3.6.37 \(\int \frac {c+d x+e x^2+f x^3}{x^4 \sqrt {a+b x^4}} \, dx\) [537]

Optimal. Leaf size=323 \[ -\frac {c \sqrt {a+b x^4}}{3 a x^3}-\frac {d \sqrt {a+b x^4}}{2 a x^2}-\frac {e \sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} e x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt [4]{b} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{3/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} \left (\sqrt {b} c-3 \sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+b x^4}} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1847, 1296, 1212, 226, 1210, 1266, 821, 272, 65, 214} \begin {gather*} -\frac {\sqrt [4]{b} \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 {b} c-3 \sqrt {a} e\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} e \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 )}{a^{3/4} \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a x^3}-\frac {d \sqrt {a+b x^4}}{2 a x^2}-\frac {e \sqrt {a+b x^4}}{a x}+\frac {\sqrt {b} e x \sqrt {a+b x^4}}{a \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(c*Sqrt[a + b*x^4])/(a*x^3) - (d*Sqrt[a + b*x^4])/(2*a*x^2) - (e*Sqrt[a + b*x^4])/(a*x) + (Sqrt[b]*e*x*Sq
rt[a + b*x^4])/(a*(Sqrt[a] + Sqrt[b]*x^2)) - (f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(2*Sqrt[a]) - (b^(1/4)*e*(Sq
rt[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]
)/(a^(3/4)*Sqrt[a + b*x^4]) - (b^(1/4)*(Sqrt[b]*c - 3*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqr
t[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(6*a^(5/4)*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 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 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 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(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
] && EqQ[Simplify[m + 2*p + 3], 0]

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 1296

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

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Mathematica [C] Result contains complex when optimal does not.
time = 10.28, size = 249, normalized size = 0.77 \begin {gather*} \frac {-\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} \left (\left (a+b x^4\right ) (2 c+3 x (d+2 e x))+3 \sqrt {a} f x^3 \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+6 \sqrt {a} \sqrt {b} e x^3 \sqrt {1+\frac {b x^4}{a}} E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right )\right |-1\right )-2 \sqrt {b} \left (-i \sqrt {b} c+3 \sqrt {a} e\right ) x^3 \sqrt {1+\frac {b x^4}{a}} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right )\right |-1\right )}{6 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x^3 \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(2*c + 3*x*(d + 2*e*x)) + 3*Sqrt[a]*f*x^3*Sqrt[a + b*x^4]*ArcTanh[Sq
rt[a + b*x^4]/Sqrt[a]])) + 6*Sqrt[a]*Sqrt[b]*e*x^3*Sqrt[1 + (b*x^4)/a]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sq
rt[a]]*x], -1] - 2*Sqrt[b]*((-I)*Sqrt[b]*c + 3*Sqrt[a]*e)*x^3*Sqrt[1 + (b*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*
Sqrt[b])/Sqrt[a]]*x], -1])/(6*a*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^3*Sqrt[a + b*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.38, size = 259, normalized size = 0.80

method	result	size
elliptic	\(-\frac {c \sqrt {b \,x^{4}+a}}{3 a \,x^{3}}-\frac {d \sqrt {b \,x^{4}+a}}{2 a \,x^{2}}-\frac {e \sqrt {b \,x^{4}+a}}{a x}-\frac {b c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {i \sqrt {b}\, e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {f \arctanh \left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2 \sqrt {a}}\)	\(248\)
default	\(-\frac {d \sqrt {b \,x^{4}+a}}{2 a \,x^{2}}+c \left (-\frac {\sqrt {b \,x^{4}+a}}{3 a \,x^{3}}-\frac {b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {f \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 \sqrt {a}}+e \left (-\frac {\sqrt {b \,x^{4}+a}}{a x}+\frac {i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(259\)
risch	\(-\frac {\sqrt {b \,x^{4}+a}\, \left (6 e \,x^{2}+3 d x +2 c \right )}{6 a \,x^{3}}+\frac {i \sqrt {b}\, e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {i \sqrt {b}\, e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{\sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {b c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {f \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 \sqrt {a}}\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.12, size = 136, normalized size = 0.42 \begin {gather*} -\frac {12 \, \sqrt {a} e x^{3} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 4 \, \sqrt {a} {\left (c + 3 \, e\right )} x^{3} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 3 \, \sqrt {a} f x^{3} \log \left (-\frac {b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) + 2 \, \sqrt {b x^{4} + a} {\left (6 \, e x^{2} + 3 \, d x + 2 \, c\right )}}{12 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.92, size = 131, normalized size = 0.41 \begin {gather*} - \frac {\sqrt {b} d \sqrt {\frac {a}{b x^{4}} + 1}}{2 a} + \frac {c \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 \sqrt {a} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {e \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x \Gamma \left (\frac {3}{4}\right )} - \frac {f \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{2 \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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