Integrand size = 23, antiderivative size = 101 \[ \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {3 b x \sqrt {b \sec (c+d x)}}{8 \sqrt {\sec (c+d x)}}+\frac {b \sqrt {b \sec (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac {7}{2}}(c+d x)}+\frac {3 b \sqrt {b \sec (c+d x)} \sin (c+d x)}{8 d \sec ^{\frac {3}{2}}(c+d x)} \]
1/4*b*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(7/2)+3/8*b*sin(d*x+c)* (b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(3/2)+3/8*b*x*(b*sec(d*x+c))^(1/2)/sec(d *x+c)^(1/2)
Time = 0.48 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.54 \[ \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {(b \sec (c+d x))^{3/2} (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{32 d \sec ^{\frac {3}{2}}(c+d x)} \]
((b*Sec[c + d*x])^(3/2)*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d* x)]))/(32*d*Sec[c + d*x]^(3/2))
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2031, 3042, 3115, 3042, 3115, 24}
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.
\(\displaystyle \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \int \cos ^4(c+d x)dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {b \sqrt {b \sec (c+d x)} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{\sqrt {\sec (c+d x)}}\) |
(b*Sqrt[b*Sec[c + d*x]]*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + ( Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/Sqrt[Sec[c + d*x]]
3.2.51.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Time = 0.44 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.64
method | result | size |
default | \(\frac {b \sqrt {b \sec \left (d x +c \right )}\, \left (2 \tan \left (d x +c \right )+3 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}+3 \left (d x +c \right ) \sec \left (d x +c \right )^{4}\right )}{8 d \sec \left (d x +c \right )^{\frac {9}{2}}}\) | \(65\) |
risch | \(\frac {3 b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, x}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sin \left (4 d x +4 c \right )}{32 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(257\) |
1/8*b/d*(b*sec(d*x+c))^(1/2)/sec(d*x+c)^(9/2)*(2*tan(d*x+c)+3*tan(d*x+c)*s ec(d*x+c)^2+3*(d*x+c)*sec(d*x+c)^4)
Time = 0.33 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.05 \[ \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\left [\frac {3 \, \sqrt {-b} b \log \left (-2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right ) + \frac {2 \, {\left (2 \, b \cos \left (d x + c\right )^{4} + 3 \, b \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, d}, \frac {3 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {b} \sqrt {\cos \left (d x + c\right )}}\right ) + \frac {{\left (2 \, b \cos \left (d x + c\right )^{4} + 3 \, b \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, d}\right ] \]
[1/16*(3*sqrt(-b)*b*log(-2*sqrt(-b)*sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2 )*sin(d*x + c) + 2*b*cos(d*x + c)^2 - b) + 2*(2*b*cos(d*x + c)^4 + 3*b*cos (d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d, 1/8* (3*b^(3/2)*arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(d*x + c)))) + (2*b*cos(d*x + c)^4 + 3*b*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*s in(d*x + c)/sqrt(cos(d*x + c)))/d]
Timed out. \[ \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]
Time = 0.40 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.52 \[ \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {{\left (12 \, {\left (d x + c\right )} b + b \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} \sqrt {b}}{32 \, d} \]
1/32*(12*(d*x + c)*b + b*sin(4*d*x + 4*c) + 8*b*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*sqrt(b)/d
\[ \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\int { \frac {\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {11}{2}}} \,d x } \]
Time = 0.52 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.52 \[ \int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx=\frac {b\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\left (8\,\sin \left (2\,c+2\,d\,x\right )+\sin \left (4\,c+4\,d\,x\right )+12\,d\,x\right )}{32\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \]