\(\int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx\) [161]

Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 a (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 d} \] 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
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\frac {2 (a (1+\sec (c+d x)))^{5/2} \left (-15 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} \left (23+11 \sec (c+d x)+3 \sec ^2(c+d x)\right )\right )}{15 d (1+\sec (c+d x))^{5/2}} \] Input:

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

Output:

(2*(a*(1 + Sec[c + d*x]))^(5/2)*(-15*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqr 
t[1 + Sec[c + d*x]]*(23 + 11*Sec[c + d*x] + 3*Sec[c + d*x]^2)))/(15*d*(1 + 
 Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 25, 4368, 60, 60, 60, 73, 220}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4368

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

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 = 6.83 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {a +a \sec \left (d x +c \right )}\, a^{2}-2 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\)	\(74\)
default	\(\frac {\frac {2 \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {a +a \sec \left (d x +c \right )}\, a^{2}-2 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d}\)	\(74\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.91 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\left [\frac {15 \, a^{\frac {5}{2}} \cos \left (d x + c\right )^{2} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (23 \, a^{2} \cos \left (d x + c\right )^{2} + 11 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{30 \, d \cos \left (d x + c\right )^{2}}, \frac {15 \, \sqrt {-a} a^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{2} + 2 \, {\left (23 \, a^{2} \cos \left (d x + c\right )^{2} + 11 \, a^{2} \cos \left (d x + c\right ) + 3 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}}\right ] \] Input:

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

Output:

[1/30*(15*a^(5/2)*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*(23*a^2*cos(d*x + c)^2 + 11*a^2*cos(d*x + c) + 3*a^ 
2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2), 1/15*(15*s 
qrt(-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)^2 + 2*(23*a^2*cos(d*x + c)^2 
 + 11*a^2*cos(d*x + c) + 3*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/( 
d*cos(d*x + c)^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\frac {15 \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 6 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} + 10 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a + 30 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}}{15 \, d} \] Input:

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

Output:

1/15*(15*a^(5/2)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos( 
d*x + c)) + sqrt(a))) + 6*(a + a/cos(d*x + c))^(5/2) + 10*(a + a/cos(d*x + 
 c))^(3/2)*a + 30*sqrt(a + a/cos(d*x + c))*a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.53 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a - 10 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{2} + 12 \, a^{3}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{15 \, d} \] Input:

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

Output:

1/15*sqrt(2)*(15*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*(15*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*a - 10 
*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^2 + 12*a^3)/((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^2*sgn(cos(d*x + c))/d
 

Mupad [B] (verification not implemented)

Time = 12.91 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.95 \[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\frac {2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{5\,d}+\frac {2\,a\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{3\,d}+\frac {2\,a^2\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{d}+\frac {a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i}}{d} \] Input:

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

Output:

(2*(a + a/cos(c + d*x))^(5/2))/(5*d) + (a^(5/2)*atan(((a + a/cos(c + d*x)) 
^(1/2)*1i)/a^(1/2))*2i)/d + (2*a*(a + a/cos(c + d*x))^(3/2))/(3*d) + (2*a^ 
2*(a + a/cos(c + d*x))^(1/2))/d
 

Reduce [F]

\[ \int (a+a \sec (c+d x))^{5/2} \tan (c+d x) \, dx=\frac {\sqrt {a}\, a^{2} \left (6 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}+22 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )+46 \sqrt {\sec \left (d x +c \right )+1}+15 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) d \right )}{15 d} \] Input:

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

Output:

(sqrt(a)*a**2*(6*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2 + 22*sqrt(sec(c + 
d*x) + 1)*sec(c + d*x) + 46*sqrt(sec(c + d*x) + 1) + 15*int((sqrt(sec(c + 
d*x) + 1)*tan(c + d*x))/(sec(c + d*x) + 1),x)*d))/(15*d)