3.505 \(\int \frac{(c+d x+e x^2+f x^3) \sqrt{a+b x^4}}{x^6} \, dx\)

Optimal. Leaf size=360 \[ \frac{b^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} e+3 \sqrt{b} c\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}-\frac{2 b^{5/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} \sqrt{a+b x^4}}+\frac{2 b^{3/2} c x \sqrt{a+b x^4}}{5 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{1}{60} \sqrt{a+b x^4} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right )-\frac{2 b c \sqrt{a+b x^4}}{5 a x}-\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{1}{2} \sqrt{b} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right ) \]

[Out]

-(((12*c)/x^5 + (15*d)/x^4 + (20*e)/x^3 + (30*f)/x^2)*Sqrt[a + b*x^4])/60 - (2*b*c*Sqrt[a + b*x^4])/(5*a*x) +
(2*b^(3/2)*c*x*Sqrt[a + b*x^4])/(5*a*(Sqrt[a] + Sqrt[b]*x^2)) + (Sqrt[b]*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^
4]])/2 - (b*d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(4*Sqrt[a]) - (2*b^(5/4)*c*(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*a^(3/4)*Sqrt[a + b*x^4]) +
(b^(3/4)*(3*Sqrt[b]*c + 5*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellip
ticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*a^(3/4)*Sqrt[a + b*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.323537, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 14, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {14, 1825, 1833, 1282, 1198, 220, 1196, 1252, 844, 217, 206, 266, 63, 208} \[ \frac{b^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} e+3 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}-\frac{2 b^{5/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{3/4} \sqrt{a+b x^4}}+\frac{2 b^{3/2} c x \sqrt{a+b x^4}}{5 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{1}{60} \sqrt{a+b x^4} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right )-\frac{2 b c \sqrt{a+b x^4}}{5 a x}-\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{1}{2} \sqrt{b} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^6,x]

[Out]

-(((12*c)/x^5 + (15*d)/x^4 + (20*e)/x^3 + (30*f)/x^2)*Sqrt[a + b*x^4])/60 - (2*b*c*Sqrt[a + b*x^4])/(5*a*x) +
(2*b^(3/2)*c*x*Sqrt[a + b*x^4])/(5*a*(Sqrt[a] + Sqrt[b]*x^2)) + (Sqrt[b]*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^
4]])/2 - (b*d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(4*Sqrt[a]) - (2*b^(5/4)*c*(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*a^(3/4)*Sqrt[a + b*x^4]) +
(b^(3/4)*(3*Sqrt[b]*c + 5*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ellip
ticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*a^(3/4)*Sqrt[a + b*x^4])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1825

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{u = IntHide[x^m*Pq, x]}, Simp[u*(a +
b*x^n)^p, x] - Dist[b*n*p, Int[x^(m + n)*(a + b*x^n)^(p - 1)*ExpandToSum[u/x^(m + 1), x], x], x]] /; FreeQ[{a
, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && GtQ[p, 0] && LtQ[m + Expon[Pq, x] + 1, 0]

Rule 1833

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)*Sum[Coeff[Pq, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p)/c^j, {
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)]

Rule 1282

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(f*x)^(m + 1)*(a
+ c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + c*x^4)^p*(a*e*(m + 1) -
c*d*(m + 4*p + 5)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p]
|| IntegerQ[m])

Rule 1198

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 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1252

