\(\int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [183]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \sqrt {a+a \sec (c+d x)}}{a^2 d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{a^3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (1-2 \sec (c+d x)-2 \sec ^2(c+d x)+\sec ^3(c+d x)-5 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \sqrt {1+\sec (c+d x)}\right )}{5 a d \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[Tan[c + d*x]^5/(a + a*Sec[c + d*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.92, 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.

\(\displaystyle \int \frac {\tan ^5(c+d x)}{(a \sec (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4368

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

Int[Tan[c + d*x]^5/(a + a*Sec[c + d*x])^(3/2),x]
 

Output:

(-2*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]] + 2*Sqrt[a + a*Sec[c 
 + d*x]] - (2*(a + a*Sec[c + d*x])^(3/2))/a + (2*(a + a*Sec[c + d*x])^(5/2 
))/(5*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 = 1.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {\left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\frac {2}{5}-\frac {6 \sec \left (d x +c \right )}{5}+\frac {2 \sec \left (d x +c \right )^{2}}{5}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{2}}\)	\(88\)

Input:

int(tan(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),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)+2/5-6/5*sec(d*x+c)+2/5*sec(d*x+c)^2)/a^2*(a*(1+se 
c(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.61 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {5 \, \sqrt {a} \cos \left (d x + c\right )^{2} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{10 \, a^{2} d \cos \left (d x + c\right )^{2}}, \frac {5 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{2} + 2 \, {\left (\cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{5 \, a^{2} d \cos \left (d x + c\right )^{2}}\right ] \] Input:

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

[1/10*(5*sqrt(a)*cos(d*x + c)^2*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*(cos(d*x + c)^2 - 3*cos(d*x + c) + 1)*sqrt((a*cos(d* 
x + c) + a)/cos(d*x + c)))/(a^2*d*cos(d*x + c)^2), 1/5*(5*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)^2 + 2*(cos(d*x + c)^2 - 3*cos(d*x + c) + 1)*sqr 
t((a*cos(d*x + c) + a)/cos(d*x + c)))/(a^2*d*cos(d*x + c)^2)]
 

Sympy [F]

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

integrate(tan(d*x+c)**5/(a+a*sec(d*x+c))**(3/2),x)
 

Output:

Integral(tan(c + d*x)**5/(a*(sec(c + d*x) + 1))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {5 \, \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a^{4}} - \frac {10 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a^{3}} + \frac {10 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{a^{2}}}{5 \, d} \] Input:

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

1/5*(5*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) 
+ sqrt(a)))/a^(3/2) + 2*(a + a/cos(d*x + c))^(5/2)/a^4 - 10*(a + a/cos(d*x 
 + c))^(3/2)/a^3 + 10*sqrt(a + a/cos(d*x + c))/a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.54 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (\frac {5 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {\sqrt {2} {\left (5 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} + 10 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a + 4 \, a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )}}{5 \, d} \] Input:

integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

2/5*(5*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(s 
qrt(-a)*a*sgn(cos(d*x + c))) + sqrt(2)*(5*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2 
 + 10*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a + 4*a^2)/((a*tan(1/2*d*x + 1/2*c)^2 
 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(tan(c + d*x)^5/(a + a/cos(c + d*x))^(3/2),x)
 

Output:

int(tan(c + d*x)^5/(a + a/cos(c + d*x))^(3/2), x)
 

Reduce [F]

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

int(tan(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),x)
 

Output:

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