3.6.48 \(\int \frac {c+d x+e x^2+f x^3}{x^2 (a+b x^4)^{3/2}} \, dx\) [548]

Optimal. Leaf size=344 \[ \frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{2 a^2}-\frac {c \sqrt {a+b x^4}}{a^2 x}+\frac {3 \sqrt {b} c x \sqrt {a+b x^4}}{2 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {3 \sqrt [4]{b} c \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{7/4} \sqrt {a+b x^4}}+\frac {\left (3 \sqrt {b} c+\sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{7/4} \sqrt [4]{b} \sqrt {a+b x^4}} \]

[Out]

-1/2*d*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(3/2)+1/2*x*(-b*d*x^3-b*c*x^2+a*f*x+a*e)/a^2/(b*x^4+a)^(1/2)+1/2*d*(
b*x^4+a)^(1/2)/a^2-c*(b*x^4+a)^(1/2)/a^2/x+3/2*c*x*b^(1/2)*(b*x^4+a)^(1/2)/a^2/(a^(1/2)+x^2*b^(1/2))-3/2*b^(1/
4)*c*(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^(7/4)/(b*x^4+a)^(
1/2)+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))*(e*a^(1/2)+3*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^(7/4)/b^(1/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.25, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1843, 1847, 1849, 1598, 1212, 226, 1210, 21, 272, 52, 65, 214} \begin {gather*} \frac {\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} e+3 \sqrt {b} c\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{7/4} \sqrt [4]{b} \sqrt {a+b x^4}}-\frac {3 \sqrt [4]{b} 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 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{7/4} \sqrt {a+b x^4}}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{a^2 x}+\frac {3 \sqrt {b} c x \sqrt {a+b x^4}}{2 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {d \sqrt {a+b x^4}}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a*e + a*f*x - b*c*x^2 - b*d*x^3))/(2*a^2*Sqrt[a + b*x^4]) + (d*Sqrt[a + b*x^4])/(2*a^2) - (c*Sqrt[a + b*x^
4])/(a^2*x) + (3*Sqrt[b]*c*x*Sqrt[a + b*x^4])/(2*a^2*(Sqrt[a] + Sqrt[b]*x^2)) - (d*ArcTanh[Sqrt[a + b*x^4]/Sqr
t[a]])/(2*a^(3/2)) - (3*b^(1/4)*c*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elliptic
E[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(7/4)*Sqrt[a + b*x^4]) + ((3*Sqrt[b]*c + Sqrt[a]*e)*(Sqrt[a] + Sqr
t[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])/(4*a^(7/4
)*b^(1/4)*Sqrt[a + b*x^4])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 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 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1843

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = Polynomi
alQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1
)*x^m*Pq, a + b*x^n, x], i}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[x^m*(a + b*x^n)^(p + 1)*Expand
ToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)/a)*Coeff[R, x, i]*x^(i - m), {i, 0, n - 1}], x], x], x] + S
imp[(-x)*R*((a + b*x^n)^(p + 1)/(a^2*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x]]] /; FreeQ[{a, b}, x] && PolyQ[P
q, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

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)]

Rule 1849

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{Pq0 = Coeff[Pq, x, 0]}, Simp[Pq0
*(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(2*a*c*(m + 1)), Int[(c*x)^(m + 1)*ExpandToSum
[2*a*(m + 1)*((Pq - Pq0)/x) - 2*b*Pq0*(m + n*(p + 1) + 1)*x^(n - 1), x]*(a + b*x^n)^p, x], x] /; NeQ[Pq0, 0]]
/; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[m, -1] && LeQ[n - 1, Expon[Pq, x]]

