3.1381 \(\int \frac {(a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\)

Optimal. Leaf size=335 \[ \frac {2 a \left (8 a^2 C+99 A b^2+77 b^2 C\right ) \sin (c+d x)}{165 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (8 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 a \left (a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^3}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a C \sin (c+d x) (a+b \cos (c+d x))^2}{33 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/231*b*(8*a^2*C+3*b^2*(11*A+9*C))*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/165*a*(99*A*b^2+8*C*a^2+77*C*b^2)*sin(d*x+c
)/d/sec(d*x+c)^(3/2)+4/33*a*C*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/11*C*(a+b*cos(d*x+c))^3*sin(d
*x+c)/d/sec(d*x+c)^(3/2)+2/231*b*(33*a^2*(7*A+5*C)+5*b^2*(11*A+9*C))*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*a*(a^2*
(5*A+3*C)+b^2*(9*A+7*C))*(cos(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+2/231*b*(33*a^2*(7*A+5*C)+5*b^2*(11*A+9*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]  time = 0.93, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4221, 3050, 3049, 3033, 3023, 2748, 2639, 2635, 2641} \[ \frac {2 a \left (8 a^2 C+99 A b^2+77 b^2 C\right ) \sin (c+d x)}{165 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (8 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {2 b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 a \left (a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^3}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a C \sin (c+d x) (a+b \cos (c+d x))^2}{33 d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Sqrt[Sec[c + d*x]],x]

[Out]

(2*a*(a^2*(5*A + 3*C) + b^2*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d
) + (2*b*(33*a^2*(7*A + 5*C) + 5*b^2*(11*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d
*x]])/(231*d) + (2*b*(8*a^2*C + 3*b^2*(11*A + 9*C))*Sin[c + d*x])/(231*d*Sec[c + d*x]^(5/2)) + (2*a*(99*A*b^2
+ 8*a^2*C + 77*b^2*C)*Sin[c + d*x])/(165*d*Sec[c + d*x]^(3/2)) + (4*a*C*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(
33*d*Sec[c + d*x]^(3/2)) + (2*C*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(11*d*Sec[c + d*x]^(3/2)) + (2*b*(33*a^2*
(7*A + 5*C) + 5*b^2*(11*A + 9*C))*Sin[c + d*x])/(231*d*Sqrt[Sec[c + d*x]])

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

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 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 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 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 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 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 4221

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*} \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 C (a+b \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{11} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2 \left (\frac {1}{2} a (11 A+3 C)+\frac {1}{2} b (11 A+9 C) \cos (c+d x)+3 a C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {4 a C (a+b \cos (c+d x))^2 \sin (c+d x)}{33 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{99} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (\frac {9}{4} a^2 (11 A+5 C)+\frac {3}{2} a b (33 A+25 C) \cos (c+d x)+\frac {3}{4} \left (8 a^2 C+3 b^2 (11 A+9 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b \left (8 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a C (a+b \cos (c+d x))^2 \sin (c+d x)}{33 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{693} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {63}{8} a^3 (11 A+5 C)+\frac {9}{8} b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \cos (c+d x)+\frac {21}{8} a \left (99 A b^2+8 a^2 C+77 b^2 C\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b \left (8 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (99 A b^2+8 a^2 C+77 b^2 C\right ) \sin (c+d x)}{165 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a C (a+b \cos (c+d x))^2 \sin (c+d x)}{33 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {693}{16} a \left (a^2 (5 A+3 C)+b^2 (9 A+7 C)\right )+\frac {45}{16} b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \cos (c+d x)\right ) \, dx}{3465}\\ &=\frac {2 b \left (8 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (99 A b^2+8 a^2 C+77 b^2 C\right ) \sin (c+d x)}{165 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a C (a+b \cos (c+d x))^2 \sin (c+d x)}{33 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{5} \left (a \left (a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{77} \left (b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 a \left (a^2 (5 A+3 C)+b^2 (9 A+7 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 b \left (8 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (99 A b^2+8 a^2 C+77 b^2 C\right ) \sin (c+d x)}{165 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a C (a+b \cos (c+d x))^2 \sin (c+d x)}{33 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}+\frac {1}{231} \left (b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a \left (a^2 (5 A+3 C)+b^2 (9 A+7 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 b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{231 d}+\frac {2 b \left (8 a^2 C+3 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a \left (99 A b^2+8 a^2 C+77 b^2 C\right ) \sin (c+d x)}{165 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 a C (a+b \cos (c+d x))^2 \sin (c+d x)}{33 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^3 \sin (c+d x)}{11 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sin (c+d x)}{231 d \sqrt {\sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.42, size = 236, normalized size = 0.70 \[ \frac {\sqrt {\sec (c+d x)} \left (\sin (2 (c+d x)) \left (154 a \left (12 a^2 C+36 A b^2+43 b^2 C\right ) \cos (c+d x)+5 b \left (12 \left (33 a^2 C+11 A b^2+16 b^2 C\right ) \cos (2 (c+d x))+1848 a^2 A+1716 a^2 C+154 a b C \cos (3 (c+d x))+572 A b^2+21 b^2 C \cos (4 (c+d x))+531 b^2 C\right )\right )+80 b \left (33 a^2 (7 A+5 C)+5 b^2 (11 A+9 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+3696 a \left (a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{9240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Sqrt[Sec[c + d*x]],x]

