\(\int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [956]

Optimal result

Integrand size = 43, antiderivative size = 652 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(a-b) \sqrt {a+b} \left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{192 a b d}+\frac {\sqrt {a+b} \left (15 A b^3+8 a^3 (9 A+16 B+12 C)+4 a^2 b (71 A+52 B+108 C)+2 a b^2 (59 A+132 B+192 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{192 a d}+\frac {\sqrt {a+b} \left (5 A b^4-160 a^3 b B-40 a b^3 B-120 a^2 b^2 (A+2 C)-16 a^4 (3 A+4 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{64 a^2 d}+\frac {\left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac {\left (5 A b^2+24 a b B+4 a^2 (3 A+4 C)\right ) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {(5 A b+8 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{4 d} \]

[Out]

1/24*(5*A*b+8*B*a)*cos(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+b*sec(d*x+c))^(5/2)*
sin(d*x+c)/d+1/192*(a-b)*(15*A*b^3+128*B*a^3+264*B*a*b^2+4*a^2*b*(71*A+108*C))*cot(d*x+c)*EllipticE((a+b*sec(d
*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a
-b))^(1/2)/a/b/d+1/192*(15*A*b^3+8*a^3*(9*A+16*B+12*C)+4*a^2*b*(71*A+52*B+108*C)+2*a*b^2*(59*A+132*B+192*C))*c
ot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b
))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a/d+1/64*(5*A*b^4-160*B*a^3*b-40*B*a*b^3-120*a^2*b^2*(A+2*C)-16*a^4*(
3*A+4*C))*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b
*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d+1/192*(15*A*b^3+128*B*a^3+264*B*a*b^2+4*a^2
*b*(71*A+108*C))*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d+1/32*(5*A*b^2+24*B*a*b+4*a^2*(3*A+4*C))*cos(d*x+c)*sin(
d*x+c)*(a+b*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 2.22 (sec) , antiderivative size = 652, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4179, 4189, 4143, 4006, 3869, 3917, 4089} \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sin (c+d x) \cos (c+d x) \left (4 a^2 (3 A+4 C)+24 a b B+5 A b^2\right ) \sqrt {a+b \sec (c+d x)}}{32 d}+\frac {\sqrt {a+b} \cot (c+d x) \left (8 a^3 (9 A+16 B+12 C)+4 a^2 b (71 A+52 B+108 C)+2 a b^2 (59 A+132 B+192 C)+15 A b^3\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{192 a d}+\frac {(a-b) \sqrt {a+b} \cot (c+d x) \left (128 a^3 B+4 a^2 b (71 A+108 C)+264 a b^2 B+15 A b^3\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{192 a b d}+\frac {\sin (c+d x) \left (128 a^3 B+4 a^2 b (71 A+108 C)+264 a b^2 B+15 A b^3\right ) \sqrt {a+b \sec (c+d x)}}{192 a d}+\frac {\sqrt {a+b} \cot (c+d x) \left (-16 a^4 (3 A+4 C)-160 a^3 b B-120 a^2 b^2 (A+2 C)-40 a b^3 B+5 A b^4\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{64 a^2 d}+\frac {(8 a B+5 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2}}{24 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2}}{4 d} \]

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((a - b)*Sqrt[a + b]*(15*A*b^3 + 128*a^3*B + 264*a*b^2*B + 4*a^2*b*(71*A + 108*C))*Cot[c + d*x]*EllipticE[ArcS
in[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 +
Sec[c + d*x]))/(a - b))])/(192*a*b*d) + (Sqrt[a + b]*(15*A*b^3 + 8*a^3*(9*A + 16*B + 12*C) + 4*a^2*b*(71*A +
52*B + 108*C) + 2*a*b^2*(59*A + 132*B + 192*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a
+ b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(192*a*d
) + (Sqrt[a + b]*(5*A*b^4 - 160*a^3*b*B - 40*a*b^3*B - 120*a^2*b^2*(A + 2*C) - 16*a^4*(3*A + 4*C))*Cot[c + d*x
]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*
x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(64*a^2*d) + ((15*A*b^3 + 128*a^3*B + 264*a*b^2*B + 4*a
^2*b*(71*A + 108*C))*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(192*a*d) + ((5*A*b^2 + 24*a*b*B + 4*a^2*(3*A + 4*
C))*Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(32*d) + ((5*A*b + 8*a*B)*Cos[c + d*x]^2*(a + b*Sec[c
+ d*x])^(3/2)*Sin[c + d*x])/(24*d) + (A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*d)

Rule 3869

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b
*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b)*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b
*Csc[c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4006

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4143

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[Csc[e + f*x
]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]

