\(\int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\) [233]

Optimal result

Integrand size = 25, antiderivative size = 200 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {163 a^{5/2} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {163 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {163 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {17 a^3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d} \] Output:

163/64*a^(5/2)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+163/64 
*a^3*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+163/96*a^3*sec(d 
*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+17/24*a^3*sec(d*x+c)^(7/2) 
*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a^2*sec(d*x+c)^(7/2)*(a+a*sec(d*x 
+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.81 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {a^3 \left (489 \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+326 \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+184 \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+48 \sqrt {1-\sec (c+d x)} \sec ^{\frac {7}{2}}(c+d x)+489 \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{192 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^3*(489*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 326*Sqrt[1 - Sec[c + d*x]]*Sec[ 
c + d*x]^(3/2) + 184*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) + 48*Sqrt[1 
 - Sec[c + d*x]]*Sec[c + d*x]^(7/2) + 489*Sqrt[-((-1 + Sec[c + d*x])*Sec[c 
 + d*x])])*Tan[c + d*x])/(192*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + 
 d*x])])
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4301, 27, 3042, 4504, 3042, 4290, 3042, 4290, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4301

\(\displaystyle \frac {1}{4} a \int \frac {1}{2} \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (17 \sec (c+d x) a+13 a)dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (17 \sec (c+d x) a+13 a)dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (17 \csc \left (c+d x+\frac {\pi }{2}\right ) a+13 a\right )dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{8} a \left (\frac {163}{6} a \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {17 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \left (\frac {163}{6} a \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {17 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {1}{8} a \left (\frac {163}{6} a \left (\frac {3}{4} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {17 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \left (\frac {163}{6} a \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {17 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {1}{8} a \left (\frac {163}{6} a \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {17 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{8} a \left (\frac {163}{6} a \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {17 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {1}{8} a \left (\frac {163}{6} a \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {17 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {1}{8} a \left (\frac {17 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {163}{6} a \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\)

Input:

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

Output:

(a^2*Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d) + (a* 
((17*a^2*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + 
 (163*a*((a*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]] 
) + (3*((Sqrt[a]*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]) 
/d + (a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4) 
)/6))/8
 

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 222

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

rule 3042

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

rule 4288

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

rule 4290

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

rule 4301

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

rule 4504

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

Maple [A] (verified)

Time = 17.62 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (489 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+489 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (489 \cos \left (d x +c \right )^{3}+326 \cos \left (d x +c \right )^{2}+184 \cos \left (d x +c \right )+48\right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \tan \left (d x +c \right )\right )}{384 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(194\)

Input:

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

Output:

1/384/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(5/2)/(cos(d*x+c)+1)/(-1/( 
cos(d*x+c)+1))^(1/2)*(489*cos(d*x+c)^3*arctan(1/2/(-1/(cos(d*x+c)+1))^(1/2 
)*(cot(d*x+c)-csc(d*x+c)+1))+489*cos(d*x+c)^3*arctan(1/2*(cot(d*x+c)-csc(d 
*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+(489*cos(d*x+c)^3+326*cos(d*x+c)^2+184 
*cos(d*x+c)+48)*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.22 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {489 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (489 \, a^{2} \cos \left (d x + c\right )^{3} + 326 \, a^{2} \cos \left (d x + c\right )^{2} + 184 \, a^{2} \cos \left (d x + c\right ) + 48 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{768 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, \frac {489 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (489 \, a^{2} \cos \left (d x + c\right )^{3} + 326 \, a^{2} \cos \left (d x + c\right )^{2} + 184 \, a^{2} \cos \left (d x + c\right ) + 48 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{384 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \] Input:

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

Output:

[1/768*(489*(a^2*cos(d*x + c)^4 + a^2*cos(d*x + c)^3)*sqrt(a)*log((a*cos(d 
*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt( 
a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) 
 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(489*a^2*cos(d*x + c)^3 + 3 
26*a^2*cos(d*x + c)^2 + 184*a^2*cos(d*x + c) + 48*a^2)*sqrt((a*cos(d*x + c 
) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4 + 
d*cos(d*x + c)^3), 1/384*(489*(a^2*cos(d*x + c)^4 + a^2*cos(d*x + c)^3)*sq 
rt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d 
*x + c) + a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d*x + c))) + 2*(489*a 
^2*cos(d*x + c)^3 + 326*a^2*cos(d*x + c)^2 + 184*a^2*cos(d*x + c) + 48*a^2 
)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c))) 
/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3860 vs. \(2 (168) = 336\).

Time = 0.46 (sec) , antiderivative size = 3860, normalized size of antiderivative = 19.30 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/768*(1956*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c 
) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(1 
5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 652*(sqrt(2)*a^2*sin(8* 
d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4* 
c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(13/4*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c))) + 6204*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin( 
6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 
2*c))*cos(11/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2060*(sqrt(2 
)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*si 
n(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(9/4*arctan2(sin(2*d*x 
 + 2*c), cos(2*d*x + 2*c))) + 2060*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt( 
2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*s 
in(2*d*x + 2*c))*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 62 
04*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt 
(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos(5/4*arctan2 
(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 652*(sqrt(2)*a^2*sin(8*d*x + 8*c) 
+ 4*sqrt(2)*a^2*sin(6*d*x + 6*c) + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt 
(2)*a^2*sin(2*d*x + 2*c))*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2* 
c))) - 1956*(sqrt(2)*a^2*sin(8*d*x + 8*c) + 4*sqrt(2)*a^2*sin(6*d*x + 6*c) 
 + 6*sqrt(2)*a^2*sin(4*d*x + 4*c) + 4*sqrt(2)*a^2*sin(2*d*x + 2*c))*cos...
 

Giac [F(-2)]

Exception generated. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{%%{[ 
%%%{%%{[2309237210123256509497344,0]:[1,0,-2]%%},[35]%%%},0]:[1,0,%%%{-1,[ 
1]%%%}]%%},[
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right )+\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) \] Input:

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

Output:

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