\(\int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx\) [2508]

Optimal result

Integrand size = 30, antiderivative size = 209 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx=-\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )+\sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )+\frac {1}{2} \text {RootSum}\left [3 a^2-2 b-5 a \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-3 a^2 \log (x)+2 b \log (x)+3 a^2 \log \left (\sqrt [4]{b x^3+a x^4}-x \text {$\#$1}\right )-2 b \log \left (\sqrt [4]{b x^3+a x^4}-x \text {$\#$1}\right )+2 a \log (x) \text {$\#$1}^4-2 a \log \left (\sqrt [4]{b x^3+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{5 a \text {$\#$1}^3-4 \text {$\#$1}^7}\&\right ] \] Output:

Unintegrable
 

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx=-\frac {x^{9/4} (b+a x)^{3/4} \left (8 \sqrt [4]{a} \left (\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )\right )+\text {RootSum}\left [3 a^2-2 b-5 a \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {3 a^2 \log (x)-2 b \log (x)-12 a^2 \log \left (\sqrt [4]{b+a x}-\sqrt [4]{x} \text {$\#$1}\right )+8 b \log \left (\sqrt [4]{b+a x}-\sqrt [4]{x} \text {$\#$1}\right )-2 a \log (x) \text {$\#$1}^4+8 a \log \left (\sqrt [4]{b+a x}-\sqrt [4]{x} \text {$\#$1}\right ) \text {$\#$1}^4}{5 a \text {$\#$1}^3-4 \text {$\#$1}^7}\&\right ]\right )}{8 \left (x^3 (b+a x)\right )^{3/4}} \] Input:

Integrate[(b*x^3 + a*x^4)^(1/4)/(-2*b + a*x + 2*x^2),x]
 

Output:

-1/8*(x^(9/4)*(b + a*x)^(3/4)*(8*a^(1/4)*(ArcTan[(a^(1/4)*x^(1/4))/(b + a* 
x)^(1/4)] - ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)]) + RootSum[3*a^2 - 
2*b - 5*a*#1^4 + 2*#1^8 & , (3*a^2*Log[x] - 2*b*Log[x] - 12*a^2*Log[(b + a 
*x)^(1/4) - x^(1/4)*#1] + 8*b*Log[(b + a*x)^(1/4) - x^(1/4)*#1] - 2*a*Log[ 
x]*#1^4 + 8*a*Log[(b + a*x)^(1/4) - x^(1/4)*#1]*#1^4)/(5*a*#1^3 - 4*#1^7) 
& ]))/(x^3*(b + a*x))^(3/4)
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1055\) vs. \(2(209)=418\).

Time = 3.10 (sec) , antiderivative size = 1055, normalized size of antiderivative = 5.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2467, 25, 1202, 25, 73, 854, 827, 216, 219, 2035, 7279, 2009}

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.

\(\displaystyle \int \frac {\sqrt [4]{a x^4+b x^3}}{a x-2 b+2 x^2} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [4]{a x^4+b x^3} \int -\frac {x^{3/4} \sqrt [4]{b+a x}}{-2 x^2-a x+2 b}dx}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \int \frac {x^{3/4} \sqrt [4]{b+a x}}{-2 x^2-a x+2 b}dx}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 1202

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (-\frac {1}{2} \int -\frac {2 a b-\left (a^2-2 b\right ) x}{\sqrt [4]{x} (b+a x)^{3/4} \left (-2 x^2-a x+2 b\right )}dx-\frac {1}{2} a \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}}dx\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (\frac {1}{2} \int \frac {2 a b-\left (a^2-2 b\right ) x}{\sqrt [4]{x} (b+a x)^{3/4} \left (-2 x^2-a x+2 b\right )}dx-\frac {1}{2} a \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}}dx\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (\frac {1}{2} \int \frac {2 a b-\left (a^2-2 b\right ) x}{\sqrt [4]{x} (b+a x)^{3/4} \left (-2 x^2-a x+2 b\right )}dx-2 a \int \frac {\sqrt {x}}{(b+a x)^{3/4}}d\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 854

