Integrand size = 25, antiderivative size = 220 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {163 a^{5/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 d}+\frac {17 a^3 \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {163 a^3 \sin (c+d x)}{96 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {163 a^3 \sin (c+d x)}{64 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]
17/24*a^3*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+163/96*a^3* sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+1/4*a^2*sin(d*x+c)*(a +a*cos(d*x+c))^(1/2)/d/sec(d*x+c)^(5/2)+163/64*a^3*sin(d*x+c)/d/(a+a*cos(d *x+c))^(1/2)/sec(d*x+c)^(1/2)+163/64*a^(5/2)*arcsin(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
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.03 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.92 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (7 (89+28 \cos (c+d x)+3 \cos (2 (c+d x))) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2},\frac {7}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )-24 (3+\cos (c+d x)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2},\frac {9}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^2(c+d x)-6 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (-\frac {1}{2},\frac {3}{2},2;1,\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^4(c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{420 d} \]
(Sqrt[Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(5/2)*Sec[(c + d*x)/2]^4*Sqrt[S ec[c + d*x]]*(7*(89 + 28*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Hypergeometric 2F1[-3/2, 1/2, 7/2, 2*Sin[(c + d*x)/2]^2] - 24*(3 + Cos[c + d*x])*Hypergeo metric2F1[-1/2, 3/2, 9/2, 2*Sin[(c + d*x)/2]^2]*Sin[c + d*x]^2 - 6*Csc[(c + d*x)/2]^2*HypergeometricPFQ[{-1/2, 3/2, 2}, {1, 9/2}, 2*Sin[(c + d*x)/2] ^2]*Sin[c + d*x]^4)*Tan[(c + d*x)/2])/(420*d)
Time = 1.11 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4710, 3042, 3242, 27, 3042, 3460, 3042, 3249, 3042, 3249, 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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4710 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{5/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx\) |
| \(\Big \downarrow \) 3242 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} \int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a} \left (17 \cos (c+d x) a^2+13 a^2\right )dx+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a} \left (17 \cos (c+d x) a^2+13 a^2\right )dx+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (17 \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+13 a^2\right )dx+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {163}{6} a^2 \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}dx+\frac {17 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {163}{6} a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {17 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {163}{6} a^2 \left (\frac {3}{4} \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {17 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {163}{6} a^2 \left (\frac {3}{4} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {17 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {163}{6} a^2 \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {17 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {163}{6} a^2 \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {17 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 3253 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{8} \left (\frac {163}{6} a^2 \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {17 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}\right )\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}+\frac {1}{8} \left (\frac {17 a^3 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {163}{6} a^2 \left (\frac {3}{4} \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\right )\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((a^2*Cos[c + d*x]^(5/2)*Sqrt[a + a* Cos[c + d*x]]*Sin[c + d*x])/(4*d) + ((17*a^3*Cos[c + d*x]^(5/2)*Sin[c + d* x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (163*a^2*((a*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((Sqrt[a]*ArcSin[(Sqrt[a]*Sin [c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Sqrt[Cos[c + d*x]]*Sin[c + d* x])/(d*Sqrt[a + a*Cos[c + d*x]])))/4))/6)/8)
3.4.60.3.1 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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* (m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[ n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 2*n + 1))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), 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] && GtQ[n, 0] && IntegerQ[2*n]
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[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[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 13.83 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.98
| method | result | size |
| default | \(\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (48 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+184 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+326 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+489 \tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+489 \sec \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) a^{2}}{192 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(216\) |
1/192/d*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/sec(d*x+c)^(3/2)/(cos(d*x+ c)/(1+cos(d*x+c)))^(1/2)*(48*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d* x+c)))^(1/2)+184*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3 26*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+489*tan(d*x+c)*(cos(d*x+c) /(1+cos(d*x+c)))^(1/2)+489*sec(d*x+c)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos (d*x+c)))^(1/2)))*a^2
Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.66 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {489 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (48 \, a^{2} \cos \left (d x + c\right )^{4} + 184 \, a^{2} \cos \left (d x + c\right )^{3} + 326 \, a^{2} \cos \left (d x + c\right )^{2} + 489 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
-1/192*(489*(a^2*cos(d*x + c) + a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (48*a^2*cos(d*x + c)^4 + 1 84*a^2*cos(d*x + c)^3 + 326*a^2*cos(d*x + c)^2 + 489*a^2*cos(d*x + c))*sqr t(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d )
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 7450 vs. \(2 (184) = 368\).
Time = 0.70 (sec) , antiderivative size = 7450, normalized size of antiderivative = 33.86 \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
1/768*(10*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/ 2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4 *d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(3/4)*((3*a^2*cos(4*d*x + 4*c)^2*sin( 4*d*x + 4*c) + 3*a^2*sin(4*d*x + 4*c)^3 + 12*(a^2*sin(4*d*x + 4*c)^3 + (a^ 2*cos(4*d*x + 4*c)^2 - 2*a^2*cos(4*d*x + 4*c) + a^2)*sin(4*d*x + 4*c))*cos (1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 12*(a^2*sin(4*d*x + 4*c)^3 + (a^2*cos(4*d*x + 4*c)^2 + 2*a^2*cos(4*d*x + 4*c) + a^2)*sin(4*d*x + 4*c))*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 3*(2*a^2 *cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))*sin(4*d*x + 4*c) + a ^2*sin(4*d*x + 4*c) - 2*(a^2*cos(4*d*x + 4*c) + a^2)*sin(1/2*arctan2(sin(4 *d*x + 4*c), cos(4*d*x + 4*c))))*cos(3/4*arctan2(sin(4*d*x + 4*c), cos(4*d *x + 4*c))) + 12*(a^2*sin(4*d*x + 4*c)^3 + (a^2*cos(4*d*x + 4*c)^2 - a^2*c os(4*d*x + 4*c))*sin(4*d*x + 4*c))*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4 *d*x + 4*c))) + (8*a^2*cos(4*d*x + 4*c)^2 + 8*a^2*sin(4*d*x + 4*c)^2 - 3*a ^2*cos(4*d*x + 4*c) + 32*(a^2*cos(4*d*x + 4*c)^2 + a^2*sin(4*d*x + 4*c)^2 - 2*a^2*cos(4*d*x + 4*c) + a^2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d* x + 4*c)))^2 + 32*(a^2*cos(4*d*x + 4*c)^2 + a^2*sin(4*d*x + 4*c)^2 + 2*a^2 *cos(4*d*x + 4*c) + a^2)*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c )))^2 + 2*(16*a^2*cos(4*d*x + 4*c)^2 + 16*a^2*sin(4*d*x + 4*c)^2 - 19*a^2* cos(4*d*x + 4*c) + 3*a^2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x +...
\[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{5/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]