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 844

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 217

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 266

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 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (c+d x+e x^2+f x^3\right ) \sqrt{a+b x^4}}{x^6} \, dx &=-\frac{1}{60} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right ) \sqrt{a+b x^4}-(2 b) \int \frac{-\frac{c}{5}-\frac{d x}{4}-\frac{e x^2}{3}-\frac{f x^3}{2}}{x^2 \sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{60} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right ) \sqrt{a+b x^4}-(2 b) \int \left (\frac{-\frac{c}{5}-\frac{e x^2}{3}}{x^2 \sqrt{a+b x^4}}+\frac{-\frac{d}{4}-\frac{f x^2}{2}}{x \sqrt{a+b x^4}}\right ) \, dx\\ &=-\frac{1}{60} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right ) \sqrt{a+b x^4}-(2 b) \int \frac{-\frac{c}{5}-\frac{e x^2}{3}}{x^2 \sqrt{a+b x^4}} \, dx-(2 b) \int \frac{-\frac{d}{4}-\frac{f x^2}{2}}{x \sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{60} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{5 a x}-b \operatorname{Subst}\left (\int \frac{-\frac{d}{4}-\frac{f x}{2}}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )+\frac{(2 b) \int \frac{\frac{a e}{3}+\frac{1}{5} b c x^2}{\sqrt{a+b x^4}} \, dx}{a}\\ &=-\frac{1}{60} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{5 a x}-\frac{\left (2 b^{3/2} c\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx}{5 \sqrt{a}}+\frac{1}{4} (b d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )+\frac{1}{15} \left (2 b \left (\frac{3 \sqrt{b} c}{\sqrt{a}}+5 e\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx+\frac{1}{2} (b f) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )\\ &=-\frac{1}{60} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{5 a x}+\frac{2 b^{3/2} c x \sqrt{a+b x^4}}{5 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{2 b^{5/4} c \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 a^{3/4} \sqrt{a+b x^4}}+\frac{b^{3/4} \left (3 \sqrt{b} c+5 \sqrt{a} e\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 a^{3/4} \sqrt{a+b x^4}}+\frac{1}{8} (b d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^4\right )+\frac{1}{2} (b f) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )\\ &=-\frac{1}{60} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{5 a x}+\frac{2 b^{3/2} c x \sqrt{a+b x^4}}{5 a \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{2} \sqrt{b} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{2 b^{5/4} c \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 a^{3/4} \sqrt{a+b x^4}}+\frac{b^{3/4} \left (3 \sqrt{b} c+5 \sqrt{a} e\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 a^{3/4} \sqrt{a+b x^4}}+\frac{1}{4} d \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^4}\right )\\ &=-\frac{1}{60} \left (\frac{12 c}{x^5}+\frac{15 d}{x^4}+\frac{20 e}{x^3}+\frac{30 f}{x^2}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{5 a x}+\frac{2 b^{3/2} c x \sqrt{a+b x^4}}{5 a \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{1}{2} \sqrt{b} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{2 b^{5/4} c \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 a^{3/4} \sqrt{a+b x^4}}+\frac{b^{3/4} \left (3 \sqrt{b} c+5 \sqrt{a} e\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 a^{3/4} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.233774, size = 179, normalized size = 0.5 \[ -\frac{\sqrt{a+b x^4} \left (12 a c \, _2F_1\left (-\frac{5}{4},-\frac{1}{2};-\frac{1}{4};-\frac{b x^4}{a}\right )+5 x \left (3 a d \sqrt{\frac{b x^4}{a}+1}+3 b d x^4 \tanh ^{-1}\left (\sqrt{\frac{b x^4}{a}+1}\right )+4 a e x \, _2F_1\left (-\frac{3}{4},-\frac{1}{2};\frac{1}{4};-\frac{b x^4}{a}\right )+6 a f x^2 \sqrt{\frac{b x^4}{a}+1}-6 \sqrt{a} \sqrt{b} f x^4 \sinh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right )\right )}{60 a x^5 \sqrt{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^6,x]

[Out]

-(Sqrt[a + b*x^4]*(12*a*c*Hypergeometric2F1[-5/4, -1/2, -1/4, -((b*x^4)/a)] + 5*x*(3*a*d*Sqrt[1 + (b*x^4)/a] +
6*a*f*x^2*Sqrt[1 + (b*x^4)/a] - 6*Sqrt[a]*Sqrt[b]*f*x^4*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]] + 3*b*d*x^4*ArcTanh[Sq
rt[1 + (b*x^4)/a]] + 4*a*e*x*Hypergeometric2F1[-3/4, -1/2, 1/4, -((b*x^4)/a)])))/(60*a*x^5*Sqrt[1 + (b*x^4)/a]
)

________________________________________________________________________________________

Maple [C]  time = 0.015, size = 404, normalized size = 1.1 \begin{align*} -{\frac{e}{3\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{2\,be}{3}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{c}{5\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{2\,bc}{5\,ax}\sqrt{b{x}^{4}+a}}+{{\frac{2\,i}{5}}c{b}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{2\,i}{5}}c{b}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{d}{4\,a{x}^{4}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}-{\frac{bd}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{bd}{4\,a}\sqrt{b{x}^{4}+a}}-{\frac{f}{2\,a{x}^{2}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{bf{x}^{2}}{2\,a}\sqrt{b{x}^{4}+a}}+{\frac{f}{2}\sqrt{b}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^6,x)

[Out]

-1/3*e/x^3*(b*x^4+a)^(1/2)+2/3*e*b/(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)-1/5*c/x^5*(b*x^4+a)^(1/2)-2/5*b*c*(b*x
^4+a)^(1/2)/a/x+2/5*I*c/a^(1/2)*b^(3/2)/(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)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-2/5*I*c/a^(1/2)*b^(3/2)/(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)*EllipticE(x*(
I/a^(1/2)*b^(1/2))^(1/2),I)-1/4*d/a/x^4*(b*x^4+a)^(3/2)-1/4*d*b/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2
)+1/4*d*b/a*(b*x^4+a)^(1/2)-1/2*f/a/x^2*(b*x^4+a)^(3/2)+1/2*f*b/a*x^2*(b*x^4+a)^(1/2)+1/2*f*b^(1/2)*ln(x^2*b^(
1/2)+(b*x^4+a)^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^6,x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^6, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^6,x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^6, x)

________________________________________________________________________________________

Sympy [C]  time = 6.0442, size = 216, normalized size = 0.6 \begin{align*} \frac{\sqrt{a} c \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} + \frac{\sqrt{a} e \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{\sqrt{a} f}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} - \frac{\sqrt{b} d \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} + \frac{\sqrt{b} f \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2} - \frac{b d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4 \sqrt{a}} - \frac{b f x^{2}}{2 \sqrt{a} \sqrt{1 + \frac{b x^{4}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**(1/2)/x**6,x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^(1/2)/x^6,x, algorithm="giac")

[Out]

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^6, x)