\(\int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx\) [41]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [B] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [F]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 32, antiderivative size = 153 \[ \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx=\frac {2 \text {arctanh}\left (\frac {d+4 e x}{\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}}\right )}{\sqrt {d^4-64 a e^3} \sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}}-\frac {2 \text {arctanh}\left (\frac {d+4 e x}{\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}}\right )}{\sqrt {d^4-64 a e^3} \sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}} \] Output:

2*arctanh((4*e*x+d)/(3*d^2-2*(-64*a*e^3+d^4)^(1/2))^(1/2))/(-64*a*e^3+d^4) 
^(1/2)/(3*d^2-2*(-64*a*e^3+d^4)^(1/2))^(1/2)-2*arctanh((4*e*x+d)/(3*d^2+2* 
(-64*a*e^3+d^4)^(1/2))^(1/2))/(-64*a*e^3+d^4)^(1/2)/(3*d^2+2*(-64*a*e^3+d^ 
4)^(1/2))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.46 \[ \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx=-\text {RootSum}\left [8 a e^2-d^3 \text {$\#$1}+8 d e^2 \text {$\#$1}^3+8 e^3 \text {$\#$1}^4\&,\frac {\log (x-\text {$\#$1})}{d^3-24 d e^2 \text {$\#$1}^2-32 e^3 \text {$\#$1}^3}\&\right ] \] Input:

Integrate[(8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4)^(-1),x]
 

Output:

-RootSum[8*a*e^2 - d^3*#1 + 8*d*e^2*#1^3 + 8*e^3*#1^4 & , Log[x - #1]/(d^3 
 - 24*d*e^2*#1^2 - 32*e^3*#1^3) & ]
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2458, 1406, 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.

\(\displaystyle \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx\)

\(\Big \downarrow \) 2458

\(\displaystyle \int \frac {1}{\frac {1}{32} \left (256 a e^2+\frac {5 d^4}{e}\right )-3 d^2 e \left (\frac {d}{4 e}+x\right )^2+8 e^3 \left (\frac {d}{4 e}+x\right )^4}d\left (\frac {d}{4 e}+x\right )\)

\(\Big \downarrow \) 1406

\(\displaystyle \frac {4 e^2 \int \frac {1}{8 e^3 \left (\frac {d}{4 e}+x\right )^2-\frac {1}{2} e \left (3 d^2+2 \sqrt {d^4-64 a e^3}\right )}d\left (\frac {d}{4 e}+x\right )}{\sqrt {d^4-64 a e^3}}-\frac {4 e^2 \int \frac {1}{8 e^3 \left (\frac {d}{4 e}+x\right )^2-\frac {1}{2} e \left (3 d^2-2 \sqrt {d^4-64 a e^3}\right )}d\left (\frac {d}{4 e}+x\right )}{\sqrt {d^4-64 a e^3}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \text {arctanh}\left (\frac {4 e \left (\frac {d}{4 e}+x\right )}{\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}}\right )}{\sqrt {d^4-64 a e^3} \sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}}-\frac {2 \text {arctanh}\left (\frac {4 e \left (\frac {d}{4 e}+x\right )}{\sqrt {2 \sqrt {d^4-64 a e^3}+3 d^2}}\right )}{\sqrt {d^4-64 a e^3} \sqrt {2 \sqrt {d^4-64 a e^3}+3 d^2}}\)

Input:

Int[(8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4)^(-1),x]
 

Output:

(2*ArcTanh[(4*e*(d/(4*e) + x))/Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]]])/(Sqr 
t[d^4 - 64*a*e^3]*Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]]) - (2*ArcTanh[(4*e* 
(d/(4*e) + x))/Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]]])/(Sqrt[d^4 - 64*a*e^3 
]*Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]])
 

Defintions of rubi rules used

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 1406
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^ 
2 - 4*a*c, 2]}, Simp[c/q   Int[1/(b/2 - q/2 + c*x^2), x], x] - Simp[c/q   I 
nt[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c 
, 0] && PosQ[b^2 - 4*a*c]
 

rule 2458
Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp 
on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x 
- S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp 
on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P 
n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
 
Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.44

method result size
default \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (8 e^{3} \textit {\_Z}^{4}+8 d \,e^{2} \textit {\_Z}^{3}-d^{3} \textit {\_Z} +8 a \,e^{2}\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{32 \textit {\_R}^{3} e^{3}+24 \textit {\_R}^{2} d \,e^{2}-d^{3}}\) \(67\)
risch \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (8 e^{3} \textit {\_Z}^{4}+8 d \,e^{2} \textit {\_Z}^{3}-d^{3} \textit {\_Z} +8 a \,e^{2}\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{32 \textit {\_R}^{3} e^{3}+24 \textit {\_R}^{2} d \,e^{2}-d^{3}}\) \(67\)

Input:

int(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2),x,method=_RETURNVERBOSE)
 

Output:

sum(1/(32*_R^3*e^3+24*_R^2*d*e^2-d^3)*ln(x-_R),_R=RootOf(8*_Z^4*e^3+8*_Z^3 
*d*e^2-_Z*d^3+8*a*e^2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1115 vs. \(2 (133) = 266\).

Time = 0.09 (sec) , antiderivative size = 1115, normalized size of antiderivative = 7.29 \[ \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx=\text {Too large to display} \] Input:

integrate(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2),x, algorithm="fricas")
 

Output:

-sqrt((3*d^2 + 2*(5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^6)/sqrt(25*d^12 + 960 
*a*d^8*e^3 - 98304*a^2*d^4*e^6 - 4194304*a^3*e^9))/(5*d^8 - 64*a*d^4*e^3 - 
 16384*a^2*e^6))*log(8*e*x + 2*(2*d^4 - 128*a*e^3 - 3*(5*d^10 - 64*a*d^6*e 
^3 - 16384*a^2*d^2*e^6)/sqrt(25*d^12 + 960*a*d^8*e^3 - 98304*a^2*d^4*e^6 - 
 4194304*a^3*e^9))*sqrt((3*d^2 + 2*(5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^6)/ 
sqrt(25*d^12 + 960*a*d^8*e^3 - 98304*a^2*d^4*e^6 - 4194304*a^3*e^9))/(5*d^ 
8 - 64*a*d^4*e^3 - 16384*a^2*e^6)) + 2*d) + sqrt((3*d^2 + 2*(5*d^8 - 64*a* 
d^4*e^3 - 16384*a^2*e^6)/sqrt(25*d^12 + 960*a*d^8*e^3 - 98304*a^2*d^4*e^6 
- 4194304*a^3*e^9))/(5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^6))*log(8*e*x - 2* 
(2*d^4 - 128*a*e^3 - 3*(5*d^10 - 64*a*d^6*e^3 - 16384*a^2*d^2*e^6)/sqrt(25 
*d^12 + 960*a*d^8*e^3 - 98304*a^2*d^4*e^6 - 4194304*a^3*e^9))*sqrt((3*d^2 
+ 2*(5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^6)/sqrt(25*d^12 + 960*a*d^8*e^3 - 
98304*a^2*d^4*e^6 - 4194304*a^3*e^9))/(5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^ 
6)) + 2*d) - sqrt((3*d^2 - 2*(5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^6)/sqrt(2 
5*d^12 + 960*a*d^8*e^3 - 98304*a^2*d^4*e^6 - 4194304*a^3*e^9))/(5*d^8 - 64 
*a*d^4*e^3 - 16384*a^2*e^6))*log(8*e*x + 2*(2*d^4 - 128*a*e^3 + 3*(5*d^10 
- 64*a*d^6*e^3 - 16384*a^2*d^2*e^6)/sqrt(25*d^12 + 960*a*d^8*e^3 - 98304*a 
^2*d^4*e^6 - 4194304*a^3*e^9))*sqrt((3*d^2 - 2*(5*d^8 - 64*a*d^4*e^3 - 163 
84*a^2*e^6)/sqrt(25*d^12 + 960*a*d^8*e^3 - 98304*a^2*d^4*e^6 - 4194304*a^3 
*e^9))/(5*d^8 - 64*a*d^4*e^3 - 16384*a^2*e^6)) + 2*d) + sqrt((3*d^2 - 2...
 

