3.961 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=223 \[ \frac{\tan (c+d x) \left (6 a^2 b (2 A+3 C)+4 a^3 B+16 a b^2 B+3 A b^3\right )}{6 d}+\frac{\left (a^3 (3 A+4 C)+12 a^2 b B+12 a b^2 (A+2 C)+8 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec (c+d x) \left (3 a^2 (3 A+4 C)+20 a b B+6 A b^2\right )}{24 d}+\frac{(4 a B+3 A b) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^3 C x \]

[Out]

b^3*C*x + ((12*a^2*b*B + 8*b^3*B + 12*a*b^2*(A + 2*C) + a^3*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*d) + ((3*A*
b^3 + 4*a^3*B + 16*a*b^2*B + 6*a^2*b*(2*A + 3*C))*Tan[c + d*x])/(6*d) + (a*(6*A*b^2 + 20*a*b*B + 3*a^2*(3*A +
4*C))*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((3*A*b + 4*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])
/(12*d) + (A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.670789, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3047, 3031, 3021, 2735, 3770} \[ \frac{\tan (c+d x) \left (6 a^2 b (2 A+3 C)+4 a^3 B+16 a b^2 B+3 A b^3\right )}{6 d}+\frac{\left (a^3 (3 A+4 C)+12 a^2 b B+12 a b^2 (A+2 C)+8 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec (c+d x) \left (3 a^2 (3 A+4 C)+20 a b B+6 A b^2\right )}{24 d}+\frac{(4 a B+3 A b) \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^3}{4 d}+b^3 C x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

b^3*C*x + ((12*a^2*b*B + 8*b^3*B + 12*a*b^2*(A + 2*C) + a^3*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*d) + ((3*A*
b^3 + 4*a^3*B + 16*a*b^2*B + 6*a^2*b*(2*A + 3*C))*Tan[c + d*x])/(6*d) + (a*(6*A*b^2 + 20*a*b*B + 3*a^2*(3*A +
4*C))*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((3*A*b + 4*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])
/(12*d) + (A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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 3031

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[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*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] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x))^2 \left (3 A b+4 a B+(3 a A+4 b B+4 a C) \cos (c+d x)+4 b C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{(3 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \cos (c+d x)) \left (6 A b^2+20 a b B+3 a^2 (3 A+4 C)+\left (15 a A b+8 a^2 B+12 b^2 B+24 a b C\right ) \cos (c+d x)+12 b^2 C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a \left (6 A b^2+20 a b B+3 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-4 \left (3 A b^3+4 a^3 B+16 a b^2 B+6 a^2 b (2 A+3 C)\right )-3 \left (12 a^2 b B+8 b^3 B+12 a b^2 (A+2 C)+a^3 (3 A+4 C)\right ) \cos (c+d x)-24 b^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{\left (3 A b^3+4 a^3 B+16 a b^2 B+6 a^2 b (2 A+3 C)\right ) \tan (c+d x)}{6 d}+\frac{a \left (6 A b^2+20 a b B+3 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{24} \int \left (-3 \left (12 a^2 b B+8 b^3 B+12 a b^2 (A+2 C)+a^3 (3 A+4 C)\right )-24 b^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^3 C x+\frac{\left (3 A b^3+4 a^3 B+16 a b^2 B+6 a^2 b (2 A+3 C)\right ) \tan (c+d x)}{6 d}+\frac{a \left (6 A b^2+20 a b B+3 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{8} \left (-12 a^2 b B-8 b^3 B-12 a b^2 (A+2 C)-a^3 (3 A+4 C)\right ) \int \sec (c+d x) \, dx\\ &=b^3 C x+\frac{\left (12 a^2 b B+8 b^3 B+12 a b^2 (A+2 C)+a^3 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\left (3 A b^3+4 a^3 B+16 a b^2 B+6 a^2 b (2 A+3 C)\right ) \tan (c+d x)}{6 d}+\frac{a \left (6 A b^2+20 a b B+3 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(3 A b+4 a B) (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac{A (a+b \cos (c+d x))^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.25263, size = 165, normalized size = 0.74 \[ \frac{3 \left (a^3 (3 A+4 C)+12 a^2 b B+12 a b^2 (A+2 C)+8 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))+3 \tan (c+d x) \left (a \sec (c+d x) \left (a^2 (3 A+4 C)+12 a b B+12 A b^2\right )+8 \left (3 a^2 b (A+C)+a^3 B+3 a b^2 B+A b^3\right )+2 a^3 A \sec ^3(c+d x)\right )+8 a^2 (a B+3 A b) \tan ^3(c+d x)+24 b^3 C d x}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(24*b^3*C*d*x + 3*(12*a^2*b*B + 8*b^3*B + 12*a*b^2*(A + 2*C) + a^3*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + 3*(8*(
A*b^3 + a^3*B + 3*a*b^2*B + 3*a^2*b*(A + C)) + a*(12*A*b^2 + 12*a*b*B + a^2*(3*A + 4*C))*Sec[c + d*x] + 2*a^3*
A*Sec[c + d*x]^3)*Tan[c + d*x] + 8*a^2*(3*A*b + a*B)*Tan[c + d*x]^3)/(24*d)

