\(\int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx\) [562]

Optimal result

Integrand size = 31, antiderivative size = 175 \[ \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c} \] Output:

-1/2*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^(1/2)+14/15*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/ 
5*sec(2*b*x+2*a)^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/15*(-c+c 
*sec(2*b*x+2*a))^(1/2)*tan(2*b*x+2*a)/b/c
 

Mathematica [A] (verified)

Time = 1.61 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {\cos (a+b x) \sec ^3(2 (a+b x)) \sin (a+b x) \left (38+4 \cos (2 (a+b x))+26 \cos (4 (a+b x))+30 \arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \cos ^2(2 (a+b x)) \sqrt {-1+\tan ^2(a+b x)}\right )}{30 b \sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \] Input:

Integrate[Sec[2*(a + b*x)]^4/Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]
 

Output:

(Cos[a + b*x]*Sec[2*(a + b*x)]^3*Sin[a + b*x]*(38 + 4*Cos[2*(a + b*x)] + 2 
6*Cos[4*(a + b*x)] + 30*ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Cos[2*(a + b*x)] 
^2*Sqrt[-1 + Tan[a + b*x]^2]))/(30*b*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]] 
)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4897, 3042, 4309, 3042, 4498, 27, 3042, 4489, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4897

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4309

\(\displaystyle \frac {\int \frac {\sec ^2(2 a+2 b x) (\sec (2 a+2 b x) c+4 c)}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )^2 \left (\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+4 c\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 4498

\(\displaystyle \frac {\frac {2 \int \frac {\sec (2 a+2 b x) \left (14 \sec (2 a+2 b x) c^2+c^2\right )}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (2 a+2 b x) \left (14 \sec (2 a+2 b x) c^2+c^2\right )}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \left (14 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c^2+c^2\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {15 c^2 \int \frac {\sec (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx+\frac {14 c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {15 c^2 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx+\frac {14 c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {\frac {14 c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {15 c^2 \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-2 c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {\frac {\frac {14 c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {15 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b}}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}\)

Input:

Int[Sec[2*(a + b*x)]^4/Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]
 

Output:

(Sec[2*a + 2*b*x]^2*Tan[2*a + 2*b*x])/(5*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) 
+ ((Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(3*b) + ((-15*c^(3/2)* 
ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]]) 
])/(Sqrt[2]*b) + (14*c^2*Tan[2*a + 2*b*x])/(b*Sqrt[-c + c*Sec[2*a + 2*b*x] 
]))/(3*c))/(5*c)
 

Defintions of rubi rules used

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 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 4309

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[-2*d^2*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 2)/( 
f*(2*n - 3)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[d^2/(b*(2*n - 3))   Int[( 
d*Csc[e + f*x])^(n - 2)*((2*b*(n - 2) - a*Csc[e + f*x])/Sqrt[a + b*Csc[e + 
f*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 
2] && IntegerQ[2*n]
 

rule 4489

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 
 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B 
, e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b 
*(m + 1), 0] &&  !LtQ[m, -2^(-1)]
 

rule 4498

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( 
csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* 
((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int 
[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) 
*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a 
*B, 0] &&  !LtQ[m, -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. \(434\) vs. \(2(154)=308\).

Time = 1.66 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.49

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {4}\, \sin \left (b x +a \right ) \left (\cos \left (b x +a \right ) \left (208 \cos \left (b x +a \right )^{5}+208 \cos \left (b x +a \right )^{4}-200 \cos \left (b x +a \right )^{3}-200 \cos \left (b x +a \right )^{2}+60 \cos \left (b x +a \right )+60\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+\left (-120 \cos \left (b x +a \right )^{6}+180 \cos \left (b x +a \right )^{4}-90 \cos \left (b x +a \right )^{2}+15\right ) \operatorname {arctanh}\left (\frac {2 \cos \left (b x +a \right )-1}{\left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right )+\left (120 \cos \left (b x +a \right )^{6}-180 \cos \left (b x +a \right )^{4}+90 \cos \left (b x +a \right )^{2}-15\right ) \ln \left (\frac {2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cos \left (b x +a \right )+2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-4 \cos \left (b x +a \right )-2}{\cos \left (b x +a \right )+1}\right )\right )}{120 b \left (-3+2 \sqrt {2}\right )^{3} \left (3+2 \sqrt {2}\right )^{3} \left (8 \cos \left (b x +a \right )^{7}+8 \cos \left (b x +a \right )^{6}-12 \cos \left (b x +a \right )^{5}-12 \cos \left (b x +a \right )^{4}+6 \cos \left (b x +a \right )^{3}+6 \cos \left (b x +a \right )^{2}-\cos \left (b x +a \right )-1\right ) \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(435\)

