\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [1355]

Optimal result

Integrand size = 45, antiderivative size = 231 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {(75 A-19 B-5 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-1/4*(A-B+C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(13*A-5*B-3*C)*sin(d*x+c)*sec(d*x+c)^(1
/2)/a/d/(a+a*cos(d*x+c))^(3/2)-1/32*(75*A-19*B-5*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)+1/16*(49*A-9*B+C)*sin(d*x+c)*sec(d*x+
c)^(1/2)/a^2/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {4306, 3120, 3057, 3063, 12, 2861, 211} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {(75 A-19 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(49 A-9 B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(13 A-5 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

-1/16*((75*A - 19*B - 5*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[2]*a^(5/2)*d) - ((A - B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(
4*d*(a + a*Cos[c + d*x])^(5/2)) - ((13*A - 5*B - 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Cos[c +
d*x])^(3/2)) + ((49*A - 9*B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(16*a^2*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 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])

Rule 3120

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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 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+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (9 A-B+C)-2 a (A-B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a 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}{4} a^2 (49 A-9 B+C)-\frac {1}{2} a^2 (13 A-5 B-3 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {a^3 (75 A-19 B-5 C)}{8 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a^5} \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((75 A-19 B-5 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}{32 a^2} \\ & = -\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((75 A-19 B-5 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 )}{16 a d} \\ & = -\frac {(75 A-19 B-5 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)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 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 = 7.47 (sec) , antiderivative size = 952, normalized size of antiderivative = 4.12 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {B \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \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 )}} \left (11-31 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+18 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )-\frac {19 \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 ) \cos ^4\left (\frac {1}{2} (c+d x)\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 )}}}\right )}{4 d (a (1+\cos (c+d x)))^{5/2}}+\frac {C \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3-\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-5 \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 ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^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 )}}\right )}{4 d (a (1+\cos (c+d x)))^{5/2} \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}}+\frac {2 A \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right )^{3/2} \left (\frac {8 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \, _4F_3\left (2,2,2,\frac {5}{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 ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{315 \left (-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {1}{120} \csc ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2 \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 )}} \left (-15 \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 ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-343+1465 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-2021 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+824 \sin ^6\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 )}} \left (-5145+33980 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-87764 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+109737 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-66122 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+15344 \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )\right )}{d (a (1+\cos (c+d x)))^{5/2}} \]

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

-1/4*(B*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*(11
- 31*Sin[c/2 + (d*x)/2]^2 + 18*Sin[c/2 + (d*x)/2]^4 - (19*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2
+ (d*x)/2]^2))]]*Cos[(c + d*x)/2]^4)/Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(d*(a*(1 + C
os[c + d*x]))^(5/2)) + (C*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*(3 - Sin[c/2 + (d*x)/2]^2
- 5*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*Cos[(c + d*x)/2]^4*Csc[c/2 + (d*x)/2]
^2*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(4*d*(a*(1 + Cos[c + d*x]))^(5/2)*Sqrt[(1 - 2*
Sin[c/2 + (d*x)/2]^2)^(-1)]) + (2*A*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*((1 - 2*Sin[c/2
+ (d*x)/2]^2)^(-1))^(3/2)*((8*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 5/2}, {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]^2)/(315*(-1 + 2*Sin[c/2 + (d*x)/2]^2)) + (Csc[c/2 +
(d*x)/2]^8*(1 - 2*Sin[c/2 + (d*x)/2]^2)^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-15*ArcTa
nh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/2]^4*(-343 + 1465*Sin[c/2 + (d*x)/2
]^2 - 2021*Sin[c/2 + (d*x)/2]^4 + 824*Sin[c/2 + (d*x)/2]^6) + Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x
)/2]^2)]*(-5145 + 33980*Sin[c/2 + (d*x)/2]^2 - 87764*Sin[c/2 + (d*x)/2]^4 + 109737*Sin[c/2 + (d*x)/2]^6 - 6612
2*Sin[c/2 + (d*x)/2]^8 + 15344*Sin[c/2 + (d*x)/2]^10)))/120))/(d*(a*(1 + Cos[c + d*x]))^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(743\) vs. \(2(196)=392\).

Time = 2.25 (sec) , antiderivative size = 744, normalized size of antiderivative = 3.22

method	result	size
default	\(-\frac {\sqrt {2}\, \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-1}\right )^{\frac {3}{2}} \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \sqrt {\frac {a}{\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}}\, \left (2 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{5} \left (\csc ^{5}\left (d x +c \right )\right )-2 B \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{5} \left (\csc ^{5}\left (d x +c \right )\right )+2 C \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{5} \left (\csc ^{5}\left (d x +c \right )\right )+17 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \left (\csc ^{3}\left (d x +c \right )\right )-9 B \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \left (\csc ^{3}\left (d x +c \right )\right )+C \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \left (\csc ^{3}\left (d x +c \right )\right )+75 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-19 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-5 C \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-83 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+11 B \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 C \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-75 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+19 B \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+5 C \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{32 a^{3} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1\right )}\)	\(744\)
parts	\(\text {Expression too large to display}\)	\(822\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/32/a^3/d*2^(1/2)*(-((1-cos(d*x+c))^2*csc(d*x+c)^2+1)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(3/2)*(-(1-cos(d*x+
c))^2*csc(d*x+c)^2+1)^(1/2)*(a/((1-cos(d*x+c))^2*csc(d*x+c)^2+1))^(1/2)*(2*A*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1
)^(1/2)*(1-cos(d*x+c))^5*csc(d*x+c)^5-2*B*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(1-cos(d*x+c))^5*csc(d*x+c)
^5+2*C*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(1-cos(d*x+c))^5*csc(d*x+c)^5+17*A*(-(1-cos(d*x+c))^2*csc(d*x+
c)^2+1)^(1/2)*(1-cos(d*x+c))^3*csc(d*x+c)^3-9*B*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(1-cos(d*x+c))^3*csc(
d*x+c)^3+C*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(1-cos(d*x+c))^3*csc(d*x+c)^3+75*A*arcsin(cot(d*x+c)-csc(d
*x+c))*(1-cos(d*x+c))^2*csc(d*x+c)^2-19*B*arcsin(cot(d*x+c)-csc(d*x+c))*(1-cos(d*x+c))^2*csc(d*x+c)^2-5*C*arcs
in(cot(d*x+c)-csc(d*x+c))*(1-cos(d*x+c))^2*csc(d*x+c)^2-83*A*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-cot(d*
x+c)+csc(d*x+c))+11*B*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))-3*C*(-(1-cos(d*x+c))^2
*csc(d*x+c)^2+1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))-75*A*arcsin(cot(d*x+c)-csc(d*x+c))+19*B*arcsin(cot(d*x+c)-csc(
d*x+c))+5*C*arcsin(cot(d*x+c)-csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 75 \, A - 19 \, B - 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (49 \, A - 9 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (85 \, A - 13 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 32 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/32*(sqrt(2)*((75*A - 19*B - 5*C)*cos(d*x + c)^3 + 3*(75*A - 19*B - 5*C)*cos(d*x + c)^2 + 3*(75*A - 19*B - 5*
C)*cos(d*x + c) + 75*A - 19*B - 5*C)*sqrt(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 - 9*B + C)*cos(d*x + c)^2 + (85*A - 13*B + 5*C)*cos(d*x + c) + 32*A)*sqrt(a*cos(d
*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x
+ c) + a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(((1/cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(5/2),x)

[Out]

int(((1/cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^(5/2), x)