Integrand size = 25, antiderivative size = 157 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {19 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {9 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \]
-1/4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2)-9/16*sin(d*x+c)/ a/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+19/32*arctan(1/2*sin(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)*se c(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)
Time = 0.68 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\left (76 \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )-\cos (c+d x) (13+9 \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {2-2 \sec (c+d x)}\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{64 \sqrt {2} a^2 d \sqrt {a (1+\cos (c+d x))} \sqrt {1-\sec (c+d x)}} \]
((76*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]] - Cos[c + d*x]*(13 + 9*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Sqrt[2 - 2*Sec[c + d*x]])*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(64*Sqrt[2]*a^2*d*Sqrt[a*(1 + Cos[c + d*x])]*Sqrt[1 - Sec[c + d*x]])
Time = 0.79 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.04, 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, 3457, 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.
\(\displaystyle \int \frac {\sqrt {\sec (c+d x)}}{(a \cos (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4710 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3245 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {7 a-2 a \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {7 a-2 a \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {7 a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {19 a^2}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {19}{4} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {19}{4} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 3261 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {-\frac {19 a \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {19 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*(Sqrt[Cos[c + d*x]]*Sin[c + d* x])/(d*(a + a*Cos[c + d*x])^(5/2)) + ((19*ArcTan[(Sqrt[a]*Sin[c + d*x])/(S qrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*Sqrt[a]*d ) - (9*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) )/(8*a^2))
3.4.81.3.1 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*(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[(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 = 6.55 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.29
method | result | size |
default | \(-\frac {\left (9 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+19 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+13 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+38 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+19 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sqrt {\sec }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {2}}{32 d \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{3}}\) | \(202\) |
-1/32/d*(9*2^(1/2)*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2) +19*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^2+13*sin(d*x+c)*2^(1/2)*(cos( d*x+c)/(1+cos(d*x+c)))^(1/2)+38*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)+1 9*arcsin(cot(d*x+c)-csc(d*x+c)))*sec(d*x+c)^(1/2)*(a*(1+cos(d*x+c)))^(1/2) *cos(d*x+c)/(1+cos(d*x+c))^3/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)/a^3
Time = 0.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {19 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (9 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
-1/32*(19*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*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)*(9*cos(d*x + c)^2 + 13*cos( d*x + c))*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. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]