Integrand size = 27, antiderivative size = 231 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx=-\frac {i (a-i b)^2 (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {i (a+i b)^2 (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {4 (b c+a d) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (2 a b c+a^2 d-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f} \]
-I*(a-I*b)^2*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f +I*(a+I*b)^2*(c+I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f +4*(a*d+b*c)*(a*c-b*d)*(c+d*tan(f*x+e))^(1/2)/f+2/3*(a^2*d+2*a*b*c-b^2*d)* (c+d*tan(f*x+e))^(3/2)/f+4/5*a*b*(c+d*tan(f*x+e))^(5/2)/f+2/7*b^2*(c+d*tan (f*x+e))^(7/2)/d/f
Time = 2.20 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.13 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx=\frac {\frac {4 b^2 (c+d \tan (e+f x))^{7/2}}{d}+7 i (a-i b)^2 \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c-i d) \left (-3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )\right )-7 i (a+i b)^2 \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c+i d) \left (-3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )\right )}{14 f} \]
((4*b^2*(c + d*Tan[e + f*x])^(7/2))/d + (7*I)*(a - I*b)^2*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c - I*d)*(-3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Ta n[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c - (3*I)*d + d*T an[e + f*x])))/3) - (7*I)*(a + I*b)^2*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c + I*d)*(-3*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))/3))/( 14*f)
Time = 1.31 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.90, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4026, 3042, 4011, 3042, 4011, 3042, 4011, 3042, 4022, 3042, 4020, 25, 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.
| \(\displaystyle \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4026 |
| \(\displaystyle \int \left (a^2+2 b \tan (e+f x) a-b^2\right ) (c+d \tan (e+f x))^{5/2}dx+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a^2+2 b \tan (e+f x) a-b^2\right ) (c+d \tan (e+f x))^{5/2}dx+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int (c+d \tan (e+f x))^{3/2} \left (c a^2-2 b d a-b^2 c+\left (d a^2+2 b c a-b^2 d\right ) \tan (e+f x)\right )dx+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d \tan (e+f x))^{3/2} \left (c a^2-2 b d a-b^2 c+\left (d a^2+2 b c a-b^2 d\right ) \tan (e+f x)\right )dx+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \sqrt {c+d \tan (e+f x)} ((a c-b c-a d-b d) (a c+b c+a d-b d)+2 (b c+a d) (a c-b d) \tan (e+f x))dx+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {c+d \tan (e+f x)} ((a c-b c-a d-b d) (a c+b c+a d-b d)+2 (b c+a d) (a c-b d) \tan (e+f x))dx+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {\left (c^3-3 c d^2\right ) a^2-2 b d \left (3 c^2-d^2\right ) a-b^2 c \left (c^2-3 d^2\right )+\left (\left (3 c^2 d-d^3\right ) a^2+2 b c \left (c^2-3 d^2\right ) a-b^2 d \left (3 c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 (a d+b c) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (c^3-3 c d^2\right ) a^2-2 b d \left (3 c^2-d^2\right ) a-b^2 c \left (c^2-3 d^2\right )+\left (\left (3 c^2 d-d^3\right ) a^2+2 b c \left (c^2-3 d^2\right ) a-b^2 d \left (3 c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 (a d+b c) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {1}{2} (a+i b)^2 (c+i d)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b)^2 (c-i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 (a d+b c) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (a+i b)^2 (c+i d)^3 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b)^2 (c-i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 (a d+b c) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i (a-i b)^2 (c-i d)^3 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b)^2 (c+i d)^3 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 (a d+b c) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i (a-i b)^2 (c-i d)^3 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i (a+i b)^2 (c+i d)^3 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 (a d+b c) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+i b)^2 (c+i d)^3 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a-i b)^2 (c-i d)^3 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 (a d+b c) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 \left (a^2 d+2 a b c-b^2 d\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {(a-i b)^2 (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b)^2 (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {4 a b (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 (a d+b c) (a c-b d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
((a - I*b)^2*(c - I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((a + I*b)^2*(c + I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f + (4*(b*c + a*d)*(a*c - b*d)*Sqrt[c + d*Tan[e + f*x]])/f + (2*(2*a*b*c + a^2*d - b^2*d )*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (4*a*b*(c + d*Tan[e + f*x])^(5/2))/( 5*f) + (2*b^2*(c + d*Tan[e + f*x])^(7/2))/(7*d*f)
3.13.42.3.1 Defintions of rubi rules used
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[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
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_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(3586\) vs. \(2(197)=394\).
