\(\int \frac {b+a x^2}{(-b+x+a x^2) \sqrt [4]{-b x^3+a x^5}} \, dx\) [1720]

Optimal result

Integrand size = 36, antiderivative size = 116 \[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} x \sqrt [4]{-b x^3+a x^5}}{-x^2+\sqrt {-b x^3+a x^5}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b x^3+a x^5}}{\sqrt {2}}}{x \sqrt [4]{-b x^3+a x^5}}\right ) \]

[Out]

-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))

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 1.36 (sec) , antiderivative size = 397, normalized size of antiderivative = 3.42, number of steps used = 21, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2081, 6860, 372, 371, 973, 477, 441, 440, 525, 524} \[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\frac {8 a x^2 \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {4 a^2 x^2}{\left (1-\sqrt {4 a b+1}\right )^2},\frac {a x^2}{b}\right )}{5 \left (1-\sqrt {4 a b+1}\right ) \sqrt [4]{a x^5-b x^3}}+\frac {8 a x^2 \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {4 a^2 x^2}{\left (\sqrt {4 a b+1}+1\right )^2},\frac {a x^2}{b}\right )}{5 \left (\sqrt {4 a b+1}+1\right ) \sqrt [4]{a x^5-b x^3}}-\frac {4 x \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {4 a^2 x^2}{\left (1-\sqrt {4 a b+1}\right )^2},\frac {a x^2}{b}\right )}{\sqrt [4]{a x^5-b x^3}}-\frac {4 x \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {4 a^2 x^2}{\left (\sqrt {4 a b+1}+1\right )^2},\frac {a x^2}{b}\right )}{\sqrt [4]{a x^5-b x^3}}+\frac {4 x \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^2}{b}\right )}{\sqrt [4]{a x^5-b x^3}} \]

[In]

Int[(b + a*x^2)/((-b + x + a*x^2)*(-(b*x^3) + a*x^5)^(1/4)),x]

[Out]

(-4*x*(1 - (a*x^2)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (4*a^2*x^2)/(1 - Sqrt[1 + 4*a*b])^2, (a*x^2)/b])/(-(b*x
^3) + a*x^5)^(1/4) - (4*x*(1 - (a*x^2)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (4*a^2*x^2)/(1 + Sqrt[1 + 4*a*b])^2
, (a*x^2)/b])/(-(b*x^3) + a*x^5)^(1/4) + (8*a*x^2*(1 - (a*x^2)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (4*a^2*x^2
)/(1 - Sqrt[1 + 4*a*b])^2, (a*x^2)/b])/(5*(1 - Sqrt[1 + 4*a*b])*(-(b*x^3) + a*x^5)^(1/4)) + (8*a*x^2*(1 - (a*x
^2)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (4*a^2*x^2)/(1 + Sqrt[1 + 4*a*b])^2, (a*x^2)/b])/(5*(1 + Sqrt[1 + 4*a
*b])*(-(b*x^3) + a*x^5)^(1/4)) + (4*x*(1 - (a*x^2)/b)^(1/4)*Hypergeometric2F1[1/8, 1/4, 9/8, (a*x^2)/b])/(-(b*
x^3) + a*x^5)^(1/4)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 973

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[d*((g*x)^n/x^n), In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[e*((g*x)^n/x^n), Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6860

