Integrand size = 17, antiderivative size = 85 \[ \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx=-\frac {4}{\sqrt [4]{a x+3 x^2}}+\frac {4 \sqrt [4]{3} \sqrt [4]{\frac {x}{a+3 x}} \sqrt {a+3 x} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {a}}{\sqrt {a+3 x}}\right )\right |2\right )}{\sqrt {a} \sqrt [4]{a x+3 x^2}} \] Output:
-4/(a*x+3*x^2)^(1/4)+4*3^(1/4)*(x/(a+3*x))^(1/4)*(a+3*x)^(1/2)*EllipticE(s in(1/2*arcsin(a^(1/2)/(a+3*x)^(1/2))),2^(1/2))/a^(1/2)/(a*x+3*x^2)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx=-\frac {4 \sqrt [4]{1+\frac {3 x}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},-\frac {3 x}{a}\right )}{\sqrt [4]{x (a+3 x)}} \] Input:
Integrate[1/(x*(a*x + 3*x^2)^(1/4)),x]
Output:
(-4*(1 + (3*x)/a)^(1/4)*Hypergeometric2F1[-1/4, 1/4, 3/4, (-3*x)/a])/(x*(a + 3*x))^(1/4)
Time = 0.50 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.58, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {1137, 61, 73, 839, 813, 27, 858, 807, 212}
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.
\(\displaystyle \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx\) |
\(\Big \downarrow \) 1137 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \int \frac {1}{x^{5/4} \sqrt [4]{a+3 x}}dx}{\sqrt [4]{a x+3 x^2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \left (\frac {6 \int \frac {1}{\sqrt [4]{x} \sqrt [4]{a+3 x}}dx}{a}-\frac {4 (a+3 x)^{3/4}}{a \sqrt [4]{x}}\right )}{\sqrt [4]{a x+3 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \left (\frac {24 \int \frac {\sqrt {x}}{\sqrt [4]{a+3 x}}d\sqrt [4]{x}}{a}-\frac {4 (a+3 x)^{3/4}}{a \sqrt [4]{x}}\right )}{\sqrt [4]{a x+3 x^2}}\) |
\(\Big \downarrow \) 839 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \left (\frac {24 \left (\frac {x^{3/4}}{2 \sqrt [4]{a+3 x}}-\frac {1}{2} a \int \frac {\sqrt {x}}{(a+3 x)^{5/4}}d\sqrt [4]{x}\right )}{a}-\frac {4 (a+3 x)^{3/4}}{a \sqrt [4]{x}}\right )}{\sqrt [4]{a x+3 x^2}}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \left (\frac {24 \left (\frac {x^{3/4}}{2 \sqrt [4]{a+3 x}}-\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{x}+3} \int \frac {3 \sqrt [4]{3}}{\left (\frac {a}{x}+3\right )^{5/4} x^{3/4}}d\sqrt [4]{x}}{6 \sqrt [4]{3} \sqrt [4]{a+3 x}}\right )}{a}-\frac {4 (a+3 x)^{3/4}}{a \sqrt [4]{x}}\right )}{\sqrt [4]{a x+3 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \left (\frac {24 \left (\frac {x^{3/4}}{2 \sqrt [4]{a+3 x}}-\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{x}+3} \int \frac {1}{\left (\frac {a}{x}+3\right )^{5/4} x^{3/4}}d\sqrt [4]{x}}{2 \sqrt [4]{a+3 x}}\right )}{a}-\frac {4 (a+3 x)^{3/4}}{a \sqrt [4]{x}}\right )}{\sqrt [4]{a x+3 x^2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \left (\frac {24 \left (\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{x}+3} \int \frac {1}{\sqrt [4]{x} (a x+3)^{5/4}}d\frac {1}{\sqrt [4]{x}}}{2 \sqrt [4]{a+3 x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+3 x}}\right )}{a}-\frac {4 (a+3 x)^{3/4}}{a \sqrt [4]{x}}\right )}{\sqrt [4]{a