3.13.54 \(\int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1254]

3.13.54.1 Optimal result

Integrand size = 27, antiderivative size = 216 \[ \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {(i a+b)^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}-\frac {(i a-b)^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{3/2} f}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d^2 \left (c^2+d^2\right ) f} \]

output

(I*a+b)^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(3/2)/f-(I 
*a-b)^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(3/2)/f-2*b* 
(a*d*(-a*d+2*b*c)-b^2*(2*c^2+d^2))*(c+d*tan(f*x+e))^(1/2)/d^2/(c^2+d^2)/f- 
2*(-a*d+b*c)^2*(a+b*tan(f*x+e))/d/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)
 

3.13.54.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.95 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {-i b \left (3 a^2-b^2\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )+\frac {4 b^2 (b c-2 a d)}{d \sqrt {c+d \tan (e+f x)}}+\frac {\left (3 a^2 b c-b^3 c-a^3 d+3 a b^2 d\right ) \left ((-i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )+(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right )}{\left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {2 b^2 (a+b \tan (e+f x))}{\sqrt {c+d \tan (e+f x)}}}{d f} \]

input

Integrate[(a + b*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(3/2),x]
 

output

((-I)*b*(3*a^2 - b^2)*(ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]]/Sqr 
t[c - I*d] - ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]]/Sqrt[c + I*d] 
) + (4*b^2*(b*c - 2*a*d))/(d*Sqrt[c + d*Tan[e + f*x]]) + ((3*a^2*b*c - b^3 
*c - a^3*d + 3*a*b^2*d)*(((-I)*c + d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + 
 d*Tan[e + f*x])/(c - I*d)] + (I*c + d)*Hypergeometric2F1[-1/2, 1, 1/2, (c 
 + d*Tan[e + f*x])/(c + I*d)]))/((c^2 + d^2)*Sqrt[c + d*Tan[e + f*x]]) + ( 
2*b^2*(a + b*Tan[e + f*x]))/Sqrt[c + d*Tan[e + f*x]])/(d*f)
 

3.13.54.3 Rubi [A] (warning: unable to verify)

Time = 1.12 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4048, 27, 3042, 4113, 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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \frac {2 \int \frac {c d a^3+4 b d^2 a^2-5 b^2 c d a+2 b^3 c^2-b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \tan ^2(e+f x)+d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)}{2 \sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {c d a^3+4 b d^2 a^2-5 b^2 c d a+2 b^3 c^2-b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \tan ^2(e+f x)+d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {c d a^3+4 b d^2 a^2-5 b^2 c d a+2 b^3 c^2-b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \tan (e+f x)^2+d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {\int \frac {d \left (c a^3+3 b d a^2-3 b^2 c a-b^3 d\right )+d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {d \left (c a^3+3 b d a^2-3 b^2 c a-b^3 d\right )+d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4022

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} d (a-i b)^3 (c+i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} d (a+i b)^3 (c-i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} d (a-i b)^3 (c+i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} d (a+i b)^3 (c-i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i d (a-i b)^3 (c+i d) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i d (a+i b)^3 (c-i d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {i d (a-i b)^3 (c+i d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i d (a+i b)^3 (c-i d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a-i b)^3 (c+i d) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {(a+i b)^3 (c-i d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {d (a-i b)^3 (c+i d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {d (a+i b)^3 (c-i d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}-\frac {2 b \left (a d (2 b c-a d)-b^2 \left (2 c^2+d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}\)

input

Int[(a + b*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(3/2),x]
 

output

(-2*(b*c - a*d)^2*(a + b*Tan[e + f*x]))/(d*(c^2 + d^2)*f*Sqrt[c + d*Tan[e 
+ f*x]]) + (((a - I*b)^3*(c + I*d)*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/( 
Sqrt[c - I*d]*f) + ((a + I*b)^3*(c - I*d)*d*ArcTan[Tan[e + f*x]/Sqrt[c + I 
*d]])/(Sqrt[c + I*d]*f) - (2*b*(a*d*(2*b*c - a*d) - b^2*(2*c^2 + d^2))*Sqr 
t[c + d*Tan[e + f*x]])/(d*f))/(d*(c^2 + d^2))
 

3.13.54.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

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]
 

rule 221

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4020

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]
 

rule 4022

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]
 

rule 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

rule 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

3.13.54.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(7393\) vs. \(2(194)=388\).

Time = 1.05 (sec) , antiderivative size = 7394, normalized size of antiderivative = 34.23

method	result	size
parts	\(\text {Expression too large to display}\)	\(7394\)
derivativedivides	\(\text {Expression too large to display}\)	\(16343\)
default	\(\text {Expression too large to display}\)	\(16343\)

input

int((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

output

result too large to display
 

3.13.54.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 9149 vs. \(2 (188) = 376\).

