\(\int \frac {\tan ^5(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\) [171]

Optimal result

Integrand size = 23, antiderivative size = 126 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {2 \sqrt {a+a \sec (c+d x)}}{a d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a^2 d}-\frac {6 (a+a \sec (c+d x))^{5/2}}{5 a^3 d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.70 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 \left (92+46 \sec (c+d x)-64 \sec ^2(c+d x)-3 \sec ^3(c+d x)+15 \sec ^4(c+d x)-105 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \sqrt {1+\sec (c+d x)}\right )}{105 d \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(2*(92 + 46*Sec[c + d*x] - 64*Sec[c + d*x]^2 - 3*Sec[c + d*x]^3 + 15*Sec[c 
 + d*x]^4 - 105*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*Sqrt[1 + Sec[c + d*x]]))/( 
105*d*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4368

\(\displaystyle \frac {\int a^2 \cos (c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{3/2}d\sec (c+d x)}{a^4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos (c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{3/2}d\sec (c+d x)}{a^2 d}\)

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-2*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]] + 2*a*Sqrt[a + a*Sec 
[c + d*x]] + (2*(a + a*Sec[c + d*x])^(3/2))/3 - (6*(a + a*Sec[c + d*x])^(5 
/2))/(5*a) + (2*(a + a*Sec[c + d*x])^(7/2))/(7*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 = 0.83 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {\left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\frac {184}{105}-\frac {92 \sec \left (d x +c \right )}{105}-\frac {12 \sec \left (d x +c \right )^{2}}{35}+\frac {2 \sec \left (d x +c \right )^{3}}{7}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d a}\)	\(98\)

Input:

int(tan(d*x+c)^5/(a+a*sec(d*x+c))^(1/2),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)+184/105-92/105*sec(d*x+c)-12/35*sec(d*x+c)^2+2/7* 
sec(d*x+c)^3)/a*(a*(1+sec(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.26 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {105 \, \sqrt {a} \cos \left (d x + c\right )^{3} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (92 \, \cos \left (d x + c\right )^{3} - 46 \, \cos \left (d x + c\right )^{2} - 18 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{210 \, a d \cos \left (d x + c\right )^{3}}, \frac {105 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{3} + 2 \, {\left (92 \, \cos \left (d x + c\right )^{3} - 46 \, \cos \left (d x + c\right )^{2} - 18 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{105 \, a d \cos \left (d x + c\right )^{3}}\right ] \] Input:

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

Output:

[1/210*(105*sqrt(a)*cos(d*x + c)^3*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*(92*cos(d*x + c)^3 - 46*cos(d*x + c)^2 - 18*cos(d 
*x + c) + 15)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a*d*cos(d*x + c)^3 
), 1/105*(105*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)^3 + 2*(92*cos(d*x 
 + c)^3 - 46*cos(d*x + c)^2 - 18*cos(d*x + c) + 15)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c)))/(a*d*cos(d*x + c)^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\frac {105 \, \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {30 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a^{4}} - \frac {126 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a^{3}} + \frac {70 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a^{2}} + \frac {210 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{a}}{105 \, d} \] Input:

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

Output:

1/105*(105*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + 
c)) + sqrt(a)))/sqrt(a) + 30*(a + a/cos(d*x + c))^(7/2)/a^4 - 126*(a + a/c 
os(d*x + c))^(5/2)/a^3 + 70*(a + a/cos(d*x + c))^(3/2)/a^2 + 210*sqrt(a + 
a/cos(d*x + c))/a)/d
 

Giac [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.33 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} {\left (\frac {105 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} - 70 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a - 252 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{2} - 120 \, a^{3}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )}}{105 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

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

Output:

1/105*sqrt(2)*(105*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c) 
^2 + a)/sqrt(-a))/sqrt(-a) + 2*(105*(a*tan(1/2*d*x + 1/2*c)^2 - a)^3 - 70* 
(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*a - 252*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^ 
2 - 120*a^3)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c 
)^2 + a)))/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

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

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

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

Output:

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