\(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [1478]

Optimal result

Integrand size = 43, antiderivative size = 413 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 \left (5 a^4 B-30 a^2 b^2 B-3 b^4 B+20 a^3 b (A-C)-4 a b^3 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}-\frac {2 b^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right ) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b \left (42 a^3 B-28 a b^2 B+3 a^2 b (49 A-13 C)-b^3 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (21 A b+7 a B-b C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \]

[Out]

-2/105*b^2*(350*A*a*b+105*B*a^2-21*B*b^2-54*C*a*b)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/3*A*(a+b*cos(d*x+c))^4*sec(
d*x+c)^(3/2)*sin(d*x+c)/d-2/21*b*(42*B*a^3-28*B*a*b^2+3*a^2*b*(49*A-13*C)-b^3*(7*A+5*C))*sin(d*x+c)/d/sec(d*x+
c)^(1/2)-2/7*b*(21*A*b+7*B*a-C*b)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/3*(8*A*b+3*B*a)*(a+b*cos(
d*x+c))^3*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*(5*B*a^4-30*B*a^2*b^2-3*B*b^4+20*a^3*b*(A-C)-4*a*b^3*(5*A+3*C))*(c
os(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c
)^(1/2)/d+2/21*(84*B*a^3*b+28*B*a*b^3+42*a^2*b^2*(3*A+C)+7*a^4*(A+3*C)+b^4*(7*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(
1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4306, 3126, 3128, 3112, 3102, 2827, 2720, 2719} \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 b^2 \sin (c+d x) \left (105 a^2 B+350 a A b-54 a b C-21 b^2 B\right )}{105 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sin (c+d x) \left (42 a^3 B+3 a^2 b (49 A-13 C)-28 a b^2 B-b^3 (7 A+5 C)\right )}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (7 a^4 (A+3 C)+84 a^3 b B+42 a^2 b^2 (3 A+C)+28 a b^3 B+b^4 (7 A+5 C)\right )}{21 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^4 B+20 a^3 b (A-C)-30 a^2 b^2 B-4 a b^3 (5 A+3 C)-3 b^4 B\right )}{5 d}-\frac {2 b \sin (c+d x) (7 a B+21 A b-b C) (a+b \cos (c+d x))^2}{7 d \sqrt {\sec (c+d x)}}+\frac {2 (3 a B+8 A b) \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^3}{3 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^4}{3 d} \]

[In]

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[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))*Sqrt[Cos[c + d*x]]*EllipticE[(
c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(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))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) - (2*b^2*(350*a*A
*b + 105*a^2*B - 21*b^2*B - 54*a*b*C)*Sin[c + d*x])/(105*d*Sec[c + d*x]^(3/2)) - (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))*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) - (2*b*(21*A*b + 7*a*B - b*C
)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Sqrt[Sec[c + d*x]]) + (2*(8*A*b + 3*a*B)*(a + b*Cos[c + d*x])^3*Sq
rt[Sec[c + d*x]]*Sin[c + d*x])/(3*d) + (2*A*(a + b*Cos[c + d*x])^4*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)

Rule 2719

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

Rule 2720

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

Rule 2827

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 3102

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 3112

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 3126

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 3128

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 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{21} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 ) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b (21 A b+7 a B-b C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{105} \left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right ) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}-\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 ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (21 A b+7 a B-b C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{315} \left (32 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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^2 \left (350 a A b+105 a^2 B-21 b^2 B-54 a b C\right ) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}-\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 ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (21 A b+7 a B-b C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\frac {1}{5} \left (\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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (\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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\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 ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{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 ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (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 ) \sin (c+d x)}{105 d \sec ^{\frac {3}{2}}(c+d x)}-\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 ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (21 A b+7 a B-b C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 (8 A b+3 a B) (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+b \cos (c+d x))^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.02 (sec) , antiderivative size = 316, normalized size of antiderivative = 0.77 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-168 \left (5 a^4 B-30 a^2 b^2 B-3 b^4 B+20 a^3 b (A-C)-4 a b^3 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+40 \left (84 a^3 b B+28 a b^3 B+42 a^2 b^2 (3 A+C)+7 a^4 (A+3 C)+b^4 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {\left (280 a^4 A+140 A b^4+560 a b^3 B+840 a^2 b^2 C+145 b^4 C+42 \left (80 a^3 A b+20 a^4 B+3 b^4 B+12 a b^3 C\right ) \cos (c+d x)+20 b^2 \left (7 A b^2+28 a b B+42 a^2 C+8 b^2 C\right ) \cos (2 (c+d x))+42 b^4 B \cos (3 (c+d x))+168 a b^3 C \cos (3 (c+d x))+15 b^4 C \cos (4 (c+d x))\right ) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{420 d} \]

