\(\int (a+a \sec (c+d x))^{3/2} \tan ^3(c+d x) \, dx\) [148]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^3(c+d x) \, dx=\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 a \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^3(c+d x) \, dx=\frac {2 (a (1+\sec (c+d x)))^{3/2} \left (105 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} \left (-146-32 \sec (c+d x)+24 \sec ^2(c+d x)+15 \sec ^3(c+d x)\right )\right )}{105 d (1+\sec (c+d x))^{3/2}} \] Input:

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

Output:

(2*(a*(1 + Sec[c + d*x]))^(3/2)*(105*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqr 
t[1 + Sec[c + d*x]]*(-146 - 32*Sec[c + d*x] + 24*Sec[c + d*x]^2 + 15*Sec[c 
 + d*x]^3)))/(105*d*(1 + Sec[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 25, 4368, 25, 27, 90, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4368

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

\(\displaystyle -\frac {a \left (a \left (2 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}+2 \sqrt {a \sec (c+d x)+a}\right )+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}\right )-\frac {2 (a \sec (c+d x)+a)^{7/2}}{7 a}+\frac {2}{5} (a \sec (c+d x)+a)^{5/2}}{a d}\)

\(\Big \downarrow \) 220

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

Input:

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

Output:

-(((2*(a + a*Sec[c + d*x])^(5/2))/5 - (2*(a + a*Sec[c + d*x])^(7/2))/(7*a) 
 + 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]])))/(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 = 1.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\left (-2 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\frac {292}{105}-\frac {64 \sec \left (d x +c \right )}{105}+\frac {16 \sec \left (d x +c \right )^{2}}{35}+\frac {2 \sec \left (d x +c \right )^{3}}{7}\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d}\)	\(96\)

Input:

int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^3,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)-292/105-64/105*sec(d*x+c)+16/35*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 = 290, normalized size of antiderivative = 2.40 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^3(c+d x) \, dx=\left [\frac {105 \, a^{\frac {3}{2}} \cos \left (d x + c\right )^{3} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, {\left (146 \, a \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} - 24 \, a \cos \left (d x + c\right ) - 15 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{210 \, d \cos \left (d x + c\right )^{3}}, -\frac {105 \, \sqrt {-a} a \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{3} + 2 \, {\left (146 \, a \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} - 24 \, a \cos \left (d x + c\right ) - 15 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )^{3}}\right ] \] Input:

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

Output:

[1/210*(105*a^(3/2)*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*(146*a*cos(d*x + c)^3 + 32*a*cos(d*x + c)^2 - 24* 
a*cos(d*x + c) - 15*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x 
 + c)^3), -1/105*(105*sqrt(-a)*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*( 
146*a*cos(d*x + c)^3 + 32*a*cos(d*x + c)^2 - 24*a*cos(d*x + c) - 15*a)*sqr 
t((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^3(c+d x) \, dx=-\frac {105 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 70 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} - \frac {30 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a^{2}} + \frac {42 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a} + 210 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a}{105 \, d} \] Input:

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

Output:

-1/105*(105*a^(3/2)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/c 
os(d*x + c)) + sqrt(a))) + 70*(a + a/cos(d*x + c))^(3/2) - 30*(a + a/cos(d 
*x + c))^(7/2)/a^2 + 42*(a + a/cos(d*x + c))^(5/2)/a + 210*sqrt(a + a/cos( 
d*x + c))*a)/d
 

Giac [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.43 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^3(c+d x) \, dx=-\frac {\sqrt {2} {\left (\frac {105 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} a^{2} - 70 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a^{3} + 84 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{4} + 120 \, a^{5}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{105 \, d} \] Input:

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

Output:

-1/105*sqrt(2)*(105*sqrt(2)*a^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* 
a^2 - 70*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*a^3 + 84*(a*tan(1/2*d*x + 1/2*c) 
^2 - a)*a^4 + 120*a^5)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d 
*x + 1/2*c)^2 + a)))*sgn(cos(d*x + c))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+a \sec (c+d x))^{3/2} \tan ^3(c+d x) \, dx=\frac {\sqrt {a}\, a \left (30 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}-40 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )+48 \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{2}-232 \sqrt {\sec \left (d x +c \right )+1}+9 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )+1}d x \right ) d -96 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) d \right )}{105 d} \] Input:

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

Output:

(sqrt(a)*a*(30*sqrt(sec(c + d*x) + 1)*sec(c + d*x)*tan(c + d*x)**2 - 40*sq 
rt(sec(c + d*x) + 1)*sec(c + d*x) + 48*sqrt(sec(c + d*x) + 1)*tan(c + d*x) 
**2 - 232*sqrt(sec(c + d*x) + 1) + 9*int((sqrt(sec(c + d*x) + 1)*tan(c + d 
*x)**3)/(sec(c + d*x) + 1),x)*d - 96*int((sqrt(sec(c + d*x) + 1)*tan(c + d 
*x))/(sec(c + d*x) + 1),x)*d))/(105*d)