Integrand size = 36, antiderivative size = 116 \[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} x \sqrt [4]{-b x^3+a x^5}}{-x^2+\sqrt {-b x^3+a x^5}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b x^3+a x^5}}{\sqrt {2}}}{x \sqrt [4]{-b x^3+a x^5}}\right ) \]
-2^(1/2)*arctan(2^(1/2)*x*(a*x^5-b*x^3)^(1/4)/(-x^2+(a*x^5-b*x^3)^(1/2)))- 2^(1/2)*arctanh((1/2*2^(1/2)*x^2+1/2*(a*x^5-b*x^3)^(1/2)*2^(1/2))/x/(a*x^5 -b*x^3)^(1/4))
\[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx \]
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.68 (sec) , antiderivative size = 482, normalized size of antiderivative = 4.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {2467, 25, 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.
\(\displaystyle \int \frac {a x^2+b}{\left (a x^2-b+x\right ) \sqrt [4]{a x^5-b x^3}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{3/4} \sqrt [4]{a x^2-b} \int -\frac {a x^2+b}{x^{3/4} \left (-a x^2-x+b\right ) \sqrt [4]{a x^2-b}}dx}{\sqrt [4]{a x^5-b x^3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {x^{3/4} \sqrt [4]{a x^2-b} \int \frac {a x^2+b}{x^{3/4} \left (-a x^2-x+b\right ) \sqrt [4]{a x^2-b}}dx}{\sqrt [4]{a x^5-b x^3}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {4 x^{3/4} \sqrt [4]{a x^2-b} \int \frac {a x^2+b}{\left (-a x^2-x+b\right ) \sqrt [4]{a x^2-b}}d\sqrt [4]{x}}{\sqrt [4]{a x^5-b x^3}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle -\frac {4 x^{3/4} \sqrt [4]{a x^2-b} \int \left (\frac {2 b-x}{\left (-a x^2-x+b\right ) \sqrt [4]{a x^2-b}}-\frac {1}{\sqrt [4]{a x^2-b}}\right )d\sqrt [4]{x}}{\sqrt [4]{a x^5-b x^3}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 x^{3/4} \sqrt [4]{a x^2-b} \left (\frac {\sqrt [4]{x} \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^2}{2 a b-\sqrt {4 a b+1}+1},\frac {a x^2}{b}\right )}{\sqrt [4]{a x^2-b}}+\frac {\sqrt [4]{x} \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^2}{2 a b+\sqrt {4 a b+1}+1},\frac {a x^2}{b}\right )}{\sqrt [4]{a x^2-b}}-\frac {a x^{5/4} \left (1-\sqrt {4 a b+1}\right ) \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^2}{2 a b-\sqrt {4 a b+1}+1},\frac {a x^2}{b}\right )}{5 \left (2 a b-\sqrt {4 a b+1}+1\right ) \sqrt [4]{a x^2-b}}-\frac {a x^{5/4} \left (\sqrt {4 a b+1}+1\right ) \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^2}{2 a b+\sqrt {4 a b+1}+1},\frac {a x^2}{b}\right )}{5 \left (2 a b+\sqrt {4 a b+1}+1\right ) \sqrt [4]{a x^2-b}}-\frac {\sqrt [4]{x} \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^2}{b}\right )}{\sqrt [4]{a x^2-b}}\right )}{\sqrt [4]{a x^5-b x^3}}\) |
(-4*x^(3/4)*(-b + a*x^2)^(1/4)*((x^(1/4)*(1 - (a*x^2)/b)^(1/4)*AppellF1[1/ 8, 1, 1/4, 9/8, (2*a^2*x^2)/(1 + 2*a*b - Sqrt[1 + 4*a*b]), (a*x^2)/b])/(-b + a*x^2)^(1/4) + (x^(1/4)*(1 - (a*x^2)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8 , (2*a^2*x^2)/(1 + 2*a*b + Sqrt[1 + 4*a*b]), (a*x^2)/b])/(-b + a*x^2)^(1/4 ) - (a*(1 - Sqrt[1 + 4*a*b])*x^(5/4)*(1 - (a*x^2)/b)^(1/4)*AppellF1[5/8, 1 , 1/4, 13/8, (2*a^2*x^2)/(1 + 2*a*b - Sqrt[1 + 4*a*b]), (a*x^2)/b])/(5*(1 + 2*a*b - Sqrt[1 + 4*a*b])*(-b + a*x^2)^(1/4)) - (a*(1 + Sqrt[1 + 4*a*b])* x^(5/4)*(1 - (a*x^2)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (2*a^2*x^2)/(1 + 2*a*b + Sqrt[1 + 4*a*b]), (a*x^2)/b])/(5*(1 + 2*a*b + Sqrt[1 + 4*a*b])*(- b + a*x^2)^(1/4)) - (x^(1/4)*(1 - (a*x^2)/b)^(1/4)*Hypergeometric2F1[1/8, 1/4, 9/8, (a*x^2)/b])/(-b + a*x^2)^(1/4)))/(-(b*x^3) + a*x^5)^(1/4)
3.18.20.3.1 Defintions of rubi rules used
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]
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]
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]
Time = 1.02 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.28
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (x^{3} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{3} \left (a \,x^{2}-b \right )}}{\left (x^{3} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{3} \left (a \,x^{2}-b \right )}}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{2}\) | \(148\) |
1/2*2^(1/2)*(ln((-(x^3*(a*x^2-b))^(1/4)*2^(1/2)*x+x^2+(x^3*(a*x^2-b))^(1/2 ))/((x^3*(a*x^2-b))^(1/4)*2^(1/2)*x+x^2+(x^3*(a*x^2-b))^(1/2)))+2*arctan(( (x^3*(a*x^2-b))^(1/4)*2^(1/2)+x)/x)+2*arctan(((x^3*(a*x^2-b))^(1/4)*2^(1/2 )-x)/x))
Timed out. \[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\text {Timed out} \]
\[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int \frac {a x^{2} + b}{\sqrt [4]{x^{3} \left (a x^{2} - b\right )} \left (a x^{2} - b + x\right )}\, dx \]
\[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int { \frac {a x^{2} + b}{{\left (a x^{5} - b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{2} - b + x\right )}} \,d x } \]
\[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int { \frac {a x^{2} + b}{{\left (a x^{5} - b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{2} - b + x\right )}} \,d x } \]
Timed out. \[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int \frac {a\,x^2+b}{{\left (a\,x^5-b\,x^3\right )}^{1/4}\,\left (a\,x^2+x-b\right )} \,d x \]