\(\int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx\) [80]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {10 b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {10 (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^2 d} \] Output:

10/21*b*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(b*sec(d*x 
+c))^(1/2)/d+10/21*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*(b*sec(d*x+c))^(7 
/2)*sin(d*x+c)/b^2/d
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {(b \sec (c+d x))^{5/2} \left (10 \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 \sin (2 (c+d x))+6 \tan (c+d x)\right )}{21 b d} \] Input:

Integrate[Sec[c + d*x]^3*(b*Sec[c + d*x])^(3/2),x]
 

Output:

((b*Sec[c + d*x])^(5/2)*(10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 
 5*Sin[2*(c + d*x)] + 6*Tan[c + d*x]))/(21*b*d)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.14, 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, 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.

\(\displaystyle \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \sec (c+d x))^{9/2}dx}{b^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}dx}{b^3}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {5}{7} b^2 \int (b \sec (c+d x))^{5/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{7} b^2 \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^3}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \sqrt {b \sec (c+d x)}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^3}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {5}{7} b^2 \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 d}}{b^3}\)

Input:

Int[Sec[c + d*x]^3*(b*Sec[c + d*x])^(3/2),x]
 

Output:

((2*b*(b*Sec[c + d*x])^(7/2)*Sin[c + d*x])/(7*d) + (5*b^2*((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)))/7)/b^3
 

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 3120

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]
 

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 = 3.79 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {\left (\frac {10 \tan \left (d x +c \right )}{21}+\frac {2 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{7}-\frac {10 i \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ), i\right )}{21}\right ) b \sqrt {b \sec \left (d x +c \right )}}{d}\)	\(104\)

Input:

int(sec(d*x+c)^3*(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(10/21*tan(d*x+c)+2/7*tan(d*x+c)*sec(d*x+c)^2-10/21*I*(cos(d*x+c)+1)*( 
1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-co 
t(d*x+c)+csc(d*x+c)),I))*b*(b*sec(d*x+c))^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.23 \[ \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx=\frac {-5 i \, \sqrt {2} b^{\frac {3}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} b^{\frac {3}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (5 \, b \cos \left (d x + c\right )^{2} + 3 \, b\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate(sec(d*x+c)^3*(b*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

1/21*(-5*I*sqrt(2)*b^(3/2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d 
*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*b^(3/2)*cos(d*x + c)^3*weierstrass 
PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(5*b*cos(d*x + c)^2 + 3 
*b)*sqrt(b/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
 

Sympy [F]

\[ \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)**3*(b*sec(d*x+c))**(3/2),x)
 

Output:

Integral((b*sec(c + d*x))**(3/2)*sec(c + d*x)**3, x)
 

Maxima [F]

\[ \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \] Input:

integrate(sec(d*x+c)^3*(b*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*sec(d*x + c))^(3/2)*sec(d*x + c)^3, x)
 

Giac [F]

\[ \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{3} \,d x } \] Input:

integrate(sec(d*x+c)^3*(b*sec(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*sec(d*x + c))^(3/2)*sec(d*x + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx=\int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:

int((b/cos(c + d*x))^(3/2)/cos(c + d*x)^3,x)
 

Output:

int((b/cos(c + d*x))^(3/2)/cos(c + d*x)^3, x)
 

Reduce [F]

\[ \int \sec ^3(c+d x) (b \sec (c+d x))^{3/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right ) b \] Input:

int(sec(d*x+c)^3*(b*sec(d*x+c))^(3/2),x)
 

Output:

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