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (\frac {1}{2} \int \frac {2 a b-\left (a^2-2 b\right ) x}{\sqrt [4]{x} (b+a x)^{3/4} \left (-2 x^2-a x+2 b\right )}dx-2 a \int \frac {\sqrt {x}}{1-a x}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (\frac {1}{2} \int \frac {2 a b-\left (a^2-2 b\right ) x}{\sqrt [4]{x} (b+a x)^{3/4} \left (-2 x^2-a x+2 b\right )}dx-2 a \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} \sqrt {x}+1}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}\right )\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (\frac {1}{2} \int \frac {2 a b-\left (a^2-2 b\right ) x}{\sqrt [4]{x} (b+a x)^{3/4} \left (-2 x^2-a x+2 b\right )}dx-2 a \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (\frac {1}{2} \int \frac {2 a b-\left (a^2-2 b\right ) x}{\sqrt [4]{x} (b+a x)^{3/4} \left (-2 x^2-a x+2 b\right )}dx-2 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (2 \int \frac {\sqrt {x} \left (2 a b-\left (a^2-2 b\right ) x\right )}{(b+a x)^{3/4} \left (-2 x^2-a x+2 b\right )}d\sqrt [4]{x}-2 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 7279

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (2 \int \left (\frac {\left (a^2-2 b\right ) x^{3/2}}{(b+a x)^{3/4} \left (2 x^2+a x-2 b\right )}-\frac {2 a b \sqrt {x}}{(b+a x)^{3/4} \left (2 x^2+a x-2 b\right )}\right )d\sqrt [4]{x}-2 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )\right )}{x^{3/4} \sqrt [4]{a x+b}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sqrt [4]{a x^4+b x^3} \left (2 \left (\frac {\left (a^2-2 b\right ) \left (a-\sqrt {a^2+16 b}\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a^2-\sqrt {a^2+16 b} a-4 b} \sqrt [4]{x}}{\sqrt [4]{a-\sqrt {a^2+16 b}} \sqrt [4]{b+a x}}\right )}{2 \sqrt {a^2+16 b} \left (a^2-\sqrt {a^2+16 b} a-4 b\right )^{3/4}}+\frac {4 a b \arctan \left (\frac {\sqrt [4]{a^2-\sqrt {a^2+16 b} a-4 b} \sqrt [4]{x}}{\sqrt [4]{a-\sqrt {a^2+16 b}} \sqrt [4]{b+a x}}\right )}{\sqrt {a^2+16 b} \sqrt [4]{a-\sqrt {a^2+16 b}} \left (a^2-\sqrt {a^2+16 b} a-4 b\right )^{3/4}}-\frac {\left (a^2-2 b\right ) \left (a+\sqrt {a^2+16 b}\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a^2+\sqrt {a^2+16 b} a-4 b} \sqrt [4]{x}}{\sqrt [4]{a+\sqrt {a^2+16 b}} \sqrt [4]{b+a x}}\right )}{2 \sqrt {a^2+16 b} \left (a^2+\sqrt {a^2+16 b} a-4 b\right )^{3/4}}-\frac {4 a b \arctan \left (\frac {\sqrt [4]{a^2+\sqrt {a^2+16 b} a-4 b} \sqrt [4]{x}}{\sqrt [4]{a+\sqrt {a^2+16 b}} \sqrt [4]{b+a x}}\right )}{\sqrt {a^2+16 b} \sqrt [4]{a+\sqrt {a^2+16 b}} \left (a^2+\sqrt {a^2+16 b} a-4 b\right )^{3/4}}-\frac {\left (a^2-2 b\right ) \left (a-\sqrt {a^2+16 b}\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a^2-\sqrt {a^2+16 b} a-4 b} \sqrt [4]{x}}{\sqrt [4]{a-\sqrt {a^2+16 b}} \sqrt [4]{b+a x}}\right )}{2 \sqrt {a^2+16 b} \left (a^2-\sqrt {a^2+16 b} a-4 b\right )^{3/4}}-\frac {4 a b \text {arctanh}\left (\frac {\sqrt [4]{a^2-\sqrt {a^2+16 b} a-4 b} \sqrt [4]{x}}{\sqrt [4]{a-\sqrt {a^2+16 b}} \sqrt [4]{b+a x}}\right )}{\sqrt {a^2+16 b} \sqrt [4]{a-\sqrt {a^2+16 b}} \left (a^2-\sqrt {a^2+16 b} a-4 b\right )^{3/4}}+\frac {\left (a^2-2 b\right ) \left (a+\sqrt {a^2+16 b}\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a^2+\sqrt {a^2+16 b} a-4 b} \sqrt [4]{x}}{\sqrt [4]{a+\sqrt {a^2+16 b}} \sqrt [4]{b+a x}}\right )}{2 \sqrt {a^2+16 b} \left (a^2+\sqrt {a^2+16 b} a-4 b\right )^{3/4}}+\frac {4 a b \text {arctanh}\left (\frac {\sqrt [4]{a^2+\sqrt {a^2+16 b} a-4 b} \sqrt [4]{x}}{\sqrt [4]{a+\sqrt {a^2+16 b}} \sqrt [4]{b+a x}}\right )}{\sqrt {a^2+16 b} \sqrt [4]{a+\sqrt {a^2+16 b}} \left (a^2+\sqrt {a^2+16 b} a-4 b\right )^{3/4}}\right )-2 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )}{2 a^{3/4}}\right )\right )}{x^{3/4} \sqrt [4]{b+a x}}\)

