\(\int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx\) [102]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {6 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 b d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^3 d} \] Output:

-6/5*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c 
))^(1/2)+6/5*(b*sec(d*x+c))^(1/2)*sin(d*x+c)/b/d+2/5*(b*sec(d*x+c))^(5/2)* 
sin(d*x+c)/b^3/d
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.63 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+2 \left (3+\sec ^2(c+d x)\right ) \tan (c+d x)}{5 d \sqrt {b \sec (c+d x)}} \] Input:

Integrate[Sec[c + d*x]^4/Sqrt[b*Sec[c + d*x]],x]
 

Output:

((-6*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*(3 + Sec[c + d*x]^2 
)*Tan[c + d*x])/(5*d*Sqrt[b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2030, 3042, 4255, 3042, 4255, 3042, 4258, 3042, 3119}

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.

\(\displaystyle \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \sec (c+d x))^{7/2}dx}{b^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx}{b^4}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {3}{5} b^2 \int (b \sec (c+d x))^{3/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}}{b^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{5} b^2 \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}}{b^4}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}}{b^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}}{b^4}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}}{b^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}}{b^4}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {3}{5} b^2 \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d}}{b^4}\)

Input:

Int[Sec[c + d*x]^4/Sqrt[b*Sec[c + d*x]],x]
 

Output:

((2*b*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d) + (3*b^2*((-2*b^2*Ellipti 
cE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*Sqr 
t[b*Sec[c + d*x]]*Sin[c + d*x])/d))/5)/b^4
 

Defintions of rubi rules used

rule 2030

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4255

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]
 

rule 4258

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.92 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.06

method	result	size
default	\(-\frac {2 \left (\left (-3 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )-1\right ) \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}-3 i \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right ) \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )+3 i \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (2+\cos \left (d x +c \right )+\sec \left (d x +c \right )\right ) \operatorname {EllipticE}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}\)	\(200\)

Input:

int(sec(d*x+c)^4/(b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/5/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)*((-3*cos(d*x+c)^2-cos(d*x+c)-1) 
*tan(d*x+c)*sec(d*x+c)^2-3*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x 
+c)+1))^(1/2)*(2+cos(d*x+c)+sec(d*x+c))*EllipticF(I*(-cot(d*x+c)+csc(d*x+c 
)),I)+3*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*(2+co 
s(d*x+c)+sec(d*x+c))*EllipticE(I*(-cot(d*x+c)+csc(d*x+c)),I))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {-3 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, b d \cos \left (d x + c\right )^{2}} \] Input:

integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/5*(-3*I*sqrt(2)*sqrt(b)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstras 
sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*sqrt(b)*cos 
(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) 
 - I*sin(d*x + c))) + 2*(3*cos(d*x + c)^2 + 1)*sqrt(b/cos(d*x + c))*sin(d* 
x + c))/(b*d*cos(d*x + c)^2)
 

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt {b \sec {\left (c + d x \right )}}}\, dx \] Input:

integrate(sec(d*x+c)**4/(b*sec(d*x+c))**(1/2),x)
 

Output:

Integral(sec(c + d*x)**4/sqrt(b*sec(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\sqrt {b \sec \left (d x + c\right )}} \,d x } \] Input:

integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sec(d*x + c)^4/sqrt(b*sec(d*x + c)), x)
 

Giac [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\sqrt {b \sec \left (d x + c\right )}} \,d x } \] Input:

integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(sec(d*x + c)^4/sqrt(b*sec(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \] Input:

int(1/(cos(c + d*x)^4*(b/cos(c + d*x))^(1/2)),x)
 

Output:

int(1/(cos(c + d*x)^4*(b/cos(c + d*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^4(c+d x)}{\sqrt {b \sec (c+d x)}} \, dx=\frac {\sqrt {b}\, \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right )}{b} \] Input:

int(sec(d*x+c)^4/(b*sec(d*x+c))^(1/2),x)
 

Output:

(sqrt(b)*int(sqrt(sec(c + d*x))*sec(c + d*x)**3,x))/b