3.308 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=216 \[ \frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {a \left (4 a^3 B+16 a^2 A b+34 a b^2 B+19 A b^3\right ) \sin (c+d x)}{6 d}+\frac {1}{8} x \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac {b^4 B \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

1/8*(3*A*a^4+24*A*a^2*b^2+8*A*b^4+16*B*a^3*b+32*B*a*b^3)*x+b^4*B*arctanh(sin(d*x+c))/d+1/6*a*(16*A*a^2*b+19*A*
b^3+4*B*a^3+34*B*a*b^2)*sin(d*x+c)/d+1/24*a^2*(9*A*a^2+26*A*b^2+32*B*a*b)*cos(d*x+c)*sin(d*x+c)/d+1/12*a*(7*A*
b+4*B*a)*cos(d*x+c)^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/4*a*A*cos(d*x+c)^3*(a+b*sec(d*x+c))^3*sin(d*x+c)/d

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Rubi [A]  time = 0.61, antiderivative size = 216, 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 = {4025, 4094, 4074, 4047, 8, 4045, 3770} \[ \frac {a \left (16 a^2 A b+4 a^3 B+34 a b^2 B+19 A b^3\right ) \sin (c+d x)}{6 d}+\frac {a^2 \left (9 a^2 A+32 a b B+26 A b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (24 a^2 A b^2+3 a^4 A+16 a^3 b B+32 a b^3 B+8 A b^4\right )+\frac {a (4 a B+7 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac {b^4 B \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*x)/8 + (b^4*B*ArcTanh[Sin[c + d*x]])/d + (a*(16*
a^2*A*b + 19*A*b^3 + 4*a^3*B + 34*a*b^2*B)*Sin[c + d*x])/(6*d) + (a^2*(9*a^2*A + 26*A*b^2 + 32*a*b*B)*Cos[c +
d*x]*Sin[c + d*x])/(24*d) + (a*(7*A*b + 4*a*B)*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(12*d) + (a
*A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*d)

