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

Optimal. Leaf size=165 \[ -\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {a^4 (12 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(11 A+9 B) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}+a^4 x (A+4 B)+\frac {(2 A+B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}+\frac {a A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

[Out]

a^4*(A+4*B)*x+1/2*a^4*(12*A+13*B)*arctanh(sin(d*x+c))/d-5/2*a^4*(2*A+B)*sin(d*x+c)/d+1/3*(11*A+9*B)*(a^4+a^4*c
os(d*x+c))*tan(d*x+c)/d+1/2*(2*A+B)*(a^2+a^2*cos(d*x+c))^2*sec(d*x+c)*tan(d*x+c)/d+1/3*a*A*(a+a*cos(d*x+c))^3*
sec(d*x+c)^2*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.51, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2975, 2968, 3023, 2735, 3770} \[ -\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {a^4 (12 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(11 A+9 B) \tan (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{3 d}+\frac {(2 A+B) \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 d}+a^4 x (A+4 B)+\frac {a A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*(A + 4*B)*x + (a^4*(12*A + 13*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(2*A + B)*Sin[c + d*x])/(2*d) + ((1
1*A + 9*B)*(a^4 + a^4*Cos[c + d*x])*Tan[c + d*x])/(3*d) + ((2*A + B)*(a^2 + a^2*Cos[c + d*x])^2*Sec[c + d*x]*T
an[c + d*x])/(2*d) + (a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 3023

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 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+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\frac {a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+a \cos (c+d x))^3 (3 a (2 A+B)-a (A-3 B) \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac {(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int (a+a \cos (c+d x))^2 \left (2 a^2 (11 A+9 B)-a^2 (8 A-3 B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int (a+a \cos (c+d x)) \left (3 a^3 (12 A+13 B)-15 a^3 (2 A+B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (3 a^4 (12 A+13 B)+\left (-15 a^4 (2 A+B)+3 a^4 (12 A+13 B)\right ) \cos (c+d x)-15 a^4 (2 A+B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (3 a^4 (12 A+13 B)+6 a^4 (A+4 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=a^4 (A+4 B) x-\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} \left (a^4 (12 A+13 B)\right ) \int \sec (c+d x) \, dx\\ &=a^4 (A+4 B) x+\frac {a^4 (12 A+13 B) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^4 (2 A+B) \sin (c+d x)}{2 d}+\frac {(11 A+9 B) \left (a^4+a^4 \cos (c+d x)\right ) \tan (c+d x)}{3 d}+\frac {(2 A+B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A (a+a \cos (c+d x))^3 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 6.22, size = 380, normalized size = 2.30 \[ a^4 \left (\frac {(A+4 B) (c+d x)}{d}+\frac {-13 A-3 B}{12 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 \left (5 A \sin \left (\frac {1}{2} (c+d x)\right )+3 B \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 \left (5 A \sin \left (\frac {1}{2} (c+d x)\right )+3 B \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {13 A+3 B}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {(-12 A-13 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {(12 A+13 B) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {B \sin (c+d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*(((A + 4*B)*(c + d*x))/d + ((-12*A - 13*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2*d) + ((12*A + 13*B
)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2*d) + (13*A + 3*B)/(12*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2
) + (A*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (A*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d
*x)/2] + Sin[(c + d*x)/2])^3) + (-13*A - 3*B)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(5*A*Sin[(c
+ d*x)/2] + 3*B*Sin[(c + d*x)/2]))/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (4*(5*A*Sin[(c + d*x)/2] + 3*
B*Sin[(c + d*x)/2]))/(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (B*Sin[c + d*x])/d)

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 159, normalized size = 0.96 \[ \frac {12 \, {\left (A + 4 \, B\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (12 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (12 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, B a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 3 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) + 2 \, A a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(A + 4*B)*a^4*d*x*cos(d*x + c)^3 + 3*(12*A + 13*B)*a^4*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(12*A
+ 13*B)*a^4*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(6*B*a^4*cos(d*x + c)^3 + 8*(5*A + 3*B)*a^4*cos(d*x + c
)^2 + 3*(4*A + B)*a^4*cos(d*x + c) + 2*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^3)

________________________________________________________________________________________

giac [A]  time = 0.49, size = 227, normalized size = 1.38 \[ \frac {\frac {12 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 6 \, {\left (A a^{4} + 4 \, B a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (12 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (12 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (30 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 76 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 48 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 54 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(12*B*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(A*a^4 + 4*B*a^4)*(d*x + c) + 3*(12*A*a^4
+ 13*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*A*a^4 + 13*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(30*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 76*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 48*B*
a^4*tan(1/2*d*x + 1/2*c)^3 + 54*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2
*c)^2 - 1)^3)/d

________________________________________________________________________________________

maple [A]  time = 0.18, size = 189, normalized size = 1.15 \[ A \,a^{4} x +\frac {A \,a^{4} c}{d}+\frac {a^{4} B \sin \left (d x +c \right )}{d}+\frac {6 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+4 a^{4} B x +\frac {4 a^{4} B c}{d}+\frac {20 A \,a^{4} \tan \left (d x +c \right )}{3 d}+\frac {13 a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {4 a^{4} B \tan \left (d x +c \right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a^4*x+1/d*A*a^4*c+1/d*a^4*B*sin(d*x+c)+6/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+4*a^4*B*x+4/d*a^4*B*c+20/3/d*A*a^
4*tan(d*x+c)+13/2/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))+2/d*A*a^4*sec(d*x+c)*tan(d*x+c)+4/d*a^4*B*tan(d*x+c)+1/3/d
*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a^4*B*sec(d*x+c)*tan(d*x+c)

________________________________________________________________________________________

maxima [A]  time = 0.59, size = 235, normalized size = 1.42 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 12 \, {\left (d x + c\right )} A a^{4} + 48 \, {\left (d x + c\right )} B a^{4} - 12 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{4} \sin \left (d x + c\right ) + 72 \, A a^{4} \tan \left (d x + c\right ) + 48 \, B a^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c))*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 + 12*(d*x + c)*A*a^4 + 48*(d*x + c)*B*a^4 - 12*A*a^4*(2*sin(d*
x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*a^4*(2*sin(d*x + c)/(sin(d*
x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x
+ c) - 1)) + 36*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*B*a^4*sin(d*x + c) + 72*A*a^4*tan(
d*x + c) + 48*B*a^4*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 0.41, size = 254, normalized size = 1.54 \[ \frac {B\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {4\,B\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*a^4*sin(c + d*x))/d + (2*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*A*a^4*atanh(sin(c/2 + (
d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13*B*a^4*atanh(sin
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (20*A*a^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (2*A*a^4*sin(c + d*x))/(
d*cos(c + d*x)^2) + (A*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (4*B*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (B*a
^4*sin(c + d*x))/(2*d*cos(c + d*x)^2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________