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

Optimal. Leaf size=229 \[ -\frac{a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (24 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right )+\frac{4 a^3 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b (4 A-C) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}-\frac{a b (12 A-7 C) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{A \tan (c+d x) (a+b \cos (c+d x))^4}{d} \]

[Out]

((8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*x)/8 + (4*a^3*A*b*ArcTanh[Sin[c + d*x]])/d - (a*b*(a^2*(12
*A - 19*C) - 8*b^2*(3*A + 2*C))*Sin[c + d*x])/(6*d) - (b^2*(a^2*(24*A - 26*C) - 3*b^2*(4*A + 3*C))*Cos[c + d*x
]*Sin[c + d*x])/(24*d) - (a*b*(12*A - 7*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) - (b*(4*A - C)*(a + b*C
os[c + d*x])^3*Sin[c + d*x])/(4*d) + (A*(a + b*Cos[c + d*x])^4*Tan[c + d*x])/d

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Rubi [A]  time = 0.802126, antiderivative size = 229, 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, 3049, 3033, 3023, 2735, 3770} \[ -\frac{a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (24 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right )+\frac{4 a^3 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b (4 A-C) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}-\frac{a b (12 A-7 C) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{A \tan (c+d x) (a+b \cos (c+d x))^4}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*x)/8 + (4*a^3*A*b*ArcTanh[Sin[c + d*x]])/d - (a*b*(a^2*(12
*A - 19*C) - 8*b^2*(3*A + 2*C))*Sin[c + d*x])/(6*d) - (b^2*(a^2*(24*A - 26*C) - 3*b^2*(4*A + 3*C))*Cos[c + d*x
]*Sin[c + d*x])/(24*d) - (a*b*(12*A - 7*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) - (b*(4*A - C)*(a + b*C
os[c + d*x])^3*Sin[c + d*x])/(4*d) + (A*(a + b*Cos[c + d*x])^4*Tan[c + d*x])/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 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 ^2(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\int (a+b \cos (c+d x))^3 \left (4 A b+a C \cos (c+d x)-b (4 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{1}{4} \int (a+b \cos (c+d x))^2 \left (16 a A b+\left (4 A b^2+4 a^2 C+3 b^2 C\right ) \cos (c+d x)-a b (12 A-7 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{1}{12} \int (a+b \cos (c+d x)) \left (48 a^2 A b+a \left (36 A b^2+12 a^2 C+23 b^2 C\right ) \cos (c+d x)-b \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac{a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{1}{24} \int \left (96 a^3 A b+3 \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \cos (c+d x)-4 a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac{a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\frac{1}{24} \int \left (96 a^3 A b+3 \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{1}{8} \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x-\frac{a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac{a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}+\left (4 a^3 A b\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x+\frac{4 a^3 A b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sin (c+d x)}{6 d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac{a b (12 A-7 C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{A (a+b \cos (c+d x))^4 \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 1.34126, size = 274, normalized size = 1.2 \[ \frac{12 (c+d x) \left (24 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right )+96 a b \left (4 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x)+24 b^2 \left (C \left (6 a^2+b^2\right )+A b^2\right ) \sin (2 (c+d x))-384 a^3 A b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+384 a^3 A b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{96 a^4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{96 a^4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+32 a b^3 C \sin (3 (c+d x))+3 b^4 C \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*(c + d*x) - 384*a^3*A*b*Log[Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]] + 384*a^3*A*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (96*a^4*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]) + (96*a^4*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 96*a*b*(4*A*b^2 +
4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 24*b^2*(A*b^2 + (6*a^2 + b^2)*C)*Sin[2*(c + d*x)] + 32*a*b^3*C*Sin[3*(c + d
*x)] + 3*b^4*C*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.061, size = 296, normalized size = 1.3 \begin{align*}{\frac{A{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{4}x}{2}}+{\frac{A{b}^{4}c}{2\,d}}+{\frac{C{b}^{4}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,C{b}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{4}Cx}{8}}+{\frac{3\,C{b}^{4}c}{8\,d}}+4\,{\frac{aA{b}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{4\,C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}a{b}^{3}}{3\,d}}+{\frac{8\,Ca{b}^{3}\sin \left ( dx+c \right ) }{3\,d}}+6\,{a}^{2}A{b}^{2}x+6\,{\frac{A{a}^{2}{b}^{2}c}{d}}+3\,{\frac{{a}^{2}{b}^{2}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+3\,{a}^{2}{b}^{2}Cx+3\,{\frac{{a}^{2}{b}^{2}Cc}{d}}+4\,{\frac{A{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{3}bC\sin \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+{a}^{4}Cx+{\frac{C{a}^{4}c}{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)^2,x)

