Integrand size = 23, antiderivative size = 76 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{b d \sqrt {b \sec (c+d x)}}-\frac {\sqrt {\sec (c+d x)} \sin ^3(c+d x)}{3 b d \sqrt {b \sec (c+d x)}} \] Output:
sec(d*x+c)^(1/2)*sin(d*x+c)/b/d/(b*sec(d*x+c))^(1/2)-1/3*sec(d*x+c)^(1/2)* sin(d*x+c)^3/b/d/(b*sec(d*x+c))^(1/2)
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx=\frac {(5+\cos (2 (c+d x))) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (b \sec (c+d x))^{3/2}} \] Input:
Integrate[1/(Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(3/2)),x]
Output:
((5 + Cos[2*(c + d*x)])*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*d*(b*Sec[c + d *x])^(3/2))
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.67, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2032, 3042, 3113, 2009}
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.
\(\displaystyle \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2032 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \int \cos ^3(c+d x)dx}{b \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx}{b \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle -\frac {\sqrt {\sec (c+d x)} \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{b d \sqrt {b \sec (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right ) \sqrt {\sec (c+d x)}}{b d \sqrt {b \sec (c+d x)}}\) |
Input:
Int[1/(Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(3/2)),x]
Output:
-((Sqrt[Sec[c + d*x]]*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/(b*d*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[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Time = 0.79 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.66
method | result | size |
default | \(\frac {\tan \left (d x +c \right )+2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{3 b d \sqrt {b \sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{\frac {5}{2}}}\) | \(50\) |
risch | \(-\frac {i {\mathrm e}^{4 i \left (d x +c \right )}}{24 b \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {3 i}{8 b \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{24 b \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(321\) |
Input:
int(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/3/b/d/(b*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2)*(tan(d*x+c)+2*tan(d*x+c)*sec (d*x+c)^2)
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx=\frac {{\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, b^{2} d \sqrt {\cos \left (d x + c\right )}} \] Input:
integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/3*(cos(d*x + c)^3 + 2*cos(d*x + c))*sqrt(b/cos(d*x + c))*sin(d*x + c)/(b ^2*d*sqrt(cos(d*x + c)))
Time = 14.87 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx=\begin {cases} \frac {2 \tan ^{3}{\left (c + d x \right )}}{3 d \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}} + \frac {\tan {\left (c + d x \right )}}{d \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{\frac {3}{2}}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x}{\left (b \sec {\left (c \right )}\right )^{\frac {3}{2}} \sec ^{\frac {3}{2}}{\left (c \right )}} & \text {otherwise} \end {cases} \] Input:
integrate(1/sec(d*x+c)**(3/2)/(b*sec(d*x+c))**(3/2),x)
Output:
Piecewise((2*tan(c + d*x)**3/(3*d*(b*sec(c + d*x))**(3/2)*sec(c + d*x)**(3 /2)) + tan(c + d*x)/(d*(b*sec(c + d*x))**(3/2)*sec(c + d*x)**(3/2)), Ne(d, 0)), (x/((b*sec(c))**(3/2)*sec(c)**(3/2)), True))
Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx=\frac {\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )}{12 \, b^{\frac {3}{2}} d} \] Input:
integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")
Output:
1/12*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3 *c))))/(b^(3/2)*d)
Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (3 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2} - 4\right )}}{3 \, b^{\frac {3}{2}} d {\left (\frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:
integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x, algorithm="giac")
Output:
2/3*(3*(1/tan(1/2*d*x + 1/2*c) + tan(1/2*d*x + 1/2*c))^2 - 4)/(b^(3/2)*d*( 1/tan(1/2*d*x + 1/2*c) + tan(1/2*d*x + 1/2*c))^3*sgn(cos(d*x + c)))
Time = 9.75 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx=\frac {\left (9\,\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{12\,b^2\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \] Input:
int(1/((b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(3/2)),x)
Output:
((9*sin(c + d*x) + sin(3*c + 3*d*x))*(b/cos(c + d*x))^(1/2))/(12*b^2*d*(1/ cos(c + d*x))^(1/2))
Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.37 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \sin \left (d x +c \right ) \left (-\sin \left (d x +c \right )^{2}+3\right )}{3 b^{2} d} \] Input:
int(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x)
Output:
(sqrt(b)*sin(c + d*x)*( - sin(c + d*x)**2 + 3))/(3*b**2*d)