{\mathrm e}^{i \left (d x +c \right )}-2 A \,a^{2}-18 A \,b^{2}-3 a^{2} C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {2 A \,a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {2 A \,a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {4 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(385\)

[In]

int((a+cos(d*x+c)*b)^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

-a^4*A/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(A*b^4+6*C*a^2*b^2)/d*(d*x+c)+(4*A*a*b^3+4*C*a^3*b)/d*ln(sec(d*x+c
)+tan(d*x+c))+(6*A*a^2*b^2+C*a^4)/d*tan(d*x+c)+C*b^4/d*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*A*a^3*b/d*(
1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+4/d*sin(d*x+c)*C*a*b^3

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.83 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (12 \, C a^{2} b^{2} + {\left (2 \, A + C\right )} b^{4}\right )} d x \cos \left (d x + c\right )^{3} + 6 \, {\left ({\left (A + 2 \, C\right )} a^{3} b + 2 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left ({\left (A + 2 \, C\right )} a^{3} b + 2 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, C b^{4} \cos \left (d x + c\right )^{4} + 24 \, C a b^{3} \cos \left (d x + c\right )^{3} + 12 \, A a^{3} b \cos \left (d x + c\right ) + 2 \, A a^{4} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{4} + 18 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

1/6*(3*(12*C*a^2*b^2 + (2*A + C)*b^4)*d*x*cos(d*x + c)^3 + 6*((A + 2*C)*a^3*b + 2*A*a*b^3)*cos(d*x + c)^3*log(
sin(d*x + c) + 1) - 6*((A + 2*C)*a^3*b + 2*A*a*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (3*C*b^4*cos(d*x +
c)^4 + 24*C*a*b^3*cos(d*x + c)^3 + 12*A*a^3*b*cos(d*x + c) + 2*A*a^4 + 2*((2*A + 3*C)*a^4 + 18*A*a^2*b^2)*cos
(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.88 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 72 \, {\left (d x + c\right )} C a^{2} b^{2} + 12 \, {\left (d x + c\right )} A b^{4} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{4} - 12 \, A a^{3} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C a b^{3} \sin \left (d x + c\right ) + 12 \, C a^{4} \tan \left (d x + c\right ) + 72 \, A a^{2} b^{2} \tan \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 72*(d*x + c)*C*a^2*b^2 + 12*(d*x + c)*A*b^4 + 3*(2*d*x + 2*c
+ sin(2*d*x + 2*c))*C*b^4 - 12*A*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)) + 24*C*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*A*a*b^3*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 48*C*a*b^3*sin(d*x + c) + 12*C*a^4*tan(d*x + c) + 72*A*a^2*b^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.58 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (12 \, C a^{2} b^{2} + 2 \, A b^{4} + C b^{4}\right )} {\left (d x + c\right )} + 12 \, {\left (A a^{3} b + 2 \, C a^{3} b + 2 \, A a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, {\left (A a^{3} b + 2 \, C a^{3} b + 2 \, A a b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {6 \, {\left (8 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {4 \, {\left (3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(12*C*a^2*b^2 + 2*A*b^4 + C*b^4)*(d*x + c) + 12*(A*a^3*b + 2*C*a^3*b + 2*A*a*b^3)*log(abs(tan(1/2*d*x +
1/2*c) + 1)) - 12*(A*a^3*b + 2*C*a^3*b + 2*A*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(8*C*a*b^3*tan(1/2
*d*x + 1/2*c)^3 - C*b^4*tan(1/2*d*x + 1/2*c)^3 + 8*C*a*b^3*tan(1/2*d*x + 1/2*c) + C*b^4*tan(1/2*d*x + 1/2*c))/
(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 4*(3*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^3*
b*tan(1/2*d*x + 1/2*c)^5 + 18*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 2*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 6*C*a^4*tan(
1/2*d*x + 1/2*c)^3 - 36*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) + 3*C*a^4*tan(1/2*d*x
+ 1/2*c) + 6*A*a^3*b*tan(1/2*d*x + 1/2*c) + 18*A*a^2*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)
/d

Mupad [B] (verification not implemented)

Time = 3.91 (sec) , antiderivative size = 2662, normalized size of antiderivative = 10.61 \[ \int (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

