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

Optimal. Leaf size=156 \[ \frac {\left (2 a^2 B+4 a A b+3 b^2 B\right ) \tan (c+d x)}{3 d}+\frac {\left (3 a^2 A+8 a b B+4 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (3 a^2 A+8 a b B+4 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2988, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {\left (2 a^2 B+4 a A b+3 b^2 B\right ) \tan (c+d x)}{3 d}+\frac {\left (3 a^2 A+8 a b B+4 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (3 a^2 A+8 a b B+4 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2988

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/
(f*d^2*(n + 1)*(c^2 - d^2)), x] - Dist[1/(d^2*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*Simp[d*(n
+ 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n +
1))) + 2*a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x
]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && LtQ[n, -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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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

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Mathematica [A]  time = 0.72, size = 120, normalized size = 0.77 \[ \frac {3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \sec (c+d x)+24 \left (a^2 B+2 a A b+b^2 B\right )+6 a^2 A \sec ^3(c+d x)+8 a (a B+2 A b) \tan ^2(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 180, normalized size = 1.15 \[ \frac {3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, B a^{2} + 4 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, A a^{2} + 3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.55, size = 478, normalized size = 3.06 \[ \frac {3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B 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 )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*log
(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*A*
a*b*tan(1/2*d*x + 1/2*c)^7 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*b^2*tan(
1/2*d*x + 1/2*c)^7 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 80*A*a*b*tan(1/2*d*x +
1/2*c)^5 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*B*b^2*tan(1/2*d*x + 1/2*c)^
5 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*B*
a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^2*tan(
1/2*d*x + 1/2*c) + 24*B*a^2*tan(1/2*d*x + 1/2*c) + 48*A*a*b*tan(1/2*d*x + 1/2*c) + 24*B*a*b*tan(1/2*d*x + 1/2*
c) + 12*A*b^2*tan(1/2*d*x + 1/2*c) + 24*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 0.13, size = 241, normalized size = 1.54 \[ \frac {a^{2} A \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 B \,a^{2} \tan \left (d x +c \right )}{3 d}+\frac {B \,a^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {4 A a b \tan \left (d x +c \right )}{3 d}+\frac {2 A a b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {B a b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{2} B \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 228, normalized size = 1.46 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b - 3 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B b^{2} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 32*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b - 3*A*a^2*(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)) - 24*B*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1
)) - 12*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)) + 48*B*b^2
*tan(d*x + c))/d

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mupad [B]  time = 3.87, size = 314, normalized size = 2.01 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,a^2}{8}+B\,a\,b+\frac {A\,b^2}{2}\right )}{\frac {3\,A\,a^2}{2}+4\,B\,a\,b+2\,A\,b^2}\right )\,\left (\frac {3\,A\,a^2}{4}+2\,B\,a\,b+A\,b^2\right )}{d}+\frac {\left (\frac {5\,A\,a^2}{4}+A\,b^2-2\,B\,a^2-2\,B\,b^2-4\,A\,a\,b+2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a^2}{4}-A\,b^2+\frac {10\,B\,a^2}{3}+6\,B\,b^2+\frac {20\,A\,a\,b}{3}-2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,A\,a^2}{4}-A\,b^2-\frac {10\,B\,a^2}{3}-6\,B\,b^2-\frac {20\,A\,a\,b}{3}-2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a^2}{4}+A\,b^2+2\,B\,a^2+2\,B\,b^2+4\,A\,a\,b+2\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((3*A*a^2)/8 + (A*b^2)/2 + B*a*b))/((3*A*a^2)/2 + 2*A*b^2 + 4*B*a*b))*((3*A*a^2)/
4 + A*b^2 + 2*B*a*b))/d + (tan(c/2 + (d*x)/2)^7*((5*A*a^2)/4 + A*b^2 - 2*B*a^2 - 2*B*b^2 - 4*A*a*b + 2*B*a*b)
- tan(c/2 + (d*x)/2)^3*(A*b^2 - (3*A*a^2)/4 + (10*B*a^2)/3 + 6*B*b^2 + (20*A*a*b)/3 + 2*B*a*b) + tan(c/2 + (d*
x)/2)^5*((3*A*a^2)/4 - A*b^2 + (10*B*a^2)/3 + 6*B*b^2 + (20*A*a*b)/3 - 2*B*a*b) + tan(c/2 + (d*x)/2)*((5*A*a^2
)/4 + A*b^2 + 2*B*a^2 + 2*B*b^2 + 4*A*a*b + 2*B*a*b))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*
tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**5,x)

[Out]

Timed out

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