Integrand size = 29, antiderivative size = 75 \[ \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=-\frac {4 c^2 \tan (2 a+2 b x)}{3 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b} \]
-4/3*c^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/3*c*(-c+c*sec(2*b* x+2*a))^(1/2)*tan(2*b*x+2*a)/b
Time = 0.48 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.68 \[ \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=-\frac {\cot (a+b x) (-1+4 \cot (a+b x) \cot (2 (a+b x))) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}{3 b} \]
-1/3*(Cot[a + b*x]*(-1 + 4*Cot[a + b*x]*Cot[2*(a + b*x)])*(c*Tan[a + b*x]* Tan[2*(a + b*x)])^(3/2))/b
Time = 0.40 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 4897, 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.
\(\displaystyle \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (2 a+2 b x) (c \tan (a+b x) \tan (2 a+2 b x))^{3/2}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \sec (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}dx\) |
\(\Big \downarrow \) 4280 |
\(\displaystyle \frac {c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {4}{3} c \int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {4}{3} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
\(\Big \downarrow \) 4279 |
\(\displaystyle \frac {c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {4 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
(-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)
3.7.14.3.1 Defintions of rubi rules used
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]
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]
Time = 5.74 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {\sqrt {2}\, \cot \left (x b +a \right ) c \left (5 \cos \left (x b +a \right )^{2}-3\right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \sqrt {4}}{3 b \left (2 \cos \left (x b +a \right )^{2}-1\right )}\) | \(71\) |
-1/3*2^(1/2)/b*cot(b*x+a)*c*(5*cos(b*x+a)^2-3)*(c*sin(b*x+a)^2/(2*cos(b*x+ a)^2-1))^(1/2)/(2*cos(b*x+a)^2-1)*4^(1/2)
Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.89 \[ \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=-\frac {2 \, \sqrt {2} {\left (3 \, c \tan \left (b x + a\right )^{2} - 2 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{3 \, {\left (b \tan \left (b x + a\right )^{3} - b \tan \left (b x + a\right )\right )}} \]
-2/3*sqrt(2)*(3*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)^3 - b*tan(b*x + a))
Timed out. \[ \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
\[ \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\int { \left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sec \left (2 \, b x + 2 \, a\right ) \,d x } \]
1/3*(6*(3*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(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(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( 8*b*x + 8*a) - 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(3/2*arctan2(sin(4*b*x + 4*a), cos(4 *b*x + 4*a))) + ((cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - 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(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(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(3/2*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 + si n(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*cos(8*b*x + 8*a)^2 + 2*cos(4*b* x + 4*a)^3 + (cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*sin(8*b*x + 8*a)^2 + (2*cos(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1) *sin(4*b*x + 4*a)^2 + 2*(cos(4*b*x + 4*a)^3 + cos(4*b*x + 4*a)*sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a)^2 + cos(4*b*x + 4*a))*cos(8*b*x + 8*a) + cos( 4*b*x + 4*a)^2 + 2*(sin(4*b*x + 4*a)^3 + (cos(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*sin(4*b*x + 4*a))*sin(8*b*x + 8*a))*cos(3/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + (cos(4*b*x + 4*a)^4 + sin(4*b*x + ...
Timed out. \[ \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
Time = 35.04 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.11 \[ \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=-\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,5{}\mathrm {i}+5{}\mathrm {i}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{3\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}-1\right )} \]