- (tan(c/2 + (d*x)/2)^5*((4*A*a^4)/3 - 4*C*a^4 + 6*C*b^4 - 24*A*a^2*b^2) + tan(c/2 + (d*x)/2)^3*((8*A*a^4)/3 -
4*C*b^4 + 8*A*a^3*b - 16*C*a*b^3) + tan(c/2 + (d*x)/2)^7*((8*A*a^4)/3 - 4*C*b^4 - 8*A*a^3*b + 16*C*a*b^3) + t
an(c/2 + (d*x)/2)*(2*A*a^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 + 4*A*a^3*b + 8*C*a*b^3) + tan(c/2 + (d*x)/2)^9*(2
*A*a^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 - 4*A*a^3*b - 8*C*a*b^3))/(d*(tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)
/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1)) - (b^2*atan(((b^2*(tan(c/2
+ (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 +
1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2) - (b^2
*(2*A*b^2 + 12*C*a^2 + C*b^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b)*1
i)/2)*(2*A*b^2 + 12*C*a^2 + C*b^2))/2 + (b^2*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 5
12*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C
*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2) + (b^2*(2*A*b^2 + 12*C*a^2 + C*b^2)*(32*A*b^4 + 16*C*b^4 + 192*
C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b)*1i)/2)*(2*A*b^2 + 12*C*a^2 + C*b^2))/2)/(256*A^3*a*b^11 -
(b^2*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C
^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^
6*b^2) - (b^2*(2*A*b^2 + 12*C*a^2 + C*b^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 1
28*C*a^3*b)*1i)/2)*(2*A*b^2 + 12*C*a^2 + C*b^2)*1i)/2 + (b^2*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512
*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A
*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2) + (b^2*(2*A*b^2 + 12*C*a^2 + C*b^2)*(32*A*b^4 +
16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b)*1i)/2)*(2*A*b^2 + 12*C*a^2 + C*b^2)*1i)/2
- 1024*A^3*a^2*b^10 + 128*A^3*a^3*b^9 - 1024*A^3*a^4*b^8 - 256*A^3*a^6*b^6 + 64*C^3*a^3*b^9 + 1536*C^3*a^5*b^7
- 512*C^3*a^6*b^6 + 9216*C^3*a^7*b^5 - 6144*C^3*a^8*b^4 + 64*A*C^2*a*b^11 + 256*A^2*C*a*b^11 + 1824*A*C^2*a^3
*b^9 - 1024*A*C^2*a^4*b^8 + 13056*A*C^2*a^5*b^7 - 13824*A*C^2*a^6*b^6 + 4608*A*C^2*a^7*b^5 - 6144*A*C^2*a^8*b^
4 - 512*A^2*C*a^2*b^10 + 3456*A^2*C*a^3*b^9 - 8704*A^2*C*a^4*b^8 + 1536*A^2*C*a^5*b^7 - 7296*A^2*C*a^6*b^6 - 1
536*A^2*C*a^8*b^4))*(2*A*b^2 + 12*C*a^2 + C*b^2))/d - (a*b*atan((a*b*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b
^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^
2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2) - 2*a*b*(A*a^2 + 2*A*b^2 + 2*C*a^2)*(32
*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b))*(A*a^2 + 2*A*b^2 + 2*C*a^2)*2i +
a*b*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^
2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6
*b^2) + 2*a*b*(A*a^2 + 2*A*b^2 + 2*C*a^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 12
8*C*a^3*b))*(A*a^2 + 2*A*b^2 + 2*C*a^2)*2i)/(256*A^3*a*b^11 - 1024*A^3*a^2*b^10 + 128*A^3*a^3*b^9 - 1024*A^3*a
^4*b^8 - 256*A^3*a^6*b^6 + 64*C^3*a^3*b^9 + 1536*C^3*a^5*b^7 - 512*C^3*a^6*b^6 + 9216*C^3*a^7*b^5 - 6144*C^3*a
^8*b^4 - 2*a*b*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b
^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 +
512*A*C*a^6*b^2) - 2*a*b*(A*a^2 + 2*A*b^2 + 2*C*a^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A
*a^3*b + 128*C*a^3*b))*(A*a^2 + 2*A*b^2 + 2*C*a^2) + 2*a*b*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A
^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C
*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2) + 2*a*b*(A*a^2 + 2*A*b^2 + 2*C*a^2)*(32*A*b^4 + 1
6*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b))*(A*a^2 + 2*A*b^2 + 2*C*a^2) + 64*A*C^2*a*b^
11 + 256*A^2*C*a*b^11 + 1824*A*C^2*a^3*b^9 - 1024*A*C^2*a^4*b^8 + 13056*A*C^2*a^5*b^7 - 13824*A*C^2*a^6*b^6 +
4608*A*C^2*a^7*b^5 - 6144*A*C^2*a^8*b^4 - 512*A^2*C*a^2*b^10 + 3456*A^2*C*a^3*b^9 - 8704*A^2*C*a^4*b^8 + 1536*
A^2*C*a^5*b^7 - 7296*A^2*C*a^6*b^6 - 1536*A^2*C*a^8*b^4))*(A*a^2 + 2*A*b^2 + 2*C*a^2)*4i)/d