\(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [1460]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 43, antiderivative size = 240 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (4 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

2/15*a*(4*A*b+5*B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/5*
(4*A*b^2+10*B*a*b+a^2*(3*A+5*C))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*(10*B*a*b+5*b^2*(A-C)+a^2*(3*A+5*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+2/3*(B*a^2+3*B*b^2+2*a*b*(A+3*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 = 0.68 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4306, 3126, 3110, 3100, 2827, 2720, 2719} \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 (3 A+5 C)+10 a b B+4 A b^2\right )}{5 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right )}{3 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )}{5 d}+\frac {2 a (5 a B+4 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[In]

Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(-2*(10*a*b*B + 5*b^2*(A - C) + a^2*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x
]])/(5*d) + (2*(a^2*B + 3*b^2*B + 2*a*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d
*x]])/(3*d) + (2*(4*A*b^2 + 10*a*b*B + a^2*(3*A + 5*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a*(4*A*b +
 5*a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])
/(5*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 3100

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

Rule 3110

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[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)
*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)
), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d +
 b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m
 + 1)*(a^2 - b^2)*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] && 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 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))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x)) \left (\frac {1}{2} (4 A b+5 a B)+\frac {1}{2} (3 a A+5 b B+5 a C) \cos (c+d x)-\frac {1}{2} b (A-5 C) \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a (4 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac {1}{15} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{4} \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right )-\frac {5}{4} \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \cos (c+d x)+\frac {3}{4} b^2 (A-5 C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (4 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac {1}{15} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {5}{8} \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right )+\frac {3}{8} \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (4 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} \left (\left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (\left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {2 \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 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 (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (4 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.27 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.80 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 \left (10 a b B+5 b^2 (A-C)+a^2 (3 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {\left (15 \left (A b^2+2 a b B+a^2 (A+C)\right )+10 a (2 A b+a B) \cos (c+d x)+3 \left (5 A b^2+10 a b B+a^2 (3 A+5 C)\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}\right )}{15 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-3*(10*a*b*B + 5*b^2*(A - C) + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/
2, 2] + 5*(a^2*B + 3*b^2*B + 2*a*b*(A + 3*C))*EllipticF[(c + d*x)/2, 2] + ((15*(A*b^2 + 2*a*b*B + a^2*(A + C))
 + 10*a*(2*A*b + a*B)*Cos[c + d*x] + 3*(5*A*b^2 + 10*a*b*B + a^2*(3*A + 5*C))*Cos[2*(c + d*x)])*Sin[c + d*x])/
(2*Cos[c + d*x]^(5/2))))/(15*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(972\) vs. \(2(268)=536\).

Time = 2358.52 (sec) , antiderivative size = 973, normalized size of antiderivative = 4.05

method result size
default \(\text {Expression too large to display}\) \(973\)
parts \(\text {Expression too large to display}\) \(1062\)

[In]

int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)-2*C*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+
1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+4*a*b*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2
*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*
b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2/5*A*a^2/(8*sin(1/2*d
*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2
*cos(1/2*d*x+1/2*c)-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)*EllipticE(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*a*(2*A*b+B*a)*(-1/6*cos(1/2*d*x+1/2*c)*(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x
+1/2*c),2^(1/2)))+2*(A*b^2+2*B*a*b+C*a^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-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)^2*cos(1/2*d*x+1/2*c)-(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d*x+1/2*c
)^2)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.13 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.33 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, {\left (A + 3 \, C\right )} a b + 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{2} - 2 i \, {\left (A + 3 \, C\right )} a b - 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{2} + 10 i \, B a b + 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{2} - 10 i \, B a b - 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, A a^{2} + 3 \, {\left ({\left (3 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(5*sqrt(2)*(I*B*a^2 + 2*I*(A + 3*C)*a*b + 3*I*B*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x +
 c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a^2 - 2*I*(A + 3*C)*a*b - 3*I*B*b^2)*cos(d*x + c)^2*weierstrassPInvers
e(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(I*(3*A + 5*C)*a^2 + 10*I*B*a*b + 5*I*(A - C)*b^2)*cos(d*x
 + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-I*(3*A
 + 5*C)*a^2 - 10*I*B*a*b - 5*I*(A - C)*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c
os(d*x + c) - I*sin(d*x + c))) - 2*(3*A*a^2 + 3*((3*A + 5*C)*a^2 + 10*B*a*b + 5*A*b^2)*cos(d*x + c)^2 + 5*(B*a
^2 + 2*A*a*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*sec(d*x + c)^(7/2), x)

Giac [F]

\[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*sec(d*x + c)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

[In]

int((1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int((1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)