\(\int \frac {1}{(2 b+a x) \sqrt [4]{b x^2+a x^3}} \, dx\) [2096]

Optimal result

Integrand size = 25, antiderivative size = 152 \[ \int \frac {1}{(2 b+a x) \sqrt [4]{b x^2+a x^3}} \, dx=-\frac {\arctan \left (\frac {-\frac {\sqrt {a} x^2}{2 \sqrt [4]{b}}+\frac {\sqrt [4]{b} \sqrt {b x^2+a x^3}}{\sqrt {a}}}{x \sqrt [4]{b x^2+a x^3}}\right )}{2 \sqrt {a} b^{3/4}}+\frac {\text {arctanh}\left (\frac {2 \sqrt {a} \sqrt [4]{b} x \sqrt [4]{b x^2+a x^3}}{a x^2+2 \sqrt {b} \sqrt {b x^2+a x^3}}\right )}{2 \sqrt {a} b^{3/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {2081, 108, 107, 504, 1225, 227, 1713, 209, 212} \[ \int \frac {1}{(2 b+a x) \sqrt [4]{b x^2+a x^3}} \, dx=\frac {\sqrt {-\frac {a x}{b}} \sqrt [4]{a x+b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x+b}}{\sqrt [4]{-b} \sqrt {-\frac {a x}{b}}}\right )}{\sqrt {2} a \sqrt [4]{-b} \sqrt [4]{a x^3+b x^2}}-\frac {\sqrt {-\frac {a x}{b}} \sqrt [4]{a x+b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a x+b}}{\sqrt [4]{-b} \sqrt {-\frac {a x}{b}}}\right )}{\sqrt {2} a \sqrt [4]{-b} \sqrt [4]{a x^3+b x^2}} \]

[In]

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

[Out]

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

Rule 107

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(1/4)), x_Symbol] :> Dist[-4, Subst[
Int[x^2/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d,
e, f}, x] && GtQ[-f/(d*e - c*f), 0]

Rule 108

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(1/4)), x_Symbol] :> Dist[Sqrt[(-f)*
((c + d*x)/(d*e - c*f))]/Sqrt[c + d*x], Int[1/((a + b*x)*Sqrt[(-c)*(f/(d*e - c*f)) - d*f*(x/(d*e - c*f))]*(e +
f*x)^(1/4)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&  !GtQ[-f/(d*e - c*f), 0]

Rule 209

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

Rule 212

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

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 504

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s
= Denominator[Rt[-a/b, 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), Int[1/
((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1225

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d
^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1713

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(2 b+a x) \sqrt [4]{b x^2+a x^3}} \, dx=\frac {\sqrt {x} \sqrt [4]{b+a x} \left (-\arctan \left (\frac {-a x+2 \sqrt {b} \sqrt {b+a x}}{2 \sqrt {a} \sqrt [4]{b} \sqrt {x} \sqrt [4]{b+a x}}\right )+\text {arctanh}\left (\frac {2 \sqrt {a} \sqrt [4]{b} \sqrt {x} \sqrt [4]{b+a x}}{a x+2 \sqrt {b} \sqrt {b+a x}}\right )\right )}{2 \sqrt {a} b^{3/4} \sqrt [4]{x^2 (b+a x)}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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