\(\int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) [554]

Optimal result

Integrand size = 31, antiderivative size = 208 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {34 c^2 \tan (2 a+2 b x)}{45 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {17 c^2 \sec ^3(2 a+2 b x) \tan (2 a+2 b x)}{63 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c^2 \sec ^4(2 a+2 b x) \tan (2 a+2 b x)}{9 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {68 c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{315 b}+\frac {34 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{105 b} \] Output:

34/45*c^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)-17/63*c^2*sec(2*b*x 
+2*a)^3*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/9*c^2*sec(2*b*x+2*a 
)^4*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+68/315*c*(-c+c*sec(2*b*x+ 
2*a))^(1/2)*tan(2*b*x+2*a)/b+34/105*(-c+c*sec(2*b*x+2*a))^(3/2)*tan(2*b*x+ 
2*a)/b
 

Mathematica [A] (verified)

Time = 1.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.41 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {\cot (a+b x) \left (-84+188 \cot (a+b x) \cot (2 (a+b x))+52 \sec (2 (a+b x))-50 \sec ^2(2 (a+b x))+35 \sec ^3(2 (a+b x))\right ) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}{315 b} \] Input:

Integrate[Sec[2*(a + b*x)]^4*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]
 

Output:

(Cot[a + b*x]*(-84 + 188*Cot[a + b*x]*Cot[2*(a + b*x)] + 52*Sec[2*(a + b*x 
)] - 50*Sec[2*(a + b*x)]^2 + 35*Sec[2*(a + b*x)]^3)*(c*Tan[a + b*x]*Tan[2* 
(a + b*x)])^(3/2))/(315*b)
 

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4897, 3042, 4301, 27, 2011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4897

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4301

\(\displaystyle \frac {2}{9} c \int \frac {17 \sec ^4(2 a+2 b x) (c-c \sec (2 a+2 b x))}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx+\frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {17}{9} c \int \frac {\sec ^4(2 a+2 b x) (c-c \sec (2 a+2 b x))}{\sqrt {c \sec (2 a+2 b x)-c}}dx+\frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \int \sec ^4(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^4 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \int \sec ^3(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^3 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\right )\)

\(\Big \downarrow \) 4287

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {2 \int \frac {1}{2} \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c} \left (2 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+3 c\right )dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7}{3} c \int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7}{3} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {17}{9} c \left (\frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\right )\)

Input:

Int[Sec[2*(a + b*x)]^4*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]
 

Output:

(c^2*Sec[2*a + 2*b*x]^4*Tan[2*a + 2*b*x])/(9*b*Sqrt[-c + c*Sec[2*a + 2*b*x 
]]) - (17*c*((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))/9
 

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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + 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 4301

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 
2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[b/(m + n - 1)   Int[(a + 
 b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n 
 - 4)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^ 
2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]
 

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 = 1.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.49

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {4}\, \left (2176 \cos \left (b x +a \right )^{8}-4896 \cos \left (b x +a \right )^{6}+4284 \cos \left (b x +a \right )^{4}-1785 \cos \left (b x +a \right )^{2}+315\right ) \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, c \cot \left (b x +a \right )}{315 b \left (2 \cos \left (b x +a \right )^{2}-1\right )^{4}}\)	\(101\)

Input:

int(sec(2*b*x+2*a)^4*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/315*2^(1/2)/b*4^(1/2)*(2176*cos(b*x+a)^8-4896*cos(b*x+a)^6+4284*cos(b*x+ 
a)^4-1785*cos(b*x+a)^2+315)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*c/(2 
*cos(b*x+a)^2-1)^4*cot(b*x+a)
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.63 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {2 \, \sqrt {2} {\left (315 \, c \tan \left (b x + a\right )^{8} - 525 \, c \tan \left (b x + a\right )^{6} + 819 \, c \tan \left (b x + a\right )^{4} - 423 \, c \tan \left (b x + a\right )^{2} + 94 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{315 \, {\left (b \tan \left (b x + a\right )^{9} - 4 \, b \tan \left (b x + a\right )^{7} + 6 \, b \tan \left (b x + a\right )^{5} - 4 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}} \] Input:

integrate(sec(2*b*x+2*a)^4*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorith 
m="fricas")
 

