Integrand size = 23, antiderivative size = 107 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx=\frac {3 x \sqrt {\sec (c+d x)}}{8 b^2 \sqrt {b \sec (c+d x)}}+\frac {\sin (c+d x)}{4 b^2 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}+\frac {3 \sin (c+d x)}{8 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}} \] Output:
3/8*x*sec(d*x+c)^(1/2)/b^2/(b*sec(d*x+c))^(1/2)+1/4*sin(d*x+c)/b^2/d/sec(d *x+c)^(5/2)/(b*sec(d*x+c))^(1/2)+3/8*sin(d*x+c)/b^2/d/sec(d*x+c)^(1/2)/(b* sec(d*x+c))^(1/2)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {\sec (c+d x)} (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{32 b^2 d \sqrt {b \sec (c+d x)}} \] Input:
Integrate[1/(Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(5/2)),x]
Output:
(Sqrt[Sec[c + d*x]]*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]) )/(32*b^2*d*Sqrt[b*Sec[c + d*x]])
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2032, 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.
\(\displaystyle \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2032 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \int \cos ^4(c+d x)dx}{b^2 \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx}{b^2 \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{b^2 \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{b^2 \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{b^2 \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{b^2 \sqrt {b \sec (c+d x)}}\) |
Input:
Int[1/(Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(5/2)),x]
Output:
(Sqrt[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))/(b^2*Sqrt[b*Sec[c + d*x]])
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m - 1/ 2)*b^(n + 1/2)*(Sqrt[a*v]/Sqrt[b*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && ILtQ[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 = 0.86 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.63
method | result | size |
default | \(\frac {2 \tan \left (d x +c \right )+3 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}+3 \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{8 b^{2} d \sec \left (d x +c \right )^{\frac {7}{2}} \sqrt {b \sec \left (d x +c \right )}}\) | \(67\) |
risch | \(\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} x}{8 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i {\mathrm e}^{5 i \left (d x +c \right )}}{64 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}-\frac {7 i \cos \left (3 d x +3 c \right )}{64 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {9 \sin \left (3 d x +3 c \right )}{64 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(408\) |
Input:
int(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/8/b^2/d/sec(d*x+c)^(7/2)/(b*sec(d*x+c))^(1/2)*(2*tan(d*x+c)+3*tan(d*x+c) *sec(d*x+c)^2+3*(d*x+c)*sec(d*x+c)^4)
Time = 0.12 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.94 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx=\left [\frac {\frac {2 \, {\left (2 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} - 3 \, \sqrt {-b} \log \left (2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right )}{16 \, b^{3} d}, \frac {\frac {{\left (2 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 3 \, \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {b} \sqrt {\cos \left (d x + c\right )}}\right )}{8 \, b^{3} d}\right ] \] Input:
integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[1/16*(2*(2*cos(d*x + c)^4 + 3*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d* x + c)/sqrt(cos(d*x + c)) - 3*sqrt(-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))/(b^3*d), 1/8*( (2*cos(d*x + c)^4 + 3*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sq rt(cos(d*x + c)) + 3*sqrt(b)*arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqr t(b)*sqrt(cos(d*x + c)))))/(b^3*d)]
Time = 109.39 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.72 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx=\begin {cases} \frac {3 x \tan ^{4}{\left (c + d x \right )}}{8 \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}} + \frac {3 x \tan ^{2}{\left (c + d x \right )}}{4 \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}} + \frac {3 x}{8 \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}} + \frac {3 \tan ^{3}{\left (c + d x \right )}}{8 d \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}} + \frac {5 \tan {\left (c + d x \right )}}{8 d \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x}{\left (b \sec {\left (c \right )}\right )^{\frac {5}{2}} \sec ^{\frac {3}{2}}{\left (c \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(1/sec(d*x+c)**(3/2)/(b*sec(d*x+c))**(5/2),x)
Output:
Piecewise((3*x*tan(c + d*x)**4/(8*(b*sec(c + d*x))**(5/2)*sec(c + d*x)**(3 /2)) + 3*x*tan(c + d*x)**2/(4*(b*sec(c + d*x))**(5/2)*sec(c + d*x)**(3/2)) + 3*x/(8*(b*sec(c + d*x))**(5/2)*sec(c + d*x)**(3/2)) + 3*tan(c + d*x)**3 /(8*d*(b*sec(c + d*x))**(5/2)*sec(c + d*x)**(3/2)) + 5*tan(c + d*x)/(8*d*( b*sec(c + d*x))**(5/2)*sec(c + d*x)**(3/2)), Ne(d, 0)), (x/((b*sec(c))**(5 /2)*sec(c)**(3/2)), True))
Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.46 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx=\frac {12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )}{32 \, b^{\frac {5}{2}} d} \] Input:
integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
1/32*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(1/2*arctan2(sin(4*d*x + 4*c ), cos(4*d*x + 4*c))))/(b^(5/2)*d)
Result contains complex when optimal does not.
Time = 0.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {2 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - 3 i \, \log \left (i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + 3 i \, \log \left (-i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{8 \, b^{\frac {5}{2}} d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:
integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
-1/8*(2*(5*tan(1/2*d*x + 1/2*c)^7 - 3*tan(1/2*d*x + 1/2*c)^5 + 3*tan(1/2*d *x + 1/2*c)^3 - 5*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^8 + 4*tan(1/ 2*d*x + 1/2*c)^6 + 6*tan(1/2*d*x + 1/2*c)^4 + 4*tan(1/2*d*x + 1/2*c)^2 + 1 ) - 3*I*log(I*tan(1/2*d*x + 1/2*c) - 1) + 3*I*log(-I*tan(1/2*d*x + 1/2*c) - 1))/(b^(5/2)*d*sgn(cos(d*x + c)))
Time = 9.78 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.51 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\left (8\,\sin \left (2\,c+2\,d\,x\right )+\sin \left (4\,c+4\,d\,x\right )+12\,d\,x\right )}{32\,b^3\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \] Input:
int(1/((b/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(3/2)),x)
Output:
((b/cos(c + d*x))^(1/2)*(8*sin(2*c + 2*d*x) + sin(4*c + 4*d*x) + 12*d*x))/ (32*b^3*d*(1/cos(c + d*x))^(1/2))
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+5 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x \right )}{8 b^{3} d} \] Input:
int(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(5/2),x)
Output:
(sqrt(b)*( - 2*cos(c + d*x)*sin(c + d*x)**3 + 5*cos(c + d*x)*sin(c + d*x) + 3*d*x))/(8*b**3*d)