3.2.13 \(\int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\) [113]

3.2.13.1 Optimal result
3.2.13.2 Mathematica [A] (verified)
3.2.13.3 Rubi [A] (warning: unable to verify)
3.2.13.4 Maple [A] (verified)
3.2.13.5 Fricas [C] (verification not implemented)
3.2.13.6 Sympy [F(-1)]
3.2.13.7 Maxima [F]
3.2.13.8 Giac [F]
3.2.13.9 Mupad [F(-1)]

3.2.13.1 Optimal result

Integrand size = 23, antiderivative size = 160 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\frac {a \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {a \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}-\frac {2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}} \]

output
a*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)+a*arctanh((e*sin(d*x+c))^ 
(1/2)/e^(1/2))/d/e^(5/2)-2/3*a/d/e/(e*sin(d*x+c))^(3/2)-2/3*a*cos(d*x+c)/d 
/e/(e*sin(d*x+c))^(3/2)-2/3*a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2* 
c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^ 
(1/2)/d/e^2/(e*sin(d*x+c))^(1/2)
 
3.2.13.2 Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.75 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-3 \arctan \left (\sqrt {\sin (c+d x)}\right )-3 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )+2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )+\csc ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sin (c+d x)}\right ) \sqrt {\sin (c+d x)}}{6 d e^2 \sqrt {e \sin (c+d x)}} \]

input
Integrate[(a + a*Sec[c + d*x])/(e*Sin[c + d*x])^(5/2),x]
 
output
-1/6*(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-3*ArcTan[Sqrt[Sin[c + d*x] 
]] - 3*ArcTanh[Sqrt[Sin[c + d*x]]] + 2*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] 
 + Csc[(c + d*x)/2]^2*Sqrt[Sin[c + d*x]])*Sqrt[Sin[c + d*x]])/(d*e^2*Sqrt[ 
e*Sin[c + d*x]])
 
3.2.13.3 Rubi [A] (warning: unable to verify)

Time = 0.77 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.98, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.913, Rules used = {3042, 4360, 25, 25, 3042, 25, 3317, 25, 3042, 3044, 27, 264, 266, 756, 216, 219, 3116, 3042, 3121, 3042, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

\(\displaystyle \int -\frac {\sec (c+d x) (a (-\cos (c+d x))-a)}{(e \sin (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -\frac {(\cos (c+d x) a+a) \sec (c+d x)}{(e \sin (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3317

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx+a \int \frac {1}{\cos (c+d x) (e \sin (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 264

\(\displaystyle \frac {a e \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)} \left (e^2-e^2 \sin ^2(c+d x)\right )}d(e \sin (c+d x))}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {a e \left (\frac {2 \int \frac {1}{e^2-e^4 \sin ^4(c+d x)}d\sqrt {e \sin (c+d x)}}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {a e \left (\frac {2 \left (\frac {\int \frac {1}{e-e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e}+\frac {\int \frac {1}{e^2 \sin ^2(c+d x)+e}d\sqrt {e \sin (c+d x)}}{2 e}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {a e \left (\frac {2 \left (\frac {\int \frac {1}{e-e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e}+\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 219

\(\displaystyle a \int \frac {1}{(e \sin (c+d x))^{5/2}}dx+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 3116

\(\displaystyle a \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 3121

\(\displaystyle a \left (\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )+\frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {a e \left (\frac {2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )}{e^2}-\frac {2}{3 e^2 (e \sin (c+d x))^{3/2}}\right )}{d}+a \left (\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}\right )\)

input
Int[(a + a*Sec[c + d*x])/(e*Sin[c + d*x])^(5/2),x]
 
output
(a*e*((2*(ArcTan[Sqrt[e]*Sin[c + d*x]]/(2*e^(3/2)) + ArcTanh[Sqrt[e]*Sin[c 
 + d*x]]/(2*e^(3/2))))/e^2 - 2/(3*e^2*(e*Sin[c + d*x])^(3/2))))/d + a*((-2 
*Cos[c + d*x])/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (2*EllipticF[(c - Pi/2 + d 
*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*Sin[c + d*x]]))
 

3.2.13.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 264
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3044
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3317
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 

rule 4360
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 
3.2.13.4 Maple [A] (verified)

