\(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [1082]

Optimal result

Integrand size = 23, antiderivative size = 80 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 (7 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)} \] Output:

2/21*(7*A+5*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/7*C*sin(d*x+c)/d 
/cos(d*x+c)^(7/2)+2/21*(7*A+5*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 (7 A+5 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(7 A+5 C) \sin (2 (c+d x))+6 C \tan (c+d x)}{21 d \cos ^{\frac {5}{2}}(c+d x)} \] Input:

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

Output:

(2*(7*A + 5*C)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + (7*A + 5*C)* 
Sin[2*(c + d*x)] + 6*C*Tan[c + d*x])/(21*d*Cos[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4554, 3042, 3491, 3042, 3116, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sec (c+d x)^2}{\cos (c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 4554

\(\displaystyle \int \frac {A \cos ^2(c+d x)+C}{\cos ^{\frac {9}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\)

\(\Big \downarrow \) 3491

\(\displaystyle \frac {1}{7} (7 A+5 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+5 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {1}{7} (7 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{7} (7 A+5 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

Input:

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

Output:

(2*C*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((7*A + 5*C)*((2*EllipticF[( 
c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))))/7
 

Defintions of rubi rules used

rule 3042

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

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 3491

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x 
_)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m 
+ 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1))   Int[(b*Sin[e + f* 
x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
 

rule 4554

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (C_.)*sec[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[b^2   Int[(b*Cos[e + f*x])^(m - 2)*(C + A*Cos[e 
+ f*x]^2), x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(375\) vs. \(2(71)=142\).

Time = 4.74 (sec) , antiderivative size = 376, normalized size of antiderivative = 4.70

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 A \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )+2 C \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{56 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{4}}-\frac {5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{42 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{21 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(376\)

Input:

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

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(-1/6*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)/(cos(1 
/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/ 
2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip 
ticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*C*(-1/56*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/ 
42*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) 
/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/ 
2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*EllipticF(cos(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 = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.55 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (-7 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (7 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left ({\left (7 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{21 \, d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

1/21*(sqrt(2)*(-7*I*A - 5*I*C)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, c 
os(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(7*I*A + 5*I*C)*cos(d*x + c)^4*wei 
erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*((7*A + 5*C)*co 
s(d*x + c)^2 + 3*C)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F(-1)]

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

integrate((A+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + A)/cos(d*x + c)^(5/2), x)
 

Giac [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + A)/cos(d*x + c)^(5/2), x)
 

Mupad [B] (verification not implemented)

Time = 13.67 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,A\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int((A + C/cos(c + d*x)^2)/cos(c + d*x)^(5/2),x)
 

Output:

(2*A*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c 
+ d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*sin(c + d*x)*hypergeom([-7/4, 
1/2], -3/4, cos(c + d*x)^2))/(7*d*cos(c + d*x)^(7/2)*(sin(c + d*x)^2)^(1/2 
))
 

Reduce [F]

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

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2),x)
 

Output:

int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x)*a + int((sqrt(cos(c + d*x))*sec( 
c + d*x)**2)/cos(c + d*x)**3,x)*c