Sympy [A] (verification not implemented)

Time = 1.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.80 \[ \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx=\operatorname {RootSum} {\left (t^{4} \cdot \left (1048576 a^{3} e^{9} - 12288 a^{2} d^{4} e^{6} - 384 a d^{8} e^{3} + 5 d^{12}\right ) + t^{2} \cdot \left (384 a d^{2} e^{3} - 6 d^{6}\right ) + 1, \left ( t \mapsto t \log {\left (x + \frac {- 49152 t^{3} a^{2} d^{2} e^{6} - 192 t^{3} a d^{6} e^{3} + 15 t^{3} d^{10} + 256 t a e^{3} - 13 t d^{4} + 2 d}{8 e} \right )} \right )\right )} \] Input:

integrate(1/(8*e**3*x**4+8*d*e**2*x**3-d**3*x+8*a*e**2),x)
 

Output:

RootSum(_t**4*(1048576*a**3*e**9 - 12288*a**2*d**4*e**6 - 384*a*d**8*e**3 
+ 5*d**12) + _t**2*(384*a*d**2*e**3 - 6*d**6) + 1, Lambda(_t, _t*log(x + ( 
-49152*_t**3*a**2*d**2*e**6 - 192*_t**3*a*d**6*e**3 + 15*_t**3*d**10 + 256 
*_t*a*e**3 - 13*_t*d**4 + 2*d)/(8*e))))
 

Maxima [F]

\[ \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx=\int { \frac {1}{8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}} \,d x } \] Input:

integrate(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2),x, algorithm="maxima")
 

Output:

integrate(1/(8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 577 vs. \(2 (133) = 266\).

Time = 0.12 (sec) , antiderivative size = 577, normalized size of antiderivative = 3.77 \[ \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx=-\frac {2 \, \log \left (x + \frac {1}{4} \, \sqrt {\frac {3 \, d^{2} e^{2} + 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} + \frac {d}{4 \, e}\right )}{e^{3} {\left (\sqrt {\frac {3 \, d^{2} e^{2} + 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} + \frac {d}{e}\right )}^{3} - 3 \, d e^{2} {\left (\sqrt {\frac {3 \, d^{2} e^{2} + 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} + \frac {d}{e}\right )}^{2} + 2 \, d^{3}} + \frac {2 \, \log \left (x - \frac {1}{4} \, \sqrt {\frac {3 \, d^{2} e^{2} + 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} + \frac {d}{4 \, e}\right )}{e^{3} {\left (\sqrt {\frac {3 \, d^{2} e^{2} + 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} - \frac {d}{e}\right )}^{3} + 3 \, d e^{2} {\left (\sqrt {\frac {3 \, d^{2} e^{2} + 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} - \frac {d}{e}\right )}^{2} - 2 \, d^{3}} - \frac {2 \, \log \left (x + \frac {1}{4} \, \sqrt {\frac {3 \, d^{2} e^{2} - 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} + \frac {d}{4 \, e}\right )}{e^{3} {\left (\sqrt {\frac {3 \, d^{2} e^{2} - 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} + \frac {d}{e}\right )}^{3} - 3 \, d e^{2} {\left (\sqrt {\frac {3 \, d^{2} e^{2} - 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} + \frac {d}{e}\right )}^{2} + 2 \, d^{3}} + \frac {2 \, \log \left (x - \frac {1}{4} \, \sqrt {\frac {3 \, d^{2} e^{2} - 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} + \frac {d}{4 \, e}\right )}{e^{3} {\left (\sqrt {\frac {3 \, d^{2} e^{2} - 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} - \frac {d}{e}\right )}^{3} + 3 \, d e^{2} {\left (\sqrt {\frac {3 \, d^{2} e^{2} - 2 \, \sqrt {d^{4} - 64 \, a e^{3}} e^{2}}{e^{4}}} - \frac {d}{e}\right )}^{2} - 2 \, d^{3}} \] Input:

integrate(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2),x, algorithm="giac")
 

Output:

-2*log(x + 1/4*sqrt((3*d^2*e^2 + 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) + 1/4*d/ 
e)/(e^3*(sqrt((3*d^2*e^2 + 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) + d/e)^3 - 3*d 
*e^2*(sqrt((3*d^2*e^2 + 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) + d/e)^2 + 2*d^3) 
 + 2*log(x - 1/4*sqrt((3*d^2*e^2 + 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) + 1/4* 
d/e)/(e^3*(sqrt((3*d^2*e^2 + 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) - d/e)^3 + 3 
*d*e^2*(sqrt((3*d^2*e^2 + 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) - d/e)^2 - 2*d^ 
3) - 2*log(x + 1/4*sqrt((3*d^2*e^2 - 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) + 1/ 
4*d/e)/(e^3*(sqrt((3*d^2*e^2 - 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) + d/e)^3 - 
 3*d*e^2*(sqrt((3*d^2*e^2 - 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) + d/e)^2 + 2* 
d^3) + 2*log(x - 1/4*sqrt((3*d^2*e^2 - 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) + 
1/4*d/e)/(e^3*(sqrt((3*d^2*e^2 - 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) - d/e)^3 
 + 3*d*e^2*(sqrt((3*d^2*e^2 - 2*sqrt(d^4 - 64*a*e^3)*e^2)/e^4) - d/e)^2 - 
2*d^3)
 