Time = 6.63 (sec) , antiderivative size = 9149, normalized size of antiderivative = 42.36 \[ \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

input

integrate((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")
 

output

Too large to include
 

3.13.54.6 Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

input

integrate((a+b*tan(f*x+e))**3/(c+d*tan(f*x+e))**(3/2),x)
 

output

Integral((a + b*tan(e + f*x))**3/(c + d*tan(e + f*x))**(3/2), x)
 

3.13.54.7 Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]

input

integrate((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")
 

output

Timed out
 

3.13.54.8 Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]

input

integrate((a+b*tan(f*x+e))^3/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")
 

output

Timed out
 

3.13.54.9 Mupad [B] (verification not implemented)

Time = 14.84 (sec) , antiderivative size = 20864, normalized size of antiderivative = 96.59 \[ \int \frac {(a+b \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]

input

int((a + b*tan(e + f*x))^3/(c + d*tan(e + f*x))^(3/2),x)
 

output

(2*b^3*(c + d*tan(e + f*x))^(1/2))/(d^2*f) - atan((((-(((8*a^6*c^3*f^2 - 8 
*b^6*c^3*f^2 - 48*a*b^5*d^3*f^2 - 48*a^5*b*d^3*f^2 - 24*a^6*c*d^2*f^2 + 24 
*b^6*c*d^2*f^2 + 120*a^2*b^4*c^3*f^2 - 120*a^4*b^2*c^3*f^2 + 160*a^3*b^3*d 
^3*f^2 + 144*a*b^5*c^2*d*f^2 + 144*a^5*b*c^2*d*f^2 - 360*a^2*b^4*c*d^2*f^2 
 - 480*a^3*b^3*c^2*d*f^2 + 360*a^4*b^2*c*d^2*f^2)^2/4 - (16*c^6*f^4 + 16*d 
^6*f^4 + 48*c^2*d^4*f^4 + 48*c^4*d^2*f^4)*(a^12 + b^12 + 6*a^2*b^10 + 15*a 
^4*b^8 + 20*a^6*b^6 + 15*a^8*b^4 + 6*a^10*b^2))^(1/2) + 4*a^6*c^3*f^2 - 4* 
b^6*c^3*f^2 - 24*a*b^5*d^3*f^2 - 24*a^5*b*d^3*f^2 - 12*a^6*c*d^2*f^2 + 12* 
b^6*c*d^2*f^2 + 60*a^2*b^4*c^3*f^2 - 60*a^4*b^2*c^3*f^2 + 80*a^3*b^3*d^3*f 
^2 + 72*a*b^5*c^2*d*f^2 + 72*a^5*b*c^2*d*f^2 - 180*a^2*b^4*c*d^2*f^2 - 240 
*a^3*b^3*c^2*d*f^2 + 180*a^4*b^2*c*d^2*f^2)/(16*(c^6*f^4 + d^6*f^4 + 3*c^2 
*d^4*f^4 + 3*c^4*d^2*f^4)))^(1/2)*((c + d*tan(e + f*x))^(1/2)*(-(((8*a^6*c 
^3*f^2 - 8*b^6*c^3*f^2 - 48*a*b^5*d^3*f^2 - 48*a^5*b*d^3*f^2 - 24*a^6*c*d^ 
2*f^2 + 24*b^6*c*d^2*f^2 + 120*a^2*b^4*c^3*f^2 - 120*a^4*b^2*c^3*f^2 + 160 
*a^3*b^3*d^3*f^2 + 144*a*b^5*c^2*d*f^2 + 144*a^5*b*c^2*d*f^2 - 360*a^2*b^4 
*c*d^2*f^2 - 480*a^3*b^3*c^2*d*f^2 + 360*a^4*b^2*c*d^2*f^2)^2/4 - (16*c^6* 
f^4 + 16*d^6*f^4 + 48*c^2*d^4*f^4 + 48*c^4*d^2*f^4)*(a^12 + b^12 + 6*a^2*b 
^10 + 15*a^4*b^8 + 20*a^6*b^6 + 15*a^8*b^4 + 6*a^10*b^2))^(1/2) + 4*a^6*c^ 
3*f^2 - 4*b^6*c^3*f^2 - 24*a*b^5*d^3*f^2 - 24*a^5*b*d^3*f^2 - 12*a^6*c*d^2 
*f^2 + 12*b^6*c*d^2*f^2 + 60*a^2*b^4*c^3*f^2 - 60*a^4*b^2*c^3*f^2 + 80*...