\(\int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \sec (c+d x)} \, dx\) [195]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 136 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}-\frac {\sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {3 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:

-3*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2) 
/a/d-cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1 
/2)/a/d+3*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d-sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a 
+a*sec(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.06 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.93 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (-\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (3 \left (1+e^{2 i (c+d x)}\right )+3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{d \left (-1+e^{2 i c}\right )}+\frac {\sqrt {\sec (c+d x)} \left (6 \cos (d x) \csc (c)-2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}\right )}{a (1+\sec (c+d x))} \] Input:

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

Output:

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 
+ E^((2*I)*(c + d*x)))]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-1 + E^((2*I)*c) 
)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I 
)*(c + d*x))] - E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + 
d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*E^(I*(c 
 + d*x))*(-1 + E^((2*I)*c))) + (Sqrt[Sec[c + d*x]]*(6*Cos[d*x]*Csc[c] - 2* 
Tan[(c + d*x)/2]))/d))/(a*(1 + Sec[c + d*x]))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4305, 27, 3042, 4274, 3042, 4255, 3042, 4258, 3042, 3119, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4305

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-3 a \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

Input:

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

Output:

-((Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) - ((2*a*Sqrt 
[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - 3*a*((-2* 
Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sq 
rt[Sec[c + d*x]]*Sin[c + d*x])/d))/(2*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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

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 4255
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4305
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[d^2*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 2)/(f*(a + 
b*Csc[e + f*x]))), x] - Simp[d^2/(a*b)   Int[(d*Csc[e + f*x])^(n - 2)*(b*(n 
 - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ 
[a^2 - b^2, 0] && GtQ[n, 1]
 
Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.86

method result size
default \(-\frac {-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+6 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(253\)

Input:

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

Output:

-(-cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2) 
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(EllipticF( 
cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+6*(-2 
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-5*( 
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2)/ 
a/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/ 
sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {{\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/2*((I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d 
*x + c) + I*sin(d*x + c)) + (-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(I*sqrt(2)*cos(d*x + 
 c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
 + c) + I*sin(d*x + c))) - 3*(-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierst 
rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) 
 + 2*(3*cos(d*x + c) + 2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + 
c) + a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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