[Out]

(Sqrt[Sec[c + d*x]]*(3696*a*(a^2*(5*A + 3*C) + b^2*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] +
80*b*(33*a^2*(7*A + 5*C) + 5*b^2*(11*A + 9*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (154*a*(36*A*b^
2 + 12*a^2*C + 43*b^2*C)*Cos[c + d*x] + 5*b*(1848*a^2*A + 572*A*b^2 + 1716*a^2*C + 531*b^2*C + 12*(11*A*b^2 +
33*a^2*C + 16*b^2*C)*Cos[2*(c + d*x)] + 154*a*b*C*Cos[3*(c + d*x)] + 21*b^2*C*Cos[4*(c + d*x)]))*Sin[2*(c + d*
x)]))/(9240*d)

________________________________________________________________________________________

fricas [F]  time = 1.24, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{3} \cos \left (d x + c\right )^{5} + 3 \, C a b^{2} \cos \left (d x + c\right )^{4} + 3 \, A a^{2} b \cos \left (d x + c\right ) + A a^{3} + {\left (3 \, C a^{2} b + A b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}}{\sqrt {\sec \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)

________________________________________________________________________________________

maple [B]  time = 2.98, size = 793, normalized size = 2.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x)

[Out]

-2/1155*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(6720*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^12+(-12320*C*a*b^2-16800*C*b^3)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(2640*A*b^3+7920*C*a^2*b+24640*C*
a*b^2+18960*C*b^3)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-5544*A*a*b^2-3960*A*b^3-1848*C*a^3-11880*C*a^2*b-
22792*C*a*b^2-11640*C*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(4620*A*a^2*b+5544*A*a*b^2+3080*A*b^3+1848*
C*a^3+9240*C*a^2*b+10472*C*a*b^2+4620*C*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-2310*A*a^2*b-1386*A*a*b
^2-880*A*b^3-462*C*a^3-2640*C*a^2*b-1848*C*a*b^2-930*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1155*A*a^2
*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))+275*A*b
^3*(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*
(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^3-2079*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^2+825
*C*a^2*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))+2
25*b^3*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))-6
93*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^3-1
617*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^
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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{3}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)/sec(d*x+c)**(1/2),x)

[Out]

Integral((A + C*cos(c + d*x)**2)*(a + b*cos(c + d*x))**3/sqrt(sec(c + d*x)), x)

________________________________________________________________________________________