\(\int (a+a \sec (c+d x))^{5/2} \tan ^5(c+d x) \, dx\) [159]

Optimal result

Integrand size = 23, antiderivative size = 193 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^5(c+d x) \, dx=-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^2 d}-\frac {6 (a+a \sec (c+d x))^{11/2}}{11 a^3 d}+\frac {2 (a+a \sec (c+d x))^{13/2}}{13 a^4 d} \] Output:

-2*a^(5/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d+2*a^2*(a+a*sec(d*x+c) 
)^(1/2)/d+2/3*a*(a+a*sec(d*x+c))^(3/2)/d+2/5*(a+a*sec(d*x+c))^(5/2)/d+2/7* 
(a+a*sec(d*x+c))^(7/2)/a/d+2/9*(a+a*sec(d*x+c))^(9/2)/a^2/d-6/11*(a+a*sec( 
d*x+c))^(11/2)/a^3/d+2/13*(a+a*sec(d*x+c))^(13/2)/a^4/d
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.81 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^5(c+d x) \, dx=\frac {(a (1+\sec (c+d x)))^{5/2} \left (-2 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+2 \sqrt {1+\sec (c+d x)}+\frac {2}{3} (1+\sec (c+d x))^{3/2}+\frac {2}{5} (1+\sec (c+d x))^{5/2}+\frac {2}{7} (1+\sec (c+d x))^{7/2}+\frac {2}{9} (1+\sec (c+d x))^{9/2}-\frac {6}{11} (1+\sec (c+d x))^{11/2}+\frac {2}{13} (1+\sec (c+d x))^{13/2}\right )}{d (1+\sec (c+d x))^{5/2}} \] Input:

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

Output:

((a*(1 + Sec[c + d*x]))^(5/2)*(-2*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + 2*Sqrt 
[1 + Sec[c + d*x]] + (2*(1 + Sec[c + d*x])^(3/2))/3 + (2*(1 + Sec[c + d*x] 
)^(5/2))/5 + (2*(1 + Sec[c + d*x])^(7/2))/7 + (2*(1 + Sec[c + d*x])^(9/2)) 
/9 - (6*(1 + Sec[c + d*x])^(11/2))/11 + (2*(1 + Sec[c + d*x])^(13/2))/13)) 
/(d*(1 + Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 25, 4368, 27, 99, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4368

\(\displaystyle \frac {\int a^2 \cos (c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{9/2}d\sec (c+d x)}{a^4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos (c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{9/2}d\sec (c+d x)}{a^2 d}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {(\sec (c+d x) a+a)^{11/2}}{a}+\cos (c+d x) (\sec (c+d x) a+a)^{9/2}-3 (\sec (c+d x) a+a)^{9/2}\right )d\sec (c+d x)}{a^2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-2 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )+2 a^4 \sqrt {a \sec (c+d x)+a}+\frac {2}{3} a^3 (a \sec (c+d x)+a)^{3/2}+\frac {2 (a \sec (c+d x)+a)^{13/2}}{13 a^2}+\frac {2}{5} a^2 (a \sec (c+d x)+a)^{5/2}-\frac {6 (a \sec (c+d x)+a)^{11/2}}{11 a}+\frac {2}{9} (a \sec (c+d x)+a)^{9/2}+\frac {2}{7} a (a \sec (c+d x)+a)^{7/2}}{a^2 d}\)

Input:

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

Output:

(-2*a^(9/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]] + 2*a^4*Sqrt[a + a*S 
ec[c + d*x]] + (2*a^3*(a + a*Sec[c + d*x])^(3/2))/3 + (2*a^2*(a + a*Sec[c 
+ d*x])^(5/2))/5 + (2*a*(a + a*Sec[c + d*x])^(7/2))/7 + (2*(a + a*Sec[c + 
d*x])^(9/2))/9 - (6*(a + a*Sec[c + d*x])^(11/2))/(11*a) + (2*(a + a*Sec[c 
+ d*x])^(13/2))/(13*a^2))/(a^2*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4368

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

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.66

\[\frac {\left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\frac {143378}{45045}+\frac {63446 \sec \left (d x +c \right )}{45045}-\frac {8354 \sec \left (d x +c \right )^{2}}{15015}-\frac {10838 \sec \left (d x +c \right )^{3}}{9009}-\frac {254 \sec \left (d x +c \right )^{4}}{1287}+\frac {54 \sec \left (d x +c \right )^{5}}{143}+\frac {2 \sec \left (d x +c \right )^{6}}{13}\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d}\]

Input:

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

Output:

1/d*(2*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(-cos(d*x+ 
c)/(1+cos(d*x+c)))^(1/2)+143378/45045+63446/45045*sec(d*x+c)-8354/15015*se 
c(d*x+c)^2-10838/9009*sec(d*x+c)^3-254/1287*sec(d*x+c)^4+54/143*sec(d*x+c) 
^5+2/13*sec(d*x+c)^6)*a^2*(a*(1+sec(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.00 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^5(c+d x) \, dx=\left [\frac {45045 \, a^{\frac {5}{2}} \cos \left (d x + c\right )^{6} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (71689 \, a^{2} \cos \left (d x + c\right )^{6} + 31723 \, a^{2} \cos \left (d x + c\right )^{5} - 12531 \, a^{2} \cos \left (d x + c\right )^{4} - 27095 \, a^{2} \cos \left (d x + c\right )^{3} - 4445 \, a^{2} \cos \left (d x + c\right )^{2} + 8505 \, a^{2} \cos \left (d x + c\right ) + 3465 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{90090 \, d \cos \left (d x + c\right )^{6}}, \frac {45045 \, \sqrt {-a} a^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{6} + 2 \, {\left (71689 \, a^{2} \cos \left (d x + c\right )^{6} + 31723 \, a^{2} \cos \left (d x + c\right )^{5} - 12531 \, a^{2} \cos \left (d x + c\right )^{4} - 27095 \, a^{2} \cos \left (d x + c\right )^{3} - 4445 \, a^{2} \cos \left (d x + c\right )^{2} + 8505 \, a^{2} \cos \left (d x + c\right ) + 3465 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{45045 \, d \cos \left (d x + c\right )^{6}}\right ] \] Input:

