Integrand size = 30, antiderivative size = 323 \[ \int \frac {c+d x+e x^2+f x^3}{x \left (a+b x^4\right )^{3/2}} \, dx=\frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {c \sqrt {a+b x^4}}{2 a^2}-\frac {f x \sqrt {a+b x^4}}{2 a \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d-\sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{5/4} b^{3/4} \sqrt {a+b x^4}} \]
-1/2*c*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(3/2)+1/2*x*(-b*c*x^3+a*f*x^2+a* e*x+a*d)/a^2/(b*x^4+a)^(1/2)+1/2*c*(b*x^4+a)^(1/2)/a^2-1/2*f*x*(b*x^4+a)^( 1/2)/a/b^(1/2)/(a^(1/2)+x^2*b^(1/2))+1/2*f*(cos(2*arctan(b^(1/4)*x/a^(1/4) ))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4 )*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b ^(1/2))^2)^(1/2)/a^(3/4)/b^(3/4)/(b*x^4+a)^(1/2)+1/4*(cos(2*arctan(b^(1/4) *x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arc tan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(-f*a^(1/2)+d*b^(1/2))*(a^(1/2)+x^2*b ^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(5/4)/b^(3/4)/(b*x^4+a )^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.39 \[ \int \frac {c+d x+e x^2+f x^3}{x \left (a+b x^4\right )^{3/2}} \, dx=\frac {3 c \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b x^4}{a}\right )+x \left (3 d+3 e x+3 d \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )+2 f x^2 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{6 a \sqrt {a+b x^4}} \]
(3*c*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*x^4)/a] + x*(3*d + 3*e*x + 3*d *Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] + 2*f* x^2*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[3/4, 3/2, 7/4, -((b*x^4)/a)]))/( 6*a*Sqrt[a + b*x^4])
Time = 0.65 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2368, 25, 2371, 798, 73, 221, 2424, 2009}
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.
\(\displaystyle \int \frac {c+d x+e x^2+f x^3}{x \left (a+b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2368 |
\(\displaystyle \frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {\int -\frac {\frac {2 b^2 c x^4}{a}-b f x^3+b d x+2 b c}{x \sqrt {b x^4+a}}dx}{2 a b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\frac {2 b^2 c x^4}{a}-b f x^3+b d x+2 b c}{x \sqrt {b x^4+a}}dx}{2 a b}+\frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 2371 |
\(\displaystyle \frac {\int \frac {\frac {2 b^2 c x^3}{a}-b f x^2+b d}{\sqrt {b x^4+a}}dx+2 b c \int \frac {1}{x \sqrt {b x^4+a}}dx}{2 a b}+\frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int \frac {\frac {2 b^2 c x^3}{a}-b f x^2+b d}{\sqrt {b x^4+a}}dx+\frac {1}{2} b c \int \frac {1}{x^4 \sqrt {b x^4+a}}dx^4}{2 a b}+\frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\int \frac {\frac {2 b^2 c x^3}{a}-b f x^2+b d}{\sqrt {b x^4+a}}dx+c \int \frac {1}{\frac {x^8}{b}-\frac {a}{b}}d\sqrt {b x^4+a}}{2 a b}+\frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\int \frac {\frac {2 b^2 c x^3}{a}-b f x^2+b d}{\sqrt {b x^4+a}}dx-\frac {b c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 a b}+\frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 2424 |
\(\displaystyle \frac {\int \left (\frac {2 b^2 c x^3}{a \sqrt {b x^4+a}}+\frac {b d-b f x^2}{\sqrt {b x^4+a}}\right )dx-\frac {b c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 a b}+\frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (a d+a e x+a f x^2-b c x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {\frac {\sqrt [4]{a} \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\frac {\sqrt {b} d}{\sqrt {a}}-f\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \sqrt [4]{b} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}-\frac {b c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {b c \sqrt {a+b x^4}}{a}-\frac {\sqrt {b} f x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}}{2 a b}\) |
(x*(a*d + a*e*x + a*f*x^2 - b*c*x^3))/(2*a^2*Sqrt[a + b*x^4]) + ((b*c*Sqrt [a + b*x^4])/a - (Sqrt[b]*f*x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2) - ( b*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/Sqrt[a] + (a^(1/4)*b^(1/4)*f*(Sqrt[a ] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*A rcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/Sqrt[a + b*x^4] + (a^(1/4)*b^(1/4)*((Sqr t[b]*d)/Sqrt[a] - f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + S qrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*Sqrt[a + b*x^4]))/(2*a*b)
3.6.47.3.1 Defintions of rubi rules used
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)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^ m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1)*x^m *Pq, a + b*x^n, x], i}, Simp[(-x)*R*((a + b*x^n)^(p + 1)/(a^2*n*(p + 1)*b^( Floor[(q - 1)/n] + 1))), x] + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)) Int[x^m*(a + b*x^n)^(p + 1)*ExpandToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)/a)*Coeff[R, x, i]*x^(i - m), {i, 0, n - 1}], x], x], x]]] /; F reeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]
Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Simp[Coeff[Pq, x, 0] Int[1/(x*Sqrt[a + b*x^n]), x], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IG tQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2 *((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && !PolyQ[Pq, x^(n/2)]
Result contains complex when optimal does not.
