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

Optimal. Leaf size=116 \[ -\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{a x^5-b x^3}}{\sqrt {a x^5-b x^3}-x^2}\right )-\sqrt {2} \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^5-b x^3}}{\sqrt {2}}+\frac {x^2}{\sqrt {2}}}{x \sqrt [4]{a x^5-b x^3}}\right ) \]

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Rubi [C]  time = 2.11, antiderivative size = 397, normalized size of antiderivative = 3.42, number of steps used = 21, number of rules used = 10, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2056, 6728, 365, 364, 959, 466, 430, 429, 511, 510} \begin {gather*} \frac {8 a x^2 \sqrt [4]{1-\frac {a x^2}{b}} F_1\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}} F_1\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}} F_1\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}} F_1\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}} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};\frac {a x^2}{b}\right )}{\sqrt [4]{a x^5-b x^3}} \end {gather*}

Warning: Unable to verify antiderivative.

[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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

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 429

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 430

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 466

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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 959

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 2056

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 6728

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*} \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx &=\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}} \, _2F_1\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}} \, _2F_1\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}} \, _2F_1\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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}} \, _2F_1\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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}} F_1\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}} F_1\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}} F_1\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}} F_1\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}} \, _2F_1\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*}

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Mathematica [F]  time = 1.23, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b+a x^2}{\left (-b+x+a x^2\right ) \sqrt [4]{-b x^3+a x^5}} \, dx \end {gather*}

Verification is not applicable to the result.

[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]

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IntegrateAlgebraic [A]  time = 2.90, size = 116, normalized size = 1.00 \begin {gather*} -\sqrt {2} \tan ^{-1}\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} \tanh ^{-1}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[2]*ArcTan[(Sqrt[2]*x*(-(b*x^3) + a*x^5)^(1/4))/(-x^2 + Sqrt[-(b*x^3) + a*x^5])]) - Sqrt[2]*ArcTanh[(x^2
/Sqrt[2] + Sqrt[-(b*x^3) + a*x^5]/Sqrt[2])/(x*(-(b*x^3) + a*x^5)^(1/4))]

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}+b}{\left (a \,x^{2}-b +x \right ) \left (a \,x^{5}-b \,x^{3}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + b}{\sqrt [4]{x^{3} \left (a x^{2} - b\right )} \left (a x^{2} - b + x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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