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

Output:

[1/90090*(45045*a^(5/2)*cos(d*x + c)^6*log(-8*a*cos(d*x + c)^2 + 4*(2*cos( 
d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) 
 - 8*a*cos(d*x + c) - a) + 4*(71689*a^2*cos(d*x + c)^6 + 31723*a^2*cos(d*x 
 + c)^5 - 12531*a^2*cos(d*x + c)^4 - 27095*a^2*cos(d*x + c)^3 - 4445*a^2*c 
os(d*x + c)^2 + 8505*a^2*cos(d*x + c) + 3465*a^2)*sqrt((a*cos(d*x + c) + a 
)/cos(d*x + c)))/(d*cos(d*x + c)^6), 1/45045*(45045*sqrt(-a)*a^2*arctan(2* 
sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x 
 + c) + a))*cos(d*x + c)^6 + 2*(71689*a^2*cos(d*x + c)^6 + 31723*a^2*cos(d 
*x + c)^5 - 12531*a^2*cos(d*x + c)^4 - 27095*a^2*cos(d*x + c)^3 - 4445*a^2 
*cos(d*x + c)^2 + 8505*a^2*cos(d*x + c) + 3465*a^2)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c)))/(d*cos(d*x + c)^6)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.94 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^5(c+d x) \, dx=\frac {45045 \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 18018 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} + \frac {6930 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {13}{2}}}{a^{4}} - \frac {24570 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {11}{2}}}{a^{3}} + \frac {10010 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {9}{2}}}{a^{2}} + \frac {12870 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a} + 30030 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a + 90090 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}}{45045 \, d} \] Input:

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

Output:

1/45045*(45045*a^(5/2)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + 
a/cos(d*x + c)) + sqrt(a))) + 18018*(a + a/cos(d*x + c))^(5/2) + 6930*(a + 
 a/cos(d*x + c))^(13/2)/a^4 - 24570*(a + a/cos(d*x + c))^(11/2)/a^3 + 1001 
0*(a + a/cos(d*x + c))^(9/2)/a^2 + 12870*(a + a/cos(d*x + c))^(7/2)/a + 30 
030*(a + a/cos(d*x + c))^(3/2)*a + 90090*sqrt(a + a/cos(d*x + c))*a^2)/d
 

Giac [A] (verification not implemented)

Time = 1.07 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.26 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^5(c+d x) \, dx=\frac {\sqrt {2} {\left (\frac {45045 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (45045 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{6} a - 30030 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} a^{2} + 36036 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} a^{3} - 51480 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} a^{4} + 80080 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a^{5} + 393120 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{6} + 221760 \, a^{7}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{6} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{45045 \, d} \] Input:

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

Output:

1/45045*sqrt(2)*(45045*sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 
1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) + 2*(45045*(a*tan(1/2*d*x + 1/2*c)^2 - a) 
^6*a - 30030*(a*tan(1/2*d*x + 1/2*c)^2 - a)^5*a^2 + 36036*(a*tan(1/2*d*x + 
 1/2*c)^2 - a)^4*a^3 - 51480*(a*tan(1/2*d*x + 1/2*c)^2 - a)^3*a^4 + 80080* 
(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*a^5 + 393120*(a*tan(1/2*d*x + 1/2*c)^2 - 
a)*a^6 + 221760*a^7)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^6*sqrt(-a*tan(1/2*d*x 
 + 1/2*c)^2 + a)))*a^2*sgn(cos(d*x + c))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+a \sec (c+d x))^{5/2} \tan ^5(c+d x) \, dx=\frac {\sqrt {a}\, a^{2} \left (6930 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2} \tan \left (d x +c \right )^{4}-6160 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2}+4928 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}+17010 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{4}-20320 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}+28736 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )+11130 \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{4}-20640 \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{2}+106368 \sqrt {\sec \left (d x +c \right )+1}+525 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{5}}{\sec \left (d x +c \right )+1}d x \right ) d -3240 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )+1}d x \right ) d +41280 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) d \right )}{45045 d} \] Input:

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

Output:

(sqrt(a)*a**2*(6930*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2*tan(c + d*x)**4 
 - 6160*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2*tan(c + d*x)**2 + 4928*sqrt 
(sec(c + d*x) + 1)*sec(c + d*x)**2 + 17010*sqrt(sec(c + d*x) + 1)*sec(c + 
d*x)*tan(c + d*x)**4 - 20320*sqrt(sec(c + d*x) + 1)*sec(c + d*x)*tan(c + d 
*x)**2 + 28736*sqrt(sec(c + d*x) + 1)*sec(c + d*x) + 11130*sqrt(sec(c + d* 
x) + 1)*tan(c + d*x)**4 - 20640*sqrt(sec(c + d*x) + 1)*tan(c + d*x)**2 + 1 
06368*sqrt(sec(c + d*x) + 1) + 525*int((sqrt(sec(c + d*x) + 1)*tan(c + d*x 
)**5)/(sec(c + d*x) + 1),x)*d - 3240*int((sqrt(sec(c + d*x) + 1)*tan(c + d 
*x)**3)/(sec(c + d*x) + 1),x)*d + 41280*int((sqrt(sec(c + d*x) + 1)*tan(c 
+ d*x))/(sec(c + d*x) + 1),x)*d))/(45045*d)