Integrand size = 21, antiderivative size = 72 \[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 b^3 d} \] Output:
2/3*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(b*sec(d*x+c)) ^(1/2)/b^2/d+2/3*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/b^3/d
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {2 \sec ^3(c+d x) \left (\cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sin (c+d x)\right )}{3 d (b \sec (c+d x))^{3/2}} \] Input:
Integrate[Sec[c + d*x]^4/(b*Sec[c + d*x])^(3/2),x]
Output:
(2*Sec[c + d*x]^3*(Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + Sin[c + d*x]))/(3*d*(b*Sec[c + d*x])^(3/2))
Time = 0.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2030, 3042, 4255, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{5/2}dx}{b^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{b^4}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {1}{3} b^2 \int \sqrt {b \sec (c+d x)}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}}{b^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} b^2 \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}}{b^4}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}}{b^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}}{b^4}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}}{b^4}\) |
Input:
Int[Sec[c + d*x]^4/(b*Sec[c + d*x])^(3/2),x]
Output:
((2*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]]) /(3*d) + (2*b*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d))/b^4
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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]
Result contains complex when optimal does not.
Time = 2.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.33
method | result | size |
default | \(\frac {-\frac {2 i \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (1+\sec \left (d x +c \right )\right )}{3}+\frac {2 \sec \left (d x +c \right ) \tan \left (d x +c \right )}{3}}{d \sqrt {b \sec \left (d x +c \right )}\, b}\) | \(96\) |
Input:
int(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/d*(-2/3*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-cot(d*x+c)+csc (d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(1+sec(d*x+c))+2/3*sec(d*x+c)*tan(d*x +c))/(b*sec(d*x+c))^(1/2)/b
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {-i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, b^{2} d \cos \left (d x + c\right )} \] Input:
integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/3*(-I*sqrt(2)*sqrt(b)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + I*sqrt(2)*sqrt(b)*cos(d*x + c)*weierstrassPInverse( -4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(b/cos(d*x + c))*sin(d*x + c ))/(b^2*d*cos(d*x + c))
\[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**4/(b*sec(d*x+c))**(3/2),x)
Output:
Integral(sec(c + d*x)**4/(b*sec(c + d*x))**(3/2), x)
\[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)^4/(b*sec(d*x + c))^(3/2), x)
\[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate(sec(d*x + c)^4/(b*sec(d*x + c))^(3/2), x)
Timed out. \[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(1/(cos(c + d*x)^4*(b/cos(c + d*x))^(3/2)),x)
Output:
int(1/(cos(c + d*x)^4*(b/cos(c + d*x))^(3/2)), x)
\[ \int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {b}\, \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}d x \right )}{b^{2}} \] Input:
int(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x)
Output:
(sqrt(b)*int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x))/b**2