Integrand size = 10, antiderivative size = 62 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx=\frac {2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right ) \sqrt {\sec (a+b x)}}{3 b}+\frac {2 \sin (a+b x)}{3 b \sqrt {\sec (a+b x)}} \] Output:
2/3*cos(b*x+a)^(1/2)*InverseJacobiAM(1/2*a+1/2*b*x,2^(1/2))*sec(b*x+a)^(1/ 2)/b+2/3*sin(b*x+a)/b/sec(b*x+a)^(1/2)
Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx=\frac {\sqrt {\sec (a+b x)} \left (2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )+\sin (2 (a+b x))\right )}{3 b} \] Input:
Integrate[Sec[a + b*x]^(-3/2),x]
Output:
(Sqrt[Sec[a + b*x]]*(2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2] + Sin[ 2*(a + b*x)]))/(3*b)
Time = 0.32 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4256, 3042, 4258, 3042, 3120}
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.
\(\displaystyle \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\csc \left (a+b x+\frac {\pi }{2}\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {1}{3} \int \sqrt {\sec (a+b x)}dx+\frac {2 \sin (a+b x)}{3 b \sqrt {\sec (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int \sqrt {\csc \left (a+b x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (a+b x)}{3 b \sqrt {\sec (a+b x)}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{3} \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx+\frac {2 \sin (a+b x)}{3 b \sqrt {\sec (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (a+b x)}{3 b \sqrt {\sec (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \sin (a+b x)}{3 b \sqrt {\sec (a+b x)}}+\frac {2 \sqrt {\cos (a+b x)} \sqrt {\sec (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b}\) |
Input:
Int[Sec[a + b*x]^(-3/2),x]
Output:
(2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[Sec[a + b*x]])/(3*b) + (2*Sin[a + b*x])/(3*b*Sqrt[Sec[a + b*x]])
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Leaf count of result is larger than twice the leaf count of optimal. \(178\) vs. \(2(53)=106\).
Time = 1.44 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.89
method | result | size |
default | \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \left (4 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1}\, b}\) | \(179\) |
Input:
int(1/sec(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/3*((2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(4*sin(1/2*b* x+1/2*a)^4*cos(1/2*b*x+1/2*a)-2*sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a)+(s in(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos( 1/2*b*x+1/2*a),2^(1/2)))/(-2*sin(1/2*b*x+1/2*a)^4+sin(1/2*b*x+1/2*a)^2)^(1 /2)/sin(1/2*b*x+1/2*a)/(2*cos(1/2*b*x+1/2*a)^2-1)^(1/2)/b
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx=\frac {2 \, \sqrt {\cos \left (b x + a\right )} \sin \left (b x + a\right ) - i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )}{3 \, b} \] Input:
integrate(1/sec(b*x+a)^(3/2),x, algorithm="fricas")
Output:
1/3*(2*sqrt(cos(b*x + a))*sin(b*x + a) - I*sqrt(2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) + I*sqrt(2)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)))/b
\[ \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {1}{\sec ^{\frac {3}{2}}{\left (a + b x \right )}}\, dx \] Input:
integrate(1/sec(b*x+a)**(3/2),x)
Output:
Integral(sec(a + b*x)**(-3/2), x)
\[ \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx=\int { \frac {1}{\sec \left (b x + a\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/sec(b*x+a)^(3/2),x, algorithm="maxima")
Output:
integrate(sec(b*x + a)^(-3/2), x)
\[ \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx=\int { \frac {1}{\sec \left (b x + a\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/sec(b*x+a)^(3/2),x, algorithm="giac")
Output:
integrate(sec(b*x + a)^(-3/2), x)
Timed out. \[ \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {1}{{\left (\frac {1}{\cos \left (a+b\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(1/(1/cos(a + b*x))^(3/2),x)
Output:
int(1/(1/cos(a + b*x))^(3/2), x)
\[ \int \frac {1}{\sec ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {\sqrt {\sec \left (b x +a \right )}}{\sec \left (b x +a \right )^{2}}d x \] Input:
int(1/sec(b*x+a)^(3/2),x)
Output:
int(sqrt(sec(a + b*x))/sec(a + b*x)**2,x)