Integrand size = 25, antiderivative size = 237 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {163 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {299 \sqrt {\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {95 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}} \] Output:
163/32*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-299/48*sec (d*x+c)^(1/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)-1/4*sec(d*x+c)^(3/2) *sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-17/16*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d /(a+a*cos(d*x+c))^(3/2)+95/48*sec(d*x+c)^(3/2)*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 = 8.39 (sec) , antiderivative size = 641, normalized size of antiderivative = 2.70 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
Integrate[Sec[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^(5/2),x]
Output:
-1/41580*(Cot[c/2 + (d*x)/2]^5*Csc[c/2 + (d*x)/2]^4*Sec[(c + d*x)/2]^4*((1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1))^(7/2)*(640*Cos[(c + d*x)/2]^8*Hypergeomet ricPFQ[{2, 2, 2, 2, 7/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Si n[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 - 1280*Cos[(c + d*x)/2]^6*Hyper geometricPFQ[{2, 2, 2, 7/2}, {1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Si n[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(-6 + 5*Sin[c/2 + (d*x)/2]^2) + 33*(1 - 2*Sin[c/2 + (d*x)/2]^2)^3*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c /2 + (d*x)/2]^2)]*(-105*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/2]^4*(-10935 + 72902*Sin[c/2 + (d*x)/2]^2 - 188110*Sin[c/2 + (d*x)/2]^4 + 234156*Sin[c/2 + (d*x)/2]^6 - 140732*Sin[c/2 + (d*x)/2]^8 + 33208*Sin[c/2 + (d*x)/2]^10) + Sqrt[Sin[c/2 + (d*x)/2]^2/( -1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-1148175 + 10333785*Sin[c/2 + (d*x)/2]^2 - 38990350*Sin[c/2 + (d*x)/2]^4 + 79946462*Sin[c/2 + (d*x)/2]^6 - 96281836*S in[c/2 + (d*x)/2]^8 + 68243596*Sin[c/2 + (d*x)/2]^10 - 26448512*Sin[c/2 + (d*x)/2]^12 + 4344400*Sin[c/2 + (d*x)/2]^14))))/(d*(a*(1 + Cos[c + d*x]))^ (5/2))
Time = 1.31 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 4710, 3042, 3245, 27, 3042, 3457, 27, 3042, 3463, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4710 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {11 a-6 a \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {11 a-6 a \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {11 a-6 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {95 a^2-68 a^2 \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {95 a^2-68 a^2 \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {95 a^2-68 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \int -\frac {299 a^3-190 a^3 \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {190 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {299 a^3-190 a^3 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {299 a^3-190 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {489 a^4}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {598 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {598 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-489 a^3 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {598 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-489 a^3 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {978 a^4 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {598 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {598 a^3 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {489 \sqrt {2} a^{5/2} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\right )\) |
Input:
Int[Sec[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^(5/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*Sin[c + d*x]/(d*Cos[c + d*x]^( 3/2)*(a + a*Cos[c + d*x])^(5/2)) + ((-17*a*Sin[c + d*x])/(2*d*Cos[c + d*x] ^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((190*a^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-489*Sqrt[2]*a^(5/2)*ArcTan[(Sqrt [a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/ d + (598*a^3*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) )/(3*a))/(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[((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*((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*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + 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] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
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[(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 = 7.22 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.78
| method | result | size |
| default | \(-\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (\cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (299 \cos \left (d x +c \right )^{3}+503 \cos \left (d x +c \right )^{2}+160 \cos \left (d x +c \right )-32\right ) \sqrt {2}+\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (489 \cos \left (d x +c \right )^{5}+1467 \cos \left (d x +c \right )^{4}+1467 \cos \left (d x +c \right )^{3}+489 \cos \left (d x +c \right )^{2}\right )\right )}{96 d \left (\cos \left (d x +c \right )+1\right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) a^{3}}\) | \(186\) |
Input:
int(sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/96/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*sec(d*x+c)^(5/2)/(cos(d*x+c)+1)/( cos(d*x+c)^2+2*cos(d*x+c)+1)*(cos(d*x+c)*sin(d*x+c)*(299*cos(d*x+c)^3+503* cos(d*x+c)^2+160*cos(d*x+c)-32)*2^(1/2)+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* arcsin(cot(d*x+c)-csc(d*x+c))*(489*cos(d*x+c)^5+1467*cos(d*x+c)^4+1467*cos (d*x+c)^3+489*cos(d*x+c)^2))/a^3
Time = 0.11 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (299 \, \cos \left (d x + c\right )^{3} + 503 \, \cos \left (d x + c\right )^{2} + 160 \, \cos \left (d x + c\right ) - 32\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-1/96*(489*sqrt(2)*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + 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*(299*cos(d*x + c)^3 + 503*cos(d*x + c) ^2 + 160*cos(d*x + c) - 32)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos (d*x + c)))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d *x + c)^2 + a^3*d*cos(d*x + c))
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(5/2), x)
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate(sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^(5/2), x)
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((1/cos(c + d*x))^(5/2)/(a + a*cos(c + d*x))^(5/2),x)
Output:
int((1/cos(c + d*x))^(5/2)/(a + a*cos(c + d*x))^(5/2), x)
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:
int(sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*sec(c + d*x)**2)/( cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x))/a**3