\(\int \frac {(c+d x+e x^2+f x^3) \sqrt {a+b x^4}}{x} \, dx\) [500]

Optimal result

Integrand size = 30, antiderivative size = 345 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\frac {2 a f x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{4} \left (2 c+e x^2\right ) \sqrt {a+b x^4}+\frac {1}{15} x \left (5 d+3 f x^2\right ) \sqrt {a+b x^4}+\frac {a e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {1}{2} \sqrt {a} c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {2 a^{5/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{3/4} \left (5 \sqrt {b} d+3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}} \]

[Out]

-1/2*c*arctanh((b*x^4+a)^(1/2)/a^(1/2))*a^(1/2)+1/4*a*e*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(1/2)+1/4*(e*x^
2+2*c)*(b*x^4+a)^(1/2)+1/15*x*(3*f*x^2+5*d)*(b*x^4+a)^(1/2)+2/5*a*f*x*(b*x^4+a)^(1/2)/b^(1/2)/(a^(1/2)+x^2*b^(
1/2))-2/5*a^(5/4)*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)/b^(3
/4)/(b*x^4+a)^(1/2)+1/15*a^(3/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*E
llipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(3*f*a^(1/2)+5*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)/b^(3/4)/(b*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {1847, 1266, 829, 858, 223, 212, 272, 65, 214, 1191, 1212, 226, 1210} \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\frac {a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} f+5 \sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}-\frac {2 a^{5/4} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}-\frac {1}{2} \sqrt {a} c \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {a e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}+\frac {1}{4} \sqrt {a+b x^4} \left (2 c+e x^2\right )+\frac {1}{15} x \sqrt {a+b x^4} \left (5 d+3 f x^2\right )+\frac {2 a f x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )} \]

[In]

Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x,x]

[Out]

(2*a*f*x*Sqrt[a + b*x^4])/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^2)) + ((2*c + e*x^2)*Sqrt[a + b*x^4])/4 + (x*(5*d +
3*f*x^2)*Sqrt[a + b*x^4])/15 + (a*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*Sqrt[b]) - (Sqrt[a]*c*ArcTanh[S
qrt[a + b*x^4]/Sqrt[a]])/2 - (2*a^(5/4)*f*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*
EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(5*b^(3/4)*Sqrt[a + b*x^4]) + (a^(3/4)*(5*Sqrt[b]*d + 3*Sqrt[a]
*f)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)
], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^4])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 214

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]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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]]

Rule 829

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*((a + c*x^2)^p/(c*e^2*(m + 2*p + 1)*(m +
2*p + 2))), x] + Dist[2*(p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1191

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*(d*(4*p + 3) + e*(4*p + 1)*x^2)*((a
+ c*x^4)^p/((4*p + 1)*(4*p + 3))), x] + Dist[2*(p/((4*p + 1)*(4*p + 3))), Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4
*p + 1))*x^2, x]*(a + c*x^4)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] &
& FractionQ[p] && IntegerQ[2*p]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 1266

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 1847

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)/c^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, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (c+e x^2\right ) \sqrt {a+b x^4}}{x}+\left (d+f x^2\right ) \sqrt {a+b x^4}\right ) \, dx \\ & = \int \frac {\left (c+e x^2\right ) \sqrt {a+b x^4}}{x} \, dx+\int \left (d+f x^2\right ) \sqrt {a+b x^4} \, dx \\ & = \frac {1}{15} x \left (5 d+3 f x^2\right ) \sqrt {a+b x^4}+\frac {1}{15} \int \frac {10 a d+6 a f x^2}{\sqrt {a+b x^4}} \, dx+\frac {1}{2} \text {Subst}\left (\int \frac {(c+e x) \sqrt {a+b x^2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{4} \left (2 c+e x^2\right ) \sqrt {a+b x^4}+\frac {1}{15} x \left (5 d+3 f x^2\right ) \sqrt {a+b x^4}+\frac {\text {Subst}\left (\int \frac {2 a b c+a b e x}{x \sqrt {a+b x^2}} \, dx,x,x^2\right )}{4 b}-\frac {\left (2 a^{3/2} f\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{5 \sqrt {b}}+\frac {1}{15} \left (2 a \left (5 d+\frac {3 \sqrt {a} f}{\sqrt {b}}\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx \\ & = \frac {2 a f x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{4} \left (2 c+e x^2\right ) \sqrt {a+b x^4}+\frac {1}{15} x \left (5 d+3 f x^2\right ) \sqrt {a+b x^4}-\frac {2 a^{5/4} 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 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{3/4} \left (5 \sqrt {b} d+3 \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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {1}{2} (a c) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x^2}} \, dx,x,x^2\right )+\frac {1}{4} (a e) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right ) \\ & = \frac {2 a f x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{4} \left (2 c+e x^2\right ) \sqrt {a+b x^4}+\frac {1}{15} x \left (5 d+3 f x^2\right ) \sqrt {a+b x^4}-\frac {2 a^{5/4} 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 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{3/4} \left (5 \sqrt {b} d+3 \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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {1}{4} (a c) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )+\frac {1}{4} (a e) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right ) \\ & = \frac {2 a f x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{4} \left (2 c+e x^2\right ) \sqrt {a+b x^4}+\frac {1}{15} x \left (5 d+3 f x^2\right ) \sqrt {a+b x^4}+\frac {a e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {2 a^{5/4} 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 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{3/4} \left (5 \sqrt {b} d+3 \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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}}+\frac {(a c) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^4}\right )}{2 b} \\ & = \frac {2 a f x \sqrt {a+b x^4}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {1}{4} \left (2 c+e x^2\right ) \sqrt {a+b x^4}+\frac {1}{15} x \left (5 d+3 f x^2\right ) \sqrt {a+b x^4}+\frac {a e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{4 \sqrt {b}}-\frac {1}{2} \sqrt {a} c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )-\frac {2 a^{5/4} 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 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^4}}+\frac {a^{3/4} \left (5 \sqrt {b} d+3 \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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{3/4} \sqrt {a+b x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.34 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.60 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\frac {3 a^{3/2} e \sqrt {1+\frac {b x^4}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+3 \sqrt {b} \left (\left (2 c+e x^2\right ) \left (a+b x^4\right )-2 \sqrt {a} c \sqrt {a+b x^4} \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+12 a \sqrt {b} d x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )+4 a \sqrt {b} f x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{12 \sqrt {b} \sqrt {a+b x^4}} \]

[In]

Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x,x]

[Out]

(3*a^(3/2)*e*Sqrt[1 + (b*x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + 3*Sqrt[b]*((2*c + e*x^2)*(a + b*x^4) - 2*Sqr
t[a]*c*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]]) + 12*a*Sqrt[b]*d*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric
2F1[-1/2, 1/4, 5/4, -((b*x^4)/a)] + 4*a*Sqrt[b]*f*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[-1/2, 3/4, 7/4, -(
(b*x^4)/a)])/(12*Sqrt[b]*Sqrt[a + b*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.65 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.79

method	result	size
elliptic	\(\frac {f \,x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {e \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {d x \sqrt {b \,x^{4}+a}}{3}+\frac {c \sqrt {b \,x^{4}+a}}{2}+\frac {2 a d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {a e \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}+\frac {2 i a^{\frac {3}{2}} f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}-\frac {\sqrt {a}\, c \,\operatorname {arctanh}\left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2}\)	\(274\)
default	\(d \left (\frac {x \sqrt {b \,x^{4}+a}}{3}+\frac {2 a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+f \left (\frac {x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {2 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+e \left (\frac {x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {a \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4 \sqrt {b}}\right )+c \left (\frac {\sqrt {b \,x^{4}+a}}{2}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}\right )\)	\(284\)

[In]

int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,x,method=_RETURNVERBOSE)

[Out]

1/5*f*x^3*(b*x^4+a)^(1/2)+1/4*e*x^2*(b*x^4+a)^(1/2)+1/3*d*x*(b*x^4+a)^(1/2)+1/2*c*(b*x^4+a)^(1/2)+2/3*a*d/(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)*Elliptic
F(x*(I/a^(1/2)*b^(1/2))^(1/2),I)+1/4*a*e*ln(2*x^2*b^(1/2)+2*(b*x^4+a)^(1/2))/b^(1/2)+2/5*I*a^(3/2)*f/(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)*(Elli
pticF(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*a^(1/2)*c*arctanh(a^(1/2)/(
b*x^4+a)^(1/2))

Fricas [F]

\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \]

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.06 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=- \frac {\sqrt {a} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {\sqrt {a} d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {a} e x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} + \frac {\sqrt {a} f x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {a c}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {a e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 \sqrt {b}} + \frac {\sqrt {b} c x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} \]

[In]

integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x,x)

[Out]

-sqrt(a)*c*asinh(sqrt(a)/(sqrt(b)*x**2))/2 + sqrt(a)*d*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**4*exp_pola
r(I*pi)/a)/(4*gamma(5/4)) + sqrt(a)*e*x**2*sqrt(1 + b*x**4/a)/4 + sqrt(a)*f*x**3*gamma(3/4)*hyper((-1/2, 3/4),
(7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + a*c/(2*sqrt(b)*x**2*sqrt(a/(b*x**4) + 1)) + a*e*asinh(sqrt
(b)*x**2/sqrt(a))/(4*sqrt(b)) + sqrt(b)*c*x**2/(2*sqrt(a/(b*x**4) + 1))

Maxima [F]

\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \]

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x, x)

Giac [F]

\[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\int { \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x} \,d x } \]

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x,x, algorithm="giac")

[Out]

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x+e x^2+f x^3\right ) \sqrt {a+b x^4}}{x} \, dx=\int \frac {\sqrt {b\,x^4+a}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x} \,d x \]

[In]

int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x,x)

[Out]

int(((a + b*x^4)^(1/2)*(c + d*x + e*x^2 + f*x^3))/x, x)