Output:

2/315*sqrt(2)*(315*c*tan(b*x + a)^8 - 525*c*tan(b*x + a)^6 + 819*c*tan(b*x 
 + a)^4 - 423*c*tan(b*x + a)^2 + 94*c)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a 
)^2 - 1))/(b*tan(b*x + a)^9 - 4*b*tan(b*x + a)^7 + 6*b*tan(b*x + a)^5 - 4* 
b*tan(b*x + a)^3 + b*tan(b*x + a))
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sec(2*b*x+2*a)**4*(c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\int { \left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (2 \, b x + 2 \, a\right )^{4} \,d x } \] Input:

integrate(sec(2*b*x+2*a)^4*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorith 
m="maxima")
 

Output:

-8/315*(630*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) 
+ 1)^(1/4)*((b*c*cos(4*b*x + 4*a)^4 + b*c*sin(4*b*x + 4*a)^4 + 4*b*c*cos(4 
*b*x + 4*a)^3 + 6*b*c*cos(4*b*x + 4*a)^2 + 4*b*c*cos(4*b*x + 4*a) + 2*(b*c 
*cos(4*b*x + 4*a)^2 + 2*b*c*cos(4*b*x + 4*a) + b*c)*sin(4*b*x + 4*a)^2 + b 
*c)*integrate(-(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4* 
a) + 1)^(1/4)*(((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(3/2*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(2 
0*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(3/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*cos(7/2*arcta 
n2(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 + ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 9119 vs. \(2 (188) = 376\).

Time = 12.57 (sec) , antiderivative size = 9119, normalized size of antiderivative = 43.84 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Too large to display} \] Input:

integrate(sec(2*b*x+2*a)^4*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorith 
m="giac")
 

Output:

-2/315*((((((((((94*sqrt(2)*c^5*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*t 
an(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^29 + 517*sqrt 
(2)*c^5*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^ 
2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^27 + 819*sqrt(2)*c^5*sgn(tan(b*x)^2*t 
an(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + ta 
n(a))*tan(a)^25 - 90*sqrt(2)*c^5*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4* 
tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^23 + 900*sqr 
t(2)*c^5*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a) 
^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^21 + 12663*sqrt(2)*c^5*sgn(tan(b*x)^ 
2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + 
 tan(a))*tan(a)^19 + 36309*sqrt(2)*c^5*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^ 
2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^17 + 5 
6388*sqrt(2)*c^5*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) 
- tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^15 + 59310*sqrt(2)*c^5*sgn(t 
an(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(ta 
n(b*x) + tan(a))*tan(a)^13 + 49315*sqrt(2)*c^5*sgn(tan(b*x)^2*tan(a)^2 - t 
an(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a 
)^11 + 35869*sqrt(2)*c^5*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x) 
*tan(a) - tan(a)^2 + 1)*sgn(tan(b*x) + tan(a))*tan(a)^9 + 21942*sqrt(2)*c^ 
5*sgn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 +...
 

Mupad [B] (verification not implemented)

Time = 23.99 (sec) , antiderivative size = 594, normalized size of antiderivative = 2.86 \[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx =\text {Too large to display} \] Input:

int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(3/2)/cos(2*a + 2*b*x)^4,x)
 

Output:

(((c*16i)/(9*b) + (c*exp(a*2i + b*x*2i)*16i)/(9*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)^4) - (((c*40i)/(7*b) + (c*exp(a*2i + b*x*2i)*88i)/(63*b))*((c*(exp(a*2 
i + 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)^3) + (((c*24i)/(5*b) - (c*exp(a*2i + b*x*2i)*176i)/(105*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) + (c*exp(a*2i + b*x*2i)*((c*(exp(a*2i + b*x*2 
i)*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)*272i)/(315*b*(exp(a*2i + b*x*2i) - 1)) + (c*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)*136i)/(315 
*b*(exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 1))
 

Reduce [F]

\[ \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\sqrt {c}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \sqrt {\tan \left (2 b x +2 a \right )}\, \sec \left (2 b x +2 a \right )^{4} \tan \left (2 b x +2 a \right ) \tan \left (b x +a \right )d x \right ) c \] Input:

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