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Maple [A]  time = 0.085, size = 389, normalized size = 1.7 \begin{align*}{\frac{A{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{b}^{3}Cx+{\frac{C{b}^{3}c}{d}}+{\frac{3\,aA{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,aA{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{a{b}^{2}B\tan \left ( dx+c \right ) }{d}}+3\,{\frac{Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{A{a}^{2}b\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{2}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,{a}^{2}bB\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}bB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{2}bC\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,{a}^{3}B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

1/d*A*b^3*tan(d*x+c)+1/d*b^3*B*ln(sec(d*x+c)+tan(d*x+c))+b^3*C*x+1/d*C*b^3*c+3/2/d*a*A*b^2*sec(d*x+c)*tan(d*x+
c)+3/2/d*a*A*b^2*ln(sec(d*x+c)+tan(d*x+c))+3/d*a*b^2*B*tan(d*x+c)+3/d*C*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+2/d*A*
a^2*b*tan(d*x+c)+1/d*A*a^2*b*tan(d*x+c)*sec(d*x+c)^2+3/2/d*a^2*b*B*sec(d*x+c)*tan(d*x+c)+3/2/d*a^2*b*B*ln(sec(
d*x+c)+tan(d*x+c))+3/d*a^2*b*C*tan(d*x+c)+1/4/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+3/8/d*A*a^3*sec(d*x+c)*tan(d*x+c
)+3/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*a^3*B*tan(d*x+c)+1/3/d*a^3*B*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a^3*C
*sec(d*x+c)*tan(d*x+c)+1/2/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.00369, size = 502, normalized size = 2.25 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 48 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b + 48 \,{\left (d x + c\right )} C b^{3} - 3 \, A a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{3}{\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 )} - 36 \, B a^{2} 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 )} - 36 \, A a b^{2}{\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 )} + 72 \, C a b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, B b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 144 \, C a^{2} b \tan \left (d x + c\right ) + 144 \, B a b^{2} \tan \left (d x + c\right ) + 48 \, A b^{3} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 48*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2*b + 48*(d*x + c)
*C*b^3 - 3*A*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*
x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1)
+ log(sin(d*x + c) - 1)) - 36*B*a^2*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(
d*x + c) - 1)) - 36*A*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) -
1)) + 72*C*a*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*B*b^3*(log(sin(d*x + c) + 1) - log(sin(d
*x + c) - 1)) + 144*C*a^2*b*tan(d*x + c) + 144*B*a*b^2*tan(d*x + c) + 48*A*b^3*tan(d*x + c))/d

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Fricas [A]  time = 2.19594, size = 624, normalized size = 2.8 \begin{align*} \frac{48 \, C b^{3} d x \cos \left (d x + c\right )^{4} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, B a^{2} b + 12 \,{\left (A + 2 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, B a^{2} b + 12 \,{\left (A + 2 \, C\right )} a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, A a^{3} + 8 \,{\left (2 \, B a^{3} + 3 \,{\left (2 \, A + 3 \, C\right )} a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(48*C*b^3*d*x*cos(d*x + c)^4 + 3*((3*A + 4*C)*a^3 + 12*B*a^2*b + 12*(A + 2*C)*a*b^2 + 8*B*b^3)*cos(d*x +
c)^4*log(sin(d*x + c) + 1) - 3*((3*A + 4*C)*a^3 + 12*B*a^2*b + 12*(A + 2*C)*a*b^2 + 8*B*b^3)*cos(d*x + c)^4*lo
g(-sin(d*x + c) + 1) + 2*(6*A*a^3 + 8*(2*B*a^3 + 3*(2*A + 3*C)*a^2*b + 9*B*a*b^2 + 3*A*b^3)*cos(d*x + c)^3 + 3
*((3*A + 4*C)*a^3 + 12*B*a^2*b + 12*A*a*b^2)*cos(d*x + c)^2 + 8*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c)
)/(d*cos(d*x + c)^4)

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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))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.29304, size = 1025, normalized size = 4.6 \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))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(24*(d*x + c)*C*b^3 + 3*(3*A*a^3 + 4*C*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 24*C*a*b^2 + 8*B*b^3)*log(abs(tan(
1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a^3 + 4*C*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 24*C*a*b^2 + 8*B*b^3)*log(abs(tan(1/
2*d*x + 1/2*c) - 1)) + 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^3*tan(1/2
*d*x + 1/2*c)^7 - 72*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 36*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 72*C*a^2*b*tan(1/2*d
*x + 1/2*c)^7 + 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 72*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*A*b^3*tan(1/2*d*x +
1/2*c)^7 + 9*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^3*tan(1/2*d*x + 1/2*c)^5
+ 120*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 36*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 216*C*a^2*b*tan(1/2*d*x + 1/2*c)^5
- 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 216*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*A*b^3*tan(1/2*d*x + 1/2*c)^5 +
9*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 120*A*a^2
*b*tan(1/2*d*x + 1/2*c)^3 - 36*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 216*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*b^
2*tan(1/2*d*x + 1/2*c)^3 - 216*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^3*tan
(1/2*d*x + 1/2*c) + 24*B*a^3*tan(1/2*d*x + 1/2*c) + 12*C*a^3*tan(1/2*d*x + 1/2*c) + 72*A*a^2*b*tan(1/2*d*x + 1
/2*c) + 36*B*a^2*b*tan(1/2*d*x + 1/2*c) + 72*C*a^2*b*tan(1/2*d*x + 1/2*c) + 36*A*a*b^2*tan(1/2*d*x + 1/2*c) +
72*B*a*b^2*tan(1/2*d*x + 1/2*c) + 24*A*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d