Rubi steps

\begin {align*} \int \frac {c+d x+e x^2+f x^3}{x^2 \left (a+b x^4\right )^{3/2}} \, dx &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {\int \frac {-2 b c-2 b d x-b e x^2-\frac {b^2 c x^4}{a}-\frac {2 b^2 d x^5}{a}}{x^2 \sqrt {a+b x^4}} \, dx}{2 a b}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {\int \left (\frac {-2 b c-b e x^2-\frac {b^2 c x^4}{a}}{x^2 \sqrt {a+b x^4}}+\frac {-2 b d-\frac {2 b^2 d x^4}{a}}{x \sqrt {a+b x^4}}\right ) \, dx}{2 a b}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {\int \frac {-2 b c-b e x^2-\frac {b^2 c x^4}{a}}{x^2 \sqrt {a+b x^4}} \, dx}{2 a b}-\frac {\int \frac {-2 b d-\frac {2 b^2 d x^4}{a}}{x \sqrt {a+b x^4}} \, dx}{2 a b}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{a^2 x}+\frac {\int \frac {2 a b e x+6 b^2 c x^3}{x \sqrt {a+b x^4}} \, dx}{4 a^2 b}+\frac {d \int \frac {\sqrt {a+b x^4}}{x} \, dx}{a^2}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{a^2 x}+\frac {\int \frac {2 a b e+6 b^2 c x^2}{\sqrt {a+b x^4}} \, dx}{4 a^2 b}+\frac {d \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^4\right )}{4 a^2}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{2 a^2}-\frac {c \sqrt {a+b x^4}}{a^2 x}-\frac {\left (3 \sqrt {b} c\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{2 a^{3/2}}+\frac {d \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )}{4 a}+\frac {\left (3 \sqrt {b} c+\sqrt {a} e\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{2 a^{3/2}}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{2 a^2}-\frac {c \sqrt {a+b x^4}}{a^2 x}+\frac {3 \sqrt {b} c x \sqrt {a+b x^4}}{2 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{b} 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 )}{2 a^{7/4} \sqrt {a+b x^4}}+\frac {\left (3 \sqrt {b} c+\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 )}{4 a^{7/4} \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {d \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^4}\right )}{2 a b}\\ &=\frac {x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {d \sqrt {a+b x^4}}{2 a^2}-\frac {c \sqrt {a+b x^4}}{a^2 x}+\frac {3 \sqrt {b} c x \sqrt {a+b x^4}}{2 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {3 \sqrt [4]{b} 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 )}{2 a^{7/4} \sqrt {a+b x^4}}+\frac {\left (3 \sqrt {b} c+\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 )}{4 a^{7/4} \sqrt [4]{b} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.30, size = 245, normalized size = 0.71 \begin {gather*} \frac {\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} \left (-2 a c-3 b c x^4+a x (d+x (e+f x))-\sqrt {a} d x \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+3 \sqrt {a} \sqrt {b} c x \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 )-i \sqrt {a} \left (-3 i \sqrt {b} c+\sqrt {a} e\right ) x \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 )}{2 a^2 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.42, size = 298, normalized size = 0.87

method	result	size
elliptic	\(-\frac {c \sqrt {b \,x^{4}+a}}{a^{2} x}-\frac {2 b \left (\frac {c \,x^{3}}{4 a^{2}}-\frac {x^{2} f}{4 a b}-\frac {x e}{4 a b}-\frac {d}{4 a b}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 i \sqrt {b}\, c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {d \arctanh \left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2 a^{\frac {3}{2}}}\)	\(268\)
default	\(\frac {f \,x^{2}}{2 a \sqrt {b \,x^{4}+a}}+e \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {1}{2 a \sqrt {b \,x^{4}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 a^{\frac {3}{2}}}\right )+c \left (-\frac {b \,x^{3}}{2 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {\sqrt {b \,x^{4}+a}}{a^{2} x}+\frac {3 i \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(298\)
risch	\(-\frac {c \sqrt {b \,x^{4}+a}}{a^{2} x}-\frac {b c \,x^{3}}{2 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {b}\, c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {3 i \sqrt {b}\, c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {f \,x^{2}}{2 a \sqrt {b \,x^{4}+a}}+\frac {e x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {d}{2 a \sqrt {b \,x^{4}+a}}-\frac {d \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2 a^{\frac {3}{2}}}\)	\(355\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.12, size = 215, normalized size = 0.62 \begin {gather*} -\frac {6 \, {\left (b^{2} c x^{5} + a b c x\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - 2 \, {\left ({\left (3 \, b^{2} c - a b e\right )} x^{5} + {\left (3 \, a b c - a^{2} e\right )} x\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (b^{2} d x^{5} + a b d x\right )} \sqrt {a} \log \left (-\frac {b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) + 2 \, {\left (3 \, b^{2} c x^{4} - a b f x^{3} - a b e x^{2} - a b d x + 2 \, a b c\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a^{2} b^{2} x^{5} + a^{3} b x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(6*(b^2*c*x^5 + a*b*c*x)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin(x*(-b/a)^(1/4)), -1) - 2*((3*b^2*c - a*b*
e)*x^5 + (3*a*b*c - a^2*e)*x)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) - (b^2*d*x^5 + a*b*d
*x)*sqrt(a)*log(-(b*x^4 - 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x^4) + 2*(3*b^2*c*x^4 - a*b*f*x^3 - a*b*e*x^2 - a*b
*d*x + 2*a*b*c)*sqrt(b*x^4 + a))/(a^2*b^2*x^5 + a^3*b*x)

