\(\int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx\) [271]

Optimal result

Integrand size = 23, antiderivative size = 134 \[ \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arcsinh}\left (\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )}{d}+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {26 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d \sqrt {1+\sec (c+d x)}} \] Output:

-2^(1/2)*arcsinh(tan(d*x+c)/(1+sec(d*x+c)))/d+2/5*sin(d*x+c)/d/sec(d*x+c)^ 
(3/2)/(1+sec(d*x+c))^(1/2)-2/15*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(1+sec(d*x+c 
))^(1/2)+26/15*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(1+sec(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx=\frac {\left (15 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^{\frac {5}{2}}(c+d x)+2 \sqrt {1-\sec (c+d x)} \left (3-\sec (c+d x)+13 \sec ^2(c+d x)\right )\right ) \sin (c+d x)}{15 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {-\tan ^2(c+d x)}} \] Input:

Integrate[1/(Sec[c + d*x]^(5/2)*Sqrt[1 + Sec[c + d*x]]),x]
 

Output:

((15*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*S 
ec[c + d*x]^(5/2) + 2*Sqrt[1 - Sec[c + d*x]]*(3 - Sec[c + d*x] + 13*Sec[c 
+ d*x]^2))*Sin[c + d*x])/(15*d*Sec[c + d*x]^(3/2)*Sqrt[-Tan[c + d*x]^2])
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4310, 3042, 4510, 27, 3042, 4501, 3042, 4294, 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.

\(\displaystyle \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x)+1}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx\)

\(\Big \downarrow \) 4310

\(\displaystyle \frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}-\frac {1}{5} \int \frac {1-4 \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {1}{5} \left (-\frac {2}{3} \int -\frac {13-2 \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}dx-\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {13-2 \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}dx-\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {13-2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx-\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 4501

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {\sec (c+d x)+1}}-15 \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x)+1}}dx\right )-\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {\sec (c+d x)+1}}-15 \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx\right )-\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 4294

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {15 \sqrt {2} \int \frac {1}{\sqrt {\frac {\tan ^2(c+d x)}{(\sec (c+d x)+1)^2}+1}}d\left (-\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d}+\frac {26 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {\sec (c+d x)+1}}\right )-\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {\sec (c+d x)+1}}-\frac {15 \sqrt {2} \text {arcsinh}\left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d}\right )-\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x)+1}}\)

Input:

Int[1/(Sec[c + d*x]^(5/2)*Sqrt[1 + Sec[c + d*x]]),x]
 

Output:

(2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[1 + Sec[c + d*x]]) + ((-2*Si 
n[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) + ((-15*Sqrt[2 
]*ArcSinh[Tan[c + d*x]/(1 + Sec[c + d*x])])/d + (26*Sqrt[Sec[c + d*x]]*Sin 
[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]))/3)/5
 

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 4294

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

rule 4310

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

rule 4501

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m 
 - b*B*n)/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] 
, x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a 
^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]
 

rule 4510

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d 
*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* 
n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {\sqrt {1+\sec \left (d x +c \right )}\, \left (-6 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )-26 \sec \left (d x +c \right ) \tan \left (d x +c \right )+\arctan \left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (15 \sec \left (d x +c \right )+15 \sec \left (d x +c \right )^{2}\right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\)	\(134\)

Input:

int(1/sec(d*x+c)^(5/2)/(1+sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15/d*(1+sec(d*x+c))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(5/2)*(-6*sin(d*x+c 
)+2*tan(d*x+c)-26*sec(d*x+c)*tan(d*x+c)+arctan(1/2*2^(1/2)/(-1/(cos(d*x+c) 
+1))^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(-2/(cos(d*x+c)+1))^(1/2)*(15*sec(d*x 
+c)+15*sec(d*x+c)^2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx=\frac {15 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (3 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \] Input:

integrate(1/sec(d*x+c)^(5/2)/(1+sec(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/30*(15*(sqrt(2)*cos(d*x + c) + sqrt(2))*log(-(2*sqrt(2)*sqrt((cos(d*x + 
c) + 1)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + cos(d*x + c)^2 - 2 
*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(3*cos(d*x + 
 c)^3 - cos(d*x + c)^2 + 13*cos(d*x + c))*sqrt((cos(d*x + c) + 1)/cos(d*x 
+ c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (114) = 228\).

Time = 0.22 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.64 \[ \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(1/sec(d*x+c)^(5/2)/(1+sec(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

1/60*sqrt(2)*(60*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c 
)))*sin(5/2*d*x + 5/2*c) - 5*cos(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2 
*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 60*cos(5/2*d*x + 5/2*c)*sin(4/5*arc 
tan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 5*cos(5/2*d*x + 5/2*c) 
*sin(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 30*log(cos 
(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*arct 
an2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + 2*sin(1/5*arctan2(sin 
(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 1) + 30*log(cos(1/5*arctan2(si 
n(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 + sin(1/5*arctan2(sin(5/2*d*x 
 + 5/2*c), cos(5/2*d*x + 5/2*c)))^2 - 2*sin(1/5*arctan2(sin(5/2*d*x + 5/2* 
c), cos(5/2*d*x + 5/2*c))) + 1) + 6*sin(5/2*d*x + 5/2*c) - 5*sin(3/5*arcta 
n2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 60*sin(1/5*arctan2(sin(5 
/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))/d
 

Giac [A] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx=\frac {\sqrt {2} {\left (\frac {2 \, {\left ({\left (17 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{\frac {5}{2}}} + 15 \, \log \left (\sqrt {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )}}{15 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

integrate(1/sec(d*x+c)^(5/2)/(1+sec(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

1/15*sqrt(2)*(2*((17*tan(1/2*d*x + 1/2*c)^2 + 20)*tan(1/2*d*x + 1/2*c)^2 + 
 15)*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1)^(5/2) + 15*log(sqrt 
(tan(1/2*d*x + 1/2*c)^2 + 1) - tan(1/2*d*x + 1/2*c)))/(d*sgn(cos(d*x + c)) 
)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}+\sec \left (d x +c \right )^{3}}d x \] Input:

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

Output:

int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/(sec(c + d*x)**4 + sec(c + 
 d*x)**3),x)