Integrand size = 31, antiderivative size = 72 \[ \int \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {c \tan (2 a+2 b x)}{3 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b} \]
-1/3*c*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/3*(-c+c*sec(2*b*x+2* a))^(1/2)*tan(2*b*x+2*a)/b
Time = 0.53 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.61 \[ \int \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {\sqrt {c \tan (a+b x) \tan (2 (a+b x))} (-\cot (a+b x)+\tan (2 (a+b x)))}{3 b} \]
Time = 0.49 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4897, 3042, 4285, 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.
\(\displaystyle \int \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (2 a+2 b x)^2 \sqrt {c \tan (a+b x) \tan (2 a+2 b x)}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \sec ^2(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^2 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
\(\Big \downarrow \) 4285 |
\(\displaystyle \frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {1}{3} \int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {1}{3} \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
\(\Big \downarrow \) 4279 |
\(\displaystyle \frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {c \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
-1/3*(c*Tan[2*a + 2*b*x])/(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)
3.7.5.3.1 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_)]^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)]
Time = 6.75 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97
method | result | size |
default | \(-\frac {\sqrt {2}\, \cot \left (x b +a \right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (4 \cos \left (x b +a \right )^{2}-3\right ) \sqrt {4}}{6 b \left (2 \cos \left (x b +a \right )^{2}-1\right )}\) | \(70\) |
-1/6*2^(1/2)/b*cot(b*x+a)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(4*cos (b*x+a)^2-3)/(2*cos(b*x+a)^2-1)*4^(1/2)
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89 \[ \int \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {\sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (3 \, \tan \left (b x + a\right )^{2} - 1\right )}}{3 \, {\left (b \tan \left (b x + a\right )^{3} - b \tan \left (b x + a\right )\right )}} \]
-1/3*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(3*tan(b*x + a)^ 2 - 1)/(b*tan(b*x + a)^3 - b*tan(b*x + a))
Timed out. \[ \int \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \]
\[ \int \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int { \sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )} \sec \left (2 \, b x + 2 \, a\right )^{2} \,d x } \]
-2/3*(6*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1) ^(3/4)*b*sqrt(c)*integrate(-(((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)*s in(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(1/2*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(1/2* 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(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))*cos(1/2*arctan2(sin(4*b*x + 4*a), -co s(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(1 /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))))/(((2*(2*cos(8*b*x + 8*a) + cos(4*b*x + 4 *a))*cos(12*b*x + 12*a) + cos(12*b*x + 12*a)^2 + 4*cos(8*b*x + 8*a)^2 + 4* cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + 2*(2*sin(8*b*x + 8*a) + sin(4*b*x + 4*a))*sin(12*b*x + 12*a) + sin(12*b*x + 12*a)^2 + 4*sin (8*b*x + 8*a)^2 + 4*sin(8*b*x + 8*a)*sin(4*b*x + 4*a) + sin(4*b*x + 4*a...
Leaf count of result is larger than twice the leaf count of optimal. 413066 vs. \(2 (64) = 128\).
Time = 142.29 (sec) , antiderivative size = 413066, normalized size of antiderivative = 5737.03 \[ \int \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Too large to display} \]
1/3*sqrt(2)*(((((((c^5*sgn(tan(1/2*b*x + 2*a)^4*tan(1/2*a)^12 - 66*tan(1/2 *b*x + 2*a)^4*tan(1/2*a)^10 + 48*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^11 - 6*ta n(1/2*b*x + 2*a)^2*tan(1/2*a)^12 + 495*tan(1/2*b*x + 2*a)^4*tan(1/2*a)^8 - 880*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^9 + 396*tan(1/2*b*x + 2*a)^2*tan(1/2* a)^10 - 48*tan(1/2*b*x + 2*a)*tan(1/2*a)^11 + tan(1/2*a)^12 - 924*tan(1/2* b*x + 2*a)^4*tan(1/2*a)^6 + 3168*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^7 - 2970* tan(1/2*b*x + 2*a)^2*tan(1/2*a)^8 + 880*tan(1/2*b*x + 2*a)*tan(1/2*a)^9 - 66*tan(1/2*a)^10 + 495*tan(1/2*b*x + 2*a)^4*tan(1/2*a)^4 - 3168*tan(1/2*b* x + 2*a)^3*tan(1/2*a)^5 + 5544*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^6 - 3168*ta n(1/2*b*x + 2*a)*tan(1/2*a)^7 + 495*tan(1/2*a)^8 - 66*tan(1/2*b*x + 2*a)^4 *tan(1/2*a)^2 + 880*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^3 - 2970*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^4 + 3168*tan(1/2*b*x + 2*a)*tan(1/2*a)^5 - 924*tan(1/2* a)^6 + tan(1/2*b*x + 2*a)^4 - 48*tan(1/2*b*x + 2*a)^3*tan(1/2*a) + 396*tan (1/2*b*x + 2*a)^2*tan(1/2*a)^2 - 880*tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 495 *tan(1/2*a)^4 - 6*tan(1/2*b*x + 2*a)^2 + 48*tan(1/2*b*x + 2*a)*tan(1/2*a) - 66*tan(1/2*a)^2 + 1)*sgn(-3*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^5 + tan(1/2* b*x + 2*a)*tan(1/2*a)^6 + 10*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^3 - 15*tan(1/ 2*b*x + 2*a)*tan(1/2*a)^4 + 3*tan(1/2*a)^5 - 3*tan(1/2*b*x + 2*a)^2*tan(1/ 2*a) + 15*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 10*tan(1/2*a)^3 - tan(1/2*b*x + 2*a) + 3*tan(1/2*a))*tan(1/2*a)^174 - 147*c^5*sgn(tan(1/2*b*x + 2*a)^...
Time = 32.89 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.79 \[ \int \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {2\,\left ({\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{3\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}-1\right )} \]