Input:

int(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

-1/120*2^(1/2)/b*4^(1/2)/(-3+2*2^(1/2))^3/(3+2*2^(1/2))^3*sin(b*x+a)*(cos( 
b*x+a)*(208*cos(b*x+a)^5+208*cos(b*x+a)^4-200*cos(b*x+a)^3-200*cos(b*x+a)^ 
2+60*cos(b*x+a)+60)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+(-120*cos( 
b*x+a)^6+180*cos(b*x+a)^4-90*cos(b*x+a)^2+15)*arctanh((2*cos(b*x+a)-1)/(co 
s(b*x+a)+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))+(120*cos(b*x+a)^6 
-180*cos(b*x+a)^4+90*cos(b*x+a)^2-15)*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)))/(8*cos(b*x+a)^7+8*cos(b*x+a)^6-12*cos(b*x+a)^5 
-12*cos(b*x+a)^4+6*cos(b*x+a)^3+6*cos(b*x+a)^2-cos(b*x+a)-1)/(c*sin(b*x+a) 
^2/(2*cos(b*x+a)^2-1))^(1/2)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.17 \[ \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\left [\frac {4 \, \sqrt {2} {\left (15 \, \tan \left (b x + a\right )^{4} - 20 \, \tan \left (b x + a\right )^{2} + 17\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} + \frac {15 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{5} - 2 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \log \left (\frac {\tan \left (b x + a\right )^{3} - \frac {2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{\sqrt {c}} - 2 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right )}{\sqrt {c}}}{60 \, {\left (b c \tan \left (b x + a\right )^{5} - 2 \, b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}, -\frac {15 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{5} - 2 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {1}{c}}}{\tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (15 \, \tan \left (b x + a\right )^{4} - 20 \, \tan \left (b x + a\right )^{2} + 17\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{30 \, {\left (b c \tan \left (b x + a\right )^{5} - 2 \, b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}\right ] \] Input:

integrate(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorith 
m="fricas")
 

Output:

[1/60*(4*sqrt(2)*(15*tan(b*x + a)^4 - 20*tan(b*x + a)^2 + 17)*sqrt(-c*tan( 
b*x + a)^2/(tan(b*x + a)^2 - 1)) + 15*sqrt(2)*(c*tan(b*x + a)^5 - 2*c*tan( 
b*x + a)^3 + c*tan(b*x + a))*log((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*tan(b*x + a))/tan 
(b*x + a)^3)/sqrt(c))/(b*c*tan(b*x + a)^5 - 2*b*c*tan(b*x + a)^3 + b*c*tan 
(b*x + a)), -1/30*(15*sqrt(2)*(c*tan(b*x + a)^5 - 2*c*tan(b*x + a)^3 + c*t 
an(b*x + a))*sqrt(-1/c)*arctan(sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1) 
)*(tan(b*x + a)^2 - 1)*sqrt(-1/c)/tan(b*x + a)) - 2*sqrt(2)*(15*tan(b*x + 
a)^4 - 20*tan(b*x + a)^2 + 17)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1) 
))/(b*c*tan(b*x + a)^5 - 2*b*c*tan(b*x + a)^3 + b*c*tan(b*x + a))]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \] Input:

integrate(sec(2*b*x+2*a)**4/(c*tan(b*x+a)*tan(2*b*x+2*a))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int { \frac {\sec \left (2 \, b x + 2 \, a\right )^{4}}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}} \,d x } \] Input:

integrate(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorith 
m="maxima")
 

Output:

integrate(sec(2*b*x + 2*a)^4/sqrt(c*tan(2*b*x + 2*a)*tan(b*x + a)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorith 
m="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^4\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )}} \,d x \] Input:

int(1/(cos(2*a + 2*b*x)^4*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2)),x)
                                                                                    
                                                                                    
 

Output:

int(1/(cos(2*a + 2*b*x)^4*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {\sqrt {c}\, \left (\int \frac {\sqrt {\tan \left (b x +a \right )}\, \sqrt {\tan \left (2 b x +2 a \right )}\, \sec \left (2 b x +2 a \right )^{4}}{\tan \left (2 b x +2 a \right ) \tan \left (b x +a \right )}d x \right )}{c} \] Input:

int(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x)
 

Output:

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