Time = 1.73 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.78
method | result | size |
elliptic | \(-\frac {2 b \left (-\frac {f \,x^{3}}{4 a b}-\frac {x^{2} e}{4 a b}-\frac {d x}{4 a b}-\frac {c}{4 b a}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {i f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}-\frac {c \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2 a^{\frac {3}{2}}}\) | \(253\) |
default | \(d \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+f \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+\frac {e \,x^{2}}{2 a \sqrt {b \,x^{4}+a}}+c \left (\frac {1}{2 a \sqrt {b \,x^{4}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 a^{\frac {3}{2}}}\right )\) | \(280\) |
-2*b*(-1/4/a/b*f*x^3-1/4/a/b*x^2*e-1/4/a/b*d*x-1/4/b/a*c)/((x^4+a/b)*b)^(1 /2)+1/2*d/a/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I /a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2) )^(1/2),I)-1/2*I*f/a^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)* x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)/b^(1/2)*(Ellipt icF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I ))-1/2*c/a^(3/2)*arctanh(a^(1/2)/(b*x^4+a)^(1/2))
Time = 0.12 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.59 \[ \int \frac {c+d x+e x^2+f x^3}{x \left (a+b x^4\right )^{3/2}} \, dx=\frac {2 \, {\left (a b f x^{4} + a^{2} f\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 2 \, {\left ({\left (a b d + a b f\right )} x^{4} + a^{2} d + a^{2} f\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (b^{2} c x^{4} + a b c\right )} \sqrt {a} \log \left (-\frac {b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) + 2 \, {\left (a b f x^{3} + a b e x^{2} + a b d x + a b c\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a^{2} b^{2} x^{4} + a^{3} b\right )}} \]
1/4*(2*(a*b*f*x^4 + a^2*f)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a) ^(1/4)), -1) - 2*((a*b*d + a*b*f)*x^4 + a^2*d + a^2*f)*sqrt(a)*(-b/a)^(3/4 )*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) + (b^2*c*x^4 + a*b*c)*sqrt(a)*log (-(b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x^4) + 2*(a*b*f*x^3 + a*b*e*x^ 2 + a*b*d*x + a*b*c)*sqrt(b*x^4 + a))/(a^2*b^2*x^4 + a^3*b)
Result contains complex when optimal does not.
Time = 7.64 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x+e x^2+f x^3}{x \left (a+b x^4\right )^{3/2}} \, dx=c \left (\frac {2 a^{3} \sqrt {1 + \frac {b x^{4}}{a}}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} + \frac {a^{3} \log {\left (\frac {b x^{4}}{a} \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{4}}{a}} + 1 \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} + \frac {a^{2} b x^{4} \log {\left (\frac {b x^{4}}{a} \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} - \frac {2 a^{2} b x^{4} \log {\left (\sqrt {1 + \frac {b x^{4}}{a}} + 1 \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}}\right ) + \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{2}}{2 a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {f x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \]
c*(2*a**3*sqrt(1 + b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**3*log(b *x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) - 2*a**3*log(sqrt(1 + b*x**4/a) + 1)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**2*b*x**4*log(b*x**4/a)/(4*a**(9 /2) + 4*a**(7/2)*b*x**4) - 2*a**2*b*x**4*log(sqrt(1 + b*x**4/a) + 1)/(4*a* *(9/2) + 4*a**(7/2)*b*x**4)) + d*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), b* x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(5/4)) + e*x**2/(2*a**(3/2)*sqrt( 1 + b*x**4/a)) + f*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), b*x**4*exp_po lar(I*pi)/a)/(4*a**(3/2)*gamma(7/4))
\[ \int \frac {c+d x+e x^2+f x^3}{x \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x} \,d x } \]
\[ \int \frac {c+d x+e x^2+f x^3}{x \left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {f x^{3} + e x^{2} + d x + c}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x} \,d x } \]
Timed out. \[ \int \frac {c+d x+e x^2+f x^3}{x \left (a+b x^4\right )^{3/2}} \, dx=\int \frac {f\,x^3+e\,x^2+d\,x+c}{x\,{\left (b\,x^4+a\right )}^{3/2}} \,d x \]