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-168*(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] + 40*(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] + ((280*a^4*A + 140*A*b^4 + 560*a*b^3*B + 840*a^2*b^2*C + 145*b^4*C
+ 42*(80*a^3*A*b + 20*a^4*B + 3*b^4*B + 12*a*b^3*C)*Cos[c + d*x] + 20*b^2*(7*A*b^2 + 28*a*b*B + 42*a^2*C + 8*
b^2*C)*Cos[2*(c + d*x)] + 42*b^4*B*Cos[3*(c + d*x)] + 168*a*b^3*C*Cos[3*(c + d*x)] + 15*b^4*C*Cos[4*(c + d*x)]
)*Sin[c + d*x])/Cos[c + d*x]^(3/2)))/(420*d)

Maple [F(-1)]

Timed out.

hanged

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.17 (sec) , antiderivative size = 447, normalized size of antiderivative = 1.08 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (7 i \, {\left (A + 3 \, C\right )} a^{4} + 84 i \, B a^{3} b + 42 i \, {\left (3 \, A + C\right )} a^{2} b^{2} + 28 i \, B a b^{3} + i \, {\left (7 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-7 i \, {\left (A + 3 \, C\right )} a^{4} - 84 i \, B a^{3} b - 42 i \, {\left (3 \, A + C\right )} a^{2} b^{2} - 28 i \, B a b^{3} - i \, {\left (7 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (5 i \, B a^{4} + 20 i \, {\left (A - C\right )} a^{3} b - 30 i \, B a^{2} b^{2} - 4 i \, {\left (5 \, A + 3 \, C\right )} a b^{3} - 3 i \, B b^{4}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-5 i \, B a^{4} - 20 i \, {\left (A - C\right )} a^{3} b + 30 i \, B a^{2} b^{2} + 4 i \, {\left (5 \, A + 3 \, C\right )} a b^{3} + 3 i \, B b^{4}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (15 \, C b^{4} \cos \left (d x + c\right )^{4} + 35 \, A a^{4} + 21 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (42 \, C a^{2} b^{2} + 28 \, B a b^{3} + {\left (7 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 105 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

-1/105*(5*sqrt(2)*(7*I*(A + 3*C)*a^4 + 84*I*B*a^3*b + 42*I*(3*A + C)*a^2*b^2 + 28*I*B*a*b^3 + I*(7*A + 5*C)*b^
4)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-7*I*(A + 3*C)*a^4 - 84
*I*B*a^3*b - 42*I*(3*A + C)*a^2*b^2 - 28*I*B*a*b^3 - I*(7*A + 5*C)*b^4)*cos(d*x + c)*weierstrassPInverse(-4, 0
, cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(5*I*B*a^4 + 20*I*(A - C)*a^3*b - 30*I*B*a^2*b^2 - 4*I*(5*A + 3*
C)*a*b^3 - 3*I*B*b^4)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x
+ c))) + 21*sqrt(2)*(-5*I*B*a^4 - 20*I*(A - C)*a^3*b + 30*I*B*a^2*b^2 + 4*I*(5*A + 3*C)*a*b^3 + 3*I*B*b^4)*cos
(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(15*C*b^4*cos(
d*x + c)^4 + 35*A*a^4 + 21*(4*C*a*b^3 + B*b^4)*cos(d*x + c)^3 + 5*(42*C*a^2*b^2 + 28*B*a*b^3 + (7*A + 5*C)*b^4
)*cos(d*x + c)^2 + 105*(B*a^4 + 4*A*a^3*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

Giac [F]

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

[In]

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

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^4\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

[In]

int((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)