Input:

Int[(b*x^3 + a*x^4)^(1/4)/(-2*b + a*x + 2*x^2),x]
 

Output:

-(((b*x^3 + a*x^4)^(1/4)*(-2*a*(-1/2*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1 
/4)]/a^(3/4) + ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)]/(2*a^(3/4))) + 2 
*((4*a*b*ArcTan[((a^2 - 4*b - a*Sqrt[a^2 + 16*b])^(1/4)*x^(1/4))/((a - Sqr 
t[a^2 + 16*b])^(1/4)*(b + a*x)^(1/4))])/(Sqrt[a^2 + 16*b]*(a - Sqrt[a^2 + 
16*b])^(1/4)*(a^2 - 4*b - a*Sqrt[a^2 + 16*b])^(3/4)) + ((a^2 - 2*b)*(a - S 
qrt[a^2 + 16*b])^(3/4)*ArcTan[((a^2 - 4*b - a*Sqrt[a^2 + 16*b])^(1/4)*x^(1 
/4))/((a - Sqrt[a^2 + 16*b])^(1/4)*(b + a*x)^(1/4))])/(2*Sqrt[a^2 + 16*b]* 
(a^2 - 4*b - a*Sqrt[a^2 + 16*b])^(3/4)) - (4*a*b*ArcTan[((a^2 - 4*b + a*Sq 
rt[a^2 + 16*b])^(1/4)*x^(1/4))/((a + Sqrt[a^2 + 16*b])^(1/4)*(b + a*x)^(1/ 
4))])/(Sqrt[a^2 + 16*b]*(a + Sqrt[a^2 + 16*b])^(1/4)*(a^2 - 4*b + a*Sqrt[a 
^2 + 16*b])^(3/4)) - ((a^2 - 2*b)*(a + Sqrt[a^2 + 16*b])^(3/4)*ArcTan[((a^ 
2 - 4*b + a*Sqrt[a^2 + 16*b])^(1/4)*x^(1/4))/((a + Sqrt[a^2 + 16*b])^(1/4) 
*(b + a*x)^(1/4))])/(2*Sqrt[a^2 + 16*b]*(a^2 - 4*b + a*Sqrt[a^2 + 16*b])^( 
3/4)) - (4*a*b*ArcTanh[((a^2 - 4*b - a*Sqrt[a^2 + 16*b])^(1/4)*x^(1/4))/(( 
a - Sqrt[a^2 + 16*b])^(1/4)*(b + a*x)^(1/4))])/(Sqrt[a^2 + 16*b]*(a - Sqrt 
[a^2 + 16*b])^(1/4)*(a^2 - 4*b - a*Sqrt[a^2 + 16*b])^(3/4)) - ((a^2 - 2*b) 
*(a - Sqrt[a^2 + 16*b])^(3/4)*ArcTanh[((a^2 - 4*b - a*Sqrt[a^2 + 16*b])^(1 
/4)*x^(1/4))/((a - Sqrt[a^2 + 16*b])^(1/4)*(b + a*x)^(1/4))])/(2*Sqrt[a^2 
+ 16*b]*(a^2 - 4*b - a*Sqrt[a^2 + 16*b])^(3/4)) + (4*a*b*ArcTanh[((a^2 - 4 
*b + a*Sqrt[a^2 + 16*b])^(1/4)*x^(1/4))/((a + Sqrt[a^2 + 16*b])^(1/4)*(...
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n 
)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 
2^(-1)] && IntegersQ[m, p + (m + 1)/n]
 