Mupad [B] (verification not implemented)

Time = 23.15 (sec) , antiderivative size = 1264, normalized size of antiderivative = 8.26 \[ \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx =\text {Too large to display} \] Input:

int(1/(8*a*e^2 - d^3*x + 8*e^3*x^4 + 8*d*e^2*x^3),x)
 

Output:

atan((d^3*(d^12 - 262144*a^3*e^9 - 192*a*d^8*e^3 + 12288*a^2*d^4*e^6)^(1/2 
)*3i - d^9*2i + a*d^5*e^3*256i - a^2*d*e^6*8192i - a^2*e^7*x*32768i - d^8* 
e*x*8i + d^2*e*x*(d^12 - 262144*a^3*e^9 - 192*a*d^8*e^3 + 12288*a^2*d^4*e^ 
6)^(1/2)*12i + a*d^4*e^4*x*1024i)/(5*d^12*(-(2*(d^12 - 262144*a^3*e^9 - 19 
2*a*d^8*e^3 + 12288*a^2*d^4*e^6)^(1/2) - 3*d^6 + 192*a*d^2*e^3)/(5*d^12 + 
1048576*a^3*e^9 - 384*a*d^8*e^3 - 12288*a^2*d^4*e^6))^(1/2) + 1048576*a^3* 
e^9*(-(2*(d^12 - 262144*a^3*e^9 - 192*a*d^8*e^3 + 12288*a^2*d^4*e^6)^(1/2) 
 - 3*d^6 + 192*a*d^2*e^3)/(5*d^12 + 1048576*a^3*e^9 - 384*a*d^8*e^3 - 1228 
8*a^2*d^4*e^6))^(1/2) - 384*a*d^8*e^3*(-(2*(d^12 - 262144*a^3*e^9 - 192*a* 
d^8*e^3 + 12288*a^2*d^4*e^6)^(1/2) - 3*d^6 + 192*a*d^2*e^3)/(5*d^12 + 1048 
576*a^3*e^9 - 384*a*d^8*e^3 - 12288*a^2*d^4*e^6))^(1/2) - 12288*a^2*d^4*e^ 
6*(-(2*(d^12 - 262144*a^3*e^9 - 192*a*d^8*e^3 + 12288*a^2*d^4*e^6)^(1/2) - 
 3*d^6 + 192*a*d^2*e^3)/(5*d^12 + 1048576*a^3*e^9 - 384*a*d^8*e^3 - 12288* 
a^2*d^4*e^6))^(1/2)))*(-(2*(d^12 - 262144*a^3*e^9 - 192*a*d^8*e^3 + 12288* 
a^2*d^4*e^6)^(1/2) - 3*d^6 + 192*a*d^2*e^3)/(5*d^12 + 1048576*a^3*e^9 - 38 
4*a*d^8*e^3 - 12288*a^2*d^4*e^6))^(1/2)*2i - atan((d^3*(d^12 - 262144*a^3* 
e^9 - 192*a*d^8*e^3 + 12288*a^2*d^4*e^6)^(1/2)*3i + d^9*2i - a*d^5*e^3*256 
i + a^2*d*e^6*8192i + a^2*e^7*x*32768i + d^8*e*x*8i + d^2*e*x*(d^12 - 2621 
44*a^3*e^9 - 192*a*d^8*e^3 + 12288*a^2*d^4*e^6)^(1/2)*12i - a*d^4*e^4*x*10 
24i)/(5*d^12*((2*(d^12 - 262144*a^3*e^9 - 192*a*d^8*e^3 + 12288*a^2*d^4...
 

Reduce [F]

\[ \int \frac {1}{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4} \, dx=\int \frac {1}{8 e^{3} x^{4}+8 d \,e^{2} x^{3}-d^{3} x +8 a \,e^{2}}d x \] Input:

int(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2),x)
 

Output:

int(1/(8*a*e**2 - d**3*x + 8*d*e**2*x**3 + 8*e**3*x**4),x)