3.7.55 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [655]

3.7.55.1 Optimal result
3.7.55.2 Mathematica [C] (verified)
3.7.55.3 Rubi [A] (verified)
3.7.55.4 Maple [C] (warning: unable to verify)
3.7.55.5 Fricas [F(-1)]
3.7.55.6 Sympy [F(-1)]
3.7.55.7 Maxima [F]
3.7.55.8 Giac [F]
3.7.55.9 Mupad [F(-1)]

3.7.55.1 Optimal result

Integrand size = 25, antiderivative size = 206 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{b d \sqrt {a+b \sec (c+d x)}}+\frac {2 a E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{b \left (a^2-b^2\right ) d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}-\frac {2 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

output
-2*a^2*sin(d*x+c)*sec(d*x+c)^(1/2)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)+2* 
(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2 
*c),2,2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^( 
1/2)/b/d/(a+b*sec(d*x+c))^(1/2)+2*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d 
*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d 
*x+c))^(1/2)/b/(a^2-b^2)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)
 
3.7.55.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.83 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.11 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {(b+a \cos (c+d x))^{3/2} \left (\frac {4 a b \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{\sqrt {b+a \cos (c+d x)}}+\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{\sqrt {b+a \cos (c+d x)}}+\frac {2 i \sqrt {-\frac {a (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {a (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (-2 b (a+b) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right )|\frac {-a+b}{a+b}\right )+a \left (2 b \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right ),\frac {-a+b}{a+b}\right )+a \operatorname {EllipticPi}\left (1-\frac {a}{b},i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (c+d x)}\right ),\frac {-a+b}{a+b}\right )\right )\right )}{\sqrt {\frac {1}{a-b}} b}\right )}{(a-b) (a+b)}+\frac {4 a^2 (b+a \cos (c+d x)) \sin (c+d x)}{-a^2+b^2}\right )}{2 b d (a+b \sec (c+d x))^{3/2}} \]

input
Integrate[Sec[c + d*x]^(5/2)/(a + b*Sec[c + d*x])^(3/2),x]
 
output
(Sec[c + d*x]^(3/2)*(((b + a*Cos[c + d*x])^(3/2)*((4*a*b*Sqrt[(b + a*Cos[c 
 + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c 
+ d*x]] + (2*(3*a^2 - 2*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi 
[2, (c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]] + ((2*I)*Sqrt[-( 
(a*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(a*(1 + Cos[c + d*x]))/(a - b)]*Csc 
[c + d*x]*(-2*b*(a + b)*EllipticE[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a* 
Cos[c + d*x]]], (-a + b)/(a + b)] + a*(2*b*EllipticF[I*ArcSinh[Sqrt[(a - b 
)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*EllipticPi[1 - a/ 
b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b 
)])))/(Sqrt[(a - b)^(-1)]*b)))/((a - b)*(a + b)) + (4*a^2*(b + a*Cos[c + d 
*x])*Sin[c + d*x])/(-a^2 + b^2)))/(2*b*d*(a + b*Sec[c + d*x])^(3/2))
 
3.7.55.3 Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 4332, 27, 3042, 4596, 2011, 3042, 4343, 3042, 3134, 3042, 3132, 4346, 3042, 3286, 3042, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4332

\(\displaystyle -\frac {2 \int -\frac {a^2+b \sec (c+d x) a+\left (a^2-b^2\right ) \sec ^2(c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a^2+b \sec (c+d x) a+\left (a^2-b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a^2+b \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4596

\(\displaystyle \frac {\left (a^2-b^2\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {a^2+b \sec (c+d x) a}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {\left (a^2-b^2\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+a \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4343

\(\displaystyle \frac {\left (a^2-b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{\sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (a^2-b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\left (a^2-b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{\sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (a^2-b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\left (a^2-b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4346

\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}}dx}{\sqrt {a+b \sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {a+b \sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\frac {2 \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

input
Int[Sec[c + d*x]^(5/2)/(a + b*Sec[c + d*x])^(3/2),x]
 
output
((2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x) 
/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(d*Sqrt[a + b*Sec[c + d*x]]) + (2*a 
*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(d*Sqrt[( 
b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]))/(b*(a^2 - b^2)) - (2*a^2 
*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x] 
])
 

3.7.55.3.1 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 2011
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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

rule 3132
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3284
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3286
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + 
 d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]
 

rule 4332
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[(-a^2)*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^( 
m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[d^3/ 
(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x]) 
^(n - 3)*Simp[a^2*(n - 3) + a*b*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*( 
m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 
- b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n 
, 2]))
 

rule 4343
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] 
*(d_.)], x_Symbol] :> Simp[Sqrt[a + b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*S 
qrt[b + a*Sin[e + f*x]])   Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[{a 
, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4346
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_)], x_Symbol] :> Simp[d*Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x 
]]/Sqrt[a + b*Csc[e + f*x]])   Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f*x]] 
), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4596
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)]), x_Symbol] :> Simp[C/d^2   Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a + b*C 
sc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[ 
a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - 
 b^2, 0]
 
3.7.55.4 Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.57 (sec) , antiderivative size = 1019, normalized size of antiderivative = 4.95

method result size
default \(\text {Expression too large to display}\) \(1019\)

input
int(sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 
output
-2/d/((a-b)/(a+b))^(1/2)/b/(a+b)*(-((1-cos(d*x+c))^2*csc(d*x+c)^2+1)/((1-c 
os(d*x+c))^2*csc(d*x+c)^2-1))^(5/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^3*(( 
a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/((1-c 
os(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(((a-b)/(a+b))^(1/2)*a*(1-cos(d*x+c))^ 
3*csc(d*x+c)^3+2*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc 
(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*Ellipt 
icF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a+( 
-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+ 
b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticF(((a-b)/(a+b))^ 
(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b-(-(a*(1-cos(d*x+c)) 
^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos( 
d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c) 
+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a-2*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2- 
b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d* 
x+c)^2+1)^(1/2)*EllipticPi(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(a 
+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a-2*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b* 
(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+ 
c)^2+1)^(1/2)*EllipticPi(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(a+b 
)/(a-b),I/((a-b)/(a+b))^(1/2))*b+((a-b)/(a+b))^(1/2)*a*(-cot(d*x+c)+csc(d* 
x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^3/(a*(1-cos(d*x+c))^2*csc(d*x+...
 
3.7.55.5 Fricas [F(-1)]

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

input
integrate(sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
 
output
Timed out
 
3.7.55.6 Sympy [F(-1)]

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

input
integrate(sec(d*x+c)**(5/2)/(a+b*sec(d*x+c))**(3/2),x)
 
output
Timed out
 
3.7.55.7 Maxima [F]

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

input
integrate(sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
 
output
integrate(sec(d*x + c)^(5/2)/(b*sec(d*x + c) + a)^(3/2), x)
 
3.7.55.8 Giac [F]

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

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

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

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