rule 1202

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [N/A] (verified)

Time = 0.00 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(\frac {a^{\frac {1}{4}} \ln \left (\frac {-x \,a^{\frac {1}{4}}-\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}-\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}\right )}{2}+a^{\frac {1}{4}} \arctan \left (\frac {\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}-5 a \,\textit {\_Z}^{4}+3 a^{2}-2 b \right )}{\sum }\frac {\left (2 \textit {\_R}^{4} a -3 a^{2}+2 b \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (4 \textit {\_R}^{4}-5 a \right )}\right )}{2}\)	\(150\)

Input:

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

Output:

1/2*a^(1/4)*ln((-x*a^(1/4)-(x^3*(a*x+b))^(1/4))/(x*a^(1/4)-(x^3*(a*x+b))^( 
1/4)))+a^(1/4)*arctan(1/a^(1/4)/x*(x^3*(a*x+b))^(1/4))+1/2*sum((2*_R^4*a-3 
*a^2+2*b)*ln((-_R*x+(x^3*(a*x+b))^(1/4))/x)/_R^3/(4*_R^4-5*a),_R=RootOf(2* 
_Z^8-5*_Z^4*a+3*a^2-2*b))
 

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.47 (sec) , antiderivative size = 3186, normalized size of antiderivative = 15.24 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [N/A]

Not integrable

Time = 1.77 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.11 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx=\int \frac {\sqrt [4]{x^{3} \left (a x + b\right )}}{a x - 2 b + 2 x^{2}}\, dx \] Input:

integrate((a*x**4+b*x**3)**(1/4)/(a*x+2*x**2-2*b),x)
 

Output:

Integral((x**3*(a*x + b))**(1/4)/(a*x - 2*b + 2*x**2), x)
 

Maxima [N/A]

Not integrable

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.14 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx=\int { \frac {{\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{a x + 2 \, x^{2} - 2 \, b} \,d x } \] Input:

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

Output:

integrate((a*x^4 + b*x^3)^(1/4)/(a*x + 2*x^2 - 2*b), x)
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 2.64 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx=\frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right ) \] Input:

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

Output:

1/2*sqrt(2)*(-a)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x 
)^(1/4))/(-a)^(1/4)) + 1/2*sqrt(2)*(-a)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2) 
*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4)) + 1/4*sqrt(2)*(-a)^(1/4)*log( 
sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x)) - 1/4*sqrt( 
2)*(-a)^(1/4)*log(-sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a 
+ b/x))
 

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.14 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx=\int \frac {{\left (a\,x^4+b\,x^3\right )}^{1/4}}{2\,x^2+a\,x-2\,b} \,d x \] Input:

int((a*x^4 + b*x^3)^(1/4)/(a*x - 2*b + 2*x^2),x)
 

Output:

int((a*x^4 + b*x^3)^(1/4)/(a*x - 2*b + 2*x^2), x)
 

Reduce [N/A]

Not integrable

Time = 200.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.14 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{-2 b+a x+2 x^2} \, dx=\int \frac {\left (a \,x^{4}+b \,x^{3}\right )^{\frac {1}{4}}}{a x +2 x^{2}-2 b}d x \] Input:

int((a*x^4+b*x^3)^(1/4)/(a*x+2*x^2-2*b),x)
 

Output:

int((a*x^4+b*x^3)^(1/4)/(a*x+2*x^2-2*b),x)