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

Optimal. Leaf size=227 \[ \frac{\left (a^2 b^2 (85 A+56 C)+6 a^4 C+2 b^4 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{15 d}+\frac{1}{2} a b x \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+\frac{a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a C \sin (c+d x) (a+b \cos (c+d x))^3}{5 d}+\frac{C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.792801, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3050, 3049, 3033, 3023, 2735, 3770} \[ \frac{\left (a^2 b^2 (85 A+56 C)+6 a^4 C+2 b^4 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{15 d}+\frac{1}{2} a b x \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+\frac{a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a C \sin (c+d x) (a+b \cos (c+d x))^3}{5 d}+\frac{C \sin (c+d x) (a+b \cos (c+d x))^4}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3033

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 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 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 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 (c+d x) \, dx &=\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cos (c+d x))^3 \left (5 a A+b (5 A+4 C) \cos (c+d x)+4 a C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{20} \int (a+b \cos (c+d x))^2 \left (20 a^2 A+4 a b (10 A+7 C) \cos (c+d x)+4 \left (3 a^2 C+b^2 (5 A+4 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d}+\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{60} \int (a+b \cos (c+d x)) \left (60 a^3 A+4 b \left (9 a^2 (5 A+3 C)+2 b^2 (5 A+4 C)\right ) \cos (c+d x)+4 a \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d}+\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{120} \int \left (120 a^4 A+60 a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \cos (c+d x)+8 \left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{\left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d}+\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac{1}{120} \int \left (120 a^4 A+60 a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{2} a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) x+\frac{\left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d}+\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}+\left (a^4 A\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) x+\frac{a^4 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{\left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{30 d}+\frac{\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{15 d}+\frac{a C (a+b \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac{C (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.03833, size = 226, normalized size = 1. \[ \frac{120 a b (c+d x) \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+240 a b \left (C \left (a^2+b^2\right )+A b^2\right ) \sin (2 (c+d x))+30 \left (12 a^2 b^2 (4 A+3 C)+8 a^4 C+b^4 (6 A+5 C)\right ) \sin (c+d x)+5 b^2 \left (24 a^2 C+4 A b^2+5 b^2 C\right ) \sin (3 (c+d x))-240 a^4 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+240 a^4 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+30 a b^3 C \sin (4 (c+d x))+3 b^4 C \sin (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*a*b*(4*a^2*(2*A + C) + b^2*(4*A + 3*C))*(c + d*x) - 240*a^4*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] +
240*a^4*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 30*(8*a^4*C + 12*a^2*b^2*(4*A + 3*C) + b^4*(6*A + 5*C))*S
in[c + d*x] + 240*a*b*(A*b^2 + (a^2 + b^2)*C)*Sin[2*(c + d*x)] + 5*b^2*(4*A*b^2 + 24*a^2*C + 5*b^2*C)*Sin[3*(c
+ d*x)] + 30*a*b^3*C*Sin[4*(c + d*x)] + 3*b^4*C*Sin[5*(c + d*x)])/(240*d)

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Maple [A]  time = 0.054, size = 364, normalized size = 1.6 \begin{align*}{\frac{A\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{b}^{4}}{3\,d}}+{\frac{2\,A{b}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{8\,C{b}^{4}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{C{b}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,C{b}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+2\,{\frac{aA{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+2\,aA{b}^{3}x+2\,{\frac{aA{b}^{3}c}{d}}+{\frac{Ca{b}^{3}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{3\,Ca{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,a{b}^{3}Cx}{2}}+{\frac{3\,Ca{b}^{3}c}{2\,d}}+6\,{\frac{{a}^{2}A{b}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{2}{b}^{2}}{d}}+4\,{\frac{{a}^{2}{b}^{2}C\sin \left ( dx+c \right ) }{d}}+4\,A{a}^{3}bx+4\,{\frac{A{a}^{3}bc}{d}}+2\,{\frac{{a}^{3}bC\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+2\,{a}^{3}bCx+2\,{\frac{{a}^{3}bCc}{d}}+{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{4}C\sin \left ( dx+c \right ) }{d}} \end{align*}

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),x)

[Out]

