3.1312 \(\int \cos ^{\frac{11}{2}}(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=404 \[ \frac{2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)+220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)\right )}{231 d}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (7 A+9 C)+54 a^2 b^2 B+7 a^4 B+12 a b^3 (3 A+5 C)+15 b^4 B\right )}{15 d}+\frac{2 a \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (2 a^2 b (673 A+891 C)+539 a^3 B+1353 a b^2 B+192 A b^3\right )}{3465 d}+\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (3 a^2 (9 A+11 C)+55 a b B+16 A b^2\right ) (a \cos (c+d x)+b)^2}{231 d}+\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (9 a^2 b^2 (101 A+143 C)+15 a^4 (9 A+11 C)+660 a^3 b B+682 a b^3 B+64 A b^4\right )}{693 d}+\frac{2 (11 a B+8 A b) \sin (c+d x) \sqrt{\cos (c+d x)} (a \cos (c+d x)+b)^3}{99 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} (a \cos (c+d x)+b)^4}{11 d} \]

[Out]

(2*(7*a^4*B + 54*a^2*b^2*B + 15*b^4*B + 12*a*b^3*(3*A + 5*C) + 4*a^3*b*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2])
/(15*d) + (2*(220*a^3*b*B + 308*a*b^3*B + 77*b^4*(A + 3*C) + 66*a^2*b^2*(5*A + 7*C) + 5*a^4*(9*A + 11*C))*Elli
pticF[(c + d*x)/2, 2])/(231*d) + (2*(64*A*b^4 + 660*a^3*b*B + 682*a*b^3*B + 15*a^4*(9*A + 11*C) + 9*a^2*b^2*(1
01*A + 143*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(693*d) + (2*a*(192*A*b^3 + 539*a^3*B + 1353*a*b^2*B + 2*a^2*b
*(673*A + 891*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d) + (2*(16*A*b^2 + 55*a*b*B + 3*a^2*(9*A + 11*C))*Sq
rt[Cos[c + d*x]]*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(231*d) + (2*(8*A*b + 11*a*B)*Sqrt[Cos[c + d*x]]*(b + a*
Cos[c + d*x])^3*Sin[c + d*x])/(99*d) + (2*A*Sqrt[Cos[c + d*x]]*(b + a*Cos[c + d*x])^4*Sin[c + d*x])/(11*d)

________________________________________________________________________________________

Rubi [A]  time = 1.32034, antiderivative size = 404, 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 = {4112, 3049, 3033, 3023, 2748, 2641, 2639} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)+220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)\right )}{231 d}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (7 A+9 C)+54 a^2 b^2 B+7 a^4 B+12 a b^3 (3 A+5 C)+15 b^4 B\right )}{15 d}+\frac{2 a \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (2 a^2 b (673 A+891 C)+539 a^3 B+1353 a b^2 B+192 A b^3\right )}{3465 d}+\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (3 a^2 (9 A+11 C)+55 a b B+16 A b^2\right ) (a \cos (c+d x)+b)^2}{231 d}+\frac{2 \sin (c+d x) \sqrt{\cos (c+d x)} \left (9 a^2 b^2 (101 A+143 C)+15 a^4 (9 A+11 C)+660 a^3 b B+682 a b^3 B+64 A b^4\right )}{693 d}+\frac{2 (11 a B+8 A b) \sin (c+d x) \sqrt{\cos (c+d x)} (a \cos (c+d x)+b)^3}{99 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} (a \cos (c+d x)+b)^4}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(11/2)*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*(7*a^4*B + 54*a^2*b^2*B + 15*b^4*B + 12*a*b^3*(3*A + 5*C) + 4*a^3*b*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2])
/(15*d) + (2*(220*a^3*b*B + 308*a*b^3*B + 77*b^4*(A + 3*C) + 66*a^2*b^2*(5*A + 7*C) + 5*a^4*(9*A + 11*C))*Elli
pticF[(c + d*x)/2, 2])/(231*d) + (2*(64*A*b^4 + 660*a^3*b*B + 682*a*b^3*B + 15*a^4*(9*A + 11*C) + 9*a^2*b^2*(1
01*A + 143*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(693*d) + (2*a*(192*A*b^3 + 539*a^3*B + 1353*a*b^2*B + 2*a^2*b
*(673*A + 891*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3465*d) + (2*(16*A*b^2 + 55*a*b*B + 3*a^2*(9*A + 11*C))*Sq
rt[Cos[c + d*x]]*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(231*d) + (2*(8*A*b + 11*a*B)*Sqrt[Cos[c + d*x]]*(b + a*
Cos[c + d*x])^3*Sin[c + d*x])/(99*d) + (2*A*Sqrt[Cos[c + d*x]]*(b + a*Cos[c + d*x])^4*Sin[c + d*x])/(11*d)

