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

Optimal. Leaf size=336 \[ -\frac{a^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (9 \sqrt{a} f+5 \sqrt{b} d\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{30 b^{7/4} \sqrt{a+b x^4}}+\frac{3 a^{5/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{7/4} \sqrt{a+b x^4}}-\frac{a e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{3/2}}-\frac{3 a f x \sqrt{a+b x^4}}{5 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{\sqrt{a+b x^4} \left (2 c+e x^2\right )}{4 b}+\frac{d x \sqrt{a+b x^4}}{3 b}+\frac{f x^3 \sqrt{a+b x^4}}{5 b} \]

[Out]

(d*x*Sqrt[a + b*x^4])/(3*b) + (f*x^3*Sqrt[a + b*x^4])/(5*b) - (3*a*f*x*Sqrt[a + b*x^4])/(5*b^(3/2)*(Sqrt[a] +
Sqrt[b]*x^2)) + ((2*c + e*x^2)*Sqrt[a + b*x^4])/(4*b) - (a*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2
)) + (3*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^(7/4)*Sqrt[a + b*x^4]) - (a^(3/4)*(5*Sqrt[b]*d + 9*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])/(30*b^(7/4)*Sqr
t[a + b*x^4])

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Rubi [A]  time = 0.265053, antiderivative size = 336, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1833, 1252, 780, 217, 206, 1280, 1198, 220, 1196} \[ -\frac{a^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (9 \sqrt{a} f+5 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{30 b^{7/4} \sqrt{a+b x^4}}+\frac{3 a^{5/4} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 b^{7/4} \sqrt{a+b x^4}}-\frac{a e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{3/2}}-\frac{3 a f x \sqrt{a+b x^4}}{5 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{\sqrt{a+b x^4} \left (2 c+e x^2\right )}{4 b}+\frac{d x \sqrt{a+b x^4}}{3 b}+\frac{f x^3 \sqrt{a+b x^4}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*Sqrt[a + b*x^4])/(3*b) + (f*x^3*Sqrt[a + b*x^4])/(5*b) - (3*a*f*x*Sqrt[a + b*x^4])/(5*b^(3/2)*(Sqrt[a] +
Sqrt[b]*x^2)) + ((2*c + e*x^2)*Sqrt[a + b*x^4])/(4*b) - (a*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4*b^(3/2
)) + (3*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^(7/4)*Sqrt[a + b*x^4]) - (a^(3/4)*(5*Sqrt[b]*d + 9*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])/(30*b^(7/4)*Sqr
t[a + b*x^4])

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 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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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 1280

