Integrand size = 31, antiderivative size = 110 \[ \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {7 c \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {2 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b}+\frac {(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b c} \]
1/5*(-c+c*sec(2*b*x+2*a))^(3/2)*tan(2*b*x+2*a)/b/c+7/15*c*tan(2*b*x+2*a)/b /(-c+c*sec(2*b*x+2*a))^(1/2)+2/15*(-c+c*sec(2*b*x+2*a))^(1/2)*tan(2*b*x+2* a)/b
Time = 0.73 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.56 \[ \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {(5 \cos (a+b x)+2 \cos (5 (a+b x))) \csc (a+b x) \sec ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{15 b} \]
((5*Cos[a + b*x] + 2*Cos[5*(a + b*x)])*Csc[a + b*x]*Sec[2*(a + b*x)]^2*Sqr t[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(15*b)
Time = 0.68 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4897, 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.
\(\displaystyle \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (2 a+2 b x)^3 \sqrt {c \tan (a+b x) \tan (2 a+2 b x)}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \sec ^3(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^3 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
\(\Big \downarrow \) 4287 |
\(\displaystyle \frac {2 \int \frac {1}{2} \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c} \left (2 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+3 c\right )dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\) |
\(\Big \downarrow \) 4489 |
\(\displaystyle \frac {\frac {7}{3} c \int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {7}{3} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\) |
\(\Big \downarrow \) 4279 |
\(\displaystyle \frac {\frac {7 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\) |
((-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)
3.7.4.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)]^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)]
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)]
Time = 6.55 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73
method | result | size |
default | \(\frac {\sqrt {2}\, \cot \left (x b +a \right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (32 \cos \left (x b +a \right )^{4}-40 \cos \left (x b +a \right )^{2}+15\right ) \sqrt {4}}{30 b \left (2 \cos \left (x b +a \right )^{2}-1\right )^{2}}\) | \(80\) |
1/30*2^(1/2)/b*cot(b*x+a)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(32*co s(b*x+a)^4-40*cos(b*x+a)^2+15)/(2*cos(b*x+a)^2-1)^2*4^(1/2)
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {\sqrt {2} {\left (15 \, \tan \left (b x + a\right )^{4} - 10 \, \tan \left (b x + a\right )^{2} + 7\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{15 \, {\left (b \tan \left (b x + a\right )^{5} - 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}} \]
1/15*sqrt(2)*(15*tan(b*x + a)^4 - 10*tan(b*x + a)^2 + 7)*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))
Timed out. \[ \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \]
\[ \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int { \sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )} \sec \left (2 \, b x + 2 \, a\right )^{3} \,d x } \]
4/15*(30*(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) ^(1/4)*sqrt(c)*integrate(-(((cos(16*b*x + 16*a)*cos(4*b*x + 4*a) + 3*cos(1 2*b*x + 12*a)*cos(4*b*x + 4*a) + 3*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos (4*b*x + 4*a)^2 + sin(16*b*x + 16*a)*sin(4*b*x + 4*a) + 3*sin(12*b*x + 12* a)*sin(4*b*x + 4*a) + 3*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(16*b*x + 16*a) + 3*cos(4*b*x + 4*a)*sin(12*b*x + 12*a) + 3*c os(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(16*b*x + 16*a)*sin(4*b*x + 4*a) - 3 *cos(12*b*x + 12*a)*sin(4*b*x + 4*a) - 3*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*arcta n2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))) + ((cos(4*b*x + 4*a)*sin(16*b*x + 16*a) + 3*cos(4*b*x + 4*a)*sin(12*b*x + 12*a) + 3*cos(4*b*x + 4*a)*sin(8*b *x + 8*a) - cos(16*b*x + 16*a)*sin(4*b*x + 4*a) - 3*cos(12*b*x + 12*a)*sin (4*b*x + 4*a) - 3*cos(8*b*x + 8*a)*sin(4*b*x + 4*a))*cos(1/2*arctan2(sin(4 *b*x + 4*a), -cos(4*b*x + 4*a) - 1)) - (cos(16*b*x + 16*a)*cos(4*b*x + 4*a ) + 3*cos(12*b*x + 12*a)*cos(4*b*x + 4*a) + 3*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(16*b*x + 16*a)*sin(4*b*x + 4*a) + 3*sin(1 2*b*x + 12*a)*sin(4*b*x + 4*a) + 3*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) - ...
Timed out. \[ \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \]
Time = 43.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.35 \[ \int \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {4\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,5{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,5{}\mathrm {i}+{\mathrm {e}}^{a\,10{}\mathrm {i}+b\,x\,10{}\mathrm {i}}\,2{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{15\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}^2} \]