x+3 x^2}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \left (\frac {24 \left (\frac {a \sqrt [4]{x} \sqrt [4]{\frac {a}{x}+3} \int \frac {1}{\left (\sqrt {x} a+3\right )^{5/4}}d\sqrt {x}}{4 \sqrt [4]{a+3 x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+3 x}}\right )}{a}-\frac {4 (a+3 x)^{3/4}}{a \sqrt [4]{x}}\right )}{\sqrt [4]{a x+3 x^2}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a+3 x} \left (\frac {24 \left (\frac {\sqrt {a} \sqrt [4]{x} \sqrt [4]{\frac {a}{x}+3} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {3}}\right )\right |2\right )}{2\ 3^{3/4} \sqrt [4]{a+3 x}}+\frac {x^{3/4}}{2 \sqrt [4]{a+3 x}}\right )}{a}-\frac {4 (a+3 x)^{3/4}}{a \sqrt [4]{x}}\right )}{\sqrt [4]{a x+3 x^2}}\) |
Input:
Int[1/(x*(a*x + 3*x^2)^(1/4)),x]
Output:
(x^(1/4)*(a + 3*x)^(1/4)*((-4*(a + 3*x)^(3/4))/(a*x^(1/4)) + (24*(x^(3/4)/ (2*(a + 3*x)^(1/4)) + (Sqrt[a]*(3 + a/x)^(1/4)*x^(1/4)*EllipticE[ArcTan[(S qrt[a]*Sqrt[x])/Sqrt[3]]/2, 2])/(2*3^(3/4)*(a + 3*x)^(1/4))))/a))/(a*x + 3 *x^2)^(1/4)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
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[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) )^(1/4)/(b*(a + b*x^4)^(1/4))) Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 )^(1/4)), x] - Simp[a/2 Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} , x] && PosQ[b/a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p)) Int[x^(m + p)*(b + c*x)^ p, x], x] /; FreeQ[{b, c, e, m}, x]
\[\int \frac {1}{x \left (a x +3 x^{2}\right )^{\frac {1}{4}}}d x\]
Input:
int(1/x/(a*x+3*x^2)^(1/4),x)
Output:
int(1/x/(a*x+3*x^2)^(1/4),x)
\[ \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx=\int { \frac {1}{{\left (a x + 3 \, x^{2}\right )}^{\frac {1}{4}} x} \,d x } \] Input:
integrate(1/x/(a*x+3*x^2)^(1/4),x, algorithm="fricas")
Output:
integral((a*x + 3*x^2)^(3/4)/(a*x^2 + 3*x^3), x)
\[ \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx=\int \frac {1}{x \sqrt [4]{x \left (a + 3 x\right )}}\, dx \] Input:
integrate(1/x/(a*x+3*x**2)**(1/4),x)
Output:
Integral(1/(x*(x*(a + 3*x))**(1/4)), x)
\[ \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx=\int { \frac {1}{{\left (a x + 3 \, x^{2}\right )}^{\frac {1}{4}} x} \,d x } \] Input:
integrate(1/x/(a*x+3*x^2)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((a*x + 3*x^2)^(1/4)*x), x)
\[ \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx=\int { \frac {1}{{\left (a x + 3 \, x^{2}\right )}^{\frac {1}{4}} x} \,d x } \] Input:
integrate(1/x/(a*x+3*x^2)^(1/4),x, algorithm="giac")
Output:
integrate(1/((a*x + 3*x^2)^(1/4)*x), x)
Timed out. \[ \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx=\int \frac {1}{x\,{\left (3\,x^2+a\,x\right )}^{1/4}} \,d x \] Input:
int(1/(x*(a*x + 3*x^2)^(1/4)),x)
Output:
int(1/(x*(a*x + 3*x^2)^(1/4)), x)
\[ \int \frac {1}{x \sqrt [4]{a x+3 x^2}} \, dx=\int \frac {1}{x^{\frac {5}{4}} \left (a +3 x \right )^{\frac {1}{4}}}d x \] Input:
int(1/x/(a*x+3*x^2)^(1/4),x)
Output:
int(1/(x**(1/4)*(a + 3*x)**(1/4)*x),x)