Integrand size = 29, antiderivative size = 63 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x \left (b+a x+x^2\right )} \, dx=-\text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{b x^3+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-a+2 \text {$\#$1}^4}\&\right ] \] Output:
Unintegrable
Time = 0.00 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.46 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x \left (b+a x+x^2\right )} \, dx=-\frac {\sqrt [4]{x^3 (b+a x)} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt [4]{x}\right ) \text {$\#$1}+\log \left (\sqrt [4]{b+a x}-\sqrt [4]{x} \text {$\#$1}\right ) \text {$\#$1}}{-a+2 \text {$\#$1}^4}\&\right ]}{x^{3/4} \sqrt [4]{b+a x}} \] Input:
Integrate[(b*x^3 + a*x^4)^(1/4)/(x*(b + a*x + x^2)),x]
Output:
-(((x^3*(b + a*x))^(1/4)*RootSum[b - a*#1^4 + #1^8 & , (-(Log[x^(1/4)]*#1) + Log[(b + a*x)^(1/4) - x^(1/4)*#1]*#1)/(-a + 2*#1^4) & ])/(x^(3/4)*(b + a*x)^(1/4)))
Leaf count is larger than twice the leaf count of optimal. \(487\) vs. \(2(63)=126\).
Time = 0.89 (sec) , antiderivative size = 487, normalized size of antiderivative = 7.73, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2467, 1205, 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.
\(\displaystyle \int \frac {\sqrt [4]{a x^4+b x^3}}{x \left (a x+b+x^2\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{a x^4+b x^3} \int \frac {\sqrt [4]{b+a x}}{\sqrt [4]{x} \left (x^2+a x+b\right )}dx}{x^{3/4} \sqrt [4]{a x+b}}\) |
\(\Big \downarrow \) 1205 |
\(\displaystyle \frac {\sqrt [4]{a x^4+b x^3} \int \left (\frac {2 \sqrt [4]{b+a x}}{\sqrt {a^2-4 b} \sqrt [4]{x} \left (a+2 x-\sqrt {a^2-4 b}\right )}-\frac {2 \sqrt [4]{b+a x}}{\sqrt {a^2-4 b} \sqrt [4]{x} \left (a+2 x+\sqrt {a^2-4 b}\right )}\right )dx}{x^{3/4} \sqrt [4]{a x+b}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt [4]{a x^4+b x^3} \left (\frac {2 \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b} \arctan \left (\frac {\sqrt [4]{x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x+b}}\right )}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt {a^2-4 b}}-\frac {2 \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b} \arctan \left (\frac {\sqrt [4]{x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x+b}}\right )}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt {a^2-4 b}}-\frac {2 \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b} \text {arctanh}\left (\frac {\sqrt [4]{x} \sqrt [4]{-a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt [4]{a x+b}}\right )}{\sqrt [4]{a-\sqrt {a^2-4 b}} \sqrt {a^2-4 b}}+\frac {2 \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b} \text {arctanh}\left (\frac {\sqrt [4]{x} \sqrt [4]{a \sqrt {a^2-4 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt [4]{a x+b}}\right )}{\sqrt [4]{\sqrt {a^2-4 b}+a} \sqrt {a^2-4 b}}\right )}{x^{3/4} \sqrt [4]{a x+b}}\) |
