\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [522]

Optimal result

Integrand size = 35, antiderivative size = 207 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} (A-B) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 (13 A-5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (A-5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-2/15*(A-5*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/5*A*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*co
s(d*x+c))^(1/2)-(A-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))*2^(1/2)*c
os(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d/a^(1/2)+2/15*(13*A-5*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2
)

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3040, 3063, 12, 2861, 211} \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {2 (A-5 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (13 A-5 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

-((Sqrt[2]*(A - B)*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*(13*A - 5*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a
+ a*Cos[c + d*x]]) - (2*(A - 5*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[
c + d*x]^(5/2)*Sin[c + d*x])/(5*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 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) \sqrt {a+a \cos (c+d x)}} \, dx \\ & = \frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 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}{2} a (A-5 B)+2 a A \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{5 a} \\ & = -\frac {2 (A-5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 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 (13 A-5 B)-\frac {1}{2} a^2 (A-5 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^2} \\ & = \frac {2 (13 A-5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (A-5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {15 a^3 (A-B)}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^3} \\ & = \frac {2 (13 A-5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (A-5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\left ((A-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 \\ & = \frac {2 (13 A-5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (A-5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (2 a (A-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 )}{d} \\ & = -\frac {\sqrt {2} (A-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)}}{\sqrt {a} d}+\frac {2 (13 A-5 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (A-5 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 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 = 7.21 (sec) , antiderivative size = 1719, normalized size of antiderivative = 8.30 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )} \left (\frac {2 B \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{5 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{5/2}}+\frac {8 B \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{15 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{3/2}}+\frac {16 B \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{15 \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-\frac {(A-B) \csc ^7\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (4725 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-48825 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+210105 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-486630 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+655812 \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-710 \operatorname {Hypergeometric2F1}\left (2,\frac {9}{2},\frac {11}{2},\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-40 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \, _4F_3\left (2,2,2,\frac {9}{2};1,1,\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-518760 \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right )+1770 \operatorname {Hypergeometric2F1}\left (2,\frac {9}{2},\frac {11}{2},\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right )+226656 \sin ^{14}\left (\frac {c}{2}+\frac {d x}{2}\right )-1500 \operatorname {Hypergeometric2F1}\left (2,\frac {9}{2},\frac {11}{2},\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{14}\left (\frac {c}{2}+\frac {d x}{2}\right )-42048 \sin ^{16}\left (\frac {c}{2}+\frac {d x}{2}\right )+440 \operatorname {Hypergeometric2F1}\left (2,\frac {9}{2},\frac {11}{2},\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{16}\left (\frac {c}{2}+\frac {d x}{2}\right )+4725 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-56700 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+291060 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-833760 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+1458000 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-1598400 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+1080000 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-414720 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^{14}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+69120 \text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \sin ^{16}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}+60 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {9}{2};1,\frac {11}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-5+4 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )}{675 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{7/2} \left (-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )}\right )}{d \sqrt {a (1+\cos (c+d x))}} \]

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2))/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]*((2*B*Sin[c/2 +
(d*x)/2])/(5*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + (8*B*Sin[c/2 + (d*x)/2])/(15*(1 - 2*Sin[c/2 + (d*x)/2]^2)^
(3/2)) + (16*B*Sin[c/2 + (d*x)/2])/(15*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - ((A - 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*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)))/(675*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)*(-1 + 2*Sin[c/2 + (
d*x)/2]^2))))/(d*Sqrt[a*(1 + Cos[c + d*x])])

Maple [A] (verified)

Time = 9.84 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.57

method	result	size
default	\(\frac {\left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (15 A \left (\cos ^{4}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-15 B \left (\cos ^{4}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+13 A \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+15 A \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-5 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}-15 B \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+5 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+3 A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{15 d \left (1+\cos \left (d x +c \right )\right ) a}\)	\(325\)
parts	\(\frac {A \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (15 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )+15 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+13 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-\sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{15 d \left (1+\cos \left (d x +c \right )\right ) a}-\frac {B \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (3 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )+\sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-\sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right ) \sqrt {2}}{3 d \left (1+\cos \left (d x +c \right )\right ) a}\)	\(360\)

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+cos(d*x+c)*a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15/d*sec(d*x+c)^(7/2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))*(15*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)-15*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)+13*A*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)+15*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))-5*B*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)-15*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcs
in(cot(d*x+c)-csc(d*x+c))-A*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+5*B*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+3*A*sin(d*x+c)
*cos(d*x+c)*2^(1/2))*2^(1/2)/a

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.79 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{3} + {\left (A - B\right )} a \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {2 \, {\left ({\left (13 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (A - 5 \, B\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/15*(15*sqrt(2)*((A - B)*a*cos(d*x + c)^3 + (A - B)*a*cos(d*x + c)^2)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)
*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + 2*((13*A - 5*B)*cos(d*x + c)^2 - (A - 5*B)*cos(d*x + c)
+ 3*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)

Sympy [F(-1)]

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

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**(7/2)/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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