3.1373 \(\int \frac{(a+b \cos (c+d x))^2 (A+C \cos ^2(c+d x))}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=245 \[ \frac{2 \left (4 a^2 C+b^2 (9 A+7 C)\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a b (7 A+5 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{4 a b (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{8 a b C \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(2*(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) + (4*a*b*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (8*a*b*C*Sin
[c + d*x])/(63*d*Sec[c + d*x]^(5/2)) + (2*(4*a^2*C + b^2*(9*A + 7*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)) + (4*a*b*(7*A + 5*C)*Sin[c + d*x])/(21*d*
Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.553583, antiderivative size = 245, normalized size of antiderivative = 1., 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, 3050, 3033, 3023, 2748, 2639, 2635, 2641} \[ \frac{2 \left (4 a^2 C+b^2 (9 A+7 C)\right ) \sin (c+d x)}{45 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{4 a b (7 A+5 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{4 a b (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{8 a b C \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^2}{9 d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(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) + (4*a*b*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (8*a*b*C*Sin
[c + d*x])/(63*d*Sec[c + d*x]^(5/2)) + (2*(4*a^2*C + b^2*(9*A + 7*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)) + (4*a*b*(7*A + 5*C)*Sin[c + d*x])/(21*d*
Sqrt[Sec[c + d*x]])

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]

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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(a+b \cos (c+d x))^2 \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))^2 \left (A+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} b (9 A+7 C) \cos (c+d x)+2 a C \cos ^2(c+d x)\right ) \, dx\\ &=\frac{8 a b 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}{2} a b (7 A+5 C) \cos (c+d x)+\frac{7}{4} \left (4 a^2 C+b^2 (9 A+7 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac{8 a b C \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (4 a^2 C+b^2 (9 A+7 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 (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right )+\frac{45}{4} a b (7 A+5 C) \cos (c+d x)\right ) \, dx\\ &=\frac{8 a b C \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (4 a^2 C+b^2 (9 A+7 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 (2 a b (7 A+5 C) \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 (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 (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{8 a b C \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (4 a^2 C+b^2 (9 A+7 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{4 a b (7 A+5 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{1}{21} \left (2 a b (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \left (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{4 a b (7 A+5 C) \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{8 a b C \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (4 a^2 C+b^2 (9 A+7 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{4 a b (7 A+5 C) \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.57533, size = 170, normalized size = 0.69 \[ \frac{\sqrt{\sec (c+d x)} \left (\sin (2 (c+d x)) \left (7 \left (36 a^2 C+36 A b^2+43 b^2 C\right ) \cos (c+d x)+5 b (168 a A+36 a C \cos (2 (c+d x))+156 a C+7 b C \cos (3 (c+d x)))\right )+168 \left (3 a^2 (5 A+3 C)+b^2 (9 A+7 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+240 a b (7 A+5 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(168*(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*a*b*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (7*(36*A*b^2 + 36*a^2*C + 43*b^2*C)*Cos[c +
d*x] + 5*b*(168*a*A + 156*a*C + 36*a*C*Cos[2*(c + d*x)] + 7*b*C*Cos[3*(c + d*x)]))*Sin[2*(c + d*x)]))/(1260*d
)

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Maple [B]  time = 1.229, size = 587, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(A+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*b^2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^10+(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^2-504*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+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^2-126*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+150*a*b*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))-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*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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C b^{2} \cos \left (d x + c\right )^{4} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} +{\left (C a^{2} + A b^{2}\right )} \cos \left (d x + c\right )^{2}}{\sqrt{\sec \left (d x + c\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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