Integrand size = 45, antiderivative size = 231 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {(75 A-19 B-5 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(13 A-5 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(49 A-9 B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \] Output:
-1/32*(75*A-19*B-5*C)*arctan(1/2*a^(1/2)*sin(d*x+c)*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)*2^(1/2)/a^(5/ 2)/d-1/4*(A-B+C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/16 *(13*A-5*B-3*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/1 6*(49*A-9*B+C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 7.78 (sec) , antiderivative size = 952, normalized size of antiderivative = 4.12 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
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]
Output:
-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 + Cos[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*Sqr t[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-15*ArcTanh[Sqrt[Si n[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/2]^4*(-34 3 + 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)]*(-5 145 + 33980*Sin[c/2 + (d*x)/2]^2 - 87764*Sin[c/2 + (d*x)/2]^4 + 109737*Sin [c/2 + (d*x)/2]^6 - 66122*Sin[c/2 + (d*x)/2]^8 + 15344*Sin[c/2 + (d*x)/...
Time = 1.37 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 4709, 3042, 3520, 27, 3042, 3457, 27, 3042, 3463, 27, 3042, 3261, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{3/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{(a \cos (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (9 A-B+C)-4 a (A-B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (49 A-9 B+C)-2 a^2 (13 A-5 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \int -\frac {a^3 (75 A-19 B-5 C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^2 (49 A-9 B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-a^2 (75 A-19 B-5 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-a^2 (75 A-19 B-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^3 (75 A-19 B-5 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (49 A-9 B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (49 A-9 B+C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} a^{3/2} (75 A-19 B-5 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (13 A-5 B-3 C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}}\right )\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + a*Co s[c + d*x])^(5/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A - B + C)*Sin[c + d*x])/(d* Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2)) + (-1/2*(a*(13*A - 5*B - 3* C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)) + (-((S qrt[2]*a^(3/2)*(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]])])/d) + (2*a^2*(49*A - 9*B + C )*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2))/ (8*a^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-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*S in[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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[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] + Simp[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])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(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] + Simp[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] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(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 ] + Simp[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)]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 2.68 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.53
| method | result | size |
| default | \(\frac {\sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (d x +c \right ) \left (\left (75 \cos \left (d x +c \right )^{3}+225 \cos \left (d x +c \right )^{2}+225 \cos \left (d x +c \right )+75\right ) A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (-19 \cos \left (d x +c \right )^{3}-57 \cos \left (d x +c \right )^{2}-57 \cos \left (d x +c \right )-19\right ) B \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (-5 \cos \left (d x +c \right )^{3}-15 \cos \left (d x +c \right )^{2}-15 \cos \left (d x +c \right )-5\right ) C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (49 \cos \left (d x +c \right )^{2}+85 \cos \left (d x +c \right )+32\right ) \sqrt {2}\, A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-9 \cos \left (d x +c \right )-13\right ) \sqrt {2}\, B +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \left (\cos \left (d x +c \right )+5\right ) C \right ) \sec \left (d x +c \right )^{\frac {3}{2}}}{16 d \,a^{3} \left (1+\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\) | \(353\) |
| parts | \(\frac {A \sqrt {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \cos \left (d x +c \right ) \left (\sin \left (d x +c \right ) \left (49 \cos \left (d x +c \right )^{2}+85 \cos \left (d x +c \right )+32\right ) \sqrt {2}+\left (75 \cos \left (d x +c \right )^{3}+225 \cos \left (d x +c \right )^{2}+225 \cos \left (d x +c \right )+75\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}{16 d \,a^{3} \left (1+\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}-\frac {B \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {2}\, \left (9 \cos \left (d x +c \right )+13\right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \left (19 \cos \left (d x +c \right )^{4}+57 \cos \left (d x +c \right )^{3}+57 \cos \left (d x +c \right )^{2}+19 \cos \left (d x +c \right )\right )\right )}{32 d \,a^{3} \left (1+\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}+\frac {C \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \sqrt {2}\, \left (\cos \left (d x +c \right )+5\right )+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \left (-5 \cos \left (d x +c \right )^{4}-15 \cos \left (d x +c \right )^{3}-15 \cos \left (d x +c \right )^{2}-5 \cos \left (d x +c \right )\right )\right )}{32 d \,a^{3} \left (1+\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\) | \(497\) |
Input:
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)
Output:
1/16/d/a^3*(cos(1/2*d*x+1/2*c)^2*a)^(1/2)*cos(d*x+c)*((75*cos(d*x+c)^3+225 *cos(d*x+c)^2+225*cos(d*x+c)+75)*A*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x +c)/(1+cos(d*x+c)))^(1/2)+(-19*cos(d*x+c)^3-57*cos(d*x+c)^2-57*cos(d*x+c)- 19)*B*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+(-5 *cos(d*x+c)^3-15*cos(d*x+c)^2-15*cos(d*x+c)-5)*C*arcsin(-csc(d*x+c)+cot(d* x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+sin(d*x+c)*(49*cos(d*x+c)^2+85*cos (d*x+c)+32)*2^(1/2)*A+sin(d*x+c)*cos(d*x+c)*(-9*cos(d*x+c)-13)*2^(1/2)*B+s in(d*x+c)*cos(d*x+c)*2^(1/2)*(cos(d*x+c)+5)*C)*sec(d*x+c)^(3/2)/(1+cos(d*x +c))/(cos(d*x+c)^2+2*cos(d*x+c)+1)
Time = 0.12 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, A - 19 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 75 \, A - 19 \, B - 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (49 \, A - 9 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (85 \, A - 13 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 32 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
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")
Output:
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)*s qrt(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)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
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)
Output:
Timed out
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
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")
Output:
Timed out
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
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")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*co s(d*x + c) + a)^(5/2), x)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
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)
Output:
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)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:
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)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec( c + d*x))/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d *x))/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*c + int ((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x))/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*a))/a**3