Integrand size = 31, antiderivative size = 129 \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {2 \tan (2 a+2 b x)}{3 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b c} \]
-1/2*arctanh(1/2*c^(1/2)*tan(2*b*x+2*a)*2^(1/2)/(-c+c*sec(2*b*x+2*a))^(1/2 ))/b*2^(1/2)/c^(1/2)+2/3*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/c
Time = 1.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {\cos ^2(a+b x) \csc (2 (a+b x)) \left (2+2 \sec (2 (a+b x))+3 \arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{3 b c} \]
(Cos[a + b*x]^2*Csc[2*(a + b*x)]*(2 + 2*Sec[2*(a + b*x)] + 3*ArcTan[Sqrt[- 1 + Tan[a + b*x]^2]]*Sqrt[-1 + Tan[a + b*x]^2])*Sqrt[c*Tan[a + b*x]*Tan[2* (a + b*x)]])/(3*b*c)
Time = 0.74 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 4897, 3042, 4287, 27, 3042, 4489, 3042, 4282, 220}
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.
| \(\displaystyle \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (2 a+2 b x)^3}{\sqrt {c \tan (a+b x) \tan (2 a+2 b x)}}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\sec ^3(2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )^3}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx\) |
| \(\Big \downarrow \) 4287 |
| \(\displaystyle \frac {2 \int \frac {\sec (2 a+2 b x) (2 \sec (2 a+2 b x) c+c)}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec (2 a+2 b x) (2 \sec (2 a+2 b x) c+c)}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \left (2 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+c\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b c}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {3 c \int \frac {\sec (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx+\frac {2 c \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 c \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx+\frac {2 c \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b c}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {2 c \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {3 c \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-2 c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b c}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {\frac {2 c \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b}}{3 c}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b c}\) |
(Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(3*b*c) + ((-3*Sqrt[c]*Ar cTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])]) /(Sqrt[2]*b) + (2*c*Tan[2*a + 2*b*x])/(b*Sqrt[-c + c*Sec[2*a + 2*b*x]]))/( 3*c)
3.7.20.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{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)]
Leaf count of result is larger than twice the leaf count of optimal. \(606\) vs. \(2(112)=224\).
Time = 9.81 (sec) , antiderivative size = 607, normalized size of antiderivative = 4.71
| method | result | size |
| default | \(\frac {\sqrt {2}\, \sin \left (x b +a \right ) \left (12 \cos \left (x b +a \right )^{4} \ln \left (\frac {2 \cos \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+2 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )-2}{1+\cos \left (x b +a \right )}\right )-12 \cos \left (x b +a \right )^{4} \operatorname {arctanh}\left (\frac {2 \cos \left (x b +a \right )-1}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right )+8 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \cos \left (x b +a \right )^{4}+8 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \cos \left (x b +a \right )^{3}-12 \cos \left (x b +a \right )^{2} \ln \left (\frac {2 \cos \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+2 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )-2}{1+\cos \left (x b +a \right )}\right )+12 \cos \left (x b +a \right )^{2} \operatorname {arctanh}\left (\frac {2 \cos \left (x b +a \right )-1}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right )+3 \ln \left (\frac {2 \cos \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+2 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )-2}{1+\cos \left (x b +a \right )}\right )-3 \,\operatorname {arctanh}\left (\frac {2 \cos \left (x b +a \right )-1}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right )\right ) \sqrt {4}}{24 b \left (2 \cos \left (x b +a \right )^{2}-1\right )^{2} \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \left (1+\cos \left (x b +a \right )\right ) \left (-3+2 \sqrt {2}\right )^{2} \left (3+2 \sqrt {2}\right )^{2}}\) | \(607\) |
1/24*2^(1/2)/b*sin(b*x+a)*(12*cos(b*x+a)^4*ln(2*(cos(b*x+a)*((2*cos(b*x+a) ^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)- 2*cos(b*x+a)-1)/(1+cos(b*x+a)))-12*cos(b*x+a)^4*arctanh((2*cos(b*x+a)-1)/( 1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2))+8*((2*cos(b*x+a )^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x+a)^4+8*((2*cos(b*x+a)^2-1)/(1+cos(b *x+a))^2)^(1/2)*cos(b*x+a)^3-12*cos(b*x+a)^2*ln(2*(cos(b*x+a)*((2*cos(b*x+ a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2 )-2*cos(b*x+a)-1)/(1+cos(b*x+a)))+12*cos(b*x+a)^2*arctanh((2*cos(b*x+a)-1) /(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2))+3*ln(2*(cos(b *x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+c os(b*x+a))^2)^(1/2)-2*cos(b*x+a)-1)/(1+cos(b*x+a)))-3*arctanh((2*cos(b*x+a )-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)))/(2*cos(b *x+a)^2-1)^2/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)/((2*cos(b*x+a)^2-1) /(1+cos(b*x+a))^2)^(1/2)/(1+cos(b*x+a))*4^(1/2)/(-3+2*2^(1/2))^2/(3+2*2^(1 /2))^2
Time = 0.29 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\left [\frac {\frac {3 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{3} - c \tan \left (b x + a\right )\right )} \log \left (\frac {\tan \left (b x + a\right )^{3} - \frac {2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{\sqrt {c}} - 2 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right )}{\sqrt {c}} - 8 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{12 \, {\left (b c \tan \left (b x + a\right )^{3} - b c \tan \left (b x + a\right )\right )}}, -\frac {3 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{3} - c \tan \left (b x + a\right )\right )} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {1}{c}}}{\tan \left (b x + a\right )}\right ) + 4 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{6 \, {\left (b c \tan \left (b x + a\right )^{3} - b c \tan \left (b x + a\right )\right )}}\right ] \]
[1/12*(3*sqrt(2)*(c*tan(b*x + a)^3 - c*tan(b*x + a))*log((tan(b*x + a)^3 - 2*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)/sqrt( c) - 2*tan(b*x + a))/tan(b*x + a)^3)/sqrt(c) - 8*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c*tan(b*x + a)^3 - b*c*tan(b*x + a)), -1/6 *(3*sqrt(2)*(c*tan(b*x + a)^3 - c*tan(b*x + a))*sqrt(-1/c)*arctan(sqrt(-c* tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-1/c)/tan(b *x + a)) + 4*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c*ta n(b*x + a)^3 - b*c*tan(b*x + a))]
Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int { \frac {\sec \left (2 \, b x + 2 \, a\right )^{3}}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sec ^3(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^3\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )}} \,d x \]