Input:
Int[(b*x^3 + a*x^4)^(1/4)/(x*(b + a*x + x^2)),x]
Output:
((b*x^3 + a*x^4)^(1/4)*((2*(a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*ArcTan[(( a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*x^(1/4))/((a - Sqrt[a^2 - 4*b])^(1/4) *(b + a*x)^(1/4))])/((a - Sqrt[a^2 - 4*b])^(1/4)*Sqrt[a^2 - 4*b]) - (2*(a^ 2 + a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*ArcTan[((a^2 + a*Sqrt[a^2 - 4*b] - 2*b) ^(1/4)*x^(1/4))/((a + Sqrt[a^2 - 4*b])^(1/4)*(b + a*x)^(1/4))])/((a + Sqrt [a^2 - 4*b])^(1/4)*Sqrt[a^2 - 4*b]) - (2*(a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^( 1/4)*ArcTanh[((a^2 - a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*x^(1/4))/((a - Sqrt[a^ 2 - 4*b])^(1/4)*(b + a*x)^(1/4))])/((a - Sqrt[a^2 - 4*b])^(1/4)*Sqrt[a^2 - 4*b]) + (2*(a^2 + a*Sqrt[a^2 - 4*b] - 2*b)^(1/4)*ArcTanh[((a^2 + a*Sqrt[a ^2 - 4*b] - 2*b)^(1/4)*x^(1/4))/((a + Sqrt[a^2 - 4*b])^(1/4)*(b + a*x)^(1/ 4))])/((a + Sqrt[a^2 - 4*b])^(1/4)*Sqrt[a^2 - 4*b])))/(x^(3/4)*(b + a*x)^( 1/4))
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x _) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^ n, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && !IntegerQ[m] && !IntegerQ[n]
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]
Time = 0.37 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81
method | result | size |
pseudoelliptic | \(-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-\textit {\_Z}^{4} a +b \right )}{\sum }\left (-\frac {\textit {\_R} \ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{x}\right )}{-2 \textit {\_R}^{4}+a}\right )\right )\) | \(51\) |
Input:
int((a*x^4+b*x^3)^(1/4)/x/(a*x+x^2+b),x,method=_RETURNVERBOSE)
Output:
-sum(-_R*ln((-_R*x+(x^3*(a*x+b))^(1/4))/x)/(-2*_R^4+a),_R=RootOf(_Z^8-_Z^4 *a+b))
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.10 (sec) , antiderivative size = 1693, normalized size of antiderivative = 26.87 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x \left (b+a x+x^2\right )} \, dx=\text {Too large to display} \] Input:
integrate((a*x^4+b*x^3)^(1/4)/x/(a*x+x^2+b),x, algorithm="fricas")
Output:
1/2*sqrt(2)*sqrt(sqrt(2)*sqrt((a + (a^4 - 8*a^2*b + 16*b^2)/sqrt(a^6 - 12* a^4*b + 48*a^2*b^2 - 64*b^3))/(a^4 - 8*a^2*b + 16*b^2)))*log((sqrt(2)*(a^4 - 8*a^2*b + 16*b^2)*x*sqrt(sqrt(2)*sqrt((a + (a^4 - 8*a^2*b + 16*b^2)/sqr t(a^6 - 12*a^4*b + 48*a^2*b^2 - 64*b^3))/(a^4 - 8*a^2*b + 16*b^2)))/sqrt(a ^6 - 12*a^4*b + 48*a^2*b^2 - 64*b^3) + 2*(a*x^4 + b*x^3)^(1/4))/x) - 1/2*s qrt(2)*sqrt(sqrt(2)*sqrt((a + (a^4 - 8*a^2*b + 16*b^2)/sqrt(a^6 - 12*a^4*b + 48*a^2*b^2 - 64*b^3))/(a^4 - 8*a^2*b + 16*b^2)))*log(-(sqrt(2)*(a^4 - 8 *a^2*b + 16*b^2)*x*sqrt(sqrt(2)*sqrt((a + (a^4 - 8*a^2*b + 16*b^2)/sqrt(a^ 6 - 12*a^4*b + 48*a^2*b^2 - 64*b^3))/(a^4 - 8*a^2*b + 16*b^2)))/sqrt(a^6 - 12*a^4*b + 48*a^2*b^2 - 64*b^3) - 2*(a*x^4 + b*x^3)^(1/4))/x) + 1/2*sqrt( 2)*sqrt(-sqrt(2)*sqrt((a + (a^4 - 8*a^2*b + 16*b^2)/sqrt(a^6 - 12*a^4*b + 48*a^2*b^2 - 64*b^3))/(a^4 - 8*a^2*b + 16*b^2)))*log((sqrt(2)*(a^4 - 8*a^2 *b + 16*b^2)*x*sqrt(-sqrt(2)*sqrt((a + (a^4 - 8*a^2*b + 16*b^2)/sqrt(a^6 - 12*a^4*b + 48*a^2*b^2 - 64*b^3))/(a^4 - 8*a^2*b + 16*b^2)))/sqrt(a^6 - 12 *a^4*b + 48*a^2*b^2 - 64*b^3) + 2*(a*x^4 + b*x^3)^(1/4))/x) - 1/2*sqrt(2)* sqrt(-sqrt(2)*sqrt((a + (a^4 - 8*a^2*b + 16*b^2)/sqrt(a^6 - 12*a^4*b + 48* a^2*b^2 - 64*b^3))/(a^4 - 8*a^2*b + 16*b^2)))*log(-(sqrt(2)*(a^4 - 8*a^2*b + 16*b^2)*x*sqrt(-sqrt(2)*sqrt((a + (a^4 - 8*a^2*b + 16*b^2)/sqrt(a^6 - 1 2*a^4*b + 48*a^2*b^2 - 64*b^3))/(a^4 - 8*a^2*b + 16*b^2)))/sqrt(a^6 - 12*a ^4*b + 48*a^2*b^2 - 64*b^3) - 2*(a*x^4 + b*x^3)^(1/4))/x) - 1/2*sqrt(2)...
Not integrable
Time = 1.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.35 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x \left (b+a x+x^2\right )} \, dx=\int \frac {\sqrt [4]{x^{3} \left (a x + b\right )}}{x \left (a x + b + x^{2}\right )}\, dx \] Input:
integrate((a*x**4+b*x**3)**(1/4)/x/(a*x+x**2+b),x)
Output:
Integral((x**3*(a*x + b))**(1/4)/(x*(a*x + b + x**2)), x)
Not integrable
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x \left (b+a x+x^2\right )} \, dx=\int { \frac {{\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{{\left (a x + x^{2} + b\right )} x} \,d x } \] Input:
integrate((a*x^4+b*x^3)^(1/4)/x/(a*x+x^2+b),x, algorithm="maxima")
Output:
integrate((a*x^4 + b*x^3)^(1/4)/((a*x + x^2 + b)*x), x)
Not integrable
Time = 3.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x \left (b+a x+x^2\right )} \, dx=\int { \frac {{\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{{\left (a x + x^{2} + b\right )} x} \,d x } \] Input:
integrate((a*x^4+b*x^3)^(1/4)/x/(a*x+x^2+b),x, algorithm="giac")
Output:
integrate((a*x^4 + b*x^3)^(1/4)/((a*x + x^2 + b)*x), x)
Not integrable
Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x \left (b+a x+x^2\right )} \, dx=\int \frac {{\left (a\,x^4+b\,x^3\right )}^{1/4}}{x\,\left (x^2+a\,x+b\right )} \,d x \] Input:
int((a*x^4 + b*x^3)^(1/4)/(x*(b + a*x + x^2)),x)
Output:
int((a*x^4 + b*x^3)^(1/4)/(x*(b + a*x + x^2)), x)
Not integrable
Time = 0.97 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt [4]{b x^3+a x^4}}{x \left (b+a x+x^2\right )} \, dx=\int \frac {\left (a x +b \right )^{\frac {1}{4}}}{x^{\frac {5}{4}} a +x^{\frac {1}{4}} b +x^{\frac {9}{4}}}d x \] Input:
int((a*x^4+b*x^3)^(1/4)/x/(a*x+x^2+b),x)
Output:
int((a*x + b)**(1/4)/(x**(1/4)*a*x + x**(1/4)*b + x**(1/4)*x**2),x)