Integrand size = 23, antiderivative size = 107 \[ \int \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {3 \text {arctanh}(\sin (c+d x)) \sqrt {b \sec (c+d x)}}{8 d \sqrt {\sec (c+d x)}}+\frac {3 \sec ^{\frac {3}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin (c+d x)}{8 d}+\frac {\sec ^{\frac {7}{2}}(c+d x) \sqrt {b \sec (c+d x)} \sin (c+d x)}{4 d} \]
3/8*sec(d*x+c)^(3/2)*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d+1/4*sec(d*x+c)^(7/2 )*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d+3/8*arctanh(sin(d*x+c))*(b*sec(d*x+c)) ^(1/2)/d/sec(d*x+c)^(1/2)
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.60 \[ \int \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\frac {\sqrt {b \sec (c+d x)} \left (3 \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (3+2 \sec ^2(c+d x)\right ) \tan (c+d x)\right )}{8 d \sqrt {\sec (c+d x)}} \]
(Sqrt[b*Sec[c + d*x]]*(3*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(3 + 2*Sec[c + d*x]^2)*Tan[c + d*x]))/(8*d*Sqrt[Sec[c + d*x]])
Time = 0.37 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2031, 3042, 4255, 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.
\(\displaystyle \int \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \int \sec ^5(c+d x)dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\sec (c+d x)}}\) |
(Sqrt[b*Sec[c + d*x]]*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(ArcTanh[S in[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/Sqrt[Sec[c + d*x]]
3.2.32.3.1 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 = 1.02 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97
method | result | size |
default | \(\frac {\sqrt {b \sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{\frac {9}{2}} \left (\left (3 \cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 \cos \left (d x +c \right )^{5} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+1\right )-3 \cos \left (d x +c \right )^{5} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )\right )}{8 d}\) | \(104\) |
risch | \(-\frac {i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (3 \,{\mathrm e}^{6 i \left (d x +c \right )}+11 \,{\mathrm e}^{4 i \left (d x +c \right )}-11 \,{\mathrm e}^{2 i \left (d x +c \right )}-3\right )}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}+\frac {3 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \cos \left (d x +c \right )}{4 d}-\frac {3 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \cos \left (d x +c \right )}{4 d}\) | \(257\) |
1/8/d*(b*sec(d*x+c))^(1/2)*sec(d*x+c)^(9/2)*((3*cos(d*x+c)^2+2)*sin(d*x+c) *cos(d*x+c)+3*cos(d*x+c)^5*ln(csc(d*x+c)-cot(d*x+c)+1)-3*cos(d*x+c)^5*ln(- cot(d*x+c)+csc(d*x+c)-1))
Time = 0.32 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.14 \[ \int \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\left [\frac {3 \, \sqrt {b} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, d \cos \left (d x + c\right )^{3}}, -\frac {3 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b}\right ) \cos \left (d x + c\right )^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, d \cos \left (d x + c\right )^{3}}\right ] \]
[1/16*(3*sqrt(b)*cos(d*x + c)^3*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*( 3*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)^3), -1/8*(3*sqrt(-b)*arctan(sqrt(-b)*sqrt(b/cos(d*x + c) )*sqrt(cos(d*x + c))*sin(d*x + c)/b)*cos(d*x + c)^3 - (3*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)^3 )]
Timed out. \[ \int \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1656 vs. \(2 (89) = 178\).
Time = 0.48 (sec) , antiderivative size = 1656, normalized size of antiderivative = 15.48 \[ \int \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\text {Too large to display} \]
-1/16*(12*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4* sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4 4*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d* x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(sin(8 *d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c) )*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 12*(sin(8*d*x + 8 *c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(1/ 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d* x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6 *c) + 16*cos(6*d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8 *c)^2 + 16*(3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16 *sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c)*sin(2*d* x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arc tan2(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*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d *x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x +...
\[ \int \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\int { \sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{\frac {9}{2}} \,d x } \]
Timed out. \[ \int \sec ^{\frac {9}{2}}(c+d x) \sqrt {b \sec (c+d x)} \, dx=\int \sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2} \,d x \]