\(\int \frac {\sec (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\) [572]

Optimal result

Integrand size = 29, antiderivative size = 93 \[ \int \frac {\sec (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}} \] Output:

1/8*arctanh(1/2*c^(1/2)*tan(2*b*x+2*a)*2^(1/2)/(-c+c*sec(2*b*x+2*a))^(1/2) 
)*2^(1/2)/b/c^(3/2)-1/4*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(3/2)
 

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {\left (1+\arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sec (2 (a+b x)) \sin ^2(a+b x) \sqrt {-1+\tan ^2(a+b x)}\right ) \tan (2 (a+b x))}{4 b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \] Input:

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

Output:

-1/4*((1 + ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sec[2*(a + b*x)]*Sin[a + b*x] 
^2*Sqrt[-1 + Tan[a + b*x]^2])*Tan[2*(a + b*x)])/(b*(c*Tan[a + b*x]*Tan[2*( 
a + b*x)])^(3/2))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 4897, 3042, 4283, 3042, 4282, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4897

\(\displaystyle \int \frac {\sec (2 a+2 b x)}{(c \sec (2 a+2 b x)-c)^{3/2}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4283

\(\displaystyle -\frac {\int \frac {\sec (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{4 c}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{4 c}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-2 c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 b c}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\)

Input:

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

Output:

ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]]) 
]/(4*Sqrt[2]*b*c^(3/2)) - Tan[2*a + 2*b*x]/(4*b*(-c + c*Sec[2*a + 2*b*x])^ 
(3/2))
 

Defintions of rubi rules used

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 4282

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4283

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] 
 + Simp[(m + 1)/(a*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 
1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) 
] && IntegerQ[2*m]
 

rule 4897

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(392\) vs. \(2(78)=156\).

Time = 1.46 (sec) , antiderivative size = 393, normalized size of antiderivative = 4.23

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {4}\, \left (-2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \left (1-\cos \left (b x +a \right )\right )^{4} \csc \left (b x +a \right )^{3}+2 \ln \left (\frac {2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cos \left (b x +a \right )+2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-4 \cos \left (b x +a \right )-2}{\cos \left (b x +a \right )+1}\right ) \left (1-\cos \left (b x +a \right )\right )^{2} \csc \left (b x +a \right )-2 \,\operatorname {arctanh}\left (\frac {2 \cos \left (b x +a \right )-1}{\left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \left (1-\cos \left (b x +a \right )\right )^{2} \csc \left (b x +a \right )+\frac {8 \left (2 \cos \left (b x +a \right )^{2}-1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}+10 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \left (1-\cos \left (b x +a \right )\right )^{2} \csc \left (b x +a \right )\right )}{64 c b \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \left (1-\cos \left (b x +a \right )\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(393\)

Input:

int(sec(2*b*x+2*a)/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x,method=_RETURNVER 
BOSE)
 

Output:

-1/64*2^(1/2)/c/b*4^(1/2)/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)/(1-cos 
(b*x+a))/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*(-2*((2*cos(b*x+a)^2- 
1)/(cos(b*x+a)+1)^2)^(1/2)*(1-cos(b*x+a))^4*csc(b*x+a)^3+2*ln(2*(((2*cos(b 
*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)+((2*cos(b*x+a)^2-1)/(cos(b*x 
+a)+1)^2)^(1/2)-2*cos(b*x+a)-1)/(cos(b*x+a)+1))*(1-cos(b*x+a))^2*csc(b*x+a 
)-2*arctanh((2*cos(b*x+a)-1)/(cos(b*x+a)+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a 
)+1)^2)^(1/2))*(1-cos(b*x+a))^2*csc(b*x+a)+8*(2*cos(b*x+a)^2-1)/(cos(b*x+a 
)+1)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*sin(b*x+a)+10*((2*cos(b 
*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*(1-cos(b*x+a))^2*csc(b*x+a))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.89 \[ \int \frac {\sec (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\left [\frac {\sqrt {2} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} + 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right )^{3} + 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{16 \, b c^{2} \tan \left (b x + a\right )^{3}}, \frac {\sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} + \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{8 \, b c^{2} \tan \left (b x + a\right )^{3}}\right ] \] Input:

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

Output:

[1/16*(sqrt(2)*sqrt(c)*log((c*tan(b*x + a)^3 + 2*sqrt(-c*tan(b*x + a)^2/(t 
an(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(c) - 2*c*tan(b*x + a))/tan(b 
*x + a)^3)*tan(b*x + a)^3 + 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a) 
^2 - 1))*(tan(b*x + a)^2 - 1))/(b*c^2*tan(b*x + a)^3), 1/8*(sqrt(2)*sqrt(- 
c)*arctan(sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1 
)*sqrt(-c)/(c*tan(b*x + a)))*tan(b*x + a)^3 + sqrt(2)*sqrt(-c*tan(b*x + a) 
^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1))/(b*c^2*tan(b*x + a)^3)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{%%{poly1[16,0]:[1,0,-2]%%},[2,8]%%%}+%%%{%%{poly1[64,0 
]:[1,0,-2
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*int((sqrt(tan(a + b*x))*sqrt(tan(2*a + 2*b*x))*sec(2*a + 2*b*x))/ 
(tan(2*a + 2*b*x)**2*tan(a + b*x)**2),x))/c**2