3.553 \(\int (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=219 \[ -\frac{b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (a^2 (A+2 C)+12 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+2 a b x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )-\frac{b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{6 d}-\frac{a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{3 d}+\frac{2 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}+\frac{A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d} \]

[Out]

2*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*x + (a^2*(12*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*d) - (b^2*(a^2
*(39*A - 34*C) - 2*b^2*(3*A + 2*C))*Sin[c + d*x])/(6*d) - (a*b^3*(9*A - 4*C)*Cos[c + d*x]*Sin[c + d*x])/(3*d)
- (b^2*(15*A - 2*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(6*d) + (2*A*b*(a + b*Cos[c + d*x])^3*Tan[c + d*x])/d
+ (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.902528, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3048, 3047, 3049, 3033, 3023, 2735, 3770} \[ -\frac{b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}+\frac{a^2 \left (a^2 (A+2 C)+12 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+2 a b x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )-\frac{b^2 (15 A-2 C) \sin (c+d x) (a+b \cos (c+d x))^2}{6 d}-\frac{a b^3 (9 A-4 C) \sin (c+d x) \cos (c+d x)}{3 d}+\frac{2 A b \tan (c+d x) (a+b \cos (c+d x))^3}{d}+\frac{A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^4}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

2*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*x + (a^2*(12*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*d) - (b^2*(a^2
*(39*A - 34*C) - 2*b^2*(3*A + 2*C))*Sin[c + d*x])/(6*d) - (a*b^3*(9*A - 4*C)*Cos[c + d*x]*Sin[c + d*x])/(3*d)
- (b^2*(15*A - 2*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(6*d) + (2*A*b*(a + b*Cos[c + d*x])^3*Tan[c + d*x])/d
+ (A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]*Tan[c + d*x])/(2*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 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 ^3(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+b \cos (c+d x))^3 \left (4 A b+a (A+2 C) \cos (c+d x)-b (3 A-2 C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+b \cos (c+d x))^2 \left (12 A b^2+a^2 (A+2 C)-2 a b (A-2 C) \cos (c+d x)-b^2 (15 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{6} \int (a+b \cos (c+d x)) \left (3 a \left (12 A b^2+a^2 (A+2 C)\right )-b \left (3 a^2 (A-6 C)-2 b^2 (3 A+2 C)\right ) \cos (c+d x)-4 a b^2 (9 A-4 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac{b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{12} \int \left (6 a^2 \left (12 A b^2+a^2 (A+2 C)\right )+24 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)-2 b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac{a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac{b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{12} \int \left (6 a^2 \left (12 A b^2+a^2 (A+2 C)\right )+24 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x-\frac{b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac{a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac{b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a^2 \left (12 A b^2+a^2 (A+2 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x+\frac{a^2 \left (12 A b^2+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac{a b^3 (9 A-4 C) \cos (c+d x) \sin (c+d x)}{3 d}-\frac{b^2 (15 A-2 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{2 A b (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac{A (a+b \cos (c+d x))^4 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 3.09963, size = 323, normalized size = 1.47 \[ \frac{24 a b (c+d x) \left (C \left (2 a^2+b^2\right )+2 A b^2\right )+3 b^2 \left (3 C \left (8 a^2+b^2\right )+4 A b^2\right ) \sin (c+d x)-6 a^2 \left (a^2 (A+2 C)+12 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 a^2 \left (a^2 (A+2 C)+12 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{48 a^3 A b \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{48 a^3 A b \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{3 a^4 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{3 a^4 A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+12 a b^3 C \sin (2 (c+d x))+b^4 C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

(24*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*(c + d*x) - 6*a^2*(12*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] + 6*a^2*(12*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*a^4*A)/(Cos[(c +
d*x)/2] - Sin[(c + d*x)/2])^2 + (48*a^3*A*b*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (3*a^4*A
)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (48*a^3*A*b*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
) + 3*b^2*(4*A*b^2 + 3*(8*a^2 + b^2)*C)*Sin[c + d*x] + 12*a*b^3*C*Sin[2*(c + d*x)] + b^4*C*Sin[3*(c + d*x)])/(
12*d)

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Maple [A]  time = 0.065, size = 259, normalized size = 1.2 \begin{align*}{\frac{A{b}^{4}\sin \left ( dx+c \right ) }{d}}+{\frac{C{b}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{2\,C{b}^{4}\sin \left ( dx+c \right ) }{3\,d}}+4\,aA{b}^{3}x+4\,{\frac{Aa{b}^{3}c}{d}}+2\,{\frac{Ca{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+2\,a{b}^{3}Cx+2\,{\frac{Ca{b}^{3}c}{d}}+6\,{\frac{{a}^{2}A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}C\sin \left ( dx+c \right ) }{d}}+4\,{\frac{A{a}^{3}b\tan \left ( dx+c \right ) }{d}}+4\,{a}^{3}bCx+4\,{\frac{C{a}^{3}bc}{d}}+{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \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)^3,x)

[Out]

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

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Maxima [A]  time = 1.0631, size = 298, normalized size = 1.36 \begin{align*} \frac{48 \,{\left (d x + c\right )} C a^{3} b + 48 \,{\left (d x + c\right )} A a b^{3} + 12 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{3} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{4} - 3 \, 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 )} + 6 \, C 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 \, A a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 72 \, C a^{2} b^{2} \sin \left (d x + c\right ) + 12 \, A b^{4} \sin \left (d x + c\right ) + 48 \, A a^{3} b \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)^3,x, algorithm="maxima")

[Out]

1/12*(48*(d*x + c)*C*a^3*b + 48*(d*x + c)*A*a*b^3 + 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a*b^3 - 4*(sin(d*x +
c)^3 - 3*sin(d*x + c))*C*b^4 - 3*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)) + 6*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*A*a^2*b^2*(log(sin(d*x + c) + 1
) - log(sin(d*x + c) - 1)) + 72*C*a^2*b^2*sin(d*x + c) + 12*A*b^4*sin(d*x + c) + 48*A*a^3*b*tan(d*x + c))/d

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Fricas [A]  time = 1.65472, size = 516, normalized size = 2.36 \begin{align*} \frac{24 \,{\left (2 \, C a^{3} b +{\left (2 \, A + C\right )} a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 3 \,{\left ({\left (A + 2 \, C\right )} a^{4} + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (A + 2 \, C\right )} a^{4} + 12 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, C b^{4} \cos \left (d x + c\right )^{4} + 12 \, C a b^{3} \cos \left (d x + c\right )^{3} + 24 \, A a^{3} b \cos \left (d x + c\right ) + 3 \, A a^{4} + 2 \,{\left (18 \, C a^{2} b^{2} +{\left (3 \, A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \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)^3,x, algorithm="fricas")

[Out]

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

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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.38456, size = 535, normalized size = 2.44 \begin{align*} \frac{12 \,{\left (2 \, C a^{3} b + 2 \, A a b^{3} + C a b^{3}\right )}{\left (d x + c\right )} + 3 \,{\left (A a^{4} + 2 \, C a^{4} + 12 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (A a^{4} + 2 \, C a^{4} + 12 \, A a^{2} b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{6 \,{\left (A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, A a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 8 \, A a^{3} b \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 (18 \, C a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, C a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 36 \, C a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, C a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b^{4} \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 )\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)^3,x, algorithm="giac")

[Out]

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