Integrand size = 23, antiderivative size = 72 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x)) \sqrt {b \sec (c+d x)}}{2 d \sqrt {\sec (c+d x)}}+\frac {\sec ^{\frac {3}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin (c+d x)}{2 d} \]
1/2*sec(d*x+c)^(3/2)*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d+1/2*arctanh(sin(d*x +c))*(b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)
Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.69 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {\sqrt {b \sec (c+d x)} (\text {arctanh}(\sin (c+d x))+\sec (c+d x) \tan (c+d x))}{2 d \sqrt {\sec (c+d x)}} \]
(Sqrt[b*Sec[c + d*x]]*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x])) /(2*d*Sqrt[Sec[c + d*x]])
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2031, 3042, 4255, 3042, 4257}
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.
| \(\displaystyle \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {\sqrt {b \sec (c+d x)} \int \sec ^3(c+d x)dx}{\sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{\sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\sec (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\sec (c+d x)}}\) |
(Sqrt[b*Sec[c + d*x]]*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/Sqrt[Sec[c + d*x]]
3.2.34.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.57 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.28
| method | result | size |
| default | \(-\frac {\sqrt {b \sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (\cos \left (d x +c \right )^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )-\cos \left (d x +c \right )^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+1\right )-\cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{2 d}\) | \(92\) |
| risch | \(-\frac {i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \cos \left (d x +c \right )}{d}+\frac {\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \cos \left (d x +c \right )}{d}\) | \(232\) |
-1/2/d*(b*sec(d*x+c))^(1/2)*sec(d*x+c)^(5/2)*(cos(d*x+c)^3*ln(-cot(d*x+c)+ csc(d*x+c)-1)-cos(d*x+c)^3*ln(csc(d*x+c)-cot(d*x+c)+1)-cos(d*x+c)*sin(d*x+ c))
Time = 0.30 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.76 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\left [\frac {\sqrt {b} \cos \left (d x + c\right ) \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, d \cos \left (d x + c\right )}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b}\right ) \cos \left (d x + c\right ) - \frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, d \cos \left (d x + c\right )}\right ] \]
[1/4*(sqrt(b)*cos(d*x + c)*log(-(b*cos(d*x + c)^2 - 2*sqrt(b)*sqrt(b/cos(d *x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b)/cos(d*x + c)^2) + 2*sqrt(b /cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)), -1/2*(sq rt(-b)*arctan(sqrt(-b)*sqrt(b/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c )/b)*cos(d*x + c) - sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/ (d*cos(d*x + c))]
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 661 vs. \(2 (60) = 120\).
Time = 0.46 (sec) , antiderivative size = 661, normalized size of antiderivative = 9.18 \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=-\frac {{\left (4 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + 2 \, \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - 4 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + 2 \, \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) + {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) - 4 \, {\left (\cos \left (4 \, d x + 4 \, c\right ) + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 4 \, {\left (\cos \left (4 \, d x + 4 \, c\right ) + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )\right )} \sqrt {b}}{4 \, {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} d} \]
-1/4*(4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos (1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4 *d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*s in(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), co s(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^ 2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(cos(4 *d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), co s(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*a rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sqrt(b)/((2*(2*cos(2*d*x + 2* c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin (4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c) ^2 + 4*cos(2*d*x + 2*c) + 1)*d)
\[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\int { \sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\int \sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]