3.2.39 \(\int \frac {1}{(a+a \sec (c+d x))^{5/2}} \, dx\) [139]

3.2.39.1 Optimal result

Integrand size = 14, antiderivative size = 144 \[ \int \frac {1}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {43 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {11 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]

output

2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-43/32*arctan 
(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)- 
1/4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-11/16*tan(d*x+c)/a/d/(a+a*sec(d*x+ 
c))^(3/2)
 

3.2.39.2 Mathematica [A] (warning: unable to verify)

Time = 6.45 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (-64 \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {1}{1+\sec (c+d x)}}}\right ) \sqrt {\frac {\cos (c+d x)}{(1+\cos (c+d x))^2}} \sqrt {1+\sec (c+d x)}+43 \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {1+\sec (c+d x)}+\left (30-19 \sec ^2\left (\frac {1}{2} (c+d x)\right )+2 \sec ^4\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\sec (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d (a (1+\sec (c+d x)))^{5/2}} \]

input

Integrate[(a + a*Sec[c + d*x])^(-5/2),x]
 

output

-1/4*(Cos[(c + d*x)/2]^6*Sec[c + d*x]^(5/2)*(-64*ArcTan[Tan[(c + d*x)/2]/S 
qrt[(1 + Sec[c + d*x])^(-1)]]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])^2]*Sqrt 
[1 + Sec[c + d*x]] + 43*ArcSin[Tan[(c + d*x)/2]]*Sqrt[Sec[(c + d*x)/2]^2]* 
Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[1 + Sec[c + d*x]] + (30 - 19*Sec[(c + d 
*x)/2]^2 + 2*Sec[(c + d*x)/2]^4)*Sqrt[Sec[c + d*x]]*Tan[(c + d*x)/2]))/(d* 
(a*(1 + Sec[c + d*x]))^(5/2))
 

3.2.39.3 Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {3042, 4264, 27, 3042, 4410, 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.

\(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4264

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4410

\(\displaystyle \frac {-\frac {\int -\frac {32 a^2-11 a^2 \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {32 a^2-11 a^2 \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {32 a^2-11 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {32 a \int \sqrt {\sec (c+d x) a+a}dx-43 a^2 \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {32 a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-43 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {-43 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {64 a^2 \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {64 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-43 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {\frac {86 a^2 \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {64 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {64 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {43 \sqrt {2} a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {11 a \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

input

Int[(a + a*Sec[c + d*x])^(-5/2),x]
 

output

-1/4*Tan[c + d*x]/(d*(a + a*Sec[c + d*x])^(5/2)) + (((64*a^(3/2)*ArcTan[(S 
qrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (43*Sqrt[2]*a^(3/2)*Ar 
cTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/(4*a^2 
) - (11*a*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)))/(8*a^2)
 

3.2.39.3.1 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]
 

rule 4410

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

3.2.39.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(246\) vs. \(2(119)=238\).

Time = 1.01 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.72

method	result	size
default	\(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{3}-13 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-32 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+43 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}\)	\(247\)

input

int(1/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

output

-1/32/d/a^3*(-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)*(2*(1-cos(d*x+c))^3*((1-cos(d*x+c))^2*csc(d*x+c)^ 
2-1)^(1/2)*csc(d*x+c)^3-13*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d 
*x+c)+csc(d*x+c))-32*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)))+43*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d 
*x+c))^2*csc(d*x+c)^2-1)^(1/2)))
 

3.2.39.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (119) = 238\).

Time = 0.39 (sec) , antiderivative size = 585, normalized size of antiderivative = 4.06 \[ \int \frac {1}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {43 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \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 ) + 64 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \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 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 11 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {43 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \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 ) - 64 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \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 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 11 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

input

integrate(1/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

output

[-1/64*(43*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1 
)*sqrt(-a)*log(-(2*sqrt(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*cos(d*x + c) + 1)) + 64*(cos(d*x + c)^3 + 3*cos(d*x + c 
)^2 + 3*cos(d*x + c) + 1)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sq 
rt((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*(15*cos(d*x + c)^2 + 11*cos(d*x + c))* 
sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^ 
3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(43*sqrt( 
2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*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))) - 64*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(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*(15*cos(d*x + c)^2 + 11*cos(d*x + c))*sqrt((a*cos(d* 
x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*co 
s(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
 

3.2.39.6 Sympy [F]

\[ \int \frac {1}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{\left (a \sec {\left (c + d x \right )} + a\right )^{\frac {5}{2}}}\, dx \]

input

integrate(1/(a+a*sec(d*x+c))**(5/2),x)
 

output

Integral((a*sec(c + d*x) + a)**(-5/2), x)
 

3.2.39.7 Maxima [F]

\[ \int \frac {1}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(1/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

output

integrate((a*sec(d*x + c) + a)^(-5/2), x)
 

3.2.39.8 Giac [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {13 \, \sqrt {2}}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{32 \, d} \]

input

integrate(1/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
 

output

1/32*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2 
/(a^3*sgn(cos(d*x + c))) - 13*sqrt(2)/(a^3*sgn(cos(d*x + c))))*tan(1/2*d*x 
 + 1/2*c)/d
 

3.2.39.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

input

int(1/(a + a/cos(c + d*x))^(5/2),x)
 

output

int(1/(a + a/cos(c + d*x))^(5/2), x)