3.6.32 \(\int \frac {x (c+d x+e x^2+f x^3)}{\sqrt {a+b x^4}} \, dx\) [532]

Optimal. Leaf size=299 \[ \frac {e \sqrt {a+b x^4}}{2 b}+\frac {f x \sqrt {a+b x^4}}{3 b}+\frac {d x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}-\frac {\sqrt [4]{a} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {\sqrt [4]{a} \left (3 \sqrt {b} d-\sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}} \]

[Out]

1/2*c*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(1/2)+1/2*e*(b*x^4+a)^(1/2)/b+1/3*f*x*(b*x^4+a)^(1/2)/b+d*x*(b*x^
4+a)^(1/2)/b^(1/2)/(a^(1/2)+x^2*b^(1/2))-a^(1/4)*d*(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)+1/6*a^(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*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(-f*a^(1/2)+3*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^(5/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {1847, 1262, 655, 223, 212, 1294, 1212, 226, 1210} \begin {gather*} \frac {\sqrt [4]{a} \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 {b} d-\sqrt {a} f\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{a} d \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 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{b^{3/4} \sqrt {a+b x^4}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b}}+\frac {d x \sqrt {a+b x^4}}{\sqrt {b} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \sqrt {a+b x^4}}{2 b}+\frac {f x \sqrt {a+b x^4}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + b*x^4])/(2*b) + (f*x*Sqrt[a + b*x^4])/(3*b) + (d*x*Sqrt[a + b*x^4])/(Sqrt[b]*(Sqrt[a] + Sqrt[b]*x^
2)) + (c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]) - (a^(1/4)*d*(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])/(b^(3/4)*Sqrt[a + b*x^4]) + (a^
(1/4)*(3*Sqrt[b]*d - 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])/(6*b^(5/4)*Sqrt[a + b*x^4])

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 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 655

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

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 1262

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

Rule 1294

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.08, size = 160, normalized size = 0.54 \begin {gather*} \frac {3 a e+2 a f x+3 b e x^4+2 b f x^5+3 \sqrt {b} c \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-2 a f x \sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^4}{a}\right )+2 b d x^3 \sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{6 b \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*e + 2*a*f*x + 3*b*e*x^4 + 2*b*f*x^5 + 3*Sqrt[b]*c*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]]
- 2*a*f*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)] + 2*b*d*x^3*Sqrt[1 + (b*x^4)/a]*H
ypergeometric2F1[1/2, 3/4, 7/4, -((b*x^4)/a)])/(6*b*Sqrt[a + b*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.36, size = 230, normalized size = 0.77

method result size
default \(f \left (\frac {x \sqrt {b \,x^{4}+a}}{3 b}-\frac {a \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+\frac {e \sqrt {b \,x^{4}+a}}{2 b}+\frac {i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}+\frac {c \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}\) \(230\)
elliptic \(\frac {f x \sqrt {b \,x^{4}+a}}{3 b}+\frac {e \sqrt {b \,x^{4}+a}}{2 b}-\frac {a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {c \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}+\frac {i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\) \(232\)
risch \(\frac {\left (2 f x +3 e \right ) \sqrt {b \,x^{4}+a}}{6 b}+\frac {i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {b}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {i d \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {b}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {c \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {b}}-\frac {a f \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) \(279\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.11, size = 133, normalized size = 0.44 \begin {gather*} \frac {12 \, \sqrt {b} d x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 4 \, \sqrt {b} {\left (3 \, d + f\right )} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 3 \, \sqrt {b} c x \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, \sqrt {b x^{4} + a} {\left (2 \, f x^{2} + 3 \, e x + 6 \, d\right )}}{12 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*sqrt(b)*d*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) - 4*sqrt(b)*(3*d + f)*x*(-a/b)^(3/4)*
elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + 3*sqrt(b)*c*x*log(-2*b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 2*s
qrt(b*x^4 + a)*(2*f*x^2 + 3*e*x + 6*d))/(b*x)

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Sympy [A]
time = 2.15, size = 129, normalized size = 0.43 \begin {gather*} e \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 {c \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {d 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 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {f 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*Piecewise((x**4/(4*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**4)/(2*b), True)) + c*asinh(sqrt(b)*x**2/sqrt(a))/(2*s
qrt(b)) + d*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(7/4)) + f*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))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{\sqrt {b\,x^4+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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