Integrand size = 21, antiderivative size = 107 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {3 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]
3/32*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2) /d*2^(1/2)+1/4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)+3/16*tan(d*x+c)/a/d/(a+ a*sec(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.49 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) \tan (c+d x)}{4 a^2 d \sqrt {a (1+\sec (c+d x))}} \]
(Hypergeometric2F1[1/2, 3, 3/2, (1 - Sec[c + d*x])/2]*Tan[c + d*x])/(4*a^2 *d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.47 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4283, 3042, 4283, 3042, 4282, 216}
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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4283 |
\(\displaystyle \frac {3 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a}+\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4283 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}+\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}+\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle \frac {3 \left (\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {\int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{2 a d}\right )}{8 a}+\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {3 \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
Tan[c + d*x]/(4*d*(a + a*Sec[c + d*x])^(5/2)) + (3*(ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]/(2*Sqrt[2]*a^(3/2)*d) + Tan[c + d*x]/(2*d*(a + a*Sec[c + d*x])^(3/2))))/(8*a)
3.2.38.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(m + 1)/(a*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) ] && IntegerQ[2*m]
Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(88)=176\).
Time = 0.96 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.79
method | result | size |
default | \(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}\) | \(192\) |
1/32/d/a^3*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^ 2*csc(d*x+c)^2-1)^(1/2)*(2*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(-cot(d *x+c)+csc(d*x+c))-3*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+c sc(d*x+c))+3*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1 /2)))
Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (88) = 176\).
Time = 0.33 (sec) , antiderivative size = 403, normalized size of antiderivative = 3.77 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (7 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left (7 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]
[-1/64*(3*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1) *sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(co s(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(7*cos(d*x + c)^2 + 3*cos(d*x + c) )*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(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), -1/32*(3*sqr t(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*arct an(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*s in(d*x + c))) - 2*(7*cos(d*x + c)^2 + 3*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(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)]
\[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.86 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.30 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {5 \, \sqrt {2}}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3 \, \sqrt {2} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \]
-1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*tan(1/2*d*x + 1/2*c) ^2/(a^3*sgn(cos(d*x + c))) - 5*sqrt(2)/(a^3*sgn(cos(d*x + c))))*tan(1/2*d* x + 1/2*c) + 3*sqrt(2)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*ta n(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*sgn(cos(d*x + c))))/d
Timed out. \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]