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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 2.01955, antiderivative size = 652, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4094, 4104, 4058, 3921, 3784, 3832, 4004} \[ \frac{\sin (c+d x) \left (4 a^2 b (71 A+108 C)+128 a^3 B+264 a b^2 B+15 A b^3\right ) \sqrt{a+b \sec (c+d x)}}{192 a d}+\frac{\sin (c+d x) \cos (c+d x) \left (4 a^2 (3 A+4 C)+24 a b B+5 A b^2\right ) \sqrt{a+b \sec (c+d x)}}{32 d}+\frac{\sqrt{a+b} \cot (c+d x) \left (4 a^2 b (71 A+52 B+108 C)+8 a^3 (9 A+16 B+12 C)+2 a b^2 (59 A+132 B+192 C)+15 A b^3\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{192 a d}+\frac{(a-b) \sqrt{a+b} \cot (c+d x) \left (4 a^2 b (71 A+108 C)+128 a^3 B+264 a b^2 B+15 A b^3\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{192 a b d}+\frac{\sqrt{a+b} \cot (c+d x) \left (-120 a^2 b^2 (A+2 C)-16 a^4 (3 A+4 C)-160 a^3 b B-40 a b^3 B+5 A b^4\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{64 a^2 d}+\frac{(8 a B+5 A b) \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^{3/2}}{24 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^{5/2}}{4 d} \]

Antiderivative was successfully verified.

[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 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]

Rule 4104

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]

Rule 4058

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 3921

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 3784

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

Rule 3832

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

Rule 4004

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 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \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 \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 (\sin ^{-1}\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 (\sin ^{-1}\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) F\left (\sin ^{-1}\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) \Pi \left (\frac{a+b}{a};\sin ^{-1}\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]  time = 26.1058, size = 5681, normalized size = 8.71 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[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

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Maple [B]  time = 0.706, size = 5850, normalized size = 9. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{4} +{\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{4} +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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(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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)