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] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {b+a x^2}{x^{3/4} \sqrt [4]{-b+a x^2} \left (-b+x+a x^2\right )} \, dx}{\sqrt [4]{-b x^3+a x^5}} \\ & = \frac {\left (x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \left (\frac {1}{x^{3/4} \sqrt [4]{-b+a x^2}}+\frac {2 b-x}{x^{3/4} \sqrt [4]{-b+a x^2} \left (-b+x+a x^2\right )}\right ) \, dx}{\sqrt [4]{-b x^3+a x^5}} \\ & = \frac {\left (x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {1}{x^{3/4} \sqrt [4]{-b+a x^2}} \, dx}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {2 b-x}{x^{3/4} \sqrt [4]{-b+a x^2} \left (-b+x+a x^2\right )} \, dx}{\sqrt [4]{-b x^3+a x^5}} \\ & = \frac {\left (x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \left (\frac {-1+\sqrt {1+4 a b}}{x^{3/4} \left (1-\sqrt {1+4 a b}+2 a x\right ) \sqrt [4]{-b+a x^2}}+\frac {-1-\sqrt {1+4 a b}}{x^{3/4} \left (1+\sqrt {1+4 a b}+2 a x\right ) \sqrt [4]{-b+a x^2}}\right ) \, dx}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (x^{3/4} \sqrt [4]{1-\frac {a x^2}{b}}\right ) \int \frac {1}{x^{3/4} \sqrt [4]{1-\frac {a x^2}{b}}} \, dx}{\sqrt [4]{-b x^3+a x^5}} \\ & = \frac {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]{-b x^3+a x^5}}+\frac {\left (\left (-1-\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {1}{x^{3/4} \left (1+\sqrt {1+4 a b}+2 a x\right ) \sqrt [4]{-b+a x^2}} \, dx}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (\left (-1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {1}{x^{3/4} \left (1-\sqrt {1+4 a b}+2 a x\right ) \sqrt [4]{-b+a x^2}} \, dx}{\sqrt [4]{-b x^3+a x^5}} \\ & = \frac {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]{-b x^3+a x^5}}-\frac {\left (2 a \left (-1-\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x^2} \left (\left (1+\sqrt {1+4 a b}\right )^2-4 a^2 x^2\right )} \, dx}{\sqrt [4]{-b x^3+a x^5}}-\frac {\left (2 a \left (-1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {\sqrt [4]{x}}{\sqrt [4]{-b+a x^2} \left (\left (1-\sqrt {1+4 a b}\right )^2-4 a^2 x^2\right )} \, dx}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (\left (1-\sqrt {1+4 a b}\right ) \left (-1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {1}{x^{3/4} \sqrt [4]{-b+a x^2} \left (\left (1-\sqrt {1+4 a b}\right )^2-4 a^2 x^2\right )} \, dx}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (\left (-1-\sqrt {1+4 a b}\right ) \left (1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \int \frac {1}{x^{3/4} \sqrt [4]{-b+a x^2} \left (\left (1+\sqrt {1+4 a b}\right )^2-4 a^2 x^2\right )} \, dx}{\sqrt [4]{-b x^3+a x^5}} \\ & = \frac {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]{-b x^3+a x^5}}-\frac {\left (8 a \left (-1-\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{-b+a x^8} \left (\left (1+\sqrt {1+4 a b}\right )^2-4 a^2 x^8\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^5}}-\frac {\left (8 a \left (-1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{-b+a x^8} \left (\left (1-\sqrt {1+4 a b}\right )^2-4 a^2 x^8\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (4 \left (1-\sqrt {1+4 a b}\right ) \left (-1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{-b+a x^8} \left (\left (1-\sqrt {1+4 a b}\right )^2-4 a^2 x^8\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (4 \left (-1-\sqrt {1+4 a b}\right ) \left (1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{-b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{-b+a x^8} \left (\left (1+\sqrt {1+4 a b}\right )^2-4 a^2 x^8\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^5}} \\ & = \frac {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]{-b x^3+a x^5}}-\frac {\left (8 a \left (-1-\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{1-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {x^4}{\left (\left (1+\sqrt {1+4 a b}\right )^2-4 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^5}}-\frac {\left (8 a \left (-1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{1-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {x^4}{\left (\left (1-\sqrt {1+4 a b}\right )^2-4 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (4 \left (1-\sqrt {1+4 a b}\right ) \left (-1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{1-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (\left (1-\sqrt {1+4 a b}\right )^2-4 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^5}}+\frac {\left (4 \left (-1-\sqrt {1+4 a b}\right ) \left (1+\sqrt {1+4 a b}\right ) x^{3/4} \sqrt [4]{1-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (\left (1+\sqrt {1+4 a b}\right )^2-4 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^5}} \\ & = -\frac {4 x \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {4 a^2 x^2}{\left (1-\sqrt {1+4 a b}\right )^2},\frac {a x^2}{b}\right )}{\sqrt [4]{-b x^3+a x^5}}-\frac {4 x \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {4 a^2 x^2}{\left (1+\sqrt {1+4 a b}\right )^2},\frac {a x^2}{b}\right )}{\sqrt [4]{-b x^3+a x^5}}+\frac {8 a x^2 \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {4 a^2 x^2}{\left (1-\sqrt {1+4 a b}\right )^2},\frac {a x^2}{b}\right )}{5 \left (1-\sqrt {1+4 a b}\right ) \sqrt [4]{-b x^3+a x^5}}+\frac {8 a x^2 \sqrt [4]{1-\frac {a x^2}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {4 a^2 x^2}{\left (1+\sqrt {1+4 a b}\right )^2},\frac {a x^2}{b}\right )}{5 \left (1+\sqrt {1+4 a b}\right ) \sqrt [4]{-b x^3+a x^5}}+\frac {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]{-b x^3+a x^5}} \\ \end{align*}

