\(\int \sqrt {a+a \sec (c+d x)} \tan ^5(c+d x) \, dx\) [135]

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^5(c+d x) \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac {6 (a+a \sec (c+d x))^{7/2}}{7 a^3 d}+\frac {2 (a+a \sec (c+d x))^{9/2}}{9 a^4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^5(c+d x) \, dx=\frac {2 \sqrt {a (1+\sec (c+d x))} \left (-315 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} \left (383-34 \sec (c+d x)-132 \sec ^2(c+d x)+5 \sec ^3(c+d x)+35 \sec ^4(c+d x)\right )\right )}{315 d \sqrt {1+\sec (c+d x)}} \] Input:

Integrate[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^5,x]
 

Output:

(2*Sqrt[a*(1 + Sec[c + d*x])]*(-315*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqrt 
[1 + Sec[c + d*x]]*(383 - 34*Sec[c + d*x] - 132*Sec[c + d*x]^2 + 5*Sec[c + 
 d*x]^3 + 35*Sec[c + d*x]^4)))/(315*d*Sqrt[1 + Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 134, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4368

\(\displaystyle \frac {\int a^2 \cos (c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{5/2}d\sec (c+d x)}{a^4 d}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {(\sec (c+d x) a+a)^{7/2}}{a}+\cos (c+d x) (\sec (c+d x) a+a)^{5/2}-3 (\sec (c+d x) a+a)^{5/2}\right )d\sec (c+d x)}{a^2 d}\)

\(\Big \downarrow \) 2009

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

Input:

Int[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^5,x]
 

Output:

(-2*a^(5/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]] + 2*a^2*Sqrt[a + a*S 
ec[c + d*x]] + (2*a*(a + a*Sec[c + d*x])^(3/2))/3 + (2*(a + a*Sec[c + d*x] 
)^(5/2))/5 - (6*(a + a*Sec[c + d*x])^(7/2))/(7*a) + (2*(a + a*Sec[c + d*x] 
)^(9/2))/(9*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 = 0.78 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {\left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\frac {766}{315}-\frac {68 \sec \left (d x +c \right )}{315}-\frac {88 \sec \left (d x +c \right )^{2}}{105}+\frac {2 \sec \left (d x +c \right )^{3}}{63}+\frac {2 \sec \left (d x +c \right )^{4}}{9}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d}\)	\(105\)

Input:

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

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)+766/315-68/315*sec(d*x+c)-88/105*sec(d*x+c)^2+2/6 
3*sec(d*x+c)^3+2/9*sec(d*x+c)^4)*(a*(1+sec(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.03 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^5(c+d x) \, dx=\left [\frac {315 \, \sqrt {a} \cos \left (d x + c\right )^{4} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (383 \, \cos \left (d x + c\right )^{4} - 34 \, \cos \left (d x + c\right )^{3} - 132 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 35\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{630 \, d \cos \left (d x + c\right )^{4}}, \frac {315 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{4} + 2 \, {\left (383 \, \cos \left (d x + c\right )^{4} - 34 \, \cos \left (d x + c\right )^{3} - 132 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 35\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{315 \, d \cos \left (d x + c\right )^{4}}\right ] \] Input:

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

Output:

[1/630*(315*sqrt(a)*cos(d*x + c)^4*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*(383*cos(d*x + c)^4 - 34*cos(d*x + c)^3 - 132*cos 
(d*x + c)^2 + 5*cos(d*x + c) + 35)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) 
)/(d*cos(d*x + c)^4), 1/315*(315*sqrt(-a)*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 
)^4 + 2*(383*cos(d*x + c)^4 - 34*cos(d*x + c)^3 - 132*cos(d*x + c)^2 + 5*c 
os(d*x + c) + 35)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c) 
^4)]
 

Sympy [F]

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

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

Output:

Integral(sqrt(a*(sec(c + d*x) + 1))*tan(c + d*x)**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^5(c+d x) \, dx=\frac {315 \, \sqrt {a} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 630 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \frac {70 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {9}{2}}}{a^{4}} - \frac {270 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a^{3}} + \frac {126 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a^{2}} + \frac {210 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a}}{315 \, d} \] Input:

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

Output:

1/315*(315*sqrt(a)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/co 
s(d*x + c)) + sqrt(a))) + 630*sqrt(a + a/cos(d*x + c)) + 70*(a + a/cos(d*x 
 + c))^(9/2)/a^4 - 270*(a + a/cos(d*x + c))^(7/2)/a^3 + 126*(a + a/cos(d*x 
 + c))^(5/2)/a^2 + 210*(a + a/cos(d*x + c))^(3/2)/a)/d
 

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.31 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^5(c+d x) \, dx=\frac {\sqrt {2} {\left (\frac {315 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} a - 210 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} a^{2} + 252 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a^{3} + 1080 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{4} + 560 \, a^{5}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{315 \, d} \] Input:

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

Output:

1/315*sqrt(2)*(315*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*(315*(a*tan(1/2*d*x + 1/2*c)^2 - a)^4*a - 
 210*(a*tan(1/2*d*x + 1/2*c)^2 - a)^3*a^2 + 252*(a*tan(1/2*d*x + 1/2*c)^2 
- a)^2*a^3 + 1080*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^4 + 560*a^5)/((a*tan(1/ 
2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))*sgn(cos(d*x 
+ c))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {a+a \sec (c+d x)} \tan ^5(c+d x) \, dx=\frac {\sqrt {a}\, \left (10 \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{4}-16 \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{2}+64 \sqrt {\sec \left (d x +c \right )+1}+5 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{5}}{\sec \left (d x +c \right )+1}d x \right ) d -8 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )+1}d x \right ) d +32 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) d \right )}{45 d} \] Input:

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

Output:

(sqrt(a)*(10*sqrt(sec(c + d*x) + 1)*tan(c + d*x)**4 - 16*sqrt(sec(c + d*x) 
 + 1)*tan(c + d*x)**2 + 64*sqrt(sec(c + d*x) + 1) + 5*int((sqrt(sec(c + d* 
x) + 1)*tan(c + d*x)**5)/(sec(c + d*x) + 1),x)*d - 8*int((sqrt(sec(c + d*x 
) + 1)*tan(c + d*x)**3)/(sec(c + d*x) + 1),x)*d + 32*int((sqrt(sec(c + d*x 
) + 1)*tan(c + d*x))/(sec(c + d*x) + 1),x)*d))/(45*d)