\(\int \frac {\sec ^2(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\) [571]

Optimal result

Integrand size = 31, antiderivative size = 93 \[ \int \frac {\sec ^2(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}} \] Output:

-3/8*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 = 1.42 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^2(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\frac {\left (-1+3 \arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sec (2 (a+b x)) \sin ^2(a+b x) \sqrt {-1+\tan ^2(a+b x)}\right ) \tan (2 (a+b x))}{4 b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3042, 4897, 3042, 4284, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4897

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4284

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4282

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

(-3*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 4284

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^m/(f*(2*m + 1))), x 
] + Simp[m/(b*(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)]
 

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. \(339\) vs. \(2(78)=156\).

Time = 1.23 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.66

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {4}\, \left (1-\cos \left (b x +a \right )\right ) \left (2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \left (1-\cos \left (b x +a \right )\right )^{2} \csc \left (b x +a \right )^{2}-6 \,\operatorname {arctanh}\left (\frac {2 \cos \left (b x +a \right )-1}{\left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \left (1-\cos \left (b x +a \right )\right )^{2} \csc \left (b x +a \right )^{2}+6 \ln \left (\frac {2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cos \left (b x +a \right )+2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-4 \cos \left (b x +a \right )-2}{\cos \left (b x +a \right )+1}\right ) \left (1-\cos \left (b x +a \right )\right )^{2} \csc \left (b x +a \right )^{2}-2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \csc \left (b x +a \right )^{3}}{64 b c \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(340\)

Input:

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

Output:

1/64*2^(1/2)/b*4^(1/2)/c/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(1-cos( 
b*x+a))*(2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*(1-cos(b*x+a))^2*cs 
c(b*x+a)^2-6*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)^2+6*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 
-2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))*(cos(b*x+a)+1)^2/((2*cos(b 
*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*csc(b*x+a)^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.92 \[ \int \frac {\sec ^2(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right )^{3} + 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{16 \, b c^{2} \tan \left (b x + a\right )^{3}}, -\frac {3 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} - \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{8 \, b c^{2} \tan \left (b x + a\right )^{3}}\right ] \] Input:

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

Output:

[1/16*(3*sqrt(2)*sqrt(c)*log((c*tan(b*x + a)^3 - 2*sqrt(-c*tan(b*x + a)^2/ 
(tan(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*(3*sqrt(2)*s 
qrt(-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. \[ \int \frac {\sec ^2(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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