\(\int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [134]

Optimal result

Integrand size = 23, antiderivative size = 183 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {163 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {17 \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {197 \tan (c+d x)}{24 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {95 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d} \] Output:

163/32*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^(5/2)/d-1/4*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-17/16*se 
c(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)-197/24*tan(d*x+c)/a^2/d/( 
a+a*sec(d*x+c))^(1/2)+95/48*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^3/d
 

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left (978 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x)+\sqrt {1-\sec (c+d x)} \left (-299-503 \sec (c+d x)-160 \sec ^2(c+d x)+32 \sec ^3(c+d x)\right )\right ) \tan (c+d x)}{48 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

((978*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4*S 
ec[c + d*x]^2 + Sqrt[1 - Sec[c + d*x]]*(-299 - 503*Sec[c + d*x] - 160*Sec[ 
c + d*x]^2 + 32*Sec[c + d*x]^3))*Tan[c + d*x])/(48*d*Sqrt[1 - Sec[c + d*x] 
]*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 4303, 27, 3042, 4507, 27, 3042, 4498, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4303

\(\displaystyle -\frac {\int \frac {\sec ^3(c+d x) (6 a-11 a \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sec ^3(c+d x) (6 a-11 a \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (6 a-11 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4507

\(\displaystyle -\frac {\frac {\int \frac {\sec ^2(c+d x) \left (68 a^2-95 a^2 \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {\sec ^2(c+d x) \left (68 a^2-95 a^2 \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (68 a^2-95 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4498

\(\displaystyle -\frac {\frac {\frac {2 \int -\frac {\sec (c+d x) \left (95 a^3-394 a^3 \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {190 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {-\frac {\int \frac {\sec (c+d x) \left (95 a^3-394 a^3 \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {190 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (95 a^3-394 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {190 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4489

\(\displaystyle -\frac {\frac {-\frac {489 a^3 \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {788 a^3 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {190 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {-\frac {489 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {788 a^3 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {190 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle -\frac {\frac {-\frac {-\frac {978 a^3 \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {788 a^3 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {190 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\frac {-\frac {\frac {489 \sqrt {2} a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {788 a^3 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {190 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{4 a^2}+\frac {17 a \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*(Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) - ((17*a 
*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((-190*a* 
Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((489*Sqrt[2]*a^(5/2)*ArcTa 
n[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (788*a^3 
*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(4*a^2))/(8*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 4303

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

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)]
 

rule 4498

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

rule 4507

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

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-598 \cos \left (d x +c \right )^{3}-1006 \cos \left (d x +c \right )^{2}-320 \cos \left (d x +c \right )+64\right ) \tan \left (d x +c \right )+\left (489 \cos \left (d x +c \right )^{3}+1467 \cos \left (d x +c \right )^{2}+1467 \cos \left (d x +c \right )+489\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{96 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(179\)

Input:

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

Output:

1/96/d/a^3*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x 
+c)+1)*((-598*cos(d*x+c)^3-1006*cos(d*x+c)^2-320*cos(d*x+c)+64)*tan(d*x+c) 
+(489*cos(d*x+c)^3+1467*cos(d*x+c)^2+1467*cos(d*x+c)+489)*2^(1/2)*(-cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x 
+c)+csc(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.49 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (299 \, \cos \left (d x + c\right )^{3} + 503 \, \cos \left (d x + c\right )^{2} + 160 \, \cos \left (d x + c\right ) - 32\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}}, -\frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left (299 \, \cos \left (d x + c\right )^{3} + 503 \, \cos \left (d x + c\right )^{2} + 160 \, \cos \left (d x + c\right ) - 32\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}}\right ] \] Input:

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

Output:

[-1/192*(489*sqrt(2)*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 
 + cos(d*x + c))*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)) + 4*(299*cos(d*x + c)^ 
3 + 503*cos(d*x + c)^2 + 160*cos(d*x + c) - 32)*sqrt((a*cos(d*x + c) + a)/ 
cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 
 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c)), -1/96*(489*sqrt(2)*(cos(d 
*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*ar 
ctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a) 
*sin(d*x + c))) + 2*(299*cos(d*x + c)^3 + 503*cos(d*x + c)^2 + 160*cos(d*x 
 + c) - 32)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*c 
os(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*co 
s(d*x + c))]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {{\left ({\left (3 \, {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {23 \, \sqrt {2}}{a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {668 \, \sqrt {2}}{a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {465 \, \sqrt {2}}{a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {489 \, \sqrt {2} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{96 \, d} \] Input:

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

Output:

1/96*(((3*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/(a*sgn(cos(d*x + c))) + 23*sqr 
t(2)/(a*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 668*sqrt(2)/(a*sgn(co 
s(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 465*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)*sqrt(-a*tan(1/2*d*x + 
 1/2*c)^2 + a)) - 489*sqrt(2)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqr 
t(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**5)/(sec(c + d*x)**3 + 3 
*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x))/a**3