Mathematica [F]

\[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx \]

[In]

Integrate[(b + a*x^2)/((-b + x + a*x^2)*(-(b*x^3) + a*x^5)^(1/4)),x]

[Out]

Integrate[(b + a*x^2)/((-b + x + a*x^2)*(-(b*x^3) + a*x^5)^(1/4)), x]

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.28

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (x^{3} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{3} \left (a \,x^{2}-b \right )}}{\left (x^{3} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{3} \left (a \,x^{2}-b \right )}}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{2}\)	\(148\)

[In]

int((a*x^2+b)/(a*x^2-b+x)/(a*x^5-b*x^3)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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))

Fricas [F(-1)]

Timed out. \[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\text {Timed out} \]

[In]

integrate((a*x^2+b)/(a*x^2-b+x)/(a*x^5-b*x^3)^(1/4),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int \frac {a x^{2} + b}{\sqrt [4]{x^{3} \left (a x^{2} - b\right )} \left (a x^{2} - b + x\right )}\, dx \]

[In]

integrate((a*x**2+b)/(a*x**2-b+x)/(a*x**5-b*x**3)**(1/4),x)

[Out]

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

Maxima [F]

\[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int { \frac {a x^{2} + b}{{\left (a x^{5} - b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{2} - b + x\right )}} \,d x } \]

[In]

integrate((a*x^2+b)/(a*x^2-b+x)/(a*x^5-b*x^3)^(1/4),x, algorithm="maxima")

[Out]

integrate((a*x^2 + b)/((a*x^5 - b*x^3)^(1/4)*(a*x^2 - b + x)), x)

Giac [F]

\[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int { \frac {a x^{2} + b}{{\left (a x^{5} - b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{2} - b + x\right )}} \,d x } \]

[In]

integrate((a*x^2+b)/(a*x^2-b+x)/(a*x^5-b*x^3)^(1/4),x, algorithm="giac")

[Out]

integrate((a*x^2 + b)/((a*x^5 - b*x^3)^(1/4)*(a*x^2 - b + x)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx=\int \frac {a\,x^2+b}{{\left (a\,x^5-b\,x^3\right )}^{1/4}\,\left (a\,x^2+x-b\right )} \,d x \]

[In]

int((b + a*x^2)/((a*x^5 - b*x^3)^(1/4)*(x - b + a*x^2)),x)

[Out]

int((b + a*x^2)/((a*x^5 - b*x^3)^(1/4)*(x - b + a*x^2)), x)