Integrand size = 23, antiderivative size = 63 \[ \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {x \sqrt {b \sec (c+d x)}}{2 \sqrt {\sec (c+d x)}}+\frac {\sqrt {b \sec (c+d x)} \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x)} \] Output:
1/2*x*(b*sec(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+1/2*(b*sec(d*x+c))^(1/2)*sin(d *x+c)/d/sec(d*x+c)^(3/2)
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {b \sec (c+d x)} (2 (c+d x)+\sin (2 (c+d x)))}{4 d \sqrt {\sec (c+d x)}} \] Input:
Integrate[Sqrt[b*Sec[c + d*x]]/Sec[c + d*x]^(5/2),x]
Output:
(Sqrt[b*Sec[c + d*x]]*(2*(c + d*x) + Sin[2*(c + d*x)]))/(4*d*Sqrt[Sec[c + d*x]])
Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.76, 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.
\(\displaystyle \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \int \cos ^2(c+d x)dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{\sqrt {\sec (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sqrt {b \sec (c+d x)} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )}{\sqrt {\sec (c+d x)}}\) |
Input:
Int[Sqrt[b*Sec[c + d*x]]/Sec[c + d*x]^(5/2),x]
Output:
(Sqrt[b*Sec[c + d*x]]*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/Sqrt[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 = 0.75 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.71
method | result | size |
default | \(\frac {\sqrt {b \sec \left (d x +c \right )}\, \left (\tan \left (d x +c \right )+\left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{2 d \sec \left (d x +c \right )^{\frac {5}{2}}}\) | \(45\) |
risch | \(\frac {\sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, x}{2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(188\) |
Input:
int((b*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/2/d*(b*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2)*(tan(d*x+c)+(d*x+c)*sec(d*x+c) ^2)
Time = 0.11 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.51 \[ \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + \sqrt {-b} \log \left (-2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right )}{4 \, d}, \frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {b} \sqrt {\cos \left (d x + c\right )}}\right )}{2 \, d}\right ] \] Input:
integrate((b*sec(d*x+c))^(1/2)/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*c os(d*x + c)^2 - b))/d, 1/2*(sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*sin(d* x + c) + sqrt(b)*arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(co s(d*x + c)))))/d]
Time = 10.86 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.70 \[ \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\begin {cases} \frac {x \sqrt {b \sec {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}}{2 \sec ^{\frac {5}{2}}{\left (c + d x \right )}} + \frac {x \sqrt {b \sec {\left (c + d x \right )}}}{2 \sec ^{\frac {5}{2}}{\left (c + d x \right )}} + \frac {\sqrt {b \sec {\left (c + d x \right )}} \tan {\left (c + d x \right )}}{2 d \sec ^{\frac {5}{2}}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x \sqrt {b \sec {\left (c \right )}}}{\sec ^{\frac {5}{2}}{\left (c \right )}} & \text {otherwise} \end {cases} \] Input:
integrate((b*sec(d*x+c))**(1/2)/sec(d*x+c)**(5/2),x)
Output:
Piecewise((x*sqrt(b*sec(c + d*x))*tan(c + d*x)**2/(2*sec(c + d*x)**(5/2)) + x*sqrt(b*sec(c + d*x))/(2*sec(c + d*x)**(5/2)) + sqrt(b*sec(c + d*x))*ta n(c + d*x)/(2*d*sec(c + d*x)**(5/2)), Ne(d, 0)), (x*sqrt(b*sec(c))/sec(c)* *(5/2), True))
Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.40 \[ \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {b}}{4 \, d} \] Input:
integrate((b*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="maxima")
Output:
1/4*(2*d*x + 2*c + sin(2*d*x + 2*c))*sqrt(b)/d
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (51) = 102\).
Time = 0.58 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.22 \[ \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {b} d x \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, \sqrt {b} d x \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {b} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \sqrt {b} d x \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, \sqrt {b} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \] Input:
integrate((b*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="giac")
Output:
1/2*(sqrt(b)*d*x*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^4 + 2*sqrt(b)*d*x* sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 2*sqrt(b)*sgn(cos(d*x + c))*tan (1/2*d*x + 1/2*c)^3 + sqrt(b)*d*x*sgn(cos(d*x + c)) + 2*sqrt(b)*sgn(cos(d* x + c))*tan(1/2*d*x + 1/2*c))/(d*tan(1/2*d*x + 1/2*c)^4 + 2*d*tan(1/2*d*x + 1/2*c)^2 + d)
Time = 9.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (\sin \left (2\,c+2\,d\,x\right )+2\,d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{4\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \] Input:
int((b/cos(c + d*x))^(1/2)/(1/cos(c + d*x))^(5/2),x)
Output:
((sin(2*c + 2*d*x) + 2*d*x)*(b/cos(c + d*x))^(1/2))/(4*d*(1/cos(c + d*x))^ (1/2))
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {b \sec (c+d x)}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x \right )}{2 d} \] Input:
int((b*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x)
Output:
(sqrt(b)*(cos(c + d*x)*sin(c + d*x) + d*x))/(2*d)