\(\int \frac {\tan ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\) [172]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {2 \sqrt {a+a \sec (c+d x)}}{a d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a^2 d} \] 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
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 \left (-2-\sec (c+d x)+\sec ^2(c+d x)+3 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \sqrt {1+\sec (c+d x)}\right )}{3 d \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[Tan[c + d*x]^3/Sqrt[a + a*Sec[c + d*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 25, 4368, 25, 27, 90, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4368

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 220

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

Input:

Int[Tan[c + d*x]^3/Sqrt[a + a*Sec[c + d*x]],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))/(3*a))/(a*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 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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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 = 0.75 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\left (-2 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\frac {4}{3}+\frac {2 \sec \left (d x +c \right )}{3}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d a}\)	\(78\)

Input:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 241, normalized size of antiderivative = 3.09 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {3 \, \sqrt {a} \cos \left (d x + c\right ) \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (2 \, \cos \left (d x + c\right ) - 1\right )}}{6 \, a d \cos \left (d x + c\right )}, -\frac {3 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (2 \, \cos \left (d x + c\right ) - 1\right )}}{3 \, a d \cos \left (d x + c\right )}\right ] \] Input:

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

[1/6*(3*sqrt(a)*cos(d*x + c)*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*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(2*cos(d*x + c) 
 - 1))/(a*d*cos(d*x + c)), -1/3*(3*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*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(2*cos(d*x + c) - 1))/(a*d 
*cos(d*x + c))]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {3 \, \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {2 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a^{2}} + \frac {6 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{a}}{3 \, d} \] Input:

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.62 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a + 2 \, a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )}}{3 \, a d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(tan(c + d*x)^3/(a + a/cos(c + d*x))^(1/2),x)
 

Output:

int(tan(c + d*x)^3/(a + a/cos(c + d*x))^(1/2), x)
 

Reduce [F]

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

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

Output:

(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*tan(c + d*x)**3)/(sec(c + d*x) + 1),x 
))/a