1/3/d*A*cos(d*x+c)^2*sin(d*x+c)*b^4+2/3/d*A*b^4*sin(d*x+c)+8/15/d*C*b^4*sin(d*x+c)+1/5/d*C*b^4*sin(d*x+c)*cos(
d*x+c)^4+4/15/d*C*b^4*sin(d*x+c)*cos(d*x+c)^2+2/d*a*A*b^3*cos(d*x+c)*sin(d*x+c)+2*a*A*b^3*x+2/d*a*A*b^3*c+1/d*
C*a*b^3*sin(d*x+c)*cos(d*x+c)^3+3/2/d*C*a*b^3*cos(d*x+c)*sin(d*x+c)+3/2*a*b^3*C*x+3/2/d*C*a*b^3*c+6/d*a^2*A*b^
2*sin(d*x+c)+2/d*C*cos(d*x+c)^2*sin(d*x+c)*a^2*b^2+4/d*a^2*b^2*C*sin(d*x+c)+4*A*a^3*b*x+4/d*A*a^3*b*c+2/d*a^3*
b*C*cos(d*x+c)*sin(d*x+c)+2*a^3*b*C*x+2/d*a^3*b*C*c+1/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^4*C*sin(d*x+c)

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Maxima [A]  time = 1.00588, size = 313, normalized size = 1.38 \begin{align*} \frac{480 \,{\left (d x + c\right )} A a^{3} b + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} b - 240 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{3} + 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{3} - 40 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{4} + 8 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b^{4} + 120 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 120 \, C a^{4} \sin \left (d x + c\right ) + 720 \, A a^{2} b^{2} \sin \left (d x + c\right )}{120 \, d} \end{align*}

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),x, algorithm="maxima")

[Out]

1/120*(480*(d*x + c)*A*a^3*b + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3*b - 240*(sin(d*x + c)^3 - 3*sin(d*x
+ c))*C*a^2*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(
2*d*x + 2*c))*C*a*b^3 - 40*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^4 + 8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 +
15*sin(d*x + c))*C*b^4 + 120*A*a^4*log(sec(d*x + c) + tan(d*x + c)) + 120*C*a^4*sin(d*x + c) + 720*A*a^2*b^2*
sin(d*x + c))/d

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Fricas [A]  time = 1.62882, size = 473, normalized size = 2.08 \begin{align*} \frac{15 \, A a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, A a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 15 \,{\left (4 \,{\left (2 \, A + C\right )} a^{3} b +{\left (4 \, A + 3 \, C\right )} a b^{3}\right )} d x +{\left (6 \, C b^{4} \cos \left (d x + c\right )^{4} + 30 \, C a b^{3} \cos \left (d x + c\right )^{3} + 30 \, C a^{4} + 60 \,{\left (3 \, A + 2 \, C\right )} a^{2} b^{2} + 4 \,{\left (5 \, A + 4 \, C\right )} b^{4} + 2 \,{\left (30 \, C a^{2} b^{2} +{\left (5 \, A + 4 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (4 \, C a^{3} b +{\left (4 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \, d} \end{align*}

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),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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),x)

[Out]

Timed out

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Giac [B]  time = 1.78789, size = 1017, normalized size = 4.48 \begin{align*} \text{result too large to display} \end{align*}

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),x, algorithm="giac")

[Out]

1/30*(30*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 30*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 15*(8*A*a^3*
b + 4*C*a^3*b + 4*A*a*b^3 + 3*C*a*b^3)*(d*x + c) + 2*(30*C*a^4*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 - 75*C*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 + 30*C*b^4*tan(1/2*d*x +
1/2*c)^9 + 120*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 720*A*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^7 + 480*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 - 30*C*a*b^3*tan(1/2*d*
x + 1/2*c)^7 + 80*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 40*C*b^4*tan(1/2*d*x + 1/2*c)^7 + 180*C*a^4*tan(1/2*d*x + 1/2
*c)^5 + 1080*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 600*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 100*A*b^4*tan(1/2*d*x +
1/2*c)^5 + 116*C*b^4*tan(1/2*d*x + 1/2*c)^5 + 120*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 120*C*a^3*b*tan(1/2*d*x + 1/
2*c)^3 + 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 480*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 + 30*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 80*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 40*C*b^4*tan(1/2*d*x + 1/2
*c)^3 + 30*C*a^4*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) +
180*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) + 75*C*a*b^3*tan(1/2*d*x + 1/2*c) + 30*A
*b^4*tan(1/2*d*x + 1/2*c) + 30*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d