Integrand size = 23, antiderivative size = 136 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}-\frac {\sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {3 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \]
-sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))+3*sin(d*x+c)*sec(d*x+c)^(1 /2)/a/d-3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/ 2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d-(cos(1/2*d*x+1 /2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*co s(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.60 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.88 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (3 \left (1+e^{2 i (c+d x)}\right )+3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{d \left (-1+e^{2 i c}\right )}+\frac {\sqrt {\sec (c+d x)} \left (6 \cos (d x) \csc (c)-2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}\right )}{a (1+\cos (c+d x))} \]
(Cos[(c + d*x)/2]^2*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-1 + E^((2*I)*c))*Sqrt[1 + E^ ((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyperg eometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + (Sqrt[Sec[c + d*x]]*(6*Cos[d*x]*Csc[c] - 2*Tan[(c + d*x) /2]))/d))/(a*(1 + Cos[c + d*x]))
Time = 0.76 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 3717, 3042, 4305, 27, 3042, 4274, 3042, 4255, 3042, 4258, 3042, 3119, 3120}
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.
\(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a \cos (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 3717 |
\(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a \sec (c+d x)+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 4305 |
\(\displaystyle -\frac {\int \frac {1}{2} \sqrt {\sec (c+d x)} (a-3 a \sec (c+d x))dx}{a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \sqrt {\sec (c+d x)} (a-3 a \sec (c+d x))dx}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a-3 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle -\frac {a \int \sqrt {\sec (c+d x)}dx-3 a \int \sec ^{\frac {3}{2}}(c+d x)dx}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle -\frac {a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle -\frac {a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-3 a \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
-((Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) - ((2*a*Sqrt [Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - 3*a*((-2* Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sq rt[Sec[c + d*x]]*Sin[c + d*x])/d))/(2*a^2)
3.4.18.3.1 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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p) Int[(d*Csc[e + f*x])^(m - n*p )*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && !IntegerQ[m] && IntegersQ[n, p]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[d^2*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 2)/(f*(a + b*Csc[e + f*x]))), x] - Simp[d^2/(a*b) Int[(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ [a^2 - b^2, 0] && GtQ[n, 1]
Time = 3.14 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.86
method | result | size |
default | \(-\frac {-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+6 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(253\) |
-(-cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2) *(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF( cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+6*(-2 *sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-5*( -2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2)/ a/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/ sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/2*((I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d *x + c) + I*sin(d*x + c)) + (-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstr assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierst rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*cos(d*x + c) + 2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {\sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \]
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{a \cos \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{a \cos \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{a+a\,\cos \left (c+d\,x\right )} \,d x \]