Integrand size = 13, antiderivative size = 77 \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}} \] Output:
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)
Time = 5.66 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {2 \sqrt [4]{x} \sqrt [4]{b+a x^3} \left (\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )\right )}{3 \sqrt [4]{a} \sqrt [4]{x \left (b+a x^3\right )}} \] Input:
Integrate[(b*x + a*x^4)^(-1/4),x]
Output:
(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))
Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {1917, 851, 807, 770, 756, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt [4]{a x^4+b x}} \, dx\) |
\(\Big \downarrow \) 1917 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^3+b} \int \frac {1}{\sqrt [4]{x} \sqrt [4]{a x^3+b}}dx}{\sqrt [4]{a x^4+b x}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \int \frac {\sqrt {x}}{\sqrt [4]{a x^3+b}}d\sqrt [4]{x}}{\sqrt [4]{a x^4+b x}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \int \frac {1}{\sqrt [4]{b+a x}}dx^{3/4}}{3 \sqrt [4]{a x^4+b x}}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \int \frac {1}{1-a x}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{3 \sqrt [4]{a x^4+b x}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}+\frac {1}{2} \int \frac {1}{\sqrt {a} \sqrt {x}+1}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}\right )}{3 \sqrt [4]{a x^4+b x}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}+\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 \sqrt [4]{a}}\right )}{3 \sqrt [4]{a x^4+b x}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \left (\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 \sqrt [4]{a}}\right )}{3 \sqrt [4]{a x^4+b x}}\) |
Input:
Int[(b*x + a*x^4)^(-1/4),x]
Output:
(4*x^(1/4)*(b + a*x^3)^(1/4)*(ArcTan[(a^(1/4)*x^(3/4))/(b + a*x)^(1/4)]/(2 *a^(1/4)) + ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x)^(1/4)]/(2*a^(1/4))))/(3*(b *x + a*x^4)^(1/4))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[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]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(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] && !Integ erQ[p] && NeQ[n, j] && PosQ[n - j]
Time = 0.45 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(\frac {-2 \arctan \left (\frac {{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}}\right )+\ln \left (\frac {a^{\frac {1}{4}} x +{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}\right )}{3 a^{\frac {1}{4}}}\) | \(67\) |
Input:
int(1/(a*x^4+b*x)^(1/4),x,method=_RETURNVERBOSE)
Output:
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)
Timed out. \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\text {Timed out} \] Input:
integrate(1/(a*x^4+b*x)^(1/4),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\int \frac {1}{\sqrt [4]{a x^{4} + b x}}\, dx \] Input:
integrate(1/(a*x**4+b*x)**(1/4),x)
Output:
Integral((a*x**4 + b*x)**(-1/4), x)
\[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\int { \frac {1}{{\left (a x^{4} + b x\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(a*x^4+b*x)^(1/4),x, algorithm="maxima")
Output:
integrate((a*x^4 + b*x)^(-1/4), x)
Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (61) = 122\).
Time = 0.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.42 \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, a} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{6 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{6 \, a} \] Input:
integrate(1/(a*x^4+b*x)^(1/4),x, algorithm="giac")
Output:
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)*(sqr t(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(-sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sq rt(-a) + sqrt(a + b/x^3))/a
Time = 8.39 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\frac {4\,x\,{\left (\frac {a\,x^3}{b}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {a\,x^3}{b}\right )}{3\,{\left (a\,x^4+b\,x\right )}^{1/4}} \] Input:
int(1/(b*x + a*x^4)^(1/4),x)
Output:
(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))
\[ \int \frac {1}{\sqrt [4]{b x+a x^4}} \, dx=\int \frac {1}{x^{\frac {1}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}}}d x \] Input:
int(1/(a*x^4+b*x)^(1/4),x)
Output:
int(1/(x**(1/4)*(a*x**3 + b)**(1/4)),x)