\(\int \frac {1}{(b+2 a x^3) \sqrt [4]{b x+a x^4}} \, dx\) [2025]

Optimal result

Integrand size = 24, antiderivative size = 144 \[ \int \frac {1}{\left (b+2 a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x \sqrt [4]{b x+a x^4}}{-\sqrt {a} x^2+\sqrt {b x+a x^4}}\right )}{3 \sqrt [4]{a} b}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\frac {\sqrt [4]{a} x^2}{\sqrt {2}}+\frac {\sqrt {b x+a x^4}}{\sqrt {2} \sqrt [4]{a}}}{x \sqrt [4]{b x+a x^4}}\right )}{3 \sqrt [4]{a} b} \]

[Out]

1/3*2^(1/2)*arctan(2^(1/2)*a^(1/4)*x*(a*x^4+b*x)^(1/4)/(-a^(1/2)*x^2+(a*x^4+b*x)^(1/2)))/a^(1/4)/b+1/3*2^(1/2)
*arctanh((1/2*a^(1/4)*x^2*2^(1/2)+1/2*(a*x^4+b*x)^(1/2)*2^(1/2)/a^(1/4))/x/(a*x^4+b*x)^(1/4))/a^(1/4)/b

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(351\) vs. \(2(144)=288\).

Time = 0.26 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.44, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2081, 477, 476, 385, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{\left (b+2 a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=-\frac {\sqrt {2} \sqrt [4]{x} \sqrt [4]{a x^3+b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a} b \sqrt [4]{a x^4+b x}}+\frac {\sqrt {2} \sqrt [4]{x} \sqrt [4]{a x^3+b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}+1\right )}{3 \sqrt [4]{a} b \sqrt [4]{a x^4+b x}}-\frac {\sqrt [4]{x} \sqrt [4]{a x^3+b} \log \left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b}}-\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}+1\right )}{3 \sqrt {2} \sqrt [4]{a} b \sqrt [4]{a x^4+b x}}+\frac {\sqrt [4]{x} \sqrt [4]{a x^3+b} \log \left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b}}+\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}+1\right )}{3 \sqrt {2} \sqrt [4]{a} b \sqrt [4]{a x^4+b x}} \]

[In]

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

[Out]

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

Rule 210

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

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 385

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

Rule 476

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

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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{b+a x^3} \left (b+2 a x^3\right )} \, dx}{\sqrt [4]{b x+a x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{b+a x^{12}} \left (b+2 a x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^4} \left (b+2 a x^4\right )} \, dx,x,x^{3/4}\right )}{3 \sqrt [4]{b x+a x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{b+a b x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}} \\ & = \frac {\left (2 \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1-\sqrt {a} x^2}{b+a b x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}}+\frac {\left (2 \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1+\sqrt {a} x^2}{b+a b x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{b x+a x^4}} \\ & = \frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {a}}-\frac {\sqrt {2} x}{\sqrt [4]{a}}+x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt {a} b \sqrt [4]{b x+a x^4}}+\frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {a}}+\frac {\sqrt {2} x}{\sqrt [4]{a}}+x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt {a} b \sqrt [4]{b x+a x^4}}-\frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{a}}+2 x}{-\frac {1}{\sqrt {a}}-\frac {\sqrt {2} x}{\sqrt [4]{a}}-x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt {2} \sqrt [4]{a} b \sqrt [4]{b x+a x^4}}-\frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{a}}-2 x}{-\frac {1}{\sqrt {a}}+\frac {\sqrt {2} x}{\sqrt [4]{a}}-x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt {2} \sqrt [4]{a} b \sqrt [4]{b x+a x^4}} \\ & = -\frac {\sqrt [4]{x} \sqrt [4]{b+a x^3} \log \left (1+\frac {\sqrt {a} x^{3/2}}{\sqrt {b+a x^3}}-\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt {2} \sqrt [4]{a} b \sqrt [4]{b x+a x^4}}+\frac {\sqrt [4]{x} \sqrt [4]{b+a x^3} \log \left (1+\frac {\sqrt {a} x^{3/2}}{\sqrt {b+a x^3}}+\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt {2} \sqrt [4]{a} b \sqrt [4]{b x+a x^4}}+\frac {\left (\sqrt {2} \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{a} b \sqrt [4]{b x+a x^4}}-\frac {\left (\sqrt {2} \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{a} b \sqrt [4]{b x+a x^4}} \\ & = -\frac {\sqrt {2} \sqrt [4]{x} \sqrt [4]{b+a x^3} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{a} b \sqrt [4]{b x+a x^4}}+\frac {\sqrt {2} \sqrt [4]{x} \sqrt [4]{b+a x^3} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt [4]{a} b \sqrt [4]{b x+a x^4}}-\frac {\sqrt [4]{x} \sqrt [4]{b+a x^3} \log \left (1+\frac {\sqrt {a} x^{3/2}}{\sqrt {b+a x^3}}-\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt {2} \sqrt [4]{a} b \sqrt [4]{b x+a x^4}}+\frac {\sqrt [4]{x} \sqrt [4]{b+a x^3} \log \left (1+\frac {\sqrt {a} x^{3/2}}{\sqrt {b+a x^3}}+\frac {\sqrt {2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \sqrt {2} \sqrt [4]{a} b \sqrt [4]{b x+a x^4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\left (b+2 a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\frac {4 x \sqrt [4]{1+\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\frac {a x^3}{b+2 a x^3}\right )}{3 b \sqrt [4]{x \left (b+a x^3\right )} \sqrt [4]{1+\frac {2 a x^3}{b}}} \]

