Integrand size = 25, antiderivative size = 157 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {7 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \]
-1/2*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(3/2)-7/4*arctan(1/2*s in(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^(3/2)/d*2^(1/2)+5/2*sin(d*x+c)*sec(d*x+c)^(1/ 2)/a/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 = 6.31 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.59 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {-\frac {35}{2} \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (78+108 \cos (c+d x)+80 \cos (2 (c+d x))-204 \cos (3 (c+d x))-62 \cos (4 (c+d x))+12 \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) (64+55 \cos (c+d x)+64 \cos (2 (c+d x))+17 \cos (3 (c+d x))) \sqrt {2-2 \sec (c+d x)}\right ) \sqrt {\sec (c+d x)}-768 \cos ^5\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {5}{2};1,\frac {9}{2};-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^{\frac {5}{2}}(c+d x) \sin ^3\left (\frac {1}{2} (c+d x)\right )}{3360 d (a (1+\cos (c+d x)))^{3/2}} \]
((-35*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^4*(78 + 108*Cos[c + d*x] + 80*Cos[ 2*(c + d*x)] - 204*Cos[3*(c + d*x)] - 62*Cos[4*(c + d*x)] + 12*ArcTanh[Sqr t[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]]*Cos[(c + d*x)/2]^2*(64 + 55*Cos[c + d*x] + 64*Cos[2*(c + d*x)] + 17*Cos[3*(c + d*x)])*Sqrt[2 - 2*Sec[c + d*x] ])*Sqrt[Sec[c + d*x]])/2 - 768*Cos[(c + d*x)/2]^5*HypergeometricPFQ[{2, 2, 5/2}, {1, 9/2}, -(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]*Sec[c + d*x]^(5/2)*Si n[(c + d*x)/2]^3)/(3360*d*(a*(1 + Cos[c + d*x]))^(3/2))
Time = 0.75 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4710, 3042, 3245, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4710 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {5 a-2 a \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {5 a-2 a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {5 a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \int -\frac {7 a^2}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {10 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {10 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-7 a \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {10 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-7 a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {14 a^2 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {10 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {10 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {7 \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*Sin[c + d*x]/(d*Sqrt[Cos[c + d *x]]*(a + a*Cos[c + d*x])^(3/2)) + ((-7*Sqrt[2]*Sqrt[a]*ArcTan[(Sqrt[a]*Si n[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d + (1 0*a*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2) )
3.4.74.3.1 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[((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*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 = 6.53 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.29
| method | result | size |
| default | \(\frac {\left (7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+5 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}+14 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+4 \sqrt {2}\, \sin \left (d x +c \right )+7 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} a^{2}}\) | \(202\) |
1/4/d*(7*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*arcsin(cot(d*x+c)- csc(d*x+c))+5*sin(d*x+c)*cos(d*x+c)*2^(1/2)+14*cos(d*x+c)*(cos(d*x+c)/(1+c os(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+4*2^(1/2)*sin(d*x+c)+7*(co s(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c)))*sec(d*x+c)^( 3/2)*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)/(1+cos(d*x+c))^2*2^(1/2)/a^2
Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {7 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/4*(7*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2 )*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2* sqrt(a*cos(d*x + c) + a)*(5*cos(d*x + c) + 4)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]