\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [530]

Optimal result

Integrand size = 35, antiderivative size = 270 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(15 A-11 B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}+\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-1/2*(A-B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-1/30*(39*A-35*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a
/d/(a+a*cos(d*x+c))^(1/2)+1/10*(9*A-5*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/4*(15*A-11*B
)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(
1/2)/a^(3/2)/d*2^(1/2)+1/30*(147*A-95*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3040, 3057, 3063, 12, 2861, 211} \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(15 A-11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(9 A-5 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(39 A-35 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{30 a d \sqrt {a \cos (c+d x)+a}}+\frac {(147 A-95 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{30 a d \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

-1/2*((15*A - 11*B)*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[2]*a^(3/2)*d) + ((147*A - 95*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(30*a
*d*Sqrt[a + a*Cos[c + d*x]]) - ((39*A - 35*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(30*a*d*Sqrt[a + a*Cos[c + d*x]
]) - ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((9*A - 5*B)*Sec[c + d*x]^(5
/2)*Sin[c + d*x])/(10*a*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 211

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

Rule 2861

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 3040

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Sin[e + f*x])^m*((
c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3057

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

Rule 3063

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])

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx \\ & = -\frac {(A-B) \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 B)-3 a (A-B) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A-B) \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 B) \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 {1}{4} a^2 (39 A-35 B)+a^2 (9 A-5 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{5 a^3} \\ & = -\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \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 B) \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 {1}{8} a^3 (147 A-95 B)-\frac {1}{4} a^3 (39 A-35 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^4} \\ & = \frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \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 B) \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-11 B)}{16 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^5} \\ & = \frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \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 B) \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-11 B) \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 {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \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 B) \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-11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {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-11 B) \arctan \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 {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \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 B) \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] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 7.23 (sec) , antiderivative size = 2166, normalized size of antiderivative = 8.02 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]

[In]

Integrate[((A + B*Cos[c + d*x])*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]*(-1/20*((A -
B)*(1 - 2*Sin[c/2 + (d*x)/2]))/((1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + ((A - B)*(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 - B)*(-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 - B)*(-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])*Sqr
t[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 -
3*B)*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*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]^12 + 226656*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*Hypergeometric2
F1[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*Sin[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*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 833760*ArcTanh[Sqrt[Si
n[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*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)] - 1598400*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)] +
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*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 69120*ArcTanh[Sqrt[Si
n[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 [A] (verified)

Time = 10.38 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.69

method	result	size
default	\(\frac {\left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (225 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-165 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+147 A \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}+450 A \left (\cos ^{4}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-95 B \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}-330 B \left (\cos ^{4}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+108 A \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+225 A \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-60 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}-165 B \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-12 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+20 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+12 A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{60 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2}}\)	\(457\)
parts	\(\frac {A \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (75 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )+49 \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+150 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )+36 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+75 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-4 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{20 d \left (1+\cos \left (d x +c \right )\right )^{2} a^{2}}-\frac {B \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (33 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )+19 \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+66 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )+12 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+33 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-4 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right ) \sqrt {2}}{12 d \left (1+\cos \left (d x +c \right )\right )^{2} a^{2}}\)	\(489\)

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+cos(d*x+c)*a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/60/a^2/d*sec(d*x+c)^(7/2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))^2*(225*A*arcsin(cot(d*x+c)-csc(d*x+c))*cos
(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-165*B*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^5*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)+147*A*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)+450*A*cos(d*x+c)^4*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)-95*B*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)-330*B*cos(d*x+c)^4*arcsin(cot(d*x+c)-csc(d*x+
c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+108*A*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)+225*A*cos(d*x+c)^3*(cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))-60*B*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)-165*B*cos(d*x+c)^3*(cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))-12*A*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+20*B*sin(d*x+c
)*cos(d*x+c)^2*2^(1/2)+12*A*sin(d*x+c)*cos(d*x+c)*2^(1/2))*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {15 \, \sqrt {2} {\left ({\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (147 \, A - 95 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 12 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/60*(15*sqrt(2)*((15*A - 11*B)*cos(d*x + c)^4 + 2*(15*A - 11*B)*cos(d*x + c)^3 + (15*A - 11*B)*cos(d*x + c)^2
)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((147*A - 95*
B)*cos(d*x + c)^3 + 12*(9*A - 5*B)*cos(d*x + c)^2 - 4*(3*A - 5*B)*cos(d*x + c) + 12*A)*sqrt(a*cos(d*x + c) + a
)*sin(d*x + c)/sqrt(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)

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + a*cos(c + d*x))^(3/2),x)

[Out]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + a*cos(c + d*x))^(3/2), x)