Integrand size = 27, antiderivative size = 211 \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) (c-i d)^{3/2} f}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) (c+i d)^{3/2} f}-\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) (b c-a d)^{3/2} f}+\frac {2 d^2}{(b c-a d) \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \]
arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(I*a+b)/(c-I*d)^(3/2)/f-arct anh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(I*a-b)/(c+I*d)^(3/2)/f-2*b^(5/2 )*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/(-a*d+b*c)^(1/2))/(a^2+b^2)/(-a*d +b*c)^(3/2)/f+2*d^2/(-a*d+b*c)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)
Time = 2.22 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\frac {-\frac {i \left (\frac {(a+i b) (c+i d) (-b c+a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {(a-i b) (c-i d) (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )}{a^2+b^2}+\frac {2 b^{5/2} \left (c^2+d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right ) \sqrt {b c-a d}}-\frac {2 d^2}{\sqrt {c+d \tan (e+f x)}}}{(-b c+a d) \left (c^2+d^2\right ) f} \]
(((-I)*(((a + I*b)*(c + I*d)*(-(b*c) + a*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x ]]/Sqrt[c - I*d]])/Sqrt[c - I*d] + ((a - I*b)*(c - I*d)*(b*c - a*d)*ArcTan h[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d]))/(a^2 + b^2) + ( 2*b^(5/2)*(c^2 + d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)*Sqrt[b*c - a*d]) - (2*d^2)/Sqrt[c + d*Tan[e + f*x]]) /((-(b*c) + a*d)*(c^2 + d^2)*f)
Time = 1.68 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.23, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {3042, 4052, 27, 3042, 4136, 25, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 73, 221}
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.
\(\displaystyle \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4052 |
\(\displaystyle \frac {2 \int -\frac {-b d^2 \tan ^2(e+f x)+d (b c-a d) \tan (e+f x)+a c d-b \left (c^2+d^2\right )}{2 (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}+\frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {\int \frac {-b d^2 \tan ^2(e+f x)+d (b c-a d) \tan (e+f x)+a c d-b \left (c^2+d^2\right )}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {\int \frac {-b d^2 \tan (e+f x)^2+d (b c-a d) \tan (e+f x)+a c d-b \left (c^2+d^2\right )}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 4136 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {\int -\frac {(b c-a d) (a c-b d)-(b c-a d) (b c+a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan ^2(e+f x)+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {\int \frac {(b c-a d) (a c-b d)-(b c-a d) (b c+a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan ^2(e+f x)+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {\int \frac {(b c-a d) (a c-b d)-(b c-a d) (b c+a d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {\frac {1}{2} (a-i b) (c-i d) (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c+i d) (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {\frac {1}{2} (a-i b) (c-i d) (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c+i d) (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {\frac {i (a+i b) (c+i d) (b c-a d) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a-i b) (c-i d) (b c-a d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {\frac {i (a-i b) (c-i d) (b c-a d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (a+i b) (c+i d) (b c-a d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {\frac {(a-i b) (c-i d) (b c-a d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a+i b) (c+i d) (b c-a d) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {b^3 \left (c^2+d^2\right ) \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {\frac {(a+i b) (c+i d) (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {(a-i b) (c-i d) (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 4117 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {b^3 \left (c^2+d^2\right ) \int \frac {1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f \left (a^2+b^2\right )}-\frac {\frac {(a+i b) (c+i d) (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {(a-i b) (c-i d) (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {2 b^3 \left (c^2+d^2\right ) \int \frac {1}{a+\frac {b (c+d \tan (e+f x))}{d}-\frac {b c}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (a^2+b^2\right )}-\frac {\frac {(a+i b) (c+i d) (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {(a-i b) (c-i d) (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 d^2}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {2 b^{5/2} \left (c^2+d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{f \left (a^2+b^2\right ) \sqrt {b c-a d}}-\frac {\frac {(a+i b) (c+i d) (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {(a-i b) (c-i d) (b c-a d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{a^2+b^2}}{\left (c^2+d^2\right ) (b c-a d)}\) |
-((-((((a + I*b)*(c + I*d)*(b*c - a*d)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]]) /(Sqrt[c - I*d]*f) + ((a - I*b)*(c - I*d)*(b*c - a*d)*ArcTan[Tan[e + f*x]/ Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/(a^2 + b^2)) + (2*b^(5/2)*(c^2 + d^2)*A rcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/((a^2 + b^2)*S qrt[b*c - a*d]*f))/((b*c - a*d)*(c^2 + d^2))) + (2*d^2)/((b*c - a*d)*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])
3.13.57.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 /((m + 1)*(a^2 + b^2)*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / ; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ erQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^ n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ (A*b^2 - a*b*B + a^2*C)/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & & !GtQ[n, 0] && !LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(8701\) vs. \(2(181)=362\).
