Integrand size = 20, antiderivative size = 138 \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b c^{3/2}}+\frac {5 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}} \]
-arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b/c^(3/2)+5/8 *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 /c^(3/2)*2^(1/2)-1/4*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(3/2)
Time = 3.20 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {\cot (a+b x) \left (-4 \text {arctanh}\left (\sqrt {1-\tan ^2(a+b x)}\right ) \cos (2 (a+b x)) \sec ^2(a+b x)+4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\tan ^2(a+b x)}}{\sqrt {2}}\right ) \cos (2 (a+b x)) \sec ^2(a+b x)+\cot ^2(a+b x) \left (\cos (2 (a+b x)) \sec ^2(a+b x)\right )^{3/2}+\arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {-\left (-1+\tan ^2(a+b x)\right )^2}\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{8 b c^2 \sqrt {1-\tan ^2(a+b x)}} \]
-1/8*(Cot[a + b*x]*(-4*ArcTanh[Sqrt[1 - Tan[a + b*x]^2]]*Cos[2*(a + b*x)]* Sec[a + b*x]^2 + 4*Sqrt[2]*ArcTanh[Sqrt[1 - Tan[a + b*x]^2]/Sqrt[2]]*Cos[2 *(a + b*x)]*Sec[a + b*x]^2 + Cot[a + b*x]^2*(Cos[2*(a + b*x)]*Sec[a + b*x] ^2)^(3/2) + ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sqrt[-(-1 + Tan[a + b*x]^2)^ 2])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(b*c^2*Sqrt[1 - Tan[a + b*x]^2] )
Time = 0.63 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4897, 3042, 4264, 27, 3042, 4408, 3042, 4261, 220, 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.
| \(\displaystyle \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(c \tan (a+b x) \tan (2 a+2 b x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {1}{(c \sec (2 a+2 b x)-c)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4264 |
| \(\displaystyle -\frac {\int \frac {\sec (2 a+2 b x) c+4 c}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx}{2 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec (2 a+2 b x) c+4 c}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{4 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+4 c}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{4 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle -\frac {5 c \int \frac {\sec (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx-4 \int \sqrt {c \sec (2 a+2 b x)-c}dx}{4 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 c \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx-4 \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx}{4 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle -\frac {\frac {4 c \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}+5 c \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{4 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {5 c \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx+\frac {4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{4 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle -\frac {\frac {4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}-\frac {5 c \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-2 c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{4 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {\frac {4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}-\frac {5 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b}}{4 c^2}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
-1/4*((4*Sqrt[c]*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]])/b - (5*Sqrt[c]*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[ -c + c*Sec[2*a + 2*b*x]])])/(Sqrt[2]*b))/c^2 - Tan[2*a + 2*b*x]/(4*b*(-c + c*Sec[2*a + 2*b*x])^(3/2))
3.7.30.3.1 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[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c + d*x])*((a + b*Csc[c + d*x])^n/(d*(2*n + 1))), x] + Simp[1/(a^2*(2*n + 1)) Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && Int egerQ[2*n]
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_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c/a Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[(b*c - a*d)/a Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(551\) vs. \(2(117)=234\).
Time = 4.37 (sec) , antiderivative size = 552, normalized size of antiderivative = 4.00
| method | result | size |
| default | \(\frac {\sqrt {2}\, \csc \left (x b +a \right ) \left (1-\cos \left (x b +a \right )\right ) \left (16 \csc \left (x b +a \right )^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (\csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}-1\right ) \sqrt {2}}{\sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}}\right ) \left (1-\cos \left (x b +a \right )\right )^{2}-10 \csc \left (x b +a \right )^{2} \ln \left (\csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}-3+\sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}\right ) \left (1-\cos \left (x b +a \right )\right )^{2}-10 \csc \left (x b +a \right )^{2} \operatorname {arctanh}\left (\frac {3 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}-1}{\sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}}\right ) \left (1-\cos \left (x b +a \right )\right )^{2}+\csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2} \sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}-\sqrt {\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}\right ) \sqrt {4}}{64 b \left (\frac {\csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2} c}{\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1}\right )^{\frac {3}{2}} \left (\csc \left (x b +a \right )^{4} \left (1-\cos \left (x b +a \right )\right )^{4}-6 \csc \left (x b +a \right )^{2} \left (1-\cos \left (x b +a \right )\right )^{2}+1\right )^{\frac {3}{2}}}\) | \(552\) |
1/64*2^(1/2)/b*csc(b*x+a)/(csc(b*x+a)^2*(1-cos(b*x+a))^2*c/(csc(b*x+a)^4*( 1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1))^(3/2)*(1-cos(b*x+a))*( 16*csc(b*x+a)^2*2^(1/2)*arctanh((csc(b*x+a)^2*(1-cos(b*x+a))^2-1)*2^(1/2)/ (csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*( 1-cos(b*x+a))^2-10*csc(b*x+a)^2*ln(csc(b*x+a)^2*(1-cos(b*x+a))^2-3+(csc(b* x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-cos(b *x+a))^2-10*csc(b*x+a)^2*arctanh((3*csc(b*x+a)^2*(1-cos(b*x+a))^2-1)/(csc( b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-cos (b*x+a))^2+csc(b*x+a)^2*(1-cos(b*x+a))^2*(csc(b*x+a)^4*(1-cos(b*x+a))^4-6* csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2)-(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*cs c(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc (b*x+a)^2*(1-cos(b*x+a))^2+1)^(3/2)*4^(1/2)
Time = 0.27 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.17 \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\left [\frac {5 \, \sqrt {2} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} + 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right )^{3} + 8 \, \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 3 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} + 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{16 \, b c^{2} \tan \left (b x + a\right )^{3}}, \frac {5 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} - 8 \, \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{2 \, c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} + \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{8 \, b c^{2} \tan \left (b x + a\right )^{3}}\right ] \]
[1/16*(5*sqrt(2)*sqrt(c)*log((c*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*c*tan(b*x + a))/tan (b*x + a)^3)*tan(b*x + a)^3 + 8*sqrt(c)*log((c*tan(b*x + a)^3 - 2*sqrt(2)* sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(c) - 3*c*tan(b*x + a))/(tan(b*x + a)^3 + tan(b*x + a)))*tan(b*x + a)^3 + 2*sq rt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1))/( b*c^2*tan(b*x + a)^3), 1/8*(5*sqrt(2)*sqrt(-c)*arctan(sqrt(-c*tan(b*x + a) ^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-c)/(c*tan(b*x + a)))*t an(b*x + a)^3 - 8*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan( b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-c)/(c*tan(b*x + a)))*tan(b*x + a)^3 + sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a) ^2 - 1))/(b*c^2*tan(b*x + a)^3)]
Timed out. \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\int { \frac {1}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\int \frac {1}{{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2}} \,d x \]