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

Optimal. Leaf size=544 \[ \frac {b^2 \sin (c+d x) \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2+b^2 B (m+3)^2\right ) \sec ^{m+2}(c+d x)}{d (m+2) (m+3) (m+4)}+\frac {b \sin (c+d x) \left (2 a^3 B \left (m^2+8 m+19\right )+a^2 A b \left (5 m^2+37 m+68\right )+4 a b^2 B \left (m^2+6 m+8\right )+A b^3 \left (m^2+6 m+8\right )\right ) \sec ^{m+1}(c+d x)}{d (m+1) (m+3) (m+4)}-\frac {\sin (c+d x) \left (a^4 A \left (m^2+4 m+3\right )+4 a^3 b B m (m+3)+6 a^2 A b^2 m (m+3)+4 a b^3 B m (m+2)+A b^4 m (m+2)\right ) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(c+d x)\right )}{d (1-m) (m+1) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (a^4 B \left (m^2+6 m+8\right )+4 a^3 A b \left (m^2+6 m+8\right )+6 a^2 b^2 B \left (m^2+5 m+4\right )+4 a A b^3 \left (m^2+5 m+4\right )+b^4 B \left (m^2+4 m+3\right )\right ) \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},-\frac {m}{2};\frac {2-m}{2};\cos ^2(c+d x)\right )}{d m (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) (a B (m+7)+A b (m+4)) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))^2}{d (m+3) (m+4)}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))^3}{d (m+4)} \]

[Out]

b*(A*b^3*(m^2+6*m+8)+4*a*b^2*B*(m^2+6*m+8)+2*a^3*B*(m^2+8*m+19)+a^2*A*b*(5*m^2+37*m+68))*sec(d*x+c)^(1+m)*sin(
d*x+c)/d/(4+m)/(m^2+4*m+3)+b^2*(b^2*B*(3+m)^2+2*a*A*b*(4+m)^2+a^2*B*(m^2+9*m+26))*sec(d*x+c)^(2+m)*sin(d*x+c)/
d/(4+m)/(m^2+5*m+6)+b*(A*b*(4+m)+a*B*(7+m))*sec(d*x+c)^(1+m)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/(3+m)/(4+m)+b*B*s
ec(d*x+c)^(1+m)*(a+b*sec(d*x+c))^3*sin(d*x+c)/d/(4+m)-(A*b^4*m*(2+m)+4*a*b^3*B*m*(2+m)+6*a^2*A*b^2*m*(3+m)+4*a
^3*b*B*m*(3+m)+a^4*A*(m^2+4*m+3))*hypergeom([1/2, 1/2-1/2*m],[3/2-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*sin(d
*x+c)/d/(3+m)/(-m^2+1)/(sin(d*x+c)^2)^(1/2)+(b^4*B*(m^2+4*m+3)+4*a*A*b^3*(m^2+5*m+4)+6*a^2*b^2*B*(m^2+5*m+4)+4
*a^3*A*b*(m^2+6*m+8)+a^4*B*(m^2+6*m+8))*hypergeom([1/2, -1/2*m],[1-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^m*sin(d*x+c
