3.1204 \(\int \sqrt{a+a \cos (c+d x)} (A+C \cos ^2(c+d x)) \sec ^{\frac{11}{2}}(c+d x) \, dx\)

Optimal. Leaf size=213 \[ \frac{2 a (16 A+21 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a (16 A+21 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a (16 A+21 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{9 d}+\frac{2 a A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{63 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(16*a*(16*A + 21*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a*(16*A + 21*C)*Sec
[c + d*x]^(3/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(16*A + 21*C)*Sec[c + d*x]^(5/2)*Sin[c +
d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(63*d*Sqrt[a + a*Cos[c + d*x
]]) + (2*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*d)

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Rubi [A]  time = 0.58155, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {4221, 3044, 2980, 2772, 2771} \[ \frac{2 a (16 A+21 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{8 a (16 A+21 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{16 a (16 A+21 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{9 d}+\frac{2 a A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{63 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]

[Out]

(16*a*(16*A + 21*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a*(16*A + 21*C)*Sec
[c + d*x]^(3/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(16*A + 21*C)*Sec[c + d*x]^(5/2)*Sin[c +
d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(63*d*Sqrt[a + a*Cos[c + d*x
]]) + (2*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*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 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((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]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin{align*} \int \sqrt{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac{11}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\\ &=\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{a A}{2}+\frac{3}{2} a (2 A+3 C) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx}{9 a}\\ &=\frac{2 a A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{21} \left ((16 A+21 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a (16 A+21 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{105} \left (4 (16 A+21 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{8 a (16 A+21 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (16 A+21 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{315} \left (8 (16 A+21 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{16 a (16 A+21 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{8 a (16 A+21 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (16 A+21 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.708371, size = 124, normalized size = 0.58 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{9}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} (2 (88 A+63 C) \cos (c+d x)+11 (16 A+21 C) \cos (2 (c+d x))+32 A \cos (3 (c+d x))+32 A \cos (4 (c+d x))+214 A+42 C \cos (3 (c+d x))+42 C \cos (4 (c+d x))+189 C)}{315 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(11/2),x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(214*A + 189*C + 2*(88*A + 63*C)*Cos[c + d*x] + 11*(16*A + 21*C)*Cos[2*(c + d*x)]
+ 32*A*Cos[3*(c + d*x)] + 42*C*Cos[3*(c + d*x)] + 32*A*Cos[4*(c + d*x)] + 42*C*Cos[4*(c + d*x)])*Sec[c + d*x]^
(9/2)*Tan[(c + d*x)/2])/(315*d)

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Maple [A]  time = 0.22, size = 129, normalized size = 0.6 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 128\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+168\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+64\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+84\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+48\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+63\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+40\,A\cos \left ( dx+c \right ) +35\,A \right ) \cos \left ( dx+c \right ) }{315\,d\sin \left ( dx+c \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{11}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2)*(a+a*cos(d*x+c))^(1/2),x)

[Out]

-2/315/d*(-1+cos(d*x+c))*(128*A*cos(d*x+c)^4+168*C*cos(d*x+c)^4+64*A*cos(d*x+c)^3+84*C*cos(d*x+c)^3+48*A*cos(d
*x+c)^2+63*C*cos(d*x+c)^2+40*A*cos(d*x+c)+35*A)*cos(d*x+c)*(1/cos(d*x+c))^(11/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(
d*x+c)

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Maxima [B]  time = 1.77329, size = 890, normalized size = 4.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/315*(A*(315*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 735*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 1302*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1206*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7 + 431*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 107*sqrt(2)*sqrt(a)*sin(d*x + c)
^11/(cos(d*x + c) + 1)^11)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^
(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*sin(d*x + c)^
4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + sin(
d*x + c)^10/(cos(d*x + c) + 1)^10 + 1)) + 21*C*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 55*sqrt(2
)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 82*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 66*sq
rt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 31*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 7
*sqrt(2)*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(d*x
+ c)/(cos(d*x + c) + 1) + 1)^(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(5*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 10*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*sin(d*x + c)^
8/(cos(d*x + c) + 1)^8 + sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 1)))/d

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Fricas [A]  time = 1.49153, size = 311, normalized size = 1.46 \begin{align*} \frac{2 \,{\left (8 \,{\left (16 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (16 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (16 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{2} + 40 \, A \cos \left (d x + c\right ) + 35 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \,{\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/315*(8*(16*A + 21*C)*cos(d*x + c)^4 + 4*(16*A + 21*C)*cos(d*x + c)^3 + 3*(16*A + 21*C)*cos(d*x + c)^2 + 40*A
*cos(d*x + c) + 35*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^5 + d*cos(d*x + c)^4)*sqrt(cos(d*
x + c)))

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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+C*cos(d*x+c)**2)*sec(d*x+c)**(11/2)*(a+a*cos(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{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac{11}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(11/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sqrt(a*cos(d*x + c) + a)*sec(d*x + c)^(11/2), x)