Integrand size = 37, antiderivative size = 242 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {5 a^{5/2} (8 A+5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}-\frac {a^3 (24 A-49 C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (8 A-3 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]
-1/3*a*(6*A-C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)-1/24*a ^3*(24*A-49*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-1/4*a^ 2*(8*A-3*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)+2*A*(a+a* cos(d*x+c))^(5/2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+5/8*a^(5/2)*(8*A+5*C)*arcs in(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^ (1/2)/d
Time = 0.84 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.59 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (15 \sqrt {2} (8 A+5 C) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 (48 A+17 C+3 (8 A+27 C) \cos (c+d x)+17 C \cos (2 (c+d x))+2 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(15*Sq rt[2]*(8*A + 5*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2* (48*A + 17*C + 3*(8*A + 27*C)*Cos[c + d*x] + 17*C*Cos[2*(c + d*x)] + 2*C*C os[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
Time = 1.55 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {3042, 4709, 3042, 3523, 27, 3042, 3455, 27, 3042, 3455, 27, 3042, 3460, 3042, 3253, 223}
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.
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{3/2} (a \cos (c+d x)+a)^{5/2} \left (A+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{5/2} \left (C \cos ^2(c+d x)+A\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int \frac {(\cos (c+d x) a+a)^{5/2} (5 a A-a (6 A-C) \cos (c+d x))}{2 \sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {(\cos (c+d x) a+a)^{5/2} (5 a A-a (6 A-C) \cos (c+d x))}{\sqrt {\cos (c+d x)}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a A-a (6 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{3} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (a^2 (24 A+C)-3 a^2 (8 A-3 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \int \frac {(\cos (c+d x) a+a)^{3/2} \left (a^2 (24 A+C)-3 a^2 (8 A-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a^2 (24 A+C)-3 a^2 (8 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^3 (72 A+13 C)-a^3 (24 A-49 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx-\frac {3 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\cos (c+d x) a+a} \left (a^3 (72 A+13 C)-a^3 (24 A-49 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {3 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a^3 (72 A+13 C)-a^3 (24 A-49 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {3 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {15}{2} a^3 (8 A+5 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx-\frac {a^4 (24 A-49 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {15}{2} a^3 (8 A+5 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a^4 (24 A-49 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \left (-\frac {15 a^3 (8 A+5 C) \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {a^4 (24 A-49 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {15 a^{7/2} (8 A+5 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {a^4 (24 A-49 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {3 a^3 (8 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{2 d}\right )-\frac {a^2 (6 A-C) \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d}}{a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{d \sqrt {\cos (c+d x)}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Cos[c + d*x])^(5/2)*Sin [c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (-1/3*(a^2*(6*A - C)*Sqrt[Cos[c + d*x] ]*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/d + ((-3*a^3*(8*A - 3*C)*Sqrt[C os[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + ((15*a^(7/2)*( 8*A + 5*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a ^4*(24*A - 49*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d* x]]))/4)/6)/a)
3.13.27.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-2/f Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E qQ[a^2 - b^2, 0] && EqQ[d, a/b]
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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*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] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(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*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[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])
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 = 2.70 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.35
| method | result | size |
| default | \(\frac {a^{2} \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (8 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+120 A \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+34 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+75 C \cos \left (d x +c \right ) \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+24 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+120 A \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+75 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+75 C \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+48 A \sin \left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \cos \left (d x +c \right )}{24 d \left (1+\cos \left (d x +c \right )\right )}\) | \(327\) |
| parts | \(\frac {A \,a^{2} \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (5 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \cos \left (d x +c \right )+\cos \left (d x +c \right ) \sin \left (d x +c \right )+5 \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+2 \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}}{d \left (1+\cos \left (d x +c \right )\right )}+\frac {C \,a^{2} \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (8 \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+34 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+75 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+75 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+75 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{24 d \left (1+\cos \left (d x +c \right )\right )}\) | \(368\) |
1/24*a^2/d*sec(d*x+c)^(3/2)*(8*C*cos(d*x+c)^3*sin(d*x+c)+120*A*arctan((cos (d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*cos(d*x+c)*(cos(d*x+c)/(1+cos(d* x+c)))^(1/2)+34*C*cos(d*x+c)^2*sin(d*x+c)+75*C*cos(d*x+c)*arctan((cos(d*x+ c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+24* A*cos(d*x+c)*sin(d*x+c)+120*A*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan (d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+75*C*cos(d*x+c)*sin(d*x+c)+75*C *arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*(cos(d*x+c)/(1+cos(d *x+c)))^(1/2)+48*A*sin(d*x+c))*((1+cos(d*x+c))*a)^(1/2)*cos(d*x+c)/(1+cos( d*x+c))
Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.67 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {15 \, {\left ({\left (8 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (8 \, A + 5 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (8 \, C a^{2} \cos \left (d x + c\right )^{3} + 34 \, C a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (8 \, A + 25 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
-1/24*(15*((8*A + 5*C)*a^2*cos(d*x + c) + (8*A + 5*C)*a^2)*sqrt(a)*arctan( sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (8*C *a^2*cos(d*x + c)^3 + 34*C*a^2*cos(d*x + c)^2 + 3*(8*A + 25*C)*a^2*cos(d*x + c) + 48*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)) )/(d*cos(d*x + c) + d)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]