Integrand size = 23, antiderivative size = 69 \[ \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx=\frac {x \sqrt {\sec (c+d x)}}{2 b^2 \sqrt {b \sec (c+d x)}}+\frac {\sin (c+d x)}{2 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}} \] Output:
1/2*x*sec(d*x+c)^(1/2)/b^2/(b*sec(d*x+c))^(1/2)+1/2*sin(d*x+c)/b^2/d/sec(d *x+c)^(1/2)/(b*sec(d*x+c))^(1/2)
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {\sec (c+d x)} (2 (c+d x)+\sin (2 (c+d x)))}{4 b^2 d \sqrt {b \sec (c+d x)}} \] Input:
Integrate[Sqrt[Sec[c + d*x]]/(b*Sec[c + d*x])^(5/2),x]
Output:
(Sqrt[Sec[c + d*x]]*(2*(c + d*x) + Sin[2*(c + d*x)]))/(4*b^2*d*Sqrt[b*Sec[ c + d*x]])
Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.74, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2031, 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.
\(\displaystyle \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \int \cos ^2(c+d x)dx}{b^2 \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx}{b^2 \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{b^2 \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )}{b^2 \sqrt {b \sec (c+d x)}}\) |
Input:
Int[Sqrt[Sec[c + d*x]]/(b*Sec[c + d*x])^(5/2),x]
Output:
(Sqrt[Sec[c + d*x]]*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/(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[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 = 0.82 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70
method | result | size |
default | \(\frac {\tan \left (d x +c \right )+\left (d x +c \right ) \sec \left (d x +c \right )^{2}}{2 b^{2} d \sqrt {b \sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(48\) |
risch | \(\frac {\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, x}{2 b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(197\) |
Input:
int(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/2/b^2/d/(b*sec(d*x+c))^(1/2)/sec(d*x+c)^(3/2)*(tan(d*x+c)+(d*x+c)*sec(d* x+c)^2)
Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.39 \[ \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx=\left [\frac {2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) - \sqrt {-b} \log \left (2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right )}{4 \, b^{3} d}, \frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {b} \sqrt {\cos \left (d x + c\right )}}\right )}{2 \, b^{3} d}\right ] \] Input:
integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[1/4*(2*sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*sin(d*x + c) - sqrt(-b)*lo g(2*sqrt(-b)*sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*sin(d*x + c) + 2*b*co s(d*x + c)^2 - b))/(b^3*d), 1/2*(sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*s in(d*x + c) + sqrt(b)*arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sq rt(cos(d*x + c)))))/(b^3*d)]
Time = 10.56 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.55 \[ \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx=\begin {cases} \frac {x \tan ^{2}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}}{2 \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}} + \frac {x \sqrt {\sec {\left (c + d x \right )}}}{2 \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}} + \frac {\tan {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}}}{2 d \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}} & \text {for}\: d \neq 0 \\\frac {x \sqrt {\sec {\left (c \right )}}}{\left (b \sec {\left (c \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(sec(d*x+c)**(1/2)/(b*sec(d*x+c))**(5/2),x)
Output:
Piecewise((x*tan(c + d*x)**2*sqrt(sec(c + d*x))/(2*(b*sec(c + d*x))**(5/2) ) + x*sqrt(sec(c + d*x))/(2*(b*sec(c + d*x))**(5/2)) + tan(c + d*x)*sqrt(s ec(c + d*x))/(2*d*(b*sec(c + d*x))**(5/2)), Ne(d, 0)), (x*sqrt(sec(c))/(b* sec(c))**(5/2), True))
Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )}{4 \, b^{\frac {5}{2}} d} \] Input:
integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
1/4*(2*d*x + 2*c + sin(2*d*x + 2*c))/(b^(5/2)*d)
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - i \, \log \left (i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + i \, \log \left (-i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{2 \, b^{\frac {5}{2}} d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:
integrate(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
-1/2*(2*(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2 *c)^4 + 2*tan(1/2*d*x + 1/2*c)^2 + 1) - I*log(I*tan(1/2*d*x + 1/2*c) - 1) + I*log(-I*tan(1/2*d*x + 1/2*c) - 1))/(b^(5/2)*d*sgn(cos(d*x + c)))
Time = 10.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\left (\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )+4\,d\,x\,\cos \left (c+d\,x\right )\right )}{8\,b^3\,d\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \] Input:
int((1/cos(c + d*x))^(1/2)/(b/cos(c + d*x))^(5/2),x)
Output:
((b/cos(c + d*x))^(1/2)*(sin(c + d*x) + sin(3*c + 3*d*x) + 4*d*x*cos(c + d *x)))/(8*b^3*d*cos(c + d*x)*(1/cos(c + d*x))^(1/2))
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {\sec (c+d x)}}{(b \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x \right )}{2 b^{3} d} \] Input:
int(sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x)
Output:
(sqrt(b)*(cos(c + d*x)*sin(c + d*x) + d*x))/(2*b**3*d)