3.2.31 \(\int \frac {1}{(a+a \sec (c+d x))^{3/2}} \, dx\) [131]

3.2.31.1 Optimal result

Integrand size = 14, antiderivative size = 114 \[ \int \frac {1}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {5 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}} \]

output

2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d-5/4*arctan(1 
/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/ 
2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)
 

3.2.31.2 Mathematica [A] (verified)

Time = 2.61 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\sec ^{\frac {3}{2}}(c+d x) \left (5 \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {1+\sec (c+d x)}+\sqrt {2} \left (-4 \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {1}{1+\sec (c+d x)}}}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {1+\sec (c+d x)}+\frac {\sqrt {\frac {1}{1+\cos (c+d x)}} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sec (c+d x)}}\right )\right )}{d \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a (1+\sec (c+d x)))^{3/2}} \]

input

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

output

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

3.2.31.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4264, 27, 3042, 4408, 3042, 4261, 216, 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.

\(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 4264

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4408

\(\displaystyle \frac {4 \int \sqrt {\sec (c+d x) a+a}dx-5 a \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4261

\(\displaystyle \frac {-\frac {8 a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-5 a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {8 \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-5 a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {10 a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {8 \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {8 \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {5 \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

input

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

output

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

3.2.31.3.1 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 4261

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 4264

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c 
+ d*x])*((a + b*Csc[c + d*x])^n/(d*(2*n + 1))), x] + Simp[1/(a^2*(2*n + 1)) 
   Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]), 
 x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && Int 
egerQ[2*n]
 

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 4408

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c/a   Int[Sqrt[a + b*Csc[e + f*x]], x], x] - 
Simp[(b*c - a*d)/a   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F 
reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
 

3.2.31.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(198\) vs. \(2(93)=186\).

Time = 0.91 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.75

method	result	size
default	\(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (4 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{4 d \,a^{2}}\)	\(199\)

input

int(1/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

output

1/4/d/a^2*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^2 
*csc(d*x+c)^2-1)^(1/2)*(4*2^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d* 
x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))+((1-cos(d*x+c))^2*csc(d*x+c)^2-1 
)^(1/2)*(-cot(d*x+c)+csc(d*x+c))-5*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c) 
)^2*csc(d*x+c)^2-1)^(1/2)))
 

3.2.31.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (93) = 186\).

Time = 0.34 (sec) , antiderivative size = 491, normalized size of antiderivative = 4.31 \[ \int \frac {1}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {5 \, \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 ) + 8 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 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 ) + a \cos \left (d x + c\right ) - a}{\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 )}} \cos \left (d x + c\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 {5 \, \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 ) - 8 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\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 )}} \cos \left (d x + c\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(1/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

output

[-1/8*(5*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)) + 8*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*log((2 
*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos 
(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 4*sqrt( 
(a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*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*(5*sqrt(2)*(cos(d*x + c)^2 
+ 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos 
(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 8*(cos(d*x + c)^2 + 2*co 
s(d*x + c) + 1)*sqrt(a)*arctan(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))*cos(d*x + c)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c 
) + a^2*d)]
 

3.2.31.6 Sympy [F]

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

input

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

output

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

3.2.31.7 Maxima [F]

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

input

integrate(1/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

output

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

3.2.31.8 Giac [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.41 \[ \int \frac {1}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{4 \, a^{2} d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]

input

integrate(1/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")
 

output

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

3.2.31.9 Mupad [F(-1)]

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

input

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

output

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