Integrand size = 27, antiderivative size = 121 \[ \int \frac {1}{(-b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx=\frac {\arctan \left (\frac {2^{3/4} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{b^2 x+a^2 x^3}}{b+a x}\right )}{2 \sqrt [4]{2} a^{3/4} b^{3/4}}-\frac {\text {arctanh}\left (\frac {2^{3/4} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{b^2 x+a^2 x^3}}{b+a x}\right )}{2 \sqrt [4]{2} a^{3/4} b^{3/4}} \]
1/4*arctan(2^(3/4)*a^(1/4)*b^(1/4)*(a^2*x^3+b^2*x)^(1/4)/(a*x+b))*2^(3/4)/ a^(3/4)/b^(3/4)-1/4*arctanh(2^(3/4)*a^(1/4)*b^(1/4)*(a^2*x^3+b^2*x)^(1/4)/ (a*x+b))*2^(3/4)/a^(3/4)/b^(3/4)
Time = 10.80 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(-b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx=\frac {\arctan \left (\frac {2^{3/4} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{x \left (b^2+a^2 x^2\right )}}{b+a x}\right )-\text {arctanh}\left (\frac {2^{3/4} \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{x \left (b^2+a^2 x^2\right )}}{b+a x}\right )}{2 \sqrt [4]{2} a^{3/4} b^{3/4}} \]
(ArcTan[(2^(3/4)*a^(1/4)*b^(1/4)*(x*(b^2 + a^2*x^2))^(1/4))/(b + a*x)] - A rcTanh[(2^(3/4)*a^(1/4)*b^(1/4)*(x*(b^2 + a^2*x^2))^(1/4))/(b + a*x)])/(2* 2^(1/4)*a^(3/4)*b^(3/4))
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.
\(\displaystyle \int \frac {1}{(a x-b) \sqrt [4]{a^2 x^3+b^2 x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a^2 x^2+b^2} \int -\frac {1}{\sqrt [4]{x} (b-a x) \sqrt [4]{b^2+a^2 x^2}}dx}{\sqrt [4]{a^2 x^3+b^2 x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{a^2 x^2+b^2} \int \frac {1}{\sqrt [4]{x} (b-a x) \sqrt [4]{b^2+a^2 x^2}}dx}{\sqrt [4]{a^2 x^3+b^2 x}}\) |
\(\Big \downarrow \) 616 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a^2 x^2+b^2} \int \frac {\sqrt {x}}{(b-a x) \sqrt [4]{b^2+a^2 x^2}}d\sqrt [4]{x}}{\sqrt [4]{a^2 x^3+b^2 x}}\) |
\(\Big \downarrow \) 1888 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a^2 x^2+b^2} \int \frac {\sqrt {x}}{(b-a x) \sqrt [4]{b^2+a^2 x^2}}d\sqrt [4]{x}}{\sqrt [4]{a^2 x^3+b^2 x}}\) |
3.18.88.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/e))^n*(a + b*(x^(2*k)/e^2))^p, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && ILtQ[n, 0] && FractionQ[m]
Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ (n_))^(q_.), x_Symbol] :> Unintegrable[(f*x)^m*(d + e*x^n)^q*(a + c*x^(2*n) )^p, x] /; FreeQ[{a, c, d, e, f, m, n, p, q}, x] && EqQ[n2, 2*n]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
\[\int \frac {1}{\left (a x -b \right ) \left (a^{2} x^{3}+b^{2} x \right )^{\frac {1}{4}}}d x\]
Timed out. \[ \int \frac {1}{(-b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(-b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx=\int \frac {1}{\sqrt [4]{x \left (a^{2} x^{2} + b^{2}\right )} \left (a x - b\right )}\, dx \]
\[ \int \frac {1}{(-b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx=\int { \frac {1}{{\left (a^{2} x^{3} + b^{2} x\right )}^{\frac {1}{4}} {\left (a x - b\right )}} \,d x } \]
\[ \int \frac {1}{(-b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx=\int { \frac {1}{{\left (a^{2} x^{3} + b^{2} x\right )}^{\frac {1}{4}} {\left (a x - b\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(-b+a x) \sqrt [4]{b^2 x+a^2 x^3}} \, dx=-\int \frac {1}{{\left (a^2\,x^3+b^2\,x\right )}^{1/4}\,\left (b-a\,x\right )} \,d x \]