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

Optimal. Leaf size=144 \[ \frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x \sqrt [4]{a x^4+b x}}{\sqrt {a x^4+b x}-\sqrt {a} x^2}\right )}{3 \sqrt [4]{a} b}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^4+b x}}{\sqrt {2} \sqrt [4]{a}}+\frac {\sqrt [4]{a} x^2}{\sqrt {2}}}{x \sqrt [4]{a x^4+b x}}\right )}{3 \sqrt [4]{a} b} \]

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Rubi [B]  time = 0.40, antiderivative size = 351, normalized size of antiderivative = 2.44, number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2056, 466, 465, 377, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\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}}-\frac {\sqrt {2} \sqrt [4]{x} \sqrt [4]{a x^3+b} \tan ^{-1}\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} \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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 204

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

Rule 211

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 377

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 465

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 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 617

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 628

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 1162

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 1165

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\left (b+2 a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\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} \tan ^{-1}\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*}

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Mathematica [C]  time = 0.04, size = 75, normalized size = 0.52 \begin {gather*} \frac {4 x \sqrt [4]{\frac {a x^3}{b}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\frac {a x^3}{2 a x^3+b}\right )}{3 b \sqrt [4]{x \left (a x^3+b\right )} \sqrt [4]{\frac {2 a x^3}{b}+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[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))

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IntegrateAlgebraic [A]  time = 0.46, size = 144, normalized size = 1.00 \begin {gather*} \frac {\sqrt {2} \tan ^{-1}\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} \tanh ^{-1}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 126.32, size = 355, normalized size = 2.47 \begin {gather*} -\frac {2}{3} \, \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} \arctan \left (\frac {{\left (a b^{4} \sqrt {\frac {1}{b^{2}}} x \left (-\frac {1}{a b^{4}}\right )^{\frac {3}{4}} + a b^{3} x \left (-\frac {1}{a b^{4}}\right )^{\frac {3}{4}}\right )} {\left (a x^{4} + b x\right )}^{\frac {3}{4}} - {\left (a x^{4} + b x\right )}^{\frac {1}{4}} {\left ({\left (a b^{2} x^{3} + b^{3}\right )} \sqrt {\frac {1}{b^{2}}} \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}} - {\left (a b x^{3} + b^{2}\right )} \left (-\frac {1}{a b^{4}}\right )^{\frac {1}{4}}\right )}}{2 \, {\left (a x^{4} + b x\right )}}\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} \, \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 162, normalized size = 1.12 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (2 a \,x^{3}+b \right ) \left (a \,x^{4}+b x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a x^{4} + b x\right )}^{\frac {1}{4}} {\left (2 \, a x^{3} + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a\,x^4+b\,x\right )}^{1/4}\,\left (2\,a\,x^3+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{x \left (a x^{3} + b\right )} \left (2 a x^{3} + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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