[Out]

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

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Maxima [A]  time = 1.07936, size = 275, normalized size = 1.2 \begin{align*} \frac{96 \,{\left (d x + c\right )} C a^{4} + 576 \,{\left (d x + c\right )} A a^{2} b^{2} + 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b^{2} - 128 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b^{3} + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} + 3 \,{\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 b^{4} + 192 \, A 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 )} + 384 \, C a^{3} b \sin \left (d x + c\right ) + 384 \, A a b^{3} \sin \left (d x + c\right ) + 96 \, A a^{4} \tan \left (d x + c\right )}{96 \, 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)^2,x, algorithm="maxima")

[Out]

1/96*(96*(d*x + c)*C*a^4 + 576*(d*x + c)*A*a^2*b^2 + 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2*b^2 - 128*(sin
(d*x + c)^3 - 3*sin(d*x + c))*C*a*b^3 + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^4 + 3*(12*d*x + 12*c + sin(4*d
*x + 4*c) + 8*sin(2*d*x + 2*c))*C*b^4 + 192*A*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 384*C*a^
3*b*sin(d*x + c) + 384*A*a*b^3*sin(d*x + c) + 96*A*a^4*tan(d*x + c))/d

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Fricas [A]  time = 1.69908, size = 504, normalized size = 2.2 \begin{align*} \frac{48 \, A a^{3} b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 48 \, A a^{3} b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (8 \, C a^{4} + 24 \,{\left (2 \, A + C\right )} a^{2} b^{2} +{\left (4 \, A + 3 \, C\right )} b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (6 \, C b^{4} \cos \left (d x + c\right )^{4} + 32 \, C a b^{3} \cos \left (d x + c\right )^{3} + 24 \, A a^{4} + 3 \,{\left (24 \, C a^{2} b^{2} +{\left (4 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 32 \,{\left (3 \, C a^{3} b +{\left (3 \, A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \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)^2,x, algorithm="fricas")

[Out]

1/24*(48*A*a^3*b*cos(d*x + c)*log(sin(d*x + c) + 1) - 48*A*a^3*b*cos(d*x + c)*log(-sin(d*x + c) + 1) + 3*(8*C*
a^4 + 24*(2*A + C)*a^2*b^2 + (4*A + 3*C)*b^4)*d*x*cos(d*x + c) + (6*C*b^4*cos(d*x + c)^4 + 32*C*a*b^3*cos(d*x
+ c)^3 + 24*A*a^4 + 3*(24*C*a^2*b^2 + (4*A + 3*C)*b^4)*cos(d*x + c)^2 + 32*(3*C*a^3*b + (3*A + 2*C)*a*b^3)*cos
(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))

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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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.71155, size = 753, normalized size = 3.29 \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)^2,x, algorithm="giac")

[Out]

1/24*(96*A*a^3*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 96*A*a^3*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 48*A*a^4
*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 3*(8*C*a^4 + 48*A*a^2*b^2 + 24*C*a^2*b^2 + 4*A*b^4 + 3*C*
b^4)*(d*x + c) + 2*(96*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 72*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 96*A*a*b^3*tan(1
/2*d*x + 1/2*c)^7 + 96*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 12*A*b^4*tan(1/2*d*x + 1/2*c)^7 - 15*C*b^4*tan(1/2*d*x
+ 1/2*c)^7 + 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 72*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 288*A*a*b^3*tan(1/2*d
*x + 1/2*c)^5 + 160*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 9*C*b^4*tan(1/2*d*x + 1
/2*c)^5 + 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 72*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 288*A*a*b^3*tan(1/2*d*x +
1/2*c)^3 + 160*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 9*C*b^4*tan(1/2*d*x + 1/2*c
)^3 + 96*C*a^3*b*tan(1/2*d*x + 1/2*c) + 72*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 96*A*a*b^3*tan(1/2*d*x + 1/2*c) +
96*C*a*b^3*tan(1/2*d*x + 1/2*c) + 12*A*b^4*tan(1/2*d*x + 1/2*c) + 15*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 + 1)^4)/d