3.1240 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^{\frac{7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=268 \[ -\frac{(15 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(9 A+5 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{10 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(13 A+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{10 a d \sqrt{a \cos (c+d x)+a}}+\frac{(49 A+25 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((15*A + 7*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]])/(2*Sqrt[2]*a^(3/2)*d) + ((49*A + 25*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*a*d*S
qrt[a + a*Cos[c + d*x]]) - ((13*A + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]]) -
((A + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((9*A + 5*C)*Sec[c + d*x]^(5/2)*S
in[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.881977, antiderivative size = 268, 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, 3042, 2984, 12, 2782, 205} \[ -\frac{(15 A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(9 A+5 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{10 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(13 A+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{10 a d \sqrt{a \cos (c+d x)+a}}+\frac{(49 A+25 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

-((15*A + 7*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]])/(2*Sqrt[2]*a^(3/2)*d) + ((49*A + 25*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*a*d*S
qrt[a + a*Cos[c + d*x]]) - ((13*A + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]]) -
((A + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((9*A + 5*C)*Sec[c + d*x]^(5/2)*S
in[c + d*x])/(10*a*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 3042

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

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{7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (9 A+5 C)-a (3 A+C) \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A+5 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{3}{4} a^2 (13 A+5 C)+a^2 (9 A+5 C) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=-\frac{(13 A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A+5 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{10 a 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{3}{8} a^3 (49 A+25 C)-\frac{3}{4} a^3 (13 A+5 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(49 A+25 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(13 A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A+5 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int -\frac{15 a^4 (15 A+7 C)}{16 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{15 a^5}\\ &=\frac{(49 A+25 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(13 A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A+5 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{\left ((15 A+7 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}{4 a}\\ &=\frac{(49 A+25 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(13 A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A+5 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left ((15 A+7 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 )}{2 d}\\ &=-\frac{(15 A+7 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)}}{2 \sqrt{2} a^{3/2} d}+\frac{(49 A+25 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(13 A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A+5 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 8.05772, size = 2280, normalized size = 8.51 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*((4*C*Sin[c/2
+ (d*x)/2])/(5*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) - ((A + C)*(1 - 2*Sin[c/2 + (d*x)/2]))/(20*(1 + Sin[c/2 +
(d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + ((A + C)*(1 + 2*Sin[c/2 + (d*x)/2]))/(20*(1 - Sin[c/2 + (d*x)/
2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + (16*C*(Sin[c/2 + (d*x)/2]/(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2) + (2*Si
n[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]))/15 - ((A + C)*(-105*ArcTan[(1 - 2*Sin[c/2 + (d*x)/2])/Sqr
t[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (4 + 3*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2
]^2)^(3/2)) - (19 + 29*Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - (67*S
qrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/30 + ((A + C)*(-105*ArcTan[(1 + 2*Sin[c/2 + (d*x)/
2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (4 - 3*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 +
(d*x)/2]^2)^(3/2)) - (19 - 29*Sin[c/2 + (d*x)/2])/((1 + Sin[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])
- (67*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*x)/2])))/30 + ((-A + 7*C)*Csc[c/2 + (d*x)/2]^7*(4725
*Sin[c/2 + (d*x)/2]^2 - 48825*Sin[c/2 + (d*x)/2]^4 + 210105*Sin[c/2 + (d*x)/2]^6 - 486630*Sin[c/2 + (d*x)/2]^8
+ 655812*Sin[c/2 + (d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (
d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10 - 40*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 11/2}, Sin[
c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10 - 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hy
pergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 2266
56*Sin[c/2 + (d*x)/2]^14 - 1500*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2
]^2)]*Sin[c/2 + (d*x)/2]^14 - 42048*Sin[c/2 + (d*x)/2]^16 + 440*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x
)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 4725*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*S
in[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 56700*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)] + 291060*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*Sq
rt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 833760*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)] + 1458000*Ar
cTanh[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)] - 1598400*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Si
n[c/2 + (d*x)/2]^10*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1080000*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)] - 414720*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^14*S
qrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 69120*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)] + 60*Cos[(c
+ d*x)/2]^4*HypergeometricPFQ[{2, 2, 9/2}, {1, 11/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin
[c/2 + (d*x)/2]^10*(-5 + 4*Sin[c/2 + (d*x)/2]^2)))/(1350*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)*(-1 + 2*Sin[c/2 +
(d*x)/2]^2))))/(d*(a*(1 + Cos[c + d*x]))^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.205, size = 583, normalized size = 2.2 \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)^(7/2)/(a+a*cos(d*x+c))^(3/2),x)

[Out]

1/20/d*2^(1/2)/a^2*(75*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^3*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(
d*x+c))+35*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^3*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+225*
A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^2*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+105*C*(cos(d*x+
c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^2*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+225*A*(cos(d*x+c)/(1+cos(d
*x+c)))^(5/2)*cos(d*x+c)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+105*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)
*cos(d*x+c)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+75*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)*ar
csin((-1+cos(d*x+c))/sin(d*x+c))+35*C*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(
d*x+c))-49*A*2^(1/2)*cos(d*x+c)^4-25*C*2^(1/2)*cos(d*x+c)^4+13*A*cos(d*x+c)^3*2^(1/2)+5*C*cos(d*x+c)^3*2^(1/2)
+40*A*cos(d*x+c)^2*2^(1/2)+20*C*cos(d*x+c)^2*2^(1/2)-8*A*cos(d*x+c)*2^(1/2)+4*A*2^(1/2))*cos(d*x+c)*(1/cos(d*x
+c))^(7/2)*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c)^3/(-1+cos(d*x+c))^2/(1+cos(d*x+c))^3

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Maxima [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)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.36114, size = 567, normalized size = 2.12 \begin{align*} \frac{5 \, \sqrt{2}{\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \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 ) + \frac{2 \,{\left ({\left (49 \, A + 25 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, A \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{20 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\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)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/20*(5*sqrt(2)*((15*A + 7*C)*cos(d*x + c)^4 + 2*(15*A + 7*C)*cos(d*x + c)^3 + (15*A + 7*C)*cos(d*x + c)^2)*sq
rt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((49*A + 25*C)*co
s(d*x + c)^3 + 4*(9*A + 5*C)*cos(d*x + c)^2 - 4*A*cos(d*x + c) + 4*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sq
rt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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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)**(7/2)/(a+a*cos(d*x+c))**(3/2),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{7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{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)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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