3.1231 \(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=289 \[ \frac {2 (19 A+21 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}-\frac {2 (29 A+21 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (257 A+273 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{9 d \sqrt {a \cos (c+d x)+a}}-\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-2/315*(29*A+21*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/105*(19*A+21*C)*sec(d*x+c)^(5/2)*sin
(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-2/63*A*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/9*A*sec(d*x+c)^
(9/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-(A+C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*co
s(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d/a^(1/2)+2/315*(257*A+273*C)*sin(d*x+c)*sec(d*x+c)
^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 1.02, antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4221, 3044, 2984, 12, 2782, 205} \[ \frac {2 (19 A+21 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}-\frac {2 (29 A+21 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{315 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (257 A+273 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{315 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x)}{9 d \sqrt {a \cos (c+d x)+a}}-\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[C
os[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (2*(257*A + 273*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqr
t[a + a*Cos[c + d*x]]) - (2*(29*A + 21*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) +
(2*(19*A + 21*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) - (2*A*Sec[c + d*x]^(7/2)*S
in[c + d*x])/(63*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*d*Sqrt[a + a*Cos[c + d
*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(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]

Rule 2984

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

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 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 {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {11}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {11}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {a A}{2}+\frac {1}{2} a (8 A+9 C) \cos (c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{9 a}\\ &=-\frac {2 A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{4} a^2 (19 A+21 C)-\frac {3}{2} a^2 A \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{63 a^2}\\ &=\frac {2 (19 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 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{8} a^3 (29 A+21 C)+\frac {3}{2} a^3 (19 A+21 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{315 a^3}\\ &=-\frac {2 (29 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 (19 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 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{16} a^4 (257 A+273 C)-\frac {3}{8} a^4 (29 A+21 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{945 a^4}\\ &=\frac {2 (257 A+273 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (29 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 (19 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 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (32 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {945 a^5 (A+C)}{32 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{945 a^5}\\ &=\frac {2 (257 A+273 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (29 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 (19 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 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}-\left ((A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {2 (257 A+273 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (29 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 (19 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 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (2 a (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{d}\\ &=-\frac {\sqrt {2} (A+C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 (257 A+273 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (29 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 (19 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 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 9.03, size = 271, normalized size = 0.94 \[ \frac {2 e^{-\frac {1}{2} i (c+d x)} \cos \left (\frac {1}{2} (c+d x)\right ) \left (-315 i (A+C) \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-\frac {1}{4} \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+i \sin \left (\frac {1}{2} (c+d x)\right )\right ) (2 (107 A+63 C) \cos (c+d x)-8 (157 A+168 C) \cos (2 (c+d x))+58 A \cos (3 (c+d x))-257 A \cos (4 (c+d x))-1279 A+42 C \cos (3 (c+d x))-273 C \cos (4 (c+d x))-1071 C)\right )}{315 d \sqrt {a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*((-315*I)*(A + C)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d
*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - ((-1279*A - 1071*C + 2*(107*A +
63*C)*Cos[c + d*x] - 8*(157*A + 168*C)*Cos[2*(c + d*x)] + 58*A*Cos[3*(c + d*x)] + 42*C*Cos[3*(c + d*x)] - 257
*A*Cos[4*(c + d*x)] - 273*C*Cos[4*(c + d*x)])*Sec[c + d*x]^(9/2)*(Cos[(c + d*x)/2] + I*Sin[(c + d*x)/2])*Sin[(
c + d*x)/2])/4))/(315*d*E^((I/2)*(c + d*x))*Sqrt[a*(1 + Cos[c + d*x])])

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fricas [A]  time = 0.72, size = 190, normalized size = 0.66 \[ \frac {\frac {315 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{5} + {\left (A + C\right )} a \cos \left (d x + c\right )^{4}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {2 \, {\left ({\left (257 \, A + 273 \, C\right )} \cos \left (d x + c\right )^{4} - {\left (29 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 21 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, A \cos \left (d x + c\right ) + 35 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}} \]

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]

1/315*(315*sqrt(2)*((A + C)*a*cos(d*x + c)^5 + (A + C)*a*cos(d*x + c)^4)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) +
a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + 2*((257*A + 273*C)*cos(d*x + c)^4 - (29*A + 21*C)*cos(
d*x + c)^3 + 3*(19*A + 21*C)*cos(d*x + c)^2 - 5*A*cos(d*x + c) + 35*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/s
qrt(cos(d*x + c)))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [B]  time = 0.56, size = 775, normalized size = 2.68 \[ \text {result too large to display} \]

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]

1/315/d*(315*A*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^5+315*C*(cos(d*
x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^5+1575*A*(cos(d*x+c)/(1+cos(d*x+c)))^
(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^4+1575*C*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(
d*x+c))/sin(d*x+c))*cos(d*x+c)^4+3150*A*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*c
os(d*x+c)^3+3150*C*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3+3150*A*(c
os(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2+3150*C*(cos(d*x+c)/(1+cos(d*x+
c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2+1575*A*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1
+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)+1575*C*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c)
)*cos(d*x+c)+315*A*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+315*C*(cos(d*x+c)/(1+c
os(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+257*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^4+273*C*2^(1/2)*sin(d
*x+c)*cos(d*x+c)^4-29*A*cos(d*x+c)^3*2^(1/2)*sin(d*x+c)-21*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3+57*A*cos(d*x+c)^2
*2^(1/2)*sin(d*x+c)+63*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2-5*A*cos(d*x+c)*2^(1/2)*sin(d*x+c)+35*A*2^(1/2)*sin(d*
x+c))*cos(d*x+c)*sin(d*x+c)^8*(1/cos(d*x+c))^(11/2)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))^4/(1+cos(d*x+c))^
5*2^(1/2)/a

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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