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Sympy [C] Result contains complex when optimal does not.
time = 8.67, size = 291, normalized size = 0.85 \begin {gather*} d \left (\frac {2 a^{3} \sqrt {1 + \frac {b x^{4}}{a}}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} + \frac {a^{3} \log {\left (\frac {b x^{4}}{a} \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{4}}{a}} + 1 \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} + \frac {a^{2} b x^{4} \log {\left (\frac {b x^{4}}{a} \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} - \frac {2 a^{2} b x^{4} \log {\left (\sqrt {1 + \frac {b x^{4}}{a}} + 1 \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}}\right ) + \frac {c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x \Gamma \left (\frac {3}{4}\right )} + \frac {e x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {f x^{2}}{2 a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{4}}{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**2/(b*x**4+a)**(3/2),x)

[Out]

d*(2*a**3*sqrt(1 + b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**3*log(b*x**4/a)/(4*a**(9/2) + 4*a**(7/2)*b*
x**4) - 2*a**3*log(sqrt(1 + b*x**4/a) + 1)/(4*a**(9/2) + 4*a**(7/2)*b*x**4) + a**2*b*x**4*log(b*x**4/a)/(4*a**
(9/2) + 4*a**(7/2)*b*x**4) - 2*a**2*b*x**4*log(sqrt(1 + b*x**4/a) + 1)/(4*a**(9/2) + 4*a**(7/2)*b*x**4)) + c*g
amma(-1/4)*hyper((-1/4, 3/2), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*x*gamma(3/4)) + e*x*gamma(1/4)*hyp
er((1/4, 3/2), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(5/4)) + f*x**2/(2*a**(3/2)*sqrt(1 + b*x**4/
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^2/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.94, size = 133, normalized size = 0.39 \begin {gather*} \frac {d}{2\,a\,\sqrt {b\,x^4+a}}-\frac {d\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^4+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}+\frac {f\,x^2}{2\,a\,\sqrt {b\,x^4+a}}-\frac {c\,{\left (\frac {a}{b\,x^4}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{4};\ \frac {11}{4};\ -\frac {a}{b\,x^4}\right )}{7\,x\,{\left (b\,x^4+a\right )}^{3/2}}+\frac {e\,x\,{\left (\frac {b\,x^4}{a}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{2};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (b\,x^4+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d/(2*a*(a + b*x^4)^(1/2)) - (d*atanh((a + b*x^4)^(1/2)/a^(1/2)))/(2*a^(3/2)) + (f*x^2)/(2*a*(a + b*x^4)^(1/2))
- (c*(a/(b*x^4) + 1)^(3/2)*hypergeom([3/2, 7/4], 11/4, -a/(b*x^4)))/(7*x*(a + b*x^4)^(3/2)) + (e*x*((b*x^4)/a
+ 1)^(3/2)*hypergeom([1/4, 3/2], 5/4, -(b*x^4)/a))/(a + b*x^4)^(3/2)

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