[In]

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

[Out]

(4*x*(1 + (a*x^3)/b)^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, (a*x^3)/(b + 2*a*x^3)])/(3*b*(x*(b + a*x^3))^(1/4)
*(1 + (2*a*x^3)/b)^(1/4))

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(-\frac {\sqrt {2}\, \left (\frac {\ln \left (\frac {-a^{\frac {1}{4}} {\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}\, x^{2}+\sqrt {x \left (a \,x^{3}+b \right )}}{a^{\frac {1}{4}} {\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}\, x^{2}+\sqrt {x \left (a \,x^{3}+b \right )}}\right )}{2}+\arctan \left (\frac {{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}} \sqrt {2}+a^{\frac {1}{4}} x}{x \,a^{\frac {1}{4}}}\right )+\arctan \left (\frac {{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}} \sqrt {2}-a^{\frac {1}{4}} x}{x \,a^{\frac {1}{4}}}\right )\right )}{3 a^{\frac {1}{4}} b}\)	\(155\)

[In]

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

[Out]

-1/3/a^(1/4)*2^(1/2)*(1/2*ln((-a^(1/4)*(x*(a*x^3+b))^(1/4)*2^(1/2)*x+a^(1/2)*x^2+(x*(a*x^3+b))^(1/2))/(a^(1/4)
*(x*(a*x^3+b))^(1/4)*2^(1/2)*x+a^(1/2)*x^2+(x*(a*x^3+b))^(1/2)))+arctan(((x*(a*x^3+b))^(1/4)*2^(1/2)+a^(1/4)*x
)/x/a^(1/4))+arctan(((x*(a*x^3+b))^(1/4)*2^(1/2)-a^(1/4)*x)/x/a^(1/4)))/b

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 134.33 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.96 \[ \int \frac {1}{\left (b+2 a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\frac {1}{6} \, \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, {\left (a x^{4} + b x\right )}^{\frac {3}{4}} a b^{2} \left (-\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + 2 \, \sqrt {a x^{4} + b x} a b x \sqrt {-\frac {1}{a b^{4}}} + 2 \, {\left (a x^{4} + b x\right )}^{\frac {1}{4}} a x^{2} \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} - 1}{2 \, a x^{3} + b}\right ) - \frac {1}{6} \, \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {2 \, {\left (a x^{4} + b x\right )}^{\frac {3}{4}} a b^{2} \left (-\frac {1}{a b^{4}}\right )^{\frac {3}{4}} - 2 \, \sqrt {a x^{4} + b x} a b x \sqrt {-\frac {1}{a b^{4}}} + 2 \, {\left (a x^{4} + b x\right )}^{\frac {1}{4}} a x^{2} \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} + 1}{2 \, a x^{3} + b}\right ) + \frac {1}{6} i \, \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {2 i \, {\left (a x^{4} + b x\right )}^{\frac {3}{4}} a b^{2} \left (-\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + 2 \, \sqrt {a x^{4} + b x} a b x \sqrt {-\frac {1}{a b^{4}}} - 2 i \, {\left (a x^{4} + b x\right )}^{\frac {1}{4}} a x^{2} \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} + 1}{2 \, a x^{3} + b}\right ) - \frac {1}{6} i \, \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {-2 i \, {\left (a x^{4} + b x\right )}^{\frac {3}{4}} a b^{2} \left (-\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + 2 \, \sqrt {a x^{4} + b x} a b x \sqrt {-\frac {1}{a b^{4}}} + 2 i \, {\left (a x^{4} + b x\right )}^{\frac {1}{4}} a x^{2} \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} + 1}{2 \, a x^{3} + b}\right ) \]

