3.1379 \(\int (a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=284 \[ -\frac {2 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 a b^2 (35 A-11 C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b (7 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^3}{d} \]

[Out]

-2/35*a*b^2*(35*A-11*C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)-2/21*b*(6*a^2*(7*A-3*C)-b^2*(7*A+5*C))*sin(d*x+c)/d/sec(
d*x+c)^(1/2)-2/7*b*(7*A-C)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2*A*(a+b*cos(d*x+c))^3*sin(d*x+c)*
sec(d*x+c)^(1/2)/d-2/5*a*(5*a^2*(A-C)-3*b^2*(5*A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti
cE(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*b*(21*a^2*(3*A+C)+b^2*(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]  time = 0.90, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4221, 3048, 3049, 3033, 3023, 2748, 2641, 2639} \[ -\frac {2 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}-\frac {2 a b^2 (35 A-11 C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b (7 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(5*a^2*(A - C) - 3*b^2*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*
d) + (2*b*(21*a^2*(3*A + C) + b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]]
)/(21*d) - (2*a*b^2*(35*A - 11*C)*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) - (2*b*(6*a^2*(7*A - 3*C) - b^2*(7*A
+ 5*C))*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) - (2*b*(7*A - C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Sq
rt[Sec[c + d*x]]) + (2*A*(a + b*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

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 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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 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 (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^2 \left (3 A b-\frac {1}{2} a (A-C) \cos (c+d x)-\frac {1}{2} b (7 A-C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{7} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{4} a b (35 A+C)-\frac {1}{4} \left (7 a^2 (A-C)-b^2 (7 A+5 C)\right ) \cos (c+d x)-\frac {1}{4} a b (35 A-11 C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 a b^2 (35 A-11 C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{35} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{8} a^2 b (35 A+C)-\frac {7}{8} a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \cos (c+d x)-\frac {5}{8} b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 a b^2 (35 A-11 C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{105} \left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{16} b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right )-\frac {21}{16} a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 a b^2 (35 A-11 C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {1}{5} \left (a \left (5 a^2 (A-C)-3 b^2 (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 (b \left (21 a^2 (3 A+C)+b^2 (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 a \left (5 a^2 (A-C)-3 b^2 (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 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}-\frac {2 a b^2 (35 A-11 C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 b \left (6 a^2 (7 A-3 C)-b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}-\frac {2 b (7 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.05, size = 193, normalized size = 0.68 \[ \frac {\sqrt {\sec (c+d x)} \left (40 b \left (21 a^2 (3 A+C)+b^2 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-168 a \left (5 a^2 (A-C)-3 b^2 (5 A+3 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) \left (3 \left (140 a^3 A+42 a b^2 C \cos (2 (c+d x))+42 a b^2 C+5 b^3 C \cos (3 (c+d x))\right )+5 b \left (84 a^2 C+28 A b^2+29 b^2 C\right ) \cos (c+d x)\right )\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(-168*a*(5*a^2*(A - C) - 3*b^2*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] +
40*b*(21*a^2*(3*A + C) + b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(5*b*(28*A*b^2 + 8
4*a^2*C + 29*b^2*C)*Cos[c + d*x] + 3*(140*a^3*A + 42*a*b^2*C + 42*a*b^2*C*Cos[2*(c + d*x)] + 5*b^3*C*Cos[3*(c
+ d*x)]))*Sin[c + d*x]))/(420*d)

________________________________________________________________________________________

fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{3} \cos \left (d x + c\right )^{5} + 3 \, C a b^{2} \cos \left (d x + c\right )^{4} + 3 \, A a^{2} b \cos \left (d x + c\right ) + A a^{3} + {\left (3 \, C a^{2} b + A b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sec \left (d x + c\right )^{\frac {3}{2}}, x\right ) \]

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)^(3/2),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)*sec(d*x + c)^(3/2), x)

________________________________________________________________________________________

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

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)^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 3.49, size = 943, normalized size = 3.32 \[ \text {result too large to display} \]

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)^(3/2),x)

[Out]

-2/105*(240*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8
-72*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(7*a+5*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*
c)+28*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(5*A*b^2+15*C*a^2+18*C*a*b+10*C*b^2)*sin(1/2*d*x+
1/2*c)^4*cos(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)*(105*A*a^3+35*A*b^3+105*C*a
^2*b+63*C*a*b^2+40*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+105*A*(-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)^(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^3-315*A*(-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)^(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+315*A*a^2*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(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)+35*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-105*C*(-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)^(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^3-189*C*(-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)^(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+105*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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1
/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
))*(-2*sin(1/2*d*x+1/2*c)^4+sin(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

________________________________________________________________________________________

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________