Time = 11.42 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.02

method result size
default \(\frac {-\frac {2 a}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}+\frac {a \,\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}-\frac {a \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) \(164\)
parts \(-\frac {a \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a \left (-\frac {2}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}\right )}{d}\) \(166\)

input
int((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 
output
(-2/3*a/e/(e*sin(d*x+c))^(3/2)+a/e^(5/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/ 
2))+a/e^(5/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))-1/3*a/e^2*((-sin(d*x+c 
)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((-sin(d*x+c)+ 
1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^3+2*sin(d*x+c))/sin(d*x+c)^2/cos(d*x+c) 
/(e*sin(d*x+c))^(1/2))/d
 
3.2.13.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.20 (sec) , antiderivative size = 705, normalized size of antiderivative = 4.41 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx=\left [-\frac {6 \, {\left (a \cos \left (d x + c\right ) - a\right )} \sqrt {-e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} - e \sin \left (d x + c\right ) - e\right )}}\right ) + 3 \, {\left (a \cos \left (d x + c\right ) - a\right )} \sqrt {-e} \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e} + 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) - 8 \, {\left (\sqrt {2} a \cos \left (d x + c\right ) - \sqrt {2} a\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 8 \, {\left (\sqrt {2} a \cos \left (d x + c\right ) - \sqrt {2} a\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 16 \, \sqrt {e \sin \left (d x + c\right )} a}{24 \, {\left (d e^{3} \cos \left (d x + c\right ) - d e^{3}\right )}}, \frac {6 \, {\left (a \cos \left (d x + c\right ) - a\right )} \sqrt {e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} + e \sin \left (d x + c\right ) - e\right )}}\right ) + 3 \, {\left (a \cos \left (d x + c\right ) - a\right )} \sqrt {e} \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e} - 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) + 8 \, {\left (\sqrt {2} a \cos \left (d x + c\right ) - \sqrt {2} a\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 8 \, {\left (\sqrt {2} a \cos \left (d x + c\right ) - \sqrt {2} a\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 16 \, \sqrt {e \sin \left (d x + c\right )} a}{24 \, {\left (d e^{3} \cos \left (d x + c\right ) - d e^{3}\right )}}\right ] \]

input
integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")
 
output
[-1/24*(6*(a*cos(d*x + c) - a)*sqrt(-e)*arctan(1/4*(cos(d*x + c)^2 - 6*sin 
(d*x + c) - 2)*sqrt(e*sin(d*x + c))*sqrt(-e)/(e*cos(d*x + c)^2 - e*sin(d*x 
 + c) - e)) + 3*(a*cos(d*x + c) - a)*sqrt(-e)*log((e*cos(d*x + c)^4 - 72*e 
*cos(d*x + c)^2 - 8*(7*cos(d*x + c)^2 - (cos(d*x + c)^2 - 8)*sin(d*x + c) 
- 8)*sqrt(e*sin(d*x + c))*sqrt(-e) + 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + 
 c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*si 
n(d*x + c) + 8)) - 8*(sqrt(2)*a*cos(d*x + c) - sqrt(2)*a)*sqrt(-I*e)*weier 
strassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) - 8*(sqrt(2)*a*cos(d*x 
 + c) - sqrt(2)*a)*sqrt(I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*si 
n(d*x + c)) - 16*sqrt(e*sin(d*x + c))*a)/(d*e^3*cos(d*x + c) - d*e^3), 1/2 
4*(6*(a*cos(d*x + c) - a)*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + 
 c) - 2)*sqrt(e*sin(d*x + c))*sqrt(e)/(e*cos(d*x + c)^2 + e*sin(d*x + c) - 
 e)) + 3*(a*cos(d*x + c) - a)*sqrt(e)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x 
 + c)^2 - 8*(7*cos(d*x + c)^2 + (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqr 
t(e*sin(d*x + c))*sqrt(e) - 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72* 
e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c 
) + 8)) + 8*(sqrt(2)*a*cos(d*x + c) - sqrt(2)*a)*sqrt(-I*e)*weierstrassPIn 
verse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 8*(sqrt(2)*a*cos(d*x + c) - s 
qrt(2)*a)*sqrt(I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c 
)) + 16*sqrt(e*sin(d*x + c))*a)/(d*e^3*cos(d*x + c) - d*e^3)]
 
3.2.13.6 Sympy [F(-1)]

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

input
integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))**(5/2),x)
 
output
Timed out
 
3.2.13.7 Maxima [F]

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

input
integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")
 
output
integrate((a*sec(d*x + c) + a)/(e*sin(d*x + c))^(5/2), x)
 
3.2.13.8 Giac [F]

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

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

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

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