3.1232 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^{\frac{9}{2}}(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=244 \[ \frac{2 (31 A+35 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}-\frac{2 (43 A+35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d \sqrt{a \cos (c+d x)+a}}-\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}} \]

[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[Cos
[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d) - (2*(43*A + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a +
a*Cos[c + d*x]]) + (2*(31*A + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) - (2*A*
Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d*
Sqrt[a + a*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.829455, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{2 (31 A+35 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}-\frac{2 (43 A+35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{2} (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d \sqrt{a \cos (c+d x)+a}}-\frac{2 A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2))/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[Cos
[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d) - (2*(43*A + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a +
a*Cos[c + d*x]]) + (2*(31*A + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) - (2*A*
Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d*
Sqrt[a + a*Cos[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 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 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 12

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

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 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]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac{9}{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{9}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 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 (6 A+7 C) \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{7 a}\\ &=-\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 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{1}{4} a^2 (31 A+35 C)-a^2 A \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{35 a^2}\\ &=\frac{2 (31 A+35 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 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{1}{8} a^3 (43 A+35 C)+\frac{1}{4} a^3 (31 A+35 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^3}\\ &=-\frac{2 (43 A+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (16 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{105 a^4 (A+C)}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{105 a^4}\\ &=-\frac{2 (43 A+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 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 (43 A+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 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 (43 A+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (31 A+35 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 10.2709, size = 2480, normalized size = 10.16 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2))/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(2*Cos[c/2 + (d*x)/2]*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*(-(C*Sin[c/2 +
(d*x)/2])/(3*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)) + ((A + C)*Csc[c/2 + (d*x)/2]^9*(363825*Sin[c/2 + (d*x)/2]^2
- 4729725*Sin[c/2 + (d*x)/2]^4 + 26785605*Sin[c/2 + (d*x)/2]^6 - 86790165*Sin[c/2 + (d*x)/2]^8 + 177677808*Sin
[c/2 + (d*x)/2]^10 - 239283044*Sin[c/2 + (d*x)/2]^12 + 52080*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/
2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 560*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1
, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 213120160*Sin[c/2 + (d*x)
/2]^14 - 168280*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 +
(d*x)/2]^14 - 2240*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/
2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^14 - 121497024*Sin[c/2 + (d*x)/2]^16 + 212520*Hypergeometric2F1[2, 11/2, 1
3/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 3360*HypergeometricPFQ[{2, 2,
2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 401
25184*Sin[c/2 + (d*x)/2]^18 - 124320*Hypergeometric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (
d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^18 - 2240*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x
)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^18 - 5840384*Sin[c/2 + (d*x)/2]^20 + 28000*Hypergeome
tric2F1[2, 11/2, 13/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^20 + 560*Hyperge
ometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 +
(d*x)/2]^20 + 363825*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c/2 + (d*x)/2]
^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 5336100*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*
Sin[c/2 + (d*x)/2]^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 34636140*ArcTanh[Sqrt[Sin[c/2
+ (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 +
(d*x)/2]^2)] - 131060160*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^
6*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 320535600*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 +
2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^8*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 530
671680*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^10*Sqrt[Sin[c/2 +
(d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 604296000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x
)/2]^2)]]*Sin[c/2 + (d*x)/2]^12*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 468948480*ArcTanh[S
qrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^14*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 +
2*Sin[c/2 + (d*x)/2]^2)] + 237726720*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/
2 + (d*x)/2]^16*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 70963200*ArcTanh[Sqrt[Sin[c/2 + (d*
x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^18*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x
)/2]^2)] + 9461760*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^20*Sqr
t[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 1120*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 11
/2}, {1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(-6 + 5*Sin[c/2 +
(d*x)/2]^2) + 280*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 11/2}, {1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*
Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(103 - 164*Sin[c/2 + (d*x)/2]^2 + 70*Sin[c/2 + (d*x)/2]^4)))/(404
25*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(9/2)*(-1 + 2*Sin[c/2 + (d*x)/2]^2)) + (C*((5*Sin[c/2 + (d*x)/2])/(1 - 2*Sin[c
/2 + (d*x)/2]^2)^(7/2) + 2*((3*Sin[c/2 + (d*x)/2])/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2) + 4*(Sin[c/2 + (d*x)/2]/
(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2) + (2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]))))/105))/(d*Sqrt
[a*(1 + Cos[c + d*x])])

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Maple [B]  time = 0.229, size = 639, normalized size = 2.6 \begin{align*} \text{result too large to display} \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)^(9/2)/(a+a*cos(d*x+c))^(1/2),x)

[Out]

1/105/d*2^(1/2)/a*(105*A*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+105
*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+420*A*cos(d*x+c)^3*(cos(d
*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+420*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^
(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+630*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d
*x+c))/sin(d*x+c))+630*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+420
*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+420*C*cos(d*x+c)*(cos(d*x+c
)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+105*A*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+
cos(d*x+c))/sin(d*x+c))+105*C*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+43*A*2^(1/2
)*cos(d*x+c)^3*sin(d*x+c)+35*C*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)-31*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-35*C*2^(1/
2)*cos(d*x+c)^2*sin(d*x+c)+3*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)-15*A*2^(1/2)*sin(d*x+c))*cos(d*x+c)*sin(d*x+c)^6*
(1/cos(d*x+c))^(9/2)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))^3/(1+cos(d*x+c))^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70353, size = 483, normalized size = 1.98 \begin{align*} -\frac{\frac{105 \, \sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} +{\left (A + C\right )} a \cos \left (d x + c\right )^{3}\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 (43 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{3} -{\left (31 \, A + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) - 15 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{105 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\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)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/105*(105*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A + C)*a*cos(d*x + c)^3)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) +
a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + 2*((43*A + 35*C)*cos(d*x + c)^3 - (31*A + 35*C)*cos(d
*x + c)^2 + 3*A*cos(d*x + c) - 15*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x +
c)^4 + a*d*cos(d*x + c)^3)

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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)**(9/2)/(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 \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac{9}{2}}}{\sqrt{a \cos \left (d x + c\right ) + a}}\,{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)^(9/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(9/2)/sqrt(a*cos(d*x + c) + a), x)