Rule 8

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4074

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4094

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (-a (7 A b+4 a B)-\left (3 a^2 A+4 A b^2+8 a b B\right ) \sec (c+d x)-4 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac {1}{12} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (-a \left (9 a^2 A+26 A b^2+32 a b B\right )-\left (23 a^2 A b+12 A b^3+8 a^3 B+36 a b^2 B\right ) \sec (c+d x)-12 b^3 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{24} \int \cos (c+d x) \left (4 a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right )+3 \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) \sec (c+d x)+24 b^4 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{24} \int \cos (c+d x) \left (4 a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right )+24 b^4 B \sec ^2(c+d x)\right ) \, dx+\frac {1}{8} \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) \int 1 \, dx\\ &=\frac {1}{8} \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) x+\frac {a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \sin (c+d x)}{6 d}+\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\left (b^4 B\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} \left (3 a^4 A+24 a^2 A b^2+8 A b^4+16 a^3 b B+32 a b^3 B\right ) x+\frac {b^4 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a \left (16 a^2 A b+19 A b^3+4 a^3 B+34 a b^2 B\right ) \sin (c+d x)}{6 d}+\frac {a^2 \left (9 a^2 A+26 A b^2+32 a b B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {a (7 A b+4 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{12 d}+\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 210, normalized size = 0.97 \[ \frac {3 a^4 A \sin (4 (c+d x))+8 a^3 (a B+4 A b) \sin (3 (c+d x))+24 a^2 \left (a^2 A+4 a b B+6 A b^2\right ) \sin (2 (c+d x))+24 a \left (3 a^3 B+12 a^2 A b+24 a b^2 B+16 A b^3\right ) \sin (c+d x)+12 (c+d x) \left (3 a^4 A+16 a^3 b B+24 a^2 A b^2+32 a b^3 B+8 A b^4\right )-96 b^4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+96 b^4 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(3*a^4*A + 24*a^2*A*b^2 + 8*A*b^4 + 16*a^3*b*B + 32*a*b^3*B)*(c + d*x) - 96*b^4*B*Log[Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] + 96*b^4*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 24*a*(12*a^2*A*b + 16*A*b^3 + 3*a^3*B +
24*a*b^2*B)*Sin[c + d*x] + 24*a^2*(a^2*A + 6*A*b^2 + 4*a*b*B)*Sin[2*(c + d*x)] + 8*a^3*(4*A*b + a*B)*Sin[3*(c
+ d*x)] + 3*a^4*A*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.48, size = 183, normalized size = 0.85 \[ \frac {12 \, B b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, B b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} d x + {\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 16 \, B a^{4} + 64 \, A a^{3} b + 144 \, B a^{2} b^{2} + 96 \, A a b^{3} + 8 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(12*B*b^4*log(sin(d*x + c) + 1) - 12*B*b^4*log(-sin(d*x + c) + 1) + 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^
2 + 32*B*a*b^3 + 8*A*b^4)*d*x + (6*A*a^4*cos(d*x + c)^3 + 16*B*a^4 + 64*A*a^3*b + 144*B*a^2*b^2 + 96*A*a*b^3 +
8*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^2 + 3*(3*A*a^4 + 16*B*a^3*b + 24*A*a^2*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.42, size = 603, normalized size = 2.79 \[ \frac {24 \, B b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, B b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (3 \, A a^{4} + 16 \, B a^{3} b + 24 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 144 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 432 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 288 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 432 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, A a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, B a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 144 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, A a b^{3} \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(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(24*B*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*B*b^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(3*A*a^4 +
16*B*a^3*b + 24*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*(d*x + c) - 2*(15*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^4*t
an(1/2*d*x + 1/2*c)^7 - 96*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 72*A*a^2*b^2*t
an(1/2*d*x + 1/2*c)^7 - 144*B*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 - 9*A*a^4*tan
(1/2*d*x + 1/2*c)^5 - 40*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 160*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 48*B*a^3*b*tan(1/
2*d*x + 1/2*c)^5 + 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 432*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 288*A*a*b^3*ta
n(1/2*d*x + 1/2*c)^5 + 9*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 160*A*a^3*b*tan(1/2*
d*x + 1/2*c)^3 - 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 432*B*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 - 15*A*a^4*tan(1/2*d*x + 1/2*c) - 24*B*a^4*tan(1/2*d*x
+ 1/2*c) - 96*A*a^3*b*tan(1/2*d*x + 1/2*c) - 48*B*a^3*b*tan(1/2*d*x + 1/2*c) - 72*A*a^2*b^2*tan(1/2*d*x + 1/2*
c) - 144*B*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))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 1.18, size = 319, normalized size = 1.48 \[ \frac {A \,a^{4} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {3 A \,a^{4} x}{8}+\frac {3 A \,a^{4} c}{8 d}+\frac {B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}+\frac {2 a^{4} B \sin \left (d x +c \right )}{3 d}+\frac {4 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{3} b}{3 d}+\frac {8 A \,a^{3} b \sin \left (d x +c \right )}{3 d}+\frac {2 B \,a^{3} b \sin \left (d x +c \right ) \cos \left (d x +c \right )}{d}+2 B x \,a^{3} b +\frac {2 B \,a^{3} b c}{d}+\frac {3 A \,a^{2} b^{2} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{d}+3 A x \,a^{2} b^{2}+\frac {3 A \,a^{2} b^{2} c}{d}+\frac {6 a^{2} b^{2} B \sin \left (d x +c \right )}{d}+\frac {4 a A \,b^{3} \sin \left (d x +c \right )}{d}+4 B x a \,b^{3}+\frac {4 B a \,b^{3} c}{d}+A x \,b^{4}+\frac {A \,b^{4} c}{d}+\frac {B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.70, size = 215, normalized size = 1.00 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 128 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} b + 96 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} b + 144 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b^{2} + 384 \, {\left (d x + c\right )} B a b^{3} + 96 \, {\left (d x + c\right )} A b^{4} + 48 \, B b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 576 \, B a^{2} b^{2} \sin \left (d x + c\right ) + 384 \, A a b^{3} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B
*a^4 - 128*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3*b + 96*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3*b + 144*(2*d*
x + 2*c + sin(2*d*x + 2*c))*A*a^2*b^2 + 384*(d*x + c)*B*a*b^3 + 96*(d*x + c)*A*b^4 + 48*B*b^4*(log(sin(d*x + c
) + 1) - log(sin(d*x + c) - 1)) + 576*B*a^2*b^2*sin(d*x + c) + 384*A*a*b^3*sin(d*x + c))/d

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mupad [B]  time = 3.38, size = 369, normalized size = 1.71 \[ \frac {3\,B\,a^4\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {2\,A\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,b^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {8\,B\,a\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,B\,a^3\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {B\,a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {6\,B\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {6\,A\,a^2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {4\,A\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,A\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^4,x)

[Out]

(3*B*a^4*sin(c + d*x))/(4*d) + (3*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (2*A*b^4*atan(sin
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*b^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^4*sin
(2*c + 2*d*x))/(4*d) + (A*a^4*sin(4*c + 4*d*x))/(32*d) + (B*a^4*sin(3*c + 3*d*x))/(12*d) + (8*B*a*b^3*atan(sin
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*B*a^3*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^3*b*
sin(3*c + 3*d*x))/(3*d) + (B*a^3*b*sin(2*c + 2*d*x))/d + (6*B*a^2*b^2*sin(c + d*x))/d + (6*A*a^2*b^2*atan(sin(
c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*A*a^2*b^2*sin(2*c + 2*d*x))/(2*d) + (4*A*a*b^3*sin(c + d*x))/d + (3
*A*a^3*b*sin(c + d*x))/d

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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(cos(d*x+c)**4*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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