\(\int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx\) [1017]

Optimal result

Integrand size = 13, antiderivative size = 77 \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.60, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2036, 335, 281, 246, 218, 212, 209} \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {2 \sqrt [4]{x} \sqrt [4]{a x^3+b} \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a} \sqrt [4]{a x^4+b x}}+\frac {2 \sqrt [4]{x} \sqrt [4]{a x^3+b} \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{3 \sqrt [4]{a} \sqrt [4]{a x^4+b x}} \]

[In]

Int[(b*x + a*x^4)^(-1/4),x]

[Out]

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

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 218

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 281

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

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

Mathematica [A] (verified)

Time = 5.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {2 \sqrt [4]{x} \sqrt [4]{b+a x^3} \left (\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )\right )}{3 \sqrt [4]{a} \sqrt [4]{x \left (b+a x^3\right )}} \]

[In]

Integrate[(b*x + a*x^4)^(-1/4),x]

[Out]

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

Maple [A] (verified)

Time = 2.65 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92

method	result	size
pseudoelliptic	\(\frac {-2 \arctan \left (\frac {{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}}\right )+\ln \left (\frac {-a^{\frac {1}{4}} x -{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x -{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}\right )}{3 a^{\frac {1}{4}}}\)	\(71\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\int \frac {1}{\sqrt [4]{a x^{4} + b x}}\, dx \]

[In]

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

[Out]

Integral((a*x**4 + b*x)**(-1/4), x)

Maxima [F]

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

[In]

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

[Out]

integrate((a*x^4 + b*x)^(-1/4), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (61) = 122\).

Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.42 \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, a} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{6 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{6 \, a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {4\,x\,{\left (\frac {a\,x^3}{b}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {a\,x^3}{b}\right )}{3\,{\left (a\,x^4+b\,x\right )}^{1/4}} \]

[In]

int(1/(b*x + a*x^4)^(1/4),x)

[Out]

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