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

Rubi steps

\begin{align*} \int \frac{x^3 \left (c+d x+e x^2+f x^3\right )}{\sqrt{a+b x^4}} \, dx &=\int \left (\frac{x^3 \left (c+e x^2\right )}{\sqrt{a+b x^4}}+\frac{x^4 \left (d+f x^2\right )}{\sqrt{a+b x^4}}\right ) \, dx\\ &=\int \frac{x^3 \left (c+e x^2\right )}{\sqrt{a+b x^4}} \, dx+\int \frac{x^4 \left (d+f x^2\right )}{\sqrt{a+b x^4}} \, dx\\ &=\frac{f x^3 \sqrt{a+b x^4}}{5 b}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{x (c+e x)}{\sqrt{a+b x^2}} \, dx,x,x^2\right )-\frac{\int \frac{x^2 \left (3 a f-5 b d x^2\right )}{\sqrt{a+b x^4}} \, dx}{5 b}\\ &=\frac{d x \sqrt{a+b x^4}}{3 b}+\frac{f x^3 \sqrt{a+b x^4}}{5 b}+\frac{\left (2 c+e x^2\right ) \sqrt{a+b x^4}}{4 b}+\frac{\int \frac{-5 a b d-9 a b f x^2}{\sqrt{a+b x^4}} \, dx}{15 b^2}-\frac{(a e) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )}{4 b}\\ &=\frac{d x \sqrt{a+b x^4}}{3 b}+\frac{f x^3 \sqrt{a+b x^4}}{5 b}+\frac{\left (2 c+e x^2\right ) \sqrt{a+b x^4}}{4 b}-\frac{(a e) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )}{4 b}+\frac{\left (3 a^{3/2} f\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx}{5 b^{3/2}}-\frac{\left (a \left (5 \sqrt{b} d+9 \sqrt{a} f\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx}{15 b^{3/2}}\\ &=\frac{d x \sqrt{a+b x^4}}{3 b}+\frac{f x^3 \sqrt{a+b x^4}}{5 b}-\frac{3 a f x \sqrt{a+b x^4}}{5 b^{3/2} \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{\left (2 c+e x^2\right ) \sqrt{a+b x^4}}{4 b}-\frac{a e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{3/2}}+\frac{3 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^{7/4} \sqrt{a+b x^4}}-\frac{a^{3/4} \left (5 \sqrt{b} d+9 \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 )}{30 b^{7/4} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.145449, size = 212, normalized size = 0.63 \[ \frac{30 \sqrt{b} c \left (a+b x^4\right )-20 a \sqrt{b} d x \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^4}{a}\right )+20 \sqrt{b} d x \left (a+b x^4\right )+15 \sqrt{b} e x^2 \left (a+b x^4\right )-15 a e \sqrt{a+b x^4} \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-12 a \sqrt{b} f x^3 \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^4}{a}\right )+12 \sqrt{b} f x^3 \left (a+b x^4\right )}{60 b^{3/2} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30*Sqrt[b]*c*(a + b*x^4) + 20*Sqrt[b]*d*x*(a + b*x^4) + 15*Sqrt[b]*e*x^2*(a + b*x^4) + 12*Sqrt[b]*f*x^3*(a +
b*x^4) - 15*a*e*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 20*a*Sqrt[b]*d*x*Sqrt[1 + (b*x^4)/a]*
Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] - 12*a*Sqrt[b]*f*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/2,
3/4, 7/4, -((b*x^4)/a)])/(60*b^(3/2)*Sqrt[a + b*x^4])

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Maple [C]  time = 0.007, size = 325, normalized size = 1. \begin{align*}{\frac{f{x}^{3}}{5\,b}\sqrt{b{x}^{4}+a}}-{{\frac{3\,i}{5}}f{a}^{{\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 ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{{\frac{3\,i}{5}}f{a}^{{\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 ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{e{x}^{2}}{4\,b}\sqrt{b{x}^{4}+a}}-{\frac{ae}{4}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{dx}{3\,b}\sqrt{b{x}^{4}+a}}-{\frac{ad}{3\,b}\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}{2\,b}\sqrt{b{x}^{4}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{b x^{4} + a} c}{2 \, b} + \int \frac{f x^{6} + e x^{5} + d x^{4}}{\sqrt{b x^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b*x^4 + a)*c/b + integrate((f*x^6 + e*x^5 + d*x^4)/sqrt(b*x^4 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{f x^{6} + e x^{5} + d x^{4} + c x^{3}}{\sqrt{b x^{4} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x^6 + e*x^5 + d*x^4 + c*x^3)/sqrt(b*x^4 + a), x)

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Sympy [A]  time = 5.02286, size = 156, normalized size = 0.46 \begin{align*} \frac{\sqrt{a} e x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4 b} - \frac{a e \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4 b^{\frac{3}{2}}} + c \left (\begin{cases} \frac{x^{4}}{4 \sqrt{a}} & \text{for}\: b = 0 \\\frac{\sqrt{a + b x^{4}}}{2 b} & \text{otherwise} \end{cases}\right ) + \frac{d x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{9}{4}\right )} + \frac{f x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*e*x**2*sqrt(1 + b*x**4/a)/(4*b) - a*e*asinh(sqrt(b)*x**2/sqrt(a))/(4*b**(3/2)) + c*Piecewise((x**4/(4*
sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**4)/(2*b), True)) + d*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**4*exp_
polar(I*pi)/a)/(4*sqrt(a)*gamma(9/4)) + f*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)
/(4*sqrt(a)*gamma(11/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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