Time = 1.10 (sec) , antiderivative size = 8702, normalized size of antiderivative = 41.24
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(8702\) |
default | \(\text {Expression too large to display}\) | \(8702\) |
Leaf count of result is larger than twice the leaf count of optimal. 8991 vs. \(2 (175) = 350\).
Time = 32.90 (sec) , antiderivative size = 17997, normalized size of antiderivative = 85.29 \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \tan {\left (e + f x \right )}\right ) \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Timed out. \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Time = 24.08 (sec) , antiderivative size = 115510, normalized size of antiderivative = 547.44 \[ \int \frac {1}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
(log(((((2*a^2*b^2*d^6*f^4 - b^4*d^6*f^4 - 4*a^2*b^2*c^6*f^4 - a^4*d^6*f^4 + 6*a^4*c^2*d^4*f^4 - 9*a^4*c^4*d^2*f^4 + 6*b^4*c^2*d^4*f^4 - 9*b^4*c^4*d ^2*f^4 - 48*a^2*b^2*c^2*d^4*f^4 + 42*a^2*b^2*c^4*d^2*f^4 + 12*a*b^3*c*d^5* f^4 + 12*a*b^3*c^5*d*f^4 - 12*a^3*b*c*d^5*f^4 - 12*a^3*b*c^5*d*f^4 - 40*a* b^3*c^3*d^3*f^4 + 40*a^3*b*c^3*d^3*f^4)^(1/2) - a^2*c^3*f^2 + b^2*c^3*f^2 + 3*a^2*c*d^2*f^2 - 3*b^2*c*d^2*f^2 - 2*a*b*d^3*f^2 + 6*a*b*c^2*d*f^2)/(a^ 4*c^6*f^4 + a^4*d^6*f^4 + b^4*c^6*f^4 + b^4*d^6*f^4 + 2*a^2*b^2*c^6*f^4 + 2*a^2*b^2*d^6*f^4 + 3*a^4*c^2*d^4*f^4 + 3*a^4*c^4*d^2*f^4 + 3*b^4*c^2*d^4* f^4 + 3*b^4*c^4*d^2*f^4 + 6*a^2*b^2*c^2*d^4*f^4 + 6*a^2*b^2*c^4*d^2*f^4))^ (1/2)*(((((2*a^2*b^2*d^6*f^4 - b^4*d^6*f^4 - 4*a^2*b^2*c^6*f^4 - a^4*d^6*f ^4 + 6*a^4*c^2*d^4*f^4 - 9*a^4*c^4*d^2*f^4 + 6*b^4*c^2*d^4*f^4 - 9*b^4*c^4 *d^2*f^4 - 48*a^2*b^2*c^2*d^4*f^4 + 42*a^2*b^2*c^4*d^2*f^4 + 12*a*b^3*c*d^ 5*f^4 + 12*a*b^3*c^5*d*f^4 - 12*a^3*b*c*d^5*f^4 - 12*a^3*b*c^5*d*f^4 - 40* a*b^3*c^3*d^3*f^4 + 40*a^3*b*c^3*d^3*f^4)^(1/2) - a^2*c^3*f^2 + b^2*c^3*f^ 2 + 3*a^2*c*d^2*f^2 - 3*b^2*c*d^2*f^2 - 2*a*b*d^3*f^2 + 6*a*b*c^2*d*f^2)/( a^4*c^6*f^4 + a^4*d^6*f^4 + b^4*c^6*f^4 + b^4*d^6*f^4 + 2*a^2*b^2*c^6*f^4 + 2*a^2*b^2*d^6*f^4 + 3*a^4*c^2*d^4*f^4 + 3*a^4*c^4*d^2*f^4 + 3*b^4*c^2*d^ 4*f^4 + 3*b^4*c^4*d^2*f^4 + 6*a^2*b^2*c^2*d^4*f^4 + 6*a^2*b^2*c^4*d^2*f^4) )^(1/2)*(((((2*a^2*b^2*d^6*f^4 - b^4*d^6*f^4 - 4*a^2*b^2*c^6*f^4 - a^4*d^6 *f^4 + 6*a^4*c^2*d^4*f^4 - 9*a^4*c^4*d^2*f^4 + 6*b^4*c^2*d^4*f^4 - 9*b^...