Rule 4112

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rule 3049

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

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \cos ^{\frac{11}{2}}(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \frac{(b+a \cos (c+d x))^4 \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac{2}{11} \int \frac{(b+a \cos (c+d x))^3 \left (\frac{1}{2} b (A+11 C)+\frac{1}{2} (9 a A+11 b B+11 a C) \cos (c+d x)+\frac{1}{2} (8 A b+11 a B) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (8 A b+11 a B) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac{2 A \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac{4}{99} \int \frac{(b+a \cos (c+d x))^2 \left (\frac{1}{4} b (17 A b+11 a B+99 b C)+\frac{1}{4} \left (146 a A b+77 a^2 B+99 b^2 B+198 a b C\right ) \cos (c+d x)+\frac{3}{4} \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac{2 (8 A b+11 a B) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac{2 A \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac{8}{693} \int \frac{(b+a \cos (c+d x)) \left (\frac{1}{8} b \left (242 a b B+9 a^2 (9 A+11 C)+b^2 (167 A+693 C)\right )+\frac{1}{8} \left (1441 a^2 b B+693 b^3 B+45 a^3 (9 A+11 C)+a b^2 (1381 A+2079 C)\right ) \cos (c+d x)+\frac{1}{8} \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac{2 (8 A b+11 a B) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac{2 A \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac{16 \int \frac{\frac{5}{16} b^2 \left (242 a b B+9 a^2 (9 A+11 C)+b^2 (167 A+693 C)\right )+\frac{231}{16} \left (7 a^4 B+54 a^2 b^2 B+15 b^4 B+12 a b^3 (3 A+5 C)+4 a^3 b (7 A+9 C)\right ) \cos (c+d x)+\frac{15}{16} \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{3465}\\ &=\frac{2 \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{693 d}+\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac{2 (8 A b+11 a B) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac{2 A \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac{32 \int \frac{\frac{45}{32} \left (220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)+66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)\right )+\frac{693}{32} \left (7 a^4 B+54 a^2 b^2 B+15 b^4 B+12 a b^3 (3 A+5 C)+4 a^3 b (7 A+9 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{10395}\\ &=\frac{2 \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{693 d}+\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac{2 (8 A b+11 a B) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac{2 A \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^4 \sin (c+d x)}{11 d}+\frac{1}{15} \left (7 a^4 B+54 a^2 b^2 B+15 b^4 B+12 a b^3 (3 A+5 C)+4 a^3 b (7 A+9 C)\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{231} \left (220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)+66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (7 a^4 B+54 a^2 b^2 B+15 b^4 B+12 a b^3 (3 A+5 C)+4 a^3 b (7 A+9 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 \left (220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)+66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{2 \left (64 A b^4+660 a^3 b B+682 a b^3 B+15 a^4 (9 A+11 C)+9 a^2 b^2 (101 A+143 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{693 d}+\frac{2 a \left (192 A b^3+539 a^3 B+1353 a b^2 B+2 a^2 b (673 A+891 C)\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3465 d}+\frac{2 \left (16 A b^2+55 a b B+3 a^2 (9 A+11 C)\right ) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^2 \sin (c+d x)}{231 d}+\frac{2 (8 A b+11 a B) \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^3 \sin (c+d x)}{99 d}+\frac{2 A \sqrt{\cos (c+d x)} (b+a \cos (c+d x))^4 \sin (c+d x)}{11 d}\\ \end{align*}

