3.1463 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)} \, dx\)

Optimal. Leaf size=243 \[ \frac {2 \sin (c+d x) \left (4 a^2 C+14 a b B+7 A b^2+5 b^2 C\right )}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 (3 A+C)+14 a b B+b^2 (7 A+5 C)\right )}{21 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 B+10 a A b+6 a b C+3 b^2 B\right )}{5 d}+\frac {2 b (4 a C+7 b B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \sqrt {\sec (c+d x)}} \]

[Out]

2/35*b*(7*B*b+4*C*a)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/21*(7*A*b^2+14*B*a*b+4*C*a^2+5*C*b^2)*sin(d*x+c)/d/sec(d*
x+c)^(1/2)+2/7*C*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*(10*A*a*b+5*B*a^2+3*B*b^2+6*C*a*b)*(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*b*B+7*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(s
in(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.64, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4221, 3049, 3033, 3023, 2748, 2641, 2639} \[ \frac {2 \sin (c+d x) \left (4 a^2 C+14 a b B+7 A b^2+5 b^2 C\right )}{21 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 (3 A+C)+14 a b B+b^2 (7 A+5 C)\right )}{21 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 B+10 a A b+6 a b C+3 b^2 B\right )}{5 d}+\frac {2 b (4 a C+7 b B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^2}{7 d \sqrt {\sec (c+d x)}} \]

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*(10*a*A*b + 5*a^2*B + 3*b^2*B + 6*a*b*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(
5*d) + (2*(14*a*b*B + 7*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*b*(7*b*B + 4*a*C)*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*(7*A*b^2 + 14*a*b*B + 4
*a^2*C + 5*b^2*C)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]]) + (2*C*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Sqr
t[Sec[c + d*x]])

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 (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 \frac {(a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {1}{7} \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} a (7 A+C)+\frac {1}{2} (7 A b+7 a B+5 b C) \cos (c+d x)+\frac {1}{2} (7 b B+4 a C) \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {1}{35} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{4} a^2 (7 A+C)+\frac {7}{4} \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)+\frac {5}{4} \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {1}{105} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{8} \left (14 a b B+7 a^2 (3 A+C)+b^2 (7 A+5 C)\right )+\frac {21}{8} \left (10 a A b+5 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b (7 b B+4 a C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}+\frac {1}{5} \left (\left (10 a A b+5 a^2 B+3 b^2 B+6 a b 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 (14 a b B+7 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 \left (10 a A b+5 a^2 B+3 b^2 B+6 a b 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 (14 a b B+7 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 b (7 b B+4 a C) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 A b^2+14 a b B+4 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 C (a+b \cos (c+d x))^2 \sin (c+d x)}{7 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.99, size = 183, normalized size = 0.75 \[ \frac {\sqrt {\sec (c+d x)} \left (2 \sin (2 (c+d x)) \left (5 \left (14 a^2 C+28 a b B+14 A b^2+3 b^2 C \cos (2 (c+d x))+13 b^2 C\right )+42 b (2 a C+b B) \cos (c+d x)\right )+40 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (7 a^2 (3 A+C)+14 a b B+b^2 (7 A+5 C)\right )+168 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 B+2 a b (5 A+3 C)+3 b^2 B\right )\right )}{420 d} \]

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]]*(168*(5*a^2*B + 3*b^2*B + 2*a*b*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]
+ 40*(14*a*b*B + 7*a^2*(3*A + C) + b^2*(7*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(42*b*(b*
B + 2*a*C)*Cos[c + d*x] + 5*(14*A*b^2 + 28*a*b*B + 14*a^2*C + 13*b^2*C + 3*b^2*C*Cos[2*(c + d*x)]))*Sin[2*(c +
d*x)]))/(420*d)

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fricas [F]  time = 1.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sqrt {\sec \left (d x + c\right )}, x\right ) \]

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)

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

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)

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maple [B]  time = 3.15, size = 706, normalized size = 2.91 \[ \text {result too large to display} \]

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]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8+(-168*B*b^2-336*C*a*b-360*C*b^2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b^2+280*B*a*b+168*B*b^2+14
0*C*a^2+336*C*a*b+280*C*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*b^2-140*B*a*b-42*B*b^2-70*C*a^2-84
*C*a*b-80*C*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-210*A*(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))*a*b+105*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+35*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-105*B*(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))*a^2-63*B*(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))*b^2+70*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-126*C*(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))*a*b+35*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+25*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*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

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

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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)

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