3.1090 \(\int \frac{(a+b \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=373 \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+84 a^3 b B+28 a b^3 B+b^4 (7 A+5 C)\right )}{21 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (20 a^3 b (A-C)-30 a^2 b^2 B+5 a^4 B-4 a b^3 (5 A+3 C)-3 b^4 B\right )}{5 d}-\frac{2 b^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (105 a^2 B+350 a A b-54 a b C-21 b^2 B\right )}{105 d}-\frac{2 b \sin (c+d x) \sqrt{\cos (c+d x)} \left (3 a^2 b (49 A-13 C)+42 a^3 B-28 a b^2 B-b^3 (7 A+5 C)\right )}{21 d}-\frac{2 b \sin (c+d x) \sqrt{\cos (c+d x)} (7 a B+21 A b-b C) (a+b \cos (c+d x))^2}{7 d}+\frac{2 (3 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^4}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]

[Out]

(-2*(5*a^4*B - 30*a^2*b^2*B - 3*b^4*B + 20*a^3*b*(A - C) - 4*a*b^3*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*
d) + (2*(84*a^3*b*B + 28*a*b^3*B + 42*a^2*b^2*(3*A + C) + 7*a^4*(A + 3*C) + b^4*(7*A + 5*C))*EllipticF[(c + d*
x)/2, 2])/(21*d) - (2*b*(42*a^3*B - 28*a*b^2*B + 3*a^2*b*(49*A - 13*C) - b^3*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*S
in[c + d*x])/(21*d) - (2*b^2*(350*a*A*b + 105*a^2*B - 21*b^2*B - 54*a*b*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(1
05*d) - (2*b*(21*A*b + 7*a*B - b*C)*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d) + (2*(8*A*b
+ 3*a*B)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) + (2*A*(a + b*Cos[c + d*x])^4*Sin[c + d
*x])/(3*d*Cos[c + d*x]^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 1.25774, antiderivative size = 373, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3047, 3049, 3033, 3023, 2748, 2641, 2639} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+84 a^3 b B+28 a b^3 B+b^4 (7 A+5 C)\right )}{21 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (20 a^3 b (A-C)-30 a^2 b^2 B+5 a^4 B-4 a b^3 (5 A+3 C)-3 b^4 B\right )}{5 d}-\frac{2 b^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (105 a^2 B+350 a A b-54 a b C-21 b^2 B\right )}{105 d}-\frac{2 b \sin (c+d x) \sqrt{\cos (c+d x)} \left (3 a^2 b (49 A-13 C)+42 a^3 B-28 a b^2 B-b^3 (7 A+5 C)\right )}{21 d}-\frac{2 b \sin (c+d x) \sqrt{\cos (c+d x)} (7 a B+21 A b-b C) (a+b \cos (c+d x))^2}{7 d}+\frac{2 (3 a B+8 A b) \sin (c+d x) (a+b \cos (c+d x))^3}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \cos (c+d x))^4}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*a^4*B - 30*a^2*b^2*B - 3*b^4*B + 20*a^3*b*(A - C) - 4*a*b^3*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*
d) + (2*(84*a^3*b*B + 28*a*b^3*B + 42*a^2*b^2*(3*A + C) + 7*a^4*(A + 3*C) + b^4*(7*A + 5*C))*EllipticF[(c + d*
x)/2, 2])/(21*d) - (2*b*(42*a^3*B - 28*a*b^2*B + 3*a^2*b*(49*A - 13*C) - b^3*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*S
in[c + d*x])/(21*d) - (2*b^2*(350*a*A*b + 105*a^2*B - 21*b^2*B - 54*a*b*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(1
05*d) - (2*b*(21*A*b + 7*a*B - b*C)*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d) + (2*(8*A*b
+ 3*a*B)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) + (2*A*(a + b*Cos[c + d*x])^4*Sin[c + d
*x])/(3*d*Cos[c + d*x]^(3/2))

