\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [1233]

Optimal result

Integrand size = 37, antiderivative size = 201 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} (A+C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 (13 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-2/15*A*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*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*cos(d*x+c))^(1/2))*2^(1/2)*cos(d*x
+c)^(1/2)*sec(d*x+c)^(1/2)/d/a^(1/2)+2/15*(13*A+15*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4306, 3123, 3063, 12, 2861, 211} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (13 A+15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}-\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{15 d \sqrt {a \cos (c+d x)+a}} \]

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/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*(13*A + 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[a
+ a*Cos[c + d*x]]) - (2*A*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 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])

Rule 3123

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*Si
n[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 4306

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*} \text {integral}& = \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) \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 {a A}{2}+\frac {1}{2} a (4 A+5 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{5 a} \\ & = -\frac {2 A \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+15 C)-\frac {1}{2} a^2 A \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+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 A \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+C)}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^3} \\ & = \frac {2 (13 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 A \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+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 (13 A+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 A \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+C) \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+C) \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+15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {2 A \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 = 7.96 (sec) , antiderivative size = 1717, normalized size of antiderivative = 8.54 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \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 C \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 {2 C \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 {4 C \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+C) \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 + C*Cos[c + d*x]^2)*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*C*Sin[c/2
+ (d*x)/2])/(5*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(5/2)) + (2*C*Sin[c/2 + (d*x)/2])/(15*(1 - 2*Sin[c/2 + (d*x)/2]^2)
^(3/2)) + (4*C*Sin[c/2 + (d*x)/2])/(15*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) - ((A + 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)))/(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 = 1.25 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.52

method	result	size
default	\(\frac {\sqrt {2}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (15 A \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+15 C \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+15 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+13 A \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}+15 C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+15 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}-A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}+3 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\right )}{15 d a \left (1+\cos \left (d x +c \right )\right )}\)	\(305\)
parts	\(\frac {A \sqrt {2}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (15 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )+13 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+15 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )-\sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{15 d a \left (1+\cos \left (d x +c \right )\right )}+\frac {C \sqrt {2}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+\sqrt {2}\, \sin \left (d x +c \right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\cos ^{3}\left (d x +c \right )\right )}{d a \left (1+\cos \left (d x +c \right )\right )}\)	\(328\)

[In]

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

[Out]

1/15/d*2^(1/2)/a*sec(d*x+c)^(7/2)*((1+cos(d*x+c))*a)^(1/2)/(1+cos(d*x+c))*(15*A*cos(d*x+c)^4*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+15*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(
d*x+c)-csc(d*x+c))+15*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3*arcsin(cot(d*x+c)-csc(d*x+c))+13*A*cos(
d*x+c)^3*sin(d*x+c)*2^(1/2)+15*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3*arcsin(cot(d*x+c)-csc(d*x+c))+
15*C*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)-A*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+3*A*cos(d*x+c)*sin(d*x+c)*2^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.78 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{3} + {\left (A + C\right )} a \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {2 \, {\left ({\left (13 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2} - A \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

1/15*(15*sqrt(2)*((A + C)*a*cos(d*x + c)^3 + (A + C)*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 + 15*C)*cos(d*x + c)^2 - A*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. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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