Integrand size = 25, antiderivative size = 137 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {19 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {9 \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \]
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))/a^(5/2)/d*2^(1/2)-1/4*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d* x+c))^(5/2)-9/16*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(3/2)
Time = 0.85 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-76 \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+\cos (c+d x) (13+9 \cos (c+d x)) \sqrt {2-2 \sec (c+d x)}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{32 \sqrt {2} a^2 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \]
-1/32*(Sec[(c + d*x)/2]^2*(-76*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2 ]^2)]]*Cos[(c + d*x)/2]^4 + Cos[c + d*x]*(13 + 9*Cos[c + d*x])*Sqrt[2 - 2* Sec[c + d*x]])*Tan[(c + d*x)/2])/(Sqrt[2]*a^2*d*Sqrt[-1 + Cos[c + d*x]]*Sq rt[a*(1 + Cos[c + d*x])])
Time = 0.60 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {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.
| \(\displaystyle \int \frac {1}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \frac {\int \frac {7 a-2 a \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {7 a-2 a \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {7 a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {19 a^2}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {19}{4} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {19}{4} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {-\frac {19 a \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {19 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {9 a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}\) |
-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])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Co s[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)
3.3.48.3.1 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])
Leaf count of result is larger than twice the leaf count of optimal. \(228\) vs. \(2(112)=224\).
Time = 5.67 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.67
| method | result | size |
| default | \(\frac {\sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \sqrt {\frac {a}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, \left (-2 \left (\csc ^{3}\left (d x +c \right )\right ) \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (1-\cos \left (d x +c \right )\right )^{3}-11 \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-19 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{32 d \sqrt {-\frac {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, a^{3}}\) | \(229\) |
1/32/d/(-(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+ 1))^(1/2)*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*(a/(csc(d*x+c)^2*(1-cos (d*x+c))^2+1))^(1/2)*(-2*csc(d*x+c)^3*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^( 1/2)*(1-cos(d*x+c))^3-11*(-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2)*(csc(d*x +c)-cot(d*x+c))-19*arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)/a^3
Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\frac {19 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (9 \, \cos \left (d x + c\right ) + 13\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
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(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) - 2*sqrt(a*cos(d*x + c) + a)*(9*cos(d*x + c) + 13)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*co s(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}} \sqrt {\cos {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]