)/d/m/(2+m)/(4+m)/(sin(d*x+c)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.63, antiderivative size = 544, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4026, 4096, 4076, 4047, 3772, 2643, 4046} \[ -\frac {\sin (c+d x) \left (6 a^2 A b^2 m (m+3)+a^4 A \left (m^2+4 m+3\right )+4 a^3 b B m (m+3)+4 a b^3 B m (m+2)+A b^4 m (m+2)\right ) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(c+d x)\right )}{d (1-m) (m+1) (m+3) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (4 a^3 A b \left (m^2+6 m+8\right )+6 a^2 b^2 B \left (m^2+5 m+4\right )+a^4 B \left (m^2+6 m+8\right )+4 a A b^3 \left (m^2+5 m+4\right )+b^4 B \left (m^2+4 m+3\right )\right ) \sec ^m(c+d x) \, _2F_1\left (\frac {1}{2},-\frac {m}{2};\frac {2-m}{2};\cos ^2(c+d x)\right )}{d m (m+2) (m+4) \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) \left (a^2 A b \left (5 m^2+37 m+68\right )+2 a^3 B \left (m^2+8 m+19\right )+4 a b^2 B \left (m^2+6 m+8\right )+A b^3 \left (m^2+6 m+8\right )\right ) \sec ^{m+1}(c+d x)}{d (m+1) (m+3) (m+4)}+\frac {b^2 \sin (c+d x) \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2+b^2 B (m+3)^2\right ) \sec ^{m+2}(c+d x)}{d (m+2) (m+3) (m+4)}+\frac {b \sin (c+d x) (a B (m+7)+A b (m+4)) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))^2}{d (m+3) (m+4)}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))^3}{d (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(A*b^3*(8 + 6*m + m^2) + 4*a*b^2*B*(8 + 6*m + m^2) + 2*a^3*B*(19 + 8*m + m^2) + a^2*A*b*(68 + 37*m + 5*m^2)
)*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(1 + m)*(3 + m)*(4 + m)) + (b^2*(b^2*B*(3 + m)^2 + 2*a*A*b*(4 + m)^2 +
a^2*B*(26 + 9*m + m^2))*Sec[c + d*x]^(2 + m)*Sin[c + d*x])/(d*(2 + m)*(3 + m)*(4 + m)) + (b*(A*b*(4 + m) + a*
B*(7 + m))*Sec[c + d*x]^(1 + m)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(d*(3 + m)*(4 + m)) + (b*B*Sec[c + d*x]^(
1 + m)*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(d*(4 + m)) - ((A*b^4*m*(2 + m) + 4*a*b^3*B*m*(2 + m) + 6*a^2*A*b^
2*m*(3 + m) + 4*a^3*b*B*m*(3 + m) + a^4*A*(3 + 4*m + m^2))*Hypergeometric2F1[1/2, (1 - m)/2, (3 - m)/2, Cos[c
+ d*x]^2]*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(1 - m)*(1 + m)*(3 + m)*Sqrt[Sin[c + d*x]^2]) + ((b^4*B*(3 +
4*m + m^2) + 4*a*A*b^3*(4 + 5*m + m^2) + 6*a^2*b^2*B*(4 + 5*m + m^2) + 4*a^3*A*b*(8 + 6*m + m^2) + a^4*B*(8 +
6*m + m^2))*Hypergeometric2F1[1/2, -m/2, (2 - m)/2, Cos[c + d*x]^2]*Sec[c + d*x]^m*Sin[c + d*x])/(d*m*(2 + m)*
(4 + m)*Sqrt[Sin[c + d*x]^2])