Rule 4179

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]

Rule 4189

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

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {1}{2} (5 A b+8 a B)+(3 a A+4 b B+4 a C) \sec (c+d x)+\frac {1}{2} b (A+8 C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {(5 A b+8 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {1}{12} \int \cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {3}{4} \left (5 A b^2+24 a b B+4 a^2 (3 A+4 C)\right )+\frac {1}{2} \left (16 a^2 B+24 b^2 B+a b (31 A+48 C)\right ) \sec (c+d x)+\frac {1}{4} b (11 A b+8 a B+48 b C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {\left (5 A b^2+24 a b B+4 a^2 (3 A+4 C)\right ) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {(5 A b+8 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}+\frac {1}{24} \int \frac {\cos (c+d x) \left (\frac {1}{8} \left (15 A b^3+128 a^3 B+264 a b^2 B+8 a^2 \left (\frac {71 A b}{2}+54 b C\right )\right )+\frac {1}{4} \left (152 a^2 b B+96 b^3 B+12 a^3 (3 A+4 C)+a b^2 (161 A+288 C)\right ) \sec (c+d x)+\frac {1}{8} b \left (104 a b B+12 a^2 (3 A+4 C)+b^2 (59 A+192 C)\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx \\ & = \frac {\left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac {\left (5 A b^2+24 a b B+4 a^2 (3 A+4 C)\right ) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {(5 A b+8 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}-\frac {\int \frac {\frac {3}{16} \left (5 A b^4-160 a^3 b B-40 a b^3 B-120 a^2 b^2 (A+2 C)-16 a^4 (3 A+4 C)\right )-\frac {1}{8} a b \left (104 a b B+12 a^2 (3 A+4 C)+b^2 (59 A+192 C)\right ) \sec (c+d x)+\frac {1}{16} b \left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 a} \\ & = \frac {\left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac {\left (5 A b^2+24 a b B+4 a^2 (3 A+4 C)\right ) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {(5 A b+8 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}-\frac {\int \frac {\frac {3}{16} \left (5 A b^4-160 a^3 b B-40 a b^3 B-120 a^2 b^2 (A+2 C)-16 a^4 (3 A+4 C)\right )+\left (-\frac {1}{16} b \left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right )-\frac {1}{8} a b \left (104 a b B+12 a^2 (3 A+4 C)+b^2 (59 A+192 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 a}-\frac {\left (b \left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{384 a} \\ & = \frac {(a-b) \sqrt {a+b} \left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{192 a b d}+\frac {\left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac {\left (5 A b^2+24 a b B+4 a^2 (3 A+4 C)\right ) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {(5 A b+8 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{4 d}-\frac {\left (5 A b^4-160 a^3 b B-40 a b^3 B-120 a^2 b^2 (A+2 C)-16 a^4 (3 A+4 C)\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{128 a}+\frac {\left (b \left (15 A b^3+8 a^3 (9 A+16 B+12 C)+4 a^2 b (71 A+52 B+108 C)+2 a b^2 (59 A+132 B+192 C)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{384 a} \\ & = \frac {(a-b) \sqrt {a+b} \left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{192 a b d}+\frac {\sqrt {a+b} \left (15 A b^3+8 a^3 (9 A+16 B+12 C)+4 a^2 b (71 A+52 B+108 C)+2 a b^2 (59 A+132 B+192 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{192 a d}+\frac {\sqrt {a+b} \left (5 A b^4-160 a^3 b B-40 a b^3 B-120 a^2 b^2 (A+2 C)-16 a^4 (3 A+4 C)\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{64 a^2 d}+\frac {\left (15 A b^3+128 a^3 B+264 a b^2 B+4 a^2 b (71 A+108 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 a d}+\frac {\left (5 A b^2+24 a b B+4 a^2 (3 A+4 C)\right ) \cos (c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {(5 A b+8 a B) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{4 d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(5667\) vs. \(2(652)=1304\).

Time = 37.57 (sec) , antiderivative size = 5667, normalized size of antiderivative = 8.69 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

Result too large to show

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(8027\) vs. \(2(603)=1206\).

Time = 519.44 (sec) , antiderivative size = 8028, normalized size of antiderivative = 12.31

method	result	size
default	\(\text {Expression too large to display}\)	\(8028\)

[In]

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b^2*cos(d*x + c)^4*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^4*sec(d*x + c)^3 + A*a^2*cos(d*
x + c)^4 + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^4*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c)^4*sec(d*x
+ c))*sqrt(b*sec(d*x + c) + a), x)

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^4, x)

Giac [F]

\[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

[In]

int(cos(c + d*x)^4*(a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int(cos(c + d*x)^4*(a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)