\(\int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [192]

Optimal result

Integrand size = 21, antiderivative size = 113 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {1}{5 d (a+a \sin (c+d x))^{5/2}}-\frac {1}{6 a d (a+a \sin (c+d x))^{3/2}}-\frac {1}{4 a^2 d \sqrt {a+a \sin (c+d x)}} \] Output:

1/8*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(5/2)/d- 
1/5/d/(a+a*sin(d*x+c))^(5/2)-1/6/a/d/(a+a*sin(d*x+c))^(3/2)-1/4/a^2/d/(a+a 
*sin(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.36 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{5 d (a+a \sin (c+d x))^{5/2}} \] Input:

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

Output:

-1/5*Hypergeometric2F1[-5/2, 1, -3/2, (1 + Sin[c + d*x])/2]/(d*(a + a*Sin[ 
c + d*x])^(5/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3146, 61, 61, 61, 73, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

\(\displaystyle \frac {a \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{7/2}}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(a*(-1/5*1/(a*(a + a*Sin[c + d*x])^(5/2)) + (-1/3*1/(a*(a + a*Sin[c + d*x] 
)^(3/2)) + (ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[2]]/(Sqrt[2]*a^(3/2)) - 1/ 
(a*Sqrt[a + a*Sin[c + d*x]]))/(2*a))/(2*a)))/d
 

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 3042

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

rule 3146

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x 
)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I 
ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/ 
2])
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {2 a \left (\frac {1}{8 a^{3} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{12 a^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{10 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {7}{2}}}\right )}{d}\)	\(88\)

Input:

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

Output:

-2*a*(1/8/a^3/(a+a*sin(d*x+c))^(1/2)+1/12/a^2/(a+a*sin(d*x+c))^(3/2)+1/10/ 
a/(a+a*sin(d*x+c))^(5/2)-1/16/a^(7/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c)) 
^(1/2)*2^(1/2)/a^(1/2)))/d
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.50 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {15 \, \sqrt {2} {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (15 \, \cos \left (d x + c\right )^{2} - 40 \, \sin \left (d x + c\right ) - 52\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{240 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:

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

Output:

1/240*(15*sqrt(2)*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 
4)*sqrt(a)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt( 
a) + 3*a)/(sin(d*x + c) - 1)) - 4*(15*cos(d*x + c)^2 - 40*sin(d*x + c) - 5 
2)*sqrt(a*sin(d*x + c) + a))/(3*a^3*d*cos(d*x + c)^2 - 4*a^3*d + (a^3*d*co 
s(d*x + c)^2 - 4*a^3*d)*sin(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {\frac {15 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} + \frac {4 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} + 10 \, {\left (a \sin \left (d x + c\right ) + a\right )} a + 12 \, a^{2}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a}}{240 \, a d} \] Input:

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

Output:

-1/240*(15*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt 
(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a)))/a^(3/2) + 4*(15*(a*sin(d*x + c) + 
 a)^2 + 10*(a*sin(d*x + c) + a)*a + 12*a^2)/((a*sin(d*x + c) + a)^(5/2)*a) 
)/(a*d)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.40 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {a} {\left (\frac {15 \, \sqrt {2} \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {15 \, \sqrt {2} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (15 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3\right )}}{a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{240 \, d} \] Input:

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

Output:

1/240*sqrt(a)*(15*sqrt(2)*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^3*sgn 
(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 15*sqrt(2)*log(-cos(-1/4*pi + 1/2*d*x 
+ 1/2*c) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 2*sqrt(2)*(15*co 
s(-1/4*pi + 1/2*d*x + 1/2*c)^4 + 5*cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 3)/( 
a^3*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))) 
/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*sec(c + d*x))/(sin(c + d*x)**3 + 3*si 
n(c + d*x)**2 + 3*sin(c + d*x) + 1),x))/a**3