3.1169 \(\int (a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^{\frac{7}{2}}(c+d x) \, dx\)

Optimal. Leaf size=196 \[ \frac{2 a^2 (17 A+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^2 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{8 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{15 d}-\frac{16 a^2 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

[Out]

(-16*a^2*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(A + 3*C)*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^2*(17*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c
+ d*x])/(15*d) + (8*A*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*A*(a + a*Cos[c + d
*x])^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.541903, antiderivative size = 196, 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, 3044, 2975, 2968, 3021, 2748, 2641, 2639} \[ \frac{2 a^2 (17 A+15 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^2 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{8 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{15 d}-\frac{16 a^2 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*a^2*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(A + 3*C)*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^2*(17*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c
+ d*x])/(15*d) + (8*A*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*A*(a + a*Cos[c + d
*x])^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

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 3044

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/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3021

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 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 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 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]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 A (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^2 \left (2 a A-\frac{1}{2} a (A-5 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 A (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x)) \left (\frac{1}{4} a^2 (17 A+15 C)-\frac{1}{4} a^2 (7 A-15 C) \cos (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac{8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 A (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a^3 (17 A+15 C)+\left (-\frac{1}{4} a^3 (7 A-15 C)+\frac{1}{4} a^3 (17 A+15 C)\right ) \cos (c+d x)-\frac{1}{4} a^3 (7 A-15 C) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac{2 a^2 (17 A+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 A (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{5}{4} a^3 (A+3 C)-3 a^3 A \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a}\\ &=\frac{2 a^2 (17 A+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 A (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac{1}{5} \left (8 a^2 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (2 a^2 (A+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{16 a^2 A \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{4 a^2 (A+3 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 a^2 (17 A+15 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{8 A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 A (a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 2.65281, size = 301, normalized size = 1.54 \[ \frac{1}{15} a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\tan (c) \sqrt{\sec (c+d x)} \left (3 \cot (c) \csc (c) \cos (d x) (16 A-5 C \cos (2 c)+5 C)+6 A \csc (c) \sin (d x) \sec ^2(c+d x)+2 A \sec (c+d x) (10 \csc (c) \sin (d x)+3)+20 A+30 C \cos (c) \cot (c) \sin (d x)\right )}{4 d}-\frac{i \sqrt{2} \left (5 \left (-1+e^{2 i c}\right ) (A+3 C) e^{i (c+d x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-e^{2 i (c+d x)}\right )+12 A \left (-1+e^{2 i c}\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 i (c+d x)}\right )+12 A \sqrt{1+e^{2 i (c+d x)}}\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(((-I)*Sqrt[2]*(12*A*Sqrt[1 + E^((2*I)*(c + d*x))] + 12*A*(-1 + E
^((2*I)*c))*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 5*(A + 3*C)*E^(I*(c + d*x))*(-1 + E^((2*
I)*c))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*(-1 + E^((2*I)*c))*Sqrt[E^(I*(c + d*x))/(1
+ E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]) + (Sqrt[Sec[c + d*x]]*(20*A + 3*(16*A + 5*C - 5*C*Cos[2
*c])*Cos[d*x]*Cot[c]*Csc[c] + 30*C*Cos[c]*Cot[c]*Sin[d*x] + 6*A*Csc[c]*Sec[c + d*x]^2*Sin[d*x] + 2*A*Sec[c + d
*x]*(3 + 10*Csc[c]*Sin[d*x]))*Tan[c])/(4*d)))/15

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Maple [B]  time = 3.109, size = 756, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*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)^3*(20*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4+48*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))*sin(1/2*d*x+1/2*c)^4-96*A*cos(1/2*d*x+1/2*c)*s
in(1/2*d*x+1/2*c)^6+60*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))*sin(1/2*d*x+1/2*c)^4-60*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-20*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))*sin(1/2*d*x+1/2*c)^2-48*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))*sin(1/2*d*x+1/2*c)
^2+116*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-60*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))*sin(1/2*d*x+1/2*c)^2+60*C*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+
5*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*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))-37*A*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+15*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))-15*C*sin(1/2*d*x+1/2*c)^2*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)/(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{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a^{2} \cos \left (d x + c\right )^{4} + 2 \, C a^{2} \cos \left (d x + c\right )^{3} +{\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sec \left (d x + c\right )^{\frac{7}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^2*cos(d*x + c)^4 + 2*C*a^2*cos(d*x + c)^3 + (A + C)*a^2*cos(d*x + c)^2 + 2*A*a^2*cos(d*x + c) +
A*a^2)*sec(d*x + c)^(7/2), 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+a*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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