[In]

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

[Out]

1/6*(-1/(a*b^4))^(1/4)*log(-(2*(a*x^4 + b*x)^(3/4)*a*b^2*(-1/(a*b^4))^(3/4) + 2*sqrt(a*x^4 + b*x)*a*b*x*sqrt(-
1/(a*b^4)) + 2*(a*x^4 + b*x)^(1/4)*a*x^2*(-1/(a*b^4))^(1/4) - 1)/(2*a*x^3 + b)) - 1/6*(-1/(a*b^4))^(1/4)*log((
2*(a*x^4 + b*x)^(3/4)*a*b^2*(-1/(a*b^4))^(3/4) - 2*sqrt(a*x^4 + b*x)*a*b*x*sqrt(-1/(a*b^4)) + 2*(a*x^4 + b*x)^
(1/4)*a*x^2*(-1/(a*b^4))^(1/4) + 1)/(2*a*x^3 + b)) + 1/6*I*(-1/(a*b^4))^(1/4)*log((2*I*(a*x^4 + b*x)^(3/4)*a*b
^2*(-1/(a*b^4))^(3/4) + 2*sqrt(a*x^4 + b*x)*a*b*x*sqrt(-1/(a*b^4)) - 2*I*(a*x^4 + b*x)^(1/4)*a*x^2*(-1/(a*b^4)
)^(1/4) + 1)/(2*a*x^3 + b)) - 1/6*I*(-1/(a*b^4))^(1/4)*log((-2*I*(a*x^4 + b*x)^(3/4)*a*b^2*(-1/(a*b^4))^(3/4)
+ 2*sqrt(a*x^4 + b*x)*a*b*x*sqrt(-1/(a*b^4)) + 2*I*(a*x^4 + b*x)^(1/4)*a*x^2*(-1/(a*b^4))^(1/4) + 1)/(2*a*x^3
+ b))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\left (b+2 a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, a^{\frac {1}{4}}}\right )}{3 \, a^{\frac {1}{4}} b} - \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, a^{\frac {1}{4}}}\right )}{3 \, a^{\frac {1}{4}} b} + \frac {\sqrt {2} \log \left (\sqrt {2} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} a^{\frac {1}{4}} + \sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}\right )}{6 \, a^{\frac {1}{4}} b} - \frac {\sqrt {2} \log \left (-\sqrt {2} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} a^{\frac {1}{4}} + \sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}\right )}{6 \, a^{\frac {1}{4}} b} \]

[In]

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

[Out]

-1/3*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4) + 2*(a + b/x^3)^(1/4))/a^(1/4))/(a^(1/4)*b) - 1/3*sqrt(2)*arc
tan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4) - 2*(a + b/x^3)^(1/4))/a^(1/4))/(a^(1/4)*b) + 1/6*sqrt(2)*log(sqrt(2)*(a + b
/x^3)^(1/4)*a^(1/4) + sqrt(a + b/x^3) + sqrt(a))/(a^(1/4)*b) - 1/6*sqrt(2)*log(-sqrt(2)*(a + b/x^3)^(1/4)*a^(1
/4) + sqrt(a + b/x^3) + sqrt(a))/(a^(1/4)*b)

Mupad [F(-1)]

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

[In]

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

[Out]

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