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3772

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

Rule 4026

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

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, 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 4076

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[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -1]

Rule 4096

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[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(m + n + 1), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^
n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a*C
*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] &&
!LeQ[n, -1]

Rubi steps

\begin {align*} \int \sec ^m(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{d (4+m)}+\frac {\int \sec ^m(c+d x) (a+b \sec (c+d x))^2 \left (a (b B m+a A (4+m))+\left (b^2 B (3+m)+a (2 A b+a B) (4+m)\right ) \sec (c+d x)+b (A b (4+m)+a B (7+m)) \sec ^2(c+d x)\right ) \, dx}{4+m}\\ &=\frac {b (A b (4+m)+a B (7+m)) \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{d (3+m) (4+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{d (4+m)}+\frac {\int \sec ^m(c+d x) (a+b \sec (c+d x)) \left (a \left (A b^2 m (4+m)+2 a b B m (5+m)+a^2 A \left (12+7 m+m^2\right )\right )+\left (b^2 (2+m) (A b (4+m)+a B (7+m))+a (3+m) \left (3 a A b (4+m)+a^2 B (4+m)+b^2 B (3+2 m)\right )\right ) \sec (c+d x)+b \left (b^2 B (3+m)^2+2 a A b (4+m)^2+a^2 B \left (26+9 m+m^2\right )\right ) \sec ^2(c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac {b^2 \left (b^2 B (3+m)^2+2 a A b (4+m)^2+a^2 B \left (26+9 m+m^2\right )\right ) \sec ^{2+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac {b (A b (4+m)+a B (7+m)) \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{d (3+m) (4+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{d (4+m)}+\frac {\int \sec ^m(c+d x) \left (a^2 (2+m) \left (A b^2 m (4+m)+2 a b B m (5+m)+a^2 A \left (12+7 m+m^2\right )\right )+(3+m) \left (b^4 B \left (3+4 m+m^2\right )+4 a A b^3 \left (4+5 m+m^2\right )+6 a^2 b^2 B \left (4+5 m+m^2\right )+4 a^3 A b \left (8+6 m+m^2\right )+a^4 B \left (8+6 m+m^2\right )\right ) \sec (c+d x)+b (2+m) \left (A b^3 \left (8+6 m+m^2\right )+4 a b^2 B \left (8+6 m+m^2\right )+2 a^3 B \left (19+8 m+m^2\right )+a^2 A b \left (68+37 m+5 m^2\right )\right ) \sec ^2(c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=\frac {b^2 \left (b^2 B (3+m)^2+2 a A b (4+m)^2+a^2 B \left (26+9 m+m^2\right )\right ) \sec ^{2+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac {b (A b (4+m)+a B (7+m)) \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{d (3+m) (4+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{d (4+m)}+\frac {\int \sec ^m(c+d x) \left (a^2 (2+m) \left (A b^2 m (4+m)+2 a b B m (5+m)+a^2 A \left (12+7 m+m^2\right )\right )+b (2+m) \left (A b^3 \left (8+6 m+m^2\right )+4 a b^2 B \left (8+6 m+m^2\right )+2 a^3 B \left (19+8 m+m^2\right )+a^2 A b \left (68+37 m+5 m^2\right )\right ) \sec ^2(c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}+\frac {\left (b^4 B \left (3+4 m+m^2\right )+4 a A b^3 \left (4+5 m+m^2\right )+6 a^2 b^2 B \left (4+5 m+m^2\right )+4 a^3 A b \left (8+6 m+m^2\right )+a^4 B \left (8+6 m+m^2\right )\right ) \int \sec ^{1+m}(c+d x) \, dx}{8+6 m+m^2}\\ &=\frac {b \left (A b^3 \left (8+6 m+m^2\right )+4 a b^2 B \left (8+6 m+m^2\right )+2 a^3 B \left (19+8 m+m^2\right )+a^2 A b \left (68+37 m+5 m^2\right )\right ) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) \left (12+7 m+m^2\right )}+\frac {b^2 \left (b^2 B (3+m)^2+2 a A b (4+m)^2+a^2 B \left (26+9 m+m^2\right )\right ) \sec ^{2+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac {b (A b (4+m)+a B (7+m)) \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{d (3+m) (4+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{d (4+m)}+\frac {\left (A b^4 m (2+m)+4 a b^3 B m (2+m)+6 a^2 A b^2 m (3+m)+4 a^3 b B m (3+m)+a^4 A \left (3+4 m+m^2\right )\right ) \int \sec ^m(c+d x) \, dx}{(1+m) (3+m)}+\frac {\left (\left (b^4 B \left (3+4 m+m^2\right )+4 a A b^3 \left (4+5 m+m^2\right )+6 a^2 b^2 B \left (4+5 m+m^2\right )+4 a^3 A b \left (8+6 m+m^2\right )+a^4 B \left (8+6 m+m^2\right )\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{-1-m}(c+d x) \, dx}{8+6 m+m^2}\\ &=\frac {b \left (A b^3 \left (8+6 m+m^2\right )+4 a b^2 B \left (8+6 m+m^2\right )+2 a^3 B \left (19+8 m+m^2\right )+a^2 A b \left (68+37 m+5 m^2\right )\right ) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) \left (12+7 m+m^2\right )}+\frac {b^2 \left (b^2 B (3+m)^2+2 a A b (4+m)^2+a^2 B \left (26+9 m+m^2\right )\right ) \sec ^{2+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac {b (A b (4+m)+a B (7+m)) \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{d (3+m) (4+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{d (4+m)}+\frac {\left (b^4 B \left (3+4 m+m^2\right )+4 a A b^3 \left (4+5 m+m^2\right )+6 a^2 b^2 B \left (4+5 m+m^2\right )+4 a^3 A b \left (8+6 m+m^2\right )+a^4 B \left (8+6 m+m^2\right )\right ) \, _2F_1\left (\frac {1}{2},-\frac {m}{2};\frac {2-m}{2};\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m \left (8+6 m+m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\left (\left (A b^4 m (2+m)+4 a b^3 B m (2+m)+6 a^2 A b^2 m (3+m)+4 a^3 b B m (3+m)+a^4 A \left (3+4 m+m^2\right )\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{-m}(c+d x) \, dx}{(1+m) (3+m)}\\ &=\frac {b \left (A b^3 \left (8+6 m+m^2\right )+4 a b^2 B \left (8+6 m+m^2\right )+2 a^3 B \left (19+8 m+m^2\right )+a^2 A b \left (68+37 m+5 m^2\right )\right ) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) \left (12+7 m+m^2\right )}+\frac {b^2 \left (b^2 B (3+m)^2+2 a A b (4+m)^2+a^2 B \left (26+9 m+m^2\right )\right ) \sec ^{2+m}(c+d x) \sin (c+d x)}{d (2+m) \left (12+7 m+m^2\right )}+\frac {b (A b (4+m)+a B (7+m)) \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{d (3+m) (4+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{d (4+m)}-\frac {\left (A b^4 m (2+m)+4 a b^3 B m (2+m)+6 a^2 A b^2 m (3+m)+4 a^3 b B m (3+m)+a^4 A \left (3+4 m+m^2\right )\right ) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (1-m) (1+m) (3+m) \sqrt {\sin ^2(c+d x)}}+\frac {\left (b^4 B \left (3+4 m+m^2\right )+4 a A b^3 \left (4+5 m+m^2\right )+6 a^2 b^2 B \left (4+5 m+m^2\right )+4 a^3 A b \left (8+6 m+m^2\right )+a^4 B \left (8+6 m+m^2\right )\right ) \, _2F_1\left (\frac {1}{2},-\frac {m}{2};\frac {2-m}{2};\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m \left (8+6 m+m^2\right ) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 4.92, size = 365, normalized size = 0.67 \[ \frac {\sqrt {-\tan ^2(c+d x)} \csc (c+d x) \sec ^{m-1}(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \left (\frac {a^4 A \cos ^5(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};\sec ^2(c+d x)\right )}{m}+\frac {a^3 (a B+4 A b) \cos ^4(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sec ^2(c+d x)\right )}{m+1}+b \left (\frac {2 a^2 (2 a B+3 A b) \cos ^3(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};\sec ^2(c+d x)\right )}{m+2}+b \left (\frac {2 a (3 a B+2 A b) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};\sec ^2(c+d x)\right )}{m+3}+b \left (\frac {(4 a B+A b) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m+4}{2};\frac {m+6}{2};\sec ^2(c+d x)\right )}{m+4}+\frac {b B \, _2F_1\left (\frac {1}{2},\frac {m+5}{2};\frac {m+7}{2};\sec ^2(c+d x)\right )}{m+5}\right )\right )\right )\right )}{d (a \cos (c+d x)+b)^4 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*((a^4*A*Cos[c + d*x]^5*Hypergeometric2F1[1/2, m/2, (2 + m)/2, Sec[c + d*x]^2])/m + (a^3*(4*A*b +
a*B)*Cos[c + d*x]^4*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sec[c + d*x]^2])/(1 + m) + b*((2*a^2*(3*A*b
+ 2*a*B)*Cos[c + d*x]^3*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sec[c + d*x]^2])/(2 + m) + b*((2*a*(2*A*b
+ 3*a*B)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Sec[c + d*x]^2])/(3 + m) + b*(((A*b + 4*
a*B)*Cos[c + d*x]*Hypergeometric2F1[1/2, (4 + m)/2, (6 + m)/2, Sec[c + d*x]^2])/(4 + m) + (b*B*Hypergeometric2
F1[1/2, (5 + m)/2, (7 + m)/2, Sec[c + d*x]^2])/(5 + m)))))*Sec[c + d*x]^(-1 + m)*(a + b*Sec[c + d*x])^4*(A + B
*Sec[c + d*x])*Sqrt[-Tan[c + d*x]^2])/(d*(b + a*Cos[c + d*x])^4*(B + A*Cos[c + d*x]))

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b^{4} \sec \left (d x + c\right )^{5} + A a^{4} + {\left (4 \, B a b^{3} + A b^{4}\right )} \sec \left (d x + c\right )^{4} + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \sec \left (d x + c\right )^{3} + 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} \sec \left (d x + c\right )^{2} + {\left (B a^{4} + 4 \, A a^{3} b\right )} \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^4*sec(d*x + c)^5 + A*a^4 + (4*B*a*b^3 + A*b^4)*sec(d*x + c)^4 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*sec(
d*x + c)^3 + 2*(2*B*a^3*b + 3*A*a^2*b^2)*sec(d*x + c)^2 + (B*a^4 + 4*A*a^3*b)*sec(d*x + c))*sec(d*x + c)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^4*sec(d*x + c)^m, x)

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maple [F]  time = 2.48, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{m}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right )^{4} \left (A +B \sec \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^4\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^4*(1/cos(c + d*x))^m,x)

[Out]

int((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^4*(1/cos(c + d*x))^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{4} \sec ^{m}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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