\(\int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [128]

Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {7 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {2 \tan (c+d x)}{a d \sqrt {a+a \sec (c+d x)}} \] Output:

-7/4*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2) 
/a^(3/2)/d+1/2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)+2*tan(d*x+c)/a/d/(a+a*s 
ec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-7 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) (1+\sec (c+d x))+2 \sqrt {1-\sec (c+d x)} (5+4 \sec (c+d x))\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

((-7*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*(1 + Sec[c + d*x]) + 
2*Sqrt[1 - Sec[c + d*x]]*(5 + 4*Sec[c + d*x]))*Tan[c + d*x])/(4*d*Sqrt[1 - 
 Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4286, 27, 3042, 4489, 3042, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4286

\(\displaystyle \frac {\int -\frac {\sec (c+d x) (3 a-4 a \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {\int \frac {\sec (c+d x) (3 a-4 a \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {7 a \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {8 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {7 a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {8 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {-\frac {14 a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {8 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {\frac {7 \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {8 a \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}\)

Input:

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

Output:

Tan[c + d*x]/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((7*Sqrt[2]*Sqrt[a]*ArcTan 
[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (8*a*Tan[ 
c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

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

rule 4282

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4286

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), 
x] - Simp[1/(a^2*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1) 
*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[ 
a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

rule 4489

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 
 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B 
, e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b 
*(m + 1), 0] &&  !LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.14

method	result	size
default	\(\frac {\left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{4}-\frac {7 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{4}+\frac {9 \csc \left (d x +c \right )}{4}-\frac {9 \cot \left (d x +c \right )}{4}\right ) \sqrt {-a \left (-1-\sec \left (d x +c \right )\right )}}{d \,a^{2}}\)	\(120\)

Input:

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

Output:

1/d*(-1/4*(1-cos(d*x+c))^3*csc(d*x+c)^3-7/4*ln((-2*cos(d*x+c)/(cos(d*x+c)+ 
1))^(1/2)-cot(d*x+c)+csc(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+9/4* 
csc(d*x+c)-9/4*cot(d*x+c))/a^2*(-a*(-1-sec(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 336, normalized size of antiderivative = 3.20 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {7 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {7 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:

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

Output:

[-1/8*(7*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*log(-(2*sq 
rt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d* 
x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*co 
s(d*x + c) + 1)) - 4*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(5*cos(d*x + 
c) + 4)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d 
), 1/4*(7*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqr 
t(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x 
 + c))) + 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(5*cos(d*x + c) + 4)*s 
in(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]
 

Sympy [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

Integral(sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(3/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {\sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {9 \, \sqrt {2}}{a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} + \frac {7 \, \sqrt {2} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{4 \, d} \] Input:

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

Output:

1/4*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(sqrt(2)*tan(1/2*d*x + 1/2*c)^2/( 
a*sgn(cos(d*x + c))) - 9*sqrt(2)/(a*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2* 
c)/(a*tan(1/2*d*x + 1/2*c)^2 - a) + 7*sqrt(2)*log(abs(-sqrt(-a)*tan(1/2*d* 
x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a*sgn(cos(d*x 
 + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(1/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(3/2)),x)
 

Output:

int(1/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:

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

Output:

(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3)/(sec(c + d*x)**2 + 2 
*sec(c + d*x) + 1),x))/a**2