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

Optimal. Leaf size=342 \[ -\frac {\sqrt {a+b x^4} \left (c-e x^2\right )}{2 x^2}+\frac {1}{2} \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {\sqrt {a+b x^4} \left (3 d-f x^2\right )}{3 x}+\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {a} f+3 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {2 \sqrt {b} d x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}-\frac {1}{2} \sqrt {a} e \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {1833, 1252, 813, 844, 217, 206, 266, 63, 208, 1272, 1198, 220, 1196} \[ -\frac {\sqrt {a+b x^4} \left (c-e x^2\right )}{2 x^2}+\frac {1}{2} \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-\frac {\sqrt {a+b x^4} \left (3 d-f x^2\right )}{3 x}+\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (\sqrt {a} f+3 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {2 \sqrt {b} d x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}-\frac {2 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a+b x^4}}-\frac {1}{2} \sqrt {a} e \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 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 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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 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 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 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 1272

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

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)]

Rubi steps

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

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Mathematica [C]  time = 0.26, size = 204, normalized size = 0.60 \[ \frac {\sqrt {a} \sqrt {b} c x^2 \sqrt {\frac {b x^4}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )-2 a d x \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};-\frac {b x^4}{a}\right )-\sqrt {a} e x^2 \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+2 a f x^3 \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )-a c+a e x^2-b c x^4+b e x^6}{2 x^2 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*c) + a*e*x^2 - b*c*x^4 + b*e*x^6 + Sqrt[a]*Sqrt[b]*c*x^2*Sqrt[1 + (b*x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a
]] - Sqrt[a]*e*x^2*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]] - 2*a*d*x*Sqrt[1 + (b*x^4)/a]*Hypergeometr
ic2F1[-1/2, -1/4, 3/4, -((b*x^4)/a)] + 2*a*f*x^3*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[-1/2, 1/4, 5/4, -((b*x^
4)/a)])/(2*x^2*Sqrt[a + b*x^4])

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}}, x\right ) \]

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^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}}\,{d x} \]

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^3,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.18, size = 360, normalized size = 1.05 \[ \frac {2 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a f \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, \sqrt {b}\, d \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {2 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {a}\, \sqrt {b}\, d \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {\sqrt {b \,x^{4}+a}\, b c \,x^{2}}{2 a}-\frac {\sqrt {a}\, e \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{2}+\frac {\sqrt {b}\, c \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2}+\frac {\sqrt {b \,x^{4}+a}\, f x}{3}+\frac {\sqrt {b \,x^{4}+a}\, e}{2}-\frac {\sqrt {b \,x^{4}+a}\, d}{x}-\frac {\left (b \,x^{4}+a \right )^{\frac {3}{2}} c}{2 a \,x^{2}} \]

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^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{3}}\,{d x} \]

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^3,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 6.41, size = 230, normalized size = 0.67 \[ - \frac {\sqrt {a} c}{2 x^{2} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} d \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {\sqrt {a} e \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2} + \frac {\sqrt {a} f x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a e}{2 \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {\sqrt {b} c \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2} + \frac {\sqrt {b} e x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {b c x^{2}}{2 \sqrt {a} \sqrt {1 + \frac {b x^{4}}{a}}} \]

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**3,x)

[Out]

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

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