Rule 3047

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

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 \frac{(a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{(a+b \cos (c+d x))^3 \left (\frac{1}{2} (8 A b+3 a B)+\frac{1}{2} (3 b B+a (A+3 C)) \cos (c+d x)-\frac{1}{2} b (7 A-3 C) \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4}{3} \int \frac{(a+b \cos (c+d x))^2 \left (\frac{1}{4} \left (48 A b^2+21 a b B+a^2 (A+3 C)\right )-\frac{1}{4} \left (14 a A b+3 a^2 B-3 b^2 B-6 a b C\right ) \cos (c+d x)-\frac{3}{4} b (21 A b+7 a B-b C) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b (21 A b+7 a B-b C) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{8}{21} \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{8} a \left (126 a b B+3 b^2 (91 A+C)+7 a^2 (A+3 C)\right )-\frac{1}{8} \left (21 a^3 B-63 a b^2 B-3 b^3 (7 A+5 C)+a^2 (91 A b-63 b C)\right ) \cos (c+d x)-\frac{1}{8} b \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right ) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (21 A b+7 a B-b C) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{16}{105} \int \frac{\frac{5}{16} a^2 \left (126 a b B+3 b^2 (91 A+C)+7 a^2 (A+3 C)\right )-\frac{21}{16} \left (5 a^4 B-30 a^2 b^2 B-3 b^4 B+20 a^3 b (A-C)-4 a b^3 (5 A+3 C)\right ) \cos (c+d x)-\frac{15}{16} b \left (42 a^3 B-28 a b^2 B-b^3 (7 A+5 C)+a^2 (147 A b-39 b C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (42 a^3 B-28 a b^2 B+3 a^2 b (49 A-13 C)-b^3 (7 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 b^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (21 A b+7 a B-b C) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{32}{315} \int \frac{\frac{15}{32} \left (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right )-\frac{63}{32} \left (5 a^4 B-30 a^2 b^2 B-3 b^4 B+20 a^3 b (A-C)-4 a b^3 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (42 a^3 B-28 a b^2 B+3 a^2 b (49 A-13 C)-b^3 (7 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 b^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (21 A b+7 a B-b C) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} \left (-5 a^4 B+30 a^2 b^2 B+3 b^4 B-20 a^3 b (A-C)+4 a b^3 (5 A+3 C)\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 \left (5 a^4 B-30 a^2 b^2 B-3 b^4 B+20 a^3 b (A-C)-4 a b^3 (5 A+3 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{2 b \left (42 a^3 B-28 a b^2 B+3 a^2 b (49 A-13 C)-b^3 (7 A+5 C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 b^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}-\frac{2 b (21 A b+7 a B-b C) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A (a+b \cos (c+d x))^4 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 2.59591, size = 257, normalized size = 0.69 \[ \frac{10 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+84 a^3 b B+28 a b^3 B+b^4 (7 A+5 C)\right )-42 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (20 a^3 b (A-C)-30 a^2 b^2 B+5 a^4 B-4 a b^3 (5 A+3 C)-3 b^4 B\right )+\frac{168 \sin (c+d x) \left (5 a^3 (a B+4 A b)+b^3 (4 a C+b B) \cos ^2(c+d x)\right )+5 \left (b^2 \sin (2 (c+d x)) \left (168 a^2 C+112 a b B+28 A b^2+23 b^2 C\right )+56 a^4 A \tan (c+d x)+6 b^4 C \sin (3 (c+d x)) \cos (c+d x)\right )}{4 \sqrt{\cos (c+d x)}}}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-42*(5*a^4*B - 30*a^2*b^2*B - 3*b^4*B + 20*a^3*b*(A - C) - 4*a*b^3*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2] + 1
0*(84*a^3*b*B + 28*a*b^3*B + 42*a^2*b^2*(3*A + C) + 7*a^4*(A + 3*C) + b^4*(7*A + 5*C))*EllipticF[(c + d*x)/2,
2] + (168*(5*a^3*(4*A*b + a*B) + b^3*(b*B + 4*a*C)*Cos[c + d*x]^2)*Sin[c + d*x] + 5*(b^2*(28*A*b^2 + 112*a*b*B
+ 168*a^2*C + 23*b^2*C)*Sin[2*(c + d*x)] + 6*b^4*C*Cos[c + d*x]*Sin[3*(c + d*x)] + 56*a^4*A*Tan[c + d*x]))/(4
*Sqrt[Cos[c + d*x]]))/(105*d)

________________________________________________________________________________________

Maple [B]  time = 3.875, size = 2507, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2
+1)/sin(1/2*d*x+1/2*c)^3*(420*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^3-630*a^2*A*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)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))+420*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^3*b+252*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)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3-210*a^2*b^2*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))+480*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10-336*B*b^4*cos(1/2*d
*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+280*A*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+504*B*b^4*cos(1/2*d*x+1/2*c)*
sin(1/2*d*x+1/2*c)^6+920*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-280*A*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^4-420*B*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-252*B*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4
-440*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+70*A*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+70*A*b^4*c
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+210*B*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+42*B*b^4*cos(1/2*d*x+
1/2*c)*sin(1/2*d*x+1/2*c)^2+80*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-960*C*b^4*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^8+630*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^2-105*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))-420*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))*a^3*b-140*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))*a*b^3+840*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*a^3*b*sin(1/2*d*x+1/2*c)^2+280*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^3*sin(1/2*d*x+1/2*c)^2-840*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*b*sin(1/2*d*x+1/2*c)^2-504*C*EllipticE(
cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^3*sin(1/2*d*x+1/
2*c)^2+420*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*a^2*b^2*sin(1/2*d*x+1/2*c)^2+840*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(
sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*b*sin(1/2*d*x+1/2*c)^2-840*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*
d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^3*sin(1/2*d*x+1/2*c)^2-35*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))-35*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(cos(1/2*d*x+1/2*c),2^(1/2))-105*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+63*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))*b^4-25*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))+1120*B*a*b^3*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^6+1680*C*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+2016*C*a*b^3*cos(1/2*d*x+1/2*c)*sin(1/
2*d*x+1/2*c)^6-1680*A*a^3*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-1120*B*a*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^4-1680*C*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-1008*C*a*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+
1/2*c)^4+840*A*a^3*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+280*B*a*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^2+420*C*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+168*C*a*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-1
344*C*a*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-420*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+210*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*sin(1/2*d*x+1/2*c)^2+50*C*EllipticF(cos(1/2*d*x+1/2*c),2
^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^4*sin(1/2*d*x+1/2*c)^2+70*A*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*sin(1/2*d*x+1/2*
c)^2+70*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
b^4*sin(1/2*d*x+1/2*c)^2+210*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*a^4*sin(1/2*d*x+1/2*c)^2-126*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^4*sin(1/2*d*x+1/2*c)^2+1260*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b^2*sin(1/2*d*x+1/2*c)^2-1260*B*EllipticE(
cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b^2*sin(1/2*d*x+
1/2*c)^2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^4*cos(d*x + c)^6 + (4*C*a*b^3 + B*b^4)*cos(d*x + c)^5 + A*a^4 + (6*C*a^2*b^2 + 4*B*a*b^3 + A*b^4
)*cos(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)^3 + (C*a^4 + 4*B*a^3*b + 6*A*a^2*b^2)*
cos(d*x + c)^2 + (B*a^4 + 4*A*a^3*b)*cos(d*x + c))/cos(d*x + c)^(5/2), 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((a+b*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{4}}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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