3.1464 \(\int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\)

Optimal. Leaf size=291 \[ \frac {2 \sin (c+d x) \left (4 a^2 C+18 a b B+9 A b^2+7 b^2 C\right )}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x) \left (7 a^2 B+14 a A b+10 a b C+5 b^2 B\right )}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 B+14 a A b+10 a b C+5 b^2 B\right )}{21 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 (5 A+3 C)+18 a b B+b^2 (9 A+7 C)\right )}{15 d}+\frac {2 b (4 a C+9 b B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/63*b*(9*B*b+4*C*a)*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/45*(9*A*b^2+18*B*a*b+4*C*a^2+7*C*b^2)*sin(d*x+c)/d/sec(d*
x+c)^(3/2)+2/9*C*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/21*(14*A*a*b+7*B*a^2+5*B*b^2+10*C*a*b)*sin
(d*x+c)/d/sec(d*x+c)^(1/2)+2/15*(18*a*b*B+3*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+2/21*(14*A*a*b+7*B*a^2+5*
B*b^2+10*C*a*b)*(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 = 0.69, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4221, 3049, 3033, 3023, 2748, 2639, 2635, 2641} \[ \frac {2 \sin (c+d x) \left (4 a^2 C+18 a b B+9 A b^2+7 b^2 C\right )}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x) \left (7 a^2 B+14 a A b+10 a b C+5 b^2 B\right )}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 B+14 a A b+10 a b C+5 b^2 B\right )}{21 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 (5 A+3 C)+18 a b B+b^2 (9 A+7 C)\right )}{15 d}+\frac {2 b (4 a C+9 b B) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(18*a*b*B + 3*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]])/(15*d) + (2*(14*a*A*b + 7*a^2*B + 5*b^2*B + 10*a*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt
[Sec[c + d*x]])/(21*d) + (2*b*(9*b*B + 4*a*C)*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2)) + (2*(9*A*b^2 + 18*a*b*B
+ 4*a^2*C + 7*b^2*C)*Sin[c + d*x])/(45*d*Sec[c + d*x]^(3/2)) + (2*C*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(9*d
*Sec[c + d*x]^(3/2)) + (2*(14*a*A*b + 7*a^2*B + 5*b^2*B + 10*a*b*C)*Sin[c + d*x])/(21*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 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))^2 \left (A+B \cos (c+d x)+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))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{9} \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)) \left (\frac {3}{2} a (3 A+C)+\frac {1}{2} (9 A b+9 a B+7 b C) \cos (c+d x)+\frac {1}{2} (9 b B+4 a C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b (9 b B+4 a C) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{63} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {21}{4} a^2 (3 A+C)+\frac {9}{4} \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \cos (c+d x)+\frac {7}{4} \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b (9 b B+4 a C) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{315} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \left (\frac {21}{8} \left (18 a b B+3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right )+\frac {45}{8} \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {2 b (9 b B+4 a C) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{7} \left (\left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{15} \left (\left (18 a b B+3 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 {2 \left (18 a b B+3 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)}}{15 d}+\frac {2 b (9 b B+4 a C) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {1}{21} \left (\left (14 a A b+7 a^2 B+5 b^2 B+10 a b 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 (18 a b B+3 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)}}{15 d}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B+10 a b 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 b (9 b B+4 a C) \sin (c+d x)}{63 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (9 A b^2+18 a b B+4 a^2 C+7 b^2 C\right ) \sin (c+d x)}{45 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B+10 a b C\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.40, size = 218, normalized size = 0.75 \[ \frac {\sqrt {\sec (c+d x)} \left (2 \sin (2 (c+d x)) \left (7 \cos (c+d x) \left (36 a^2 C+72 a b B+36 A b^2+43 b^2 C\right )+5 \left (84 a^2 B+168 a A b+18 b (2 a C+b B) \cos (2 (c+d x))+156 a b C+78 b^2 B+7 b^2 C \cos (3 (c+d x))\right )\right )+240 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 B+2 a b (7 A+5 C)+5 b^2 B\right )+336 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 (5 A+3 C)+18 a b B+b^2 (9 A+7 C)\right )\right )}{2520 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(336*(18*a*b*B + 3*a^2*(5*A + 3*C) + b^2*(9*A + 7*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*
x)/2, 2] + 240*(7*a^2*B + 5*b^2*B + 2*a*b*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(7*(36
*A*b^2 + 72*a*b*B + 36*a^2*C + 43*b^2*C)*Cos[c + d*x] + 5*(168*a*A*b + 84*a^2*B + 78*b^2*B + 156*a*b*C + 18*b*
(b*B + 2*a*C)*Cos[2*(c + d*x)] + 7*b^2*C*Cos[3*(c + d*x)]))*Sin[2*(c + d*x)]))/(2520*d)

________________________________________________________________________________________

fricas [F]  time = 3.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )}{\sqrt {\sec \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 3.37, size = 784, normalized size = 2.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)^(1/2),x)

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^10+(720*B*b^2+1440*C*a*b+2240*C*b^2)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-504*A*b^2-1008*B*a*b-1080*B
*b^2-504*C*a^2-2160*C*a*b-2072*C*b^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(840*A*a*b+504*A*b^2+420*B*a^2+1
008*B*a*b+840*B*b^2+504*C*a^2+1680*C*a*b+952*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-420*A*a*b-126*A*
b^2-210*B*a^2-252*B*a*b-240*B*b^2-126*C*a^2-480*C*a*b-168*C*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+210*A
*a*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))-315*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^2-189*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))*b^2+105*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))+75*b^
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))-378*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*b+150*C*
a*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))-189*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^2-147*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))*b^2)/(-2*s
in(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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\sec \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________