\(\int \sec ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) [556]

Optimal result

Integrand size = 31, antiderivative size = 110 \[ \int \sec ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {4 c^2 \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{5 b}+\frac {(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.54 \[ \int \sec ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {\cot (a+b x) (-2+4 \cot (a+b x) \cot (2 (a+b x))+\sec (2 (a+b x))) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}{5 b} \] Input:

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

Output:

(Cot[a + b*x]*(-2 + 4*Cot[a + b*x]*Cot[2*(a + b*x)] + Sec[2*(a + b*x)])*(c 
*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2))/(5*b)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 4897, 3042, 4285, 3042, 4280, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4897

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4285

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4280

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4279

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

Input:

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

Output:

((-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5*b) - (3*((-4*c^2*Tan 
[2*a + 2*b*x])/(3*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (c*Sqrt[-c + c*Sec[2* 
a + 2*b*x]]*Tan[2*a + 2*b*x])/(3*b)))/5
 

Defintions of rubi rules used

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 4280

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 - 1)/(f*m)), x] 
+ Simp[a*((2*m - 1)/m)   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] && GtQ[m, 1/2] && Intege 
rQ[2*m]
 

rule 4285

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*(m + 1))), x] 
+ Simp[a*(m/(b*(m + 1)))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] 
/; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 4897

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

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {4}\, \left (12 \cos \left (b x +a \right )^{4}-15 \cos \left (b x +a \right )^{2}+5\right ) \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, c \cot \left (b x +a \right )}{5 b \left (2 \cos \left (b x +a \right )^{2}-1\right )^{2}}\)	\(81\)

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:

1/5*2^(1/2)/b*4^(1/2)*(12*cos(b*x+a)^4-15*cos(b*x+a)^2+5)*(c*sin(b*x+a)^2/ 
(2*cos(b*x+a)^2-1))^(1/2)*c/(2*cos(b*x+a)^2-1)^2*cot(b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.80 \[ \int \sec ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {2 \, \sqrt {2} {\left (5 \, c \tan \left (b x + a\right )^{4} - 5 \, c \tan \left (b x + a\right )^{2} + 2 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{5 \, {\left (b \tan \left (b x + a\right )^{5} - 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}} \] 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:

2/5*sqrt(2)*(5*c*tan(b*x + a)^4 - 5*c*tan(b*x + a)^2 + 2*c)*sqrt(-c*tan(b* 
x + a)^2/(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^5 - 2*b*tan(b*x + a)^3 + b* 
tan(b*x + a))
 

Sympy [F(-1)]

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

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

-2/5*(10*(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)^2 + b*c*sin(4*b*x + 4*a)^2 + 2*b*c*cos(4*b* 
x + 4*a) + b*c)*integrate(-(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*co 
s(4*b*x + 4*a) + 1)^(1/4)*(((cos(12*b*x + 12*a)*cos(4*b*x + 4*a) + 2*cos(8 
*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(12*b*x + 12*a)*sin 
(4*b*x + 4*a) + 2*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(12*b*x + 12*a) + 2*cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(12*b*x 
+ 12*a)*sin(4*b*x + 4*a) - 2*cos(8*b*x + 8*a)*sin(4*b*x + 4*a))*sin(3/2*ar 
ctan2(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(12*b*x + 12*a) + 2*co 
s(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(12*b*x + 12*a)*sin(4*b*x + 4*a) - 2* 
cos(8*b*x + 8*a)*sin(4*b*x + 4*a))*cos(3/2*arctan2(sin(4*b*x + 4*a), -cos( 
4*b*x + 4*a) - 1)) - (cos(12*b*x + 12*a)*cos(4*b*x + 4*a) + 2*cos(8*b*x + 
8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(12*b*x + 12*a)*sin(4*b*x 
+ 4*a) + 2*sin(8*b*x + 8*a)*sin(4*b*x + 4*a) + sin(4*b*x + 4*a)^2)*sin(3/2 
*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*sin(5/2*arctan2(sin(4* 
b*x + 4*a), cos(4*b*x + 4*a))))/((cos(4*b*x + 4*a)^4 + sin(4*b*x + 4*a)^4 
+ (cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*cos(1 
2*b*x + 12*a)^2 + 4*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3323 vs. \(2 (98) = 196\).

Time = 7.46 (sec) , antiderivative size = 3323, normalized size of antiderivative = 30.21 \[ \int \sec ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

-2/5*((((((2*sqrt(2)*c^3*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 + 7*sqrt(2)*c^3*s 
gn(tan(b*x)^2*tan(a)^2 - tan(b*x)^2 - 4*tan(b*x)*tan(a) - tan(a)^2 + 1)*sg 
n(tan(b*x) + tan(a))*tan(a)^15 + 5*sqrt(2)*c^3*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 
)^13 - 5*sqrt(2)*c^3*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)^11 + 5*sqrt(2)*c^3*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)^9 + 37*sqrt(2)*c^3*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)^7 
+ 47*sqrt(2)*c^3*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)^5 + 25*sqrt(2)*c^3*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)^3 + 5*sqrt(2)*c^3*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))*tan(b* 
x)/(tan(a)^12 + 6*tan(a)^10 + 15*tan(a)^8 + 20*tan(a)^6 + 15*tan(a)^4 + 6* 
tan(a)^2 + 1) - 5*(4*sqrt(2)*c^3*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)^16 + 17*sqrt 
(2)*c^3*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)^14 + 19*sqrt(2)*c^3*sgn(tan(b*x)^2...
 

Mupad [B] (verification not implemented)

Time = 27.93 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.35 \[ \int \sec ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {2\,c\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,5{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,5{}\mathrm {i}+{\mathrm {e}}^{a\,10{}\mathrm {i}+b\,x\,10{}\mathrm {i}}\,3{}\mathrm {i}+3{}\mathrm {i}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{5\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}^2} \] Input:

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

Output:

(2*c*(exp(a*4i + b*x*4i)*5i + exp(a*6i + b*x*6i)*5i + exp(a*10i + b*x*10i) 
*3i + 3i)*((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))/(5*b*(exp(a*2i 
+ b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 1)^2)
 

Reduce [F]

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

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