3.3.25 \(\int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\) [225]

3.3.25.1 Optimal result
3.3.25.2 Mathematica [A] (verified)
3.3.25.3 Rubi [A] (verified)
3.3.25.4 Maple [A] (verified)
3.3.25.5 Fricas [A] (verification not implemented)
3.3.25.6 Sympy [F(-1)]
3.3.25.7 Maxima [B] (verification not implemented)
3.3.25.8 Giac [F]
3.3.25.9 Mupad [F(-1)]

3.3.25.1 Optimal result

Integrand size = 25, antiderivative size = 160 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {11 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {11 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {11 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}} \]

output
11/8*a^(3/2)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+11/8*a^2 
*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+11/12*a^2*sec(d*x+c) 
^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*a^2*sec(d*x+c)^(7/2)*sin(d* 
x+c)/d/(a+a*sec(d*x+c))^(1/2)
 
3.3.25.2 Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.85 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {a^2 \left (33 \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+22 \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+33 \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2),x]
 
output
(a^2*(33*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 22*Sqrt[1 - Sec[c + d*x]]*Sec[c 
+ d*x]^(3/2) + 8*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) + 33*Sqrt[-((-1 
 + Sec[c + d*x])*Sec[c + d*x])])*Tan[c + d*x])/(24*d*Sqrt[1 - Sec[c + d*x] 
]*Sqrt[a*(1 + Sec[c + d*x])])
 
3.3.25.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4301, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4301

\(\displaystyle \frac {1}{3} a \int \frac {11}{2} \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {11}{6} a \int \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{6} a \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {11}{6} a \left (\frac {3}{4} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{6} a \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {11}{6} a \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {11}{6} a \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {11}{6} a \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {11}{6} a \left (\frac {3}{4} \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

input
Int[Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2),x]
 
output
(a^2*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (11 
*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
 

3.3.25.3.1 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 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]
 
3.3.25.4 Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.48

method result size
default \(\frac {a \left (66 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}-33 \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-33 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+44 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+16 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}}}{48 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) \(236\)

input
int(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 
output
1/48/d*a*(66*(-1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*cos(d*x+c)^2-33*cos(d*x+ 
c)^3*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+ 
1))^(1/2))-33*cos(d*x+c)^3*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c 
)+1)/(-1/(cos(d*x+c)+1))^(1/2))+44*(-1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*co 
s(d*x+c)+16*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2))*(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)
 
3.3.25.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.50 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\left [\frac {33 \, {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (33 \, a \cos \left (d x + c\right )^{2} + 22 \, a \cos \left (d x + c\right ) + 8 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{96 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {33 \, {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (33 \, a \cos \left (d x + c\right )^{2} + 22 \, a \cos \left (d x + c\right ) + 8 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

input
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
 
output
[1/96*(33*(a*cos(d*x + c)^3 + a*cos(d*x + c)^2)*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)*sqr 
t((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*(33*a*cos(d*x + c)^2 + 22*a*cos(d 
*x + c) + 8*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(c 
os(d*x + c)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), 1/48*(33*(a*cos(d*x + 
 c)^3 + a*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) 
+ a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*c 
os(d*x + c) - 2*a)) + 2*(33*a*cos(d*x + c)^2 + 22*a*cos(d*x + c) + 8*a)*sq 
rt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d* 
cos(d*x + c)^3 + d*cos(d*x + c)^2)]
 
3.3.25.6 Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**(5/2)*(a+a*sec(d*x+c))**(3/2),x)
 
output
Timed out
 
3.3.25.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2361 vs. \(2 (134) = 268\).

Time = 0.49 (sec) , antiderivative size = 2361, normalized size of antiderivative = 14.76 \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\text {Too large to display} \]

input
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
 
output
-1/96*(132*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4*c) + 3* 
sqrt(2)*a*sin(2*d*x + 2*c))*cos(11/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
 2*c))) + 44*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4*c) + 
3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(9/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
+ 2*c))) + 216*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4*c) 
+ 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d* 
x + 2*c))) - 216*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4*c 
) + 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2* 
d*x + 2*c))) - 44*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 4* 
c) + 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2 
*d*x + 2*c))) - 132*(sqrt(2)*a*sin(6*d*x + 6*c) + 3*sqrt(2)*a*sin(4*d*x + 
4*c) + 3*sqrt(2)*a*sin(2*d*x + 2*c))*cos(1/4*arctan2(sin(2*d*x + 2*c), cos 
(2*d*x + 2*c))) - 33*(a*cos(6*d*x + 6*c)^2 + 9*a*cos(4*d*x + 4*c)^2 + 9*a* 
cos(2*d*x + 2*c)^2 + a*sin(6*d*x + 6*c)^2 + 9*a*sin(4*d*x + 4*c)^2 + 18*a* 
sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*a*sin(2*d*x + 2*c)^2 + 2*(3*a*cos(4* 
d*x + 4*c) + 3*a*cos(2*d*x + 2*c) + a)*cos(6*d*x + 6*c) + 6*(3*a*cos(2*d*x 
 + 2*c) + a)*cos(4*d*x + 4*c) + 6*a*cos(2*d*x + 2*c) + 6*(a*sin(4*d*x + 4* 
c) + a*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + a)*log(2*cos(1/4*arctan2(sin(2 
*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*...
 
3.3.25.8 Giac [F]

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

input
integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")
 
output
sage0*x
 
3.3.25.9 Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

input
int((a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(5/2),x)
 
output
int((a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(5/2), x)