Time = 0.94 (sec) , antiderivative size = 3587, normalized size of antiderivative = 15.53
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(3587\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(3700\) |
| default | \(\text {Expression too large to display}\) | \(3700\) |
a^2*(2/3/f*d*(c+d*tan(f*x+e))^(3/2)+4/f*d*(c+d*tan(f*x+e))^(1/2)*c-1/4/f/d *ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c ^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c^2+1/4/f*d*l n(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2 +d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)+1/4/f/d*ln(d*ta n(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^ (1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-3/4/f*d*ln(d*tan(f*x+e)+c+(c+d*ta n(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2 )^(1/2)+2*c)^(1/2)*c+3/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*ta n(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/ 2))*c^2-1/f*d^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^( 1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-2/f*d/( 2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2) ^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*c+1/4/f/ d*ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+( c^2+d^2)^(1/2))*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-1/4/f*d* ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^ 2+d^2)^(1/2))*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-1/4/f/d*ln(d*t an(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2) ^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3+3/4/f*d*ln(d*tan(f*x+e)+c-(c+...
Leaf count of result is larger than twice the leaf count of optimal. 8466 vs. \(2 (189) = 378\).
Time = 2.77 (sec) , antiderivative size = 8466, normalized size of antiderivative = 36.65 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]
\[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{2} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]
\[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
Time = 78.78 (sec) , antiderivative size = 23642, normalized size of antiderivative = 102.35 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]
(((4*b^2*c - 4*a*b*d)/(d*f) - (4*b^2*c)/(d*f))*(c^2 + d^2) - 2*c*(2*c*((4* b^2*c - 4*a*b*d)/(d*f) - (4*b^2*c)/(d*f)) - (2*(a*d - b*c)^2)/(d*f) + (2*b ^2*(c^2 + d^2))/(d*f)))*(c + d*tan(e + f*x))^(1/2) - (c + d*tan(e + f*x))^ (3/2)*((2*c*((4*b^2*c - 4*a*b*d)/(d*f) - (4*b^2*c)/(d*f)))/3 - (2*(a*d - b *c)^2)/(3*d*f) + (2*b^2*(c^2 + d^2))/(3*d*f)) - ((4*b^2*c - 4*a*b*d)/(5*d* f) - (4*b^2*c)/(5*d*f))*(c + d*tan(e + f*x))^(5/2) - atan(((((8*(8*a^2*c*d ^5*f^2 - 8*b^2*c*d^5*f^2 + 8*a^2*c^3*d^3*f^2 - 8*b^2*c^3*d^3*f^2 - 8*a*b*d ^6*f^2 + 8*a*b*c^4*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-( ((8*a^4*c^5*f^2 + 8*b^4*c^5*f^2 + 32*a*b^3*d^5*f^2 - 32*a^3*b*d^5*f^2 + 40 *a^4*c*d^4*f^2 + 40*b^4*c*d^4*f^2 - 48*a^2*b^2*c^5*f^2 - 80*a^4*c^3*d^2*f^ 2 - 80*b^4*c^3*d^2*f^2 + 480*a^2*b^2*c^3*d^2*f^2 + 160*a*b^3*c^4*d*f^2 - 1 60*a^3*b*c^4*d*f^2 - 320*a*b^3*c^2*d^3*f^2 - 240*a^2*b^2*c*d^4*f^2 + 320*a ^3*b*c^2*d^3*f^2)^2/64 - f^4*(a^8*c^10 + a^8*d^10 + b^8*c^10 + b^8*d^10 + 4*a^2*b^6*c^10 + 6*a^4*b^4*c^10 + 4*a^6*b^2*c^10 + 4*a^2*b^6*d^10 + 6*a^4* b^4*d^10 + 4*a^6*b^2*d^10 + 5*a^8*c^2*d^8 + 10*a^8*c^4*d^6 + 10*a^8*c^6*d^ 4 + 5*a^8*c^8*d^2 + 5*b^8*c^2*d^8 + 10*b^8*c^4*d^6 + 10*b^8*c^6*d^4 + 5*b^ 8*c^8*d^2 + 20*a^2*b^6*c^2*d^8 + 40*a^2*b^6*c^4*d^6 + 40*a^2*b^6*c^6*d^4 + 20*a^2*b^6*c^8*d^2 + 30*a^4*b^4*c^2*d^8 + 60*a^4*b^4*c^4*d^6 + 60*a^4*b^4 *c^6*d^4 + 30*a^4*b^4*c^8*d^2 + 20*a^6*b^2*c^2*d^8 + 40*a^6*b^2*c^4*d^6 + 40*a^6*b^2*c^6*d^4 + 20*a^6*b^2*c^8*d^2))^(1/2) + a^4*c^5*f^2 + b^4*c^5...