Mathematica [A]  time = 2.51623, size = 320, normalized size = 0.79 \[ \frac{10 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (66 a^2 b^2 (5 A+7 C)+5 a^4 (9 A+11 C)+220 a^3 b B+308 a b^3 B+77 b^4 (A+3 C)\right )+154 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (4 a^3 b (7 A+9 C)+54 a^2 b^2 B+7 a^4 B+12 a b^3 (3 A+5 C)+15 b^4 B\right )+\frac{1}{12} \sin (c+d x) \sqrt{\cos (c+d x)} \left (154 a \cos (c+d x) \left (4 a^2 b (43 A+36 C)+43 a^3 B+216 a b^2 B+144 A b^3\right )+5 \left (36 a^2 \cos (2 (c+d x)) \left (a^2 (16 A+11 C)+44 a b B+66 A b^2\right )+3 \left (264 a^2 b^2 (13 A+14 C)+21 a^4 A \cos (4 (c+d x))+a^4 (531 A+572 C)+2288 a^3 b B+2464 a b^3 B+616 A b^4\right )+154 a^3 (a B+4 A b) \cos (3 (c+d x))\right )\right )}{1155 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(11/2)*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(154*(7*a^4*B + 54*a^2*b^2*B + 15*b^4*B + 12*a*b^3*(3*A + 5*C) + 4*a^3*b*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2
] + 10*(220*a^3*b*B + 308*a*b^3*B + 77*b^4*(A + 3*C) + 66*a^2*b^2*(5*A + 7*C) + 5*a^4*(9*A + 11*C))*EllipticF[
(c + d*x)/2, 2] + (Sqrt[Cos[c + d*x]]*(154*a*(144*A*b^3 + 43*a^3*B + 216*a*b^2*B + 4*a^2*b*(43*A + 36*C))*Cos[
c + d*x] + 5*(36*a^2*(66*A*b^2 + 44*a*b*B + a^2*(16*A + 11*C))*Cos[2*(c + d*x)] + 154*a^3*(4*A*b + a*B)*Cos[3*
(c + d*x)] + 3*(616*A*b^4 + 2288*a^3*b*B + 2464*a*b^3*B + 264*a^2*b^2*(13*A + 14*C) + a^4*(531*A + 572*C) + 21
*a^4*A*Cos[4*(c + d*x)])))*Sin[c + d*x])/12)/(1155*d)

________________________________________________________________________________________

Maple [B]  time = 2.933, size = 1273, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(11/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

-2/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(20160*A*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-50400*A*a^4-49280*A*a^3*b-12320*B*a^4)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(56880*A*a^4+98560*A
*a^3*b+47520*A*a^2*b^2+24640*B*a^4+31680*B*a^3*b+7920*C*a^4)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-34920*A
*a^4-91168*A*a^3*b-71280*A*a^2*b^2-22176*A*a*b^3-22792*B*a^4-47520*B*a^3*b-33264*B*a^2*b^2-11880*C*a^4-22176*C
*a^3*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(13860*A*a^4+41888*A*a^3*b+55440*A*a^2*b^2+22176*A*a*b^3+4620*
A*b^4+10472*B*a^4+36960*B*a^3*b+33264*B*a^2*b^2+18480*B*a*b^3+9240*C*a^4+22176*C*a^3*b+27720*C*a^2*b^2)*sin(1/
2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-2790*A*a^4-7392*A*a^3*b-15840*A*a^2*b^2-5544*A*a*b^3-2310*A*b^4-1848*B*a^4
-10560*B*a^3*b-8316*B*a^2*b^2-9240*B*a*b^3-2640*C*a^4-5544*C*a^3*b-13860*C*a^2*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1
/2*d*x+1/2*c)-6468*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c
),2^(1/2))*a^3*b-8316*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/
2*c),2^(1/2))*a*b^3+675*A*a^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*
d*x+1/2*c),2^(1/2))+4950*A*a^2*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos
(1/2*d*x+1/2*c),2^(1/2))+1155*A*b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))-1617*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))*a^4-12474*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos
(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2-3465*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))*b^4+3300*B*a^3*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))+4620*a*b^3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-8316*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b-13860*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/
2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3+825*a^4*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-
1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6930*C*a^2*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*
c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3465*C*b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin
(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [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)^(11/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{4} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{6} +{\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{5} + A a^{4} \cos \left (d x + c\right )^{5} +{\left (6 \, C a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{4} + 2 \,{\left (2 \, C a^{3} b + 3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{3} +{\left (C a^{4} + 4 \, B a^{3} b + 6 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )^{2} +{\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{5} \sec \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*b^4*cos(d*x + c)^5*sec(d*x + c)^6 + (4*C*a*b^3 + B*b^4)*cos(d*x + c)^5*sec(d*x + c)^5 + A*a^4*cos(
d*x + c)^5 + (6*C*a^2*b^2 + 4*B*a*b^3 + A*b^4)*cos(d*x + c)^5*sec(d*x + c)^4 + 2*(2*C*a^3*b + 3*B*a^2*b^2 + 2*
A*a*b^3)*cos(d*x + c)^5*sec(d*x + c)^3 + (C*a^4 + 4*B*a^3*b + 6*A*a^2*b^2)*cos(d*x + c)^5*sec(d*x + c)^2 + (B*
a^4 + 4*A*a^3*b)*cos(d*x + c)^5*sec(d*x + c))*sqrt(cos(d*x + c)), 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)**(11/2)*(a+b*sec(d*x+c))**4*(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 )}^{4} \cos \left (d x + c\right )^{\frac{11}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(11/2)*(a+b*sec(d*x+c))^4*(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)^4*cos(d*x + c)^(11/2), x)