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

Optimal. Leaf size=251 \[ -\frac{2 a b \left (a^2 (2 A+3 C)+b^2 (11 A-6 C)\right ) \sin (c+d x)}{3 d}+\frac{2 a b \left (a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^2 \left (a^2 (4 A+6 C)+3 b^2 (6 A-C)\right ) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{\left (a^2 (2 A+3 C)+6 A b^2\right ) \tan (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac{1}{2} b^2 x \left (C \left (12 a^2+b^2\right )+2 A b^2\right )+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}+\frac{2 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{3 d} \]

[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)

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Rubi [A]  time = 0.962218, antiderivative size = 251, normalized size of antiderivative = 1., 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, 3033, 3023, 2735, 3770} \[ -\frac{2 a b \left (a^2 (2 A+3 C)+b^2 (11 A-6 C)\right ) \sin (c+d x)}{3 d}+\frac{2 a b \left (a^2 (A+2 C)+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^2 \left (a^2 (4 A+6 C)+3 b^2 (6 A-C)\right ) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{\left (a^2 (2 A+3 C)+6 A b^2\right ) \tan (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac{1}{2} b^2 x \left (C \left (12 a^2+b^2\right )+2 A b^2\right )+\frac{A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^4}{3 d}+\frac{2 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[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 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 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 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 ^4(c+d x) \, dx &=\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 ) \tanh ^{-1}(\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]  time = 6.01673, size = 412, normalized size = 1.64 \[ \frac{6 b^2 (c+d x) \left (C \left (12 a^2+b^2\right )+2 A b^2\right )+\frac{4 a^2 \left (a^2 (2 A+3 C)+18 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a^2 \left (a^2 (2 A+3 C)+18 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-24 a b \left (a^2 (A+2 C)+2 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+24 a b \left (a^2 (A+2 C)+2 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^3 A (a+12 b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^3 A (a+12 b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 a^4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 a^4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+48 a b^3 C \sin (c+d x)+3 b^4 C \sin (2 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.069, size = 258, normalized size = 1. \begin{align*} A{b}^{4}x+{\frac{A{b}^{4}c}{d}}+{\frac{C{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{4}Cx}{2}}+{\frac{C{b}^{4}c}{2\,d}}+4\,{\frac{aA{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Ca{b}^{3}\sin \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{2}A{b}^{2}\tan \left ( dx+c \right ) }{d}}+6\,{a}^{2}{b}^{2}Cx+6\,{\frac{C{a}^{2}{b}^{2}c}{d}}+2\,{\frac{A{a}^{3}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{3}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,A{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{4}C\tan \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)^4,x)

[Out]

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

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Maxima [A]  time = 1.02472, size = 298, normalized size = 1.19 \begin{align*} \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} \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)^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

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Fricas [A]  time = 1.67368, size = 509, normalized size = 2.03 \begin{align*} \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}} \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)^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)

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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)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.71736, size = 536, normalized size = 2.14 \begin{align*} \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} \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)^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