Integrand size = 31, antiderivative size = 157 \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {2 c \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c \sec ^3(2 a+2 b x) \tan (2 a+2 b x)}{7 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {4 \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{35 b}-\frac {6 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b c} \]
-6/35*(-c+c*sec(2*b*x+2*a))^(3/2)*tan(2*b*x+2*a)/b/c-2/5*c*tan(2*b*x+2*a)/ b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/7*c*sec(2*b*x+2*a)^3*tan(2*b*x+2*a)/b/(-c+ c*sec(2*b*x+2*a))^(1/2)-4/35*(-c+c*sec(2*b*x+2*a))^(1/2)*tan(2*b*x+2*a)/b
Time = 1.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.41 \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {(7 \cos (3 (a+b x))+2 \cos (7 (a+b x))) \csc (a+b x) \sec ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{35 b} \]
-1/35*((7*Cos[3*(a + b*x)] + 2*Cos[7*(a + b*x)])*Csc[a + b*x]*Sec[2*(a + b *x)]^3*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/b
Time = 0.98 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4897, 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.
| \(\displaystyle \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (2 a+2 b x)^4 \sqrt {c \tan (a+b x) \tan (2 a+2 b x)}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \sec ^4(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^4 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
| \(\Big \downarrow \) 4290 |
| \(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \int \sec ^3(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right )^3 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
| \(\Big \downarrow \) 4287 |
| \(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {2 \int \frac {1}{2} \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c} (2 \sec (2 a+2 b x) c+3 c)dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c} \left (2 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+3 c\right )dx}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7}{3} c \int \sec (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7}{3} c \int \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {c \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{7 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {6}{7} \left (\frac {\frac {7 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {2 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}}{5 c}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}\right )\) |
(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
3.7.3.3.1 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_)]*(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]
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 = 7.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.57
| method | result | size |
| default | \(-\frac {\sqrt {2}\, \cot \left (x b +a \right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (128 \cos \left (x b +a \right )^{6}-224 \cos \left (x b +a \right )^{4}+140 \cos \left (x b +a \right )^{2}-35\right ) \sqrt {4}}{70 b \left (2 \cos \left (x b +a \right )^{2}-1\right )^{3}}\) | \(90\) |
-1/70*2^(1/2)/b*cot(b*x+a)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(128* cos(b*x+a)^6-224*cos(b*x+a)^4+140*cos(b*x+a)^2-35)/(2*cos(b*x+a)^2-1)^3*4^ (1/2)
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.68 \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {\sqrt {2} {\left (35 \, \tan \left (b x + a\right )^{6} - 35 \, \tan \left (b x + a\right )^{4} + 49 \, \tan \left (b x + a\right )^{2} - 9\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{35 \, {\left (b \tan \left (b x + a\right )^{7} - 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} - b \tan \left (b x + a\right )\right )}} \]
-1/35*sqrt(2)*(35*tan(b*x + a)^6 - 35*tan(b*x + a)^4 + 49*tan(b*x + a)^2 - 9)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^7 - 3*b*t an(b*x + a)^5 + 3*b*tan(b*x + a)^3 - b*tan(b*x + a))
Timed out. \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \]
\[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int { \sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )} \sec \left (2 \, b x + 2 \, a\right )^{4} \,d x } \]
-8/35*(70*(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 )^(3/4)*sqrt(c)*integrate(-(((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(1/2*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(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 + 12*a)*sin(4*b*x + 4*a) - 4*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(5/2*arctan2(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 + 12*a)*sin(4*b*x + 4*a) - 4*cos(8*b*x + 8*a)*sin(4*b*x + 4*a))*co s(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) - (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...
Timed out. \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \]
Time = 34.13 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.95 \[ \int \sec ^4(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,\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 )}}\,16{}\mathrm {i}}{35\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}+\frac {\left (\frac {8{}\mathrm {i}}{7\,b}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,8{}\mathrm {i}}{7\,b}\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 )}}}{\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}-\frac {\left (\frac {8{}\mathrm {i}}{5\,b}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,64{}\mathrm {i}}{35\,b}\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 )}}}{\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}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,\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 )}}\,8{}\mathrm {i}}{35\,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 )} \]
((8i/(7*b) - (exp(a*2i + b*x*2i)*8i)/(7*b))*((c*(exp(a*2i + b*x*2i)*1i - 1 i)*(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) - (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)*16i)/ (35*b*(exp(a*2i + b*x*2i) - 1)) - ((8i/(5*b) - (exp(a*2i + b*x*2i)*64i)/(3 5*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) - (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)*(e xp(a*4i + b*x*4i) + 1)))^(1/2)*8i)/(35*b*(exp(a*2i + b*x*2i) - 1)*(exp(a*4 i + b*x*4i) + 1))