\(\int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\) [546]

Optimal result

Integrand size = 31, antiderivative size = 157 \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {2 c \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c \sec ^3(2 a+2 b x) \tan (2 a+2 b x)}{7 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {4 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{35 b}-\frac {6 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b c} \] Output:

-2/5*c*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/7*c*sec(2*b*x+2*a)^3 
*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)-4/35*(-c+c*sec(2*b*x+2*a))^( 
1/2)*tan(2*b*x+2*a)/b-6/35*(-c+c*sec(2*b*x+2*a))^(3/2)*tan(2*b*x+2*a)/b/c
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.41 \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {(7 \cos (3 (a+b x))+2 \cos (7 (a+b x))) \csc (a+b x) \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{35 b} \] Input:

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

Output:

-1/35*((7*Cos[3*(a + b*x)] + 2*Cos[7*(a + b*x)])*Csc[a + b*x]*Sec[2*(a + b 
*x)]^3*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/b
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4897, 3042, 4290, 3042, 4287, 27, 3042, 4489, 3042, 4279}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4897

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4290

\(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \int \sec ^3(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^3 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\)

\(\Big \downarrow \) 4287

\(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {2 \int \frac {1}{2} \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c} \left (2 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+3 c\right )dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7}{3} c \int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7}{3} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\)

Input:

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

Output:

(c*Sec[2*a + 2*b*x]^3*Tan[2*a + 2*b*x])/(7*b*Sqrt[-c + c*Sec[2*a + 2*b*x]] 
) - (6*(((-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5*b*c) + ((7*c 
^2*Tan[2*a + 2*b*x])/(3*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (2*c*Sqrt[-c + 
c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(3*b))/(5*c)))/7
 

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4279

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

rule 4287

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-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*(b*( 
m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - 
b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 4290

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( 
f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 
1)))   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre 
eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && 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 4897

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

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.57

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {4}\, \left (128 \cos \left (b x +a \right )^{6}-224 \cos \left (b x +a \right )^{4}+140 \cos \left (b x +a \right )^{2}-35\right ) \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \cot \left (b x +a \right )}{70 b \left (2 \cos \left (b x +a \right )^{2}-1\right )^{3}}\)	\(90\)

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:

-1/70*2^(1/2)/b*4^(1/2)*(128*cos(b*x+a)^6-224*cos(b*x+a)^4+140*cos(b*x+a)^ 
2-35)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)/(2*cos(b*x+a)^2-1)^3*cot(b 
*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.68 \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {\sqrt {2} {\left (35 \, \tan \left (b x + a\right )^{6} - 35 \, \tan \left (b x + a\right )^{4} + 49 \, \tan \left (b x + a\right )^{2} - 9\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{35 \, {\left (b \tan \left (b x + a\right )^{7} - 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} - b \tan \left (b x + a\right )\right )}} \] 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:

-1/35*sqrt(2)*(35*tan(b*x + a)^6 - 35*tan(b*x + a)^4 + 49*tan(b*x + a)^2 - 
 9)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^7 - 3*b*t 
an(b*x + a)^5 + 3*b*tan(b*x + a)^3 - b*tan(b*x + a))
 

Sympy [F(-1)]

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

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

-8/35*(70*(b*cos(4*b*x + 4*a)^2 + b*sin(4*b*x + 4*a)^2 + 2*b*cos(4*b*x + 4 
*a) + b)*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1 
)^(3/4)*sqrt(c)*integrate(-(((cos(20*b*x + 20*a)*cos(4*b*x + 4*a) + 4*cos( 
16*b*x + 16*a)*cos(4*b*x + 4*a) + 6*cos(12*b*x + 12*a)*cos(4*b*x + 4*a) + 
4*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(20*b*x + 20 
*a)*sin(4*b*x + 4*a) + 4*sin(16*b*x + 16*a)*sin(4*b*x + 4*a) + 6*sin(12*b* 
x + 12*a)*sin(4*b*x + 4*a) + 4*sin(8*b*x + 8*a)*sin(4*b*x + 4*a) + sin(4*b 
*x + 4*a)^2)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + ( 
cos(4*b*x + 4*a)*sin(20*b*x + 20*a) + 4*cos(4*b*x + 4*a)*sin(16*b*x + 16*a 
) + 6*cos(4*b*x + 4*a)*sin(12*b*x + 12*a) + 4*cos(4*b*x + 4*a)*sin(8*b*x + 
 8*a) - cos(20*b*x + 20*a)*sin(4*b*x + 4*a) - 4*cos(16*b*x + 16*a)*sin(4*b 
*x + 4*a) - 6*cos(12*b*x + 12*a)*sin(4*b*x + 4*a) - 4*cos(8*b*x + 8*a)*sin 
(4*b*x + 4*a))*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))* 
cos(5/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))) + ((cos(4*b*x + 4*a)* 
sin(20*b*x + 20*a) + 4*cos(4*b*x + 4*a)*sin(16*b*x + 16*a) + 6*cos(4*b*x + 
 4*a)*sin(12*b*x + 12*a) + 4*cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(20*b* 
x + 20*a)*sin(4*b*x + 4*a) - 4*cos(16*b*x + 16*a)*sin(4*b*x + 4*a) - 6*cos 
(12*b*x + 12*a)*sin(4*b*x + 4*a) - 4*cos(8*b*x + 8*a)*sin(4*b*x + 4*a))*co 
s(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) - (cos(20*b*x + 20 
*a)*cos(4*b*x + 4*a) + 4*cos(16*b*x + 16*a)*cos(4*b*x + 4*a) + 6*cos(12...
 

Giac [F(-2)]

Exception generated. \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Exception raised: TypeError} \] 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 [B] (verification not implemented)

Time = 22.30 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.95 \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx =\text {Too large to display} \] Input:

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

Output:

((8i/(7*b) - (exp(a*2i + b*x*2i)*8i)/(7*b))*((c*(exp(a*2i + b*x*2i)*1i - 1 
i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x 
*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 1)^3) - 
 (exp(a*2i + b*x*2i)*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)* 
1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2)*16i)/ 
(35*b*(exp(a*2i + b*x*2i) - 1)) - ((8i/(5*b) - (exp(a*2i + b*x*2i)*64i)/(3 
5*b))*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp 
(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i 
) - 1)*(exp(a*4i + b*x*4i) + 1)^2) - (exp(a*2i + b*x*2i)*((c*(exp(a*2i + b 
*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(e 
xp(a*4i + b*x*4i) + 1)))^(1/2)*8i)/(35*b*(exp(a*2i + b*x*2i) - 1)*(exp(a*4 
i + b*x*4i) + 1))
 

Reduce [F]

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