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

Optimal. Leaf size=387 \[ -\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 \sqrt {b} c-9 \sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+b x^4}}-\frac {3 \sqrt [4]{b} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{7/4} \sqrt {a+b x^4}}-\frac {f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}-\frac {e \sqrt {a+b x^4}}{a^2 x}+\frac {3 \sqrt {b} e x \sqrt {a+b x^4}}{2 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {f \sqrt {a+b x^4}}{2 a^2} \]

[Out]

-1/2*f*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(3/2)-1/2*x*(b*f*x^3+b*e*x^2+b*d*x+b*c)/a^2/(b*x^4+a)^(1/2)+1/2*f*(b
*x^4+a)^(1/2)/a^2-1/3*c*(b*x^4+a)^(1/2)/a^2/x^3-1/2*d*(b*x^4+a)^(1/2)/a^2/x^2-e*(b*x^4+a)^(1/2)/a^2/x+3/2*e*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)*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^(7/4)/(b*x^4+a)^(1/2)-1/12*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))*(-9*e*a^(1/
2)+5*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^(9/4)/(b*x^4+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.61, antiderivative size = 387, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {1829, 1833, 1835, 1585, 1584, 1198, 220, 1196, 21, 266, 50, 63, 208} \[ -\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (5 \sqrt {b} c-9 \sqrt {a} e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}-\frac {e \sqrt {a+b x^4}}{a^2 x}+\frac {3 \sqrt {b} e x \sqrt {a+b x^4}}{2 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{b} e \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{7/4} \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{2 a^2}-\frac {f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(b*c + b*d*x + b*e*x^2 + b*f*x^3))/(2*a^2*Sqrt[a + b*x^4]) + (f*Sqrt[a + b*x^4])/(2*a^2) - (c*Sqrt[a + b*x
^4])/(3*a^2*x^3) - (d*Sqrt[a + b*x^4])/(2*a^2*x^2) - (e*Sqrt[a + b*x^4])/(a^2*x) + (3*Sqrt[b]*e*x*Sqrt[a + b*x
^4])/(2*a^2*(Sqrt[a] + Sqrt[b]*x^2)) - (f*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(2*a^(3/2)) - (3*b^(1/4)*e*(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])/(2
*a^(7/4)*Sqrt[a + b*x^4]) - (b^(1/4)*(5*Sqrt[b]*c - 9*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])/(12*a^(9/4)*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 50

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 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 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 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 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 1584

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 1585

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

Rule 1829

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)*Coeff[R, x, i]*x^(i - m))/a, {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[Pq,
x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

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 1835

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{Pq0 = Coeff[Pq, x, 0]}, Simp[(Pq
0*(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^4 \left (a+b x^4\right )^{3/2}} \, dx &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {\int \frac {-2 b c-2 b d x-2 b e x^2-2 b f x^3+\frac {b^2 c x^4}{a}-\frac {b^2 e x^6}{a}-\frac {2 b^2 f x^7}{a}}{x^4 \sqrt {a+b x^4}} \, dx}{2 a b}\\ &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {\int \left (\frac {-2 b c-2 b e x^2+\frac {b^2 c x^4}{a}-\frac {b^2 e x^6}{a}}{x^4 \sqrt {a+b x^4}}+\frac {-2 b d-2 b f x^2-\frac {2 b^2 f x^6}{a}}{x^3 \sqrt {a+b x^4}}\right ) \, dx}{2 a b}\\ &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {\int \frac {-2 b c-2 b e x^2+\frac {b^2 c x^4}{a}-\frac {b^2 e x^6}{a}}{x^4 \sqrt {a+b x^4}} \, dx}{2 a b}-\frac {\int \frac {-2 b d-2 b f x^2-\frac {2 b^2 f x^6}{a}}{x^3 \sqrt {a+b x^4}} \, dx}{2 a b}\\ &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}+\frac {\int \frac {12 a b e x-10 b^2 c x^3+6 b^2 e x^5}{x^3 \sqrt {a+b x^4}} \, dx}{12 a^2 b}+\frac {\int \frac {8 a b f x+8 b^2 f x^5}{x^2 \sqrt {a+b x^4}} \, dx}{8 a^2 b}\\ &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}+\frac {\int \frac {12 a b e-10 b^2 c x^2+6 b^2 e x^4}{x^2 \sqrt {a+b x^4}} \, dx}{12 a^2 b}+\frac {\int \frac {8 a b f+8 b^2 f x^4}{x \sqrt {a+b x^4}} \, dx}{8 a^2 b}\\ &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}-\frac {e \sqrt {a+b x^4}}{a^2 x}-\frac {\int \frac {20 a b^2 c x-36 a b^2 e x^3}{x \sqrt {a+b x^4}} \, dx}{24 a^3 b}+\frac {f \int \frac {\sqrt {a+b x^4}}{x} \, dx}{a^2}\\ &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}-\frac {e \sqrt {a+b x^4}}{a^2 x}-\frac {\int \frac {20 a b^2 c-36 a b^2 e x^2}{\sqrt {a+b x^4}} \, dx}{24 a^3 b}+\frac {f \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^4\right )}{4 a^2}\\ &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{2 a^2}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}-\frac {e \sqrt {a+b x^4}}{a^2 x}-\frac {\left (3 \sqrt {b} e\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{2 a^{3/2}}-\frac {\left (\sqrt {b} \left (5 \sqrt {b} c-9 \sqrt {a} e\right )\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{6 a^2}+\frac {f \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^4\right )}{4 a}\\ &=-\frac {x \left (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{2 a^2}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}-\frac {e \sqrt {a+b x^4}}{a^2 x}+\frac {3 \sqrt {b} e x \sqrt {a+b x^4}}{2 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \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 )}{2 a^{7/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} \left (5 \sqrt {b} c-9 \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 )}{12 a^{9/4} \sqrt {a+b x^4}}+\frac {f \operatorname {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 (b c+b d x+b e x^2+b f x^3\right )}{2 a^2 \sqrt {a+b x^4}}+\frac {f \sqrt {a+b x^4}}{2 a^2}-\frac {c \sqrt {a+b x^4}}{3 a^2 x^3}-\frac {d \sqrt {a+b x^4}}{2 a^2 x^2}-\frac {e \sqrt {a+b x^4}}{a^2 x}+\frac {3 \sqrt {b} e x \sqrt {a+b x^4}}{2 a^2 \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {f \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {3 \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 )}{2 a^{7/4} \sqrt {a+b x^4}}-\frac {\sqrt [4]{b} \left (5 \sqrt {b} c-9 \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 )}{12 a^{9/4} \sqrt {a+b x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.14, size = 136, normalized size = 0.35 \[ \frac {-2 a c \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-\frac {b x^4}{a}\right )-3 x \left (2 a e x \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {1}{4},\frac {3}{2};\frac {3}{4};-\frac {b x^4}{a}\right )-a f x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b x^4}{a}+1\right )+a d+2 b d x^4\right )}{6 a^2 x^3 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{4} + a} {\left (f x^{3} + e x^{2} + d x + c\right )}}{b^{2} x^{12} + 2 \, a b x^{8} + a^{2} x^{4}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/(b^2*x^12 + 2*a*b*x^8 + a^2*x^4), x)

________________________________________________________________________________________

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

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)^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.19, size = 383, normalized size = 0.99 \[ -\frac {b e \,x^{3}}{2 \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, a^{2}}-\frac {3 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {b}\, e \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, a^{\frac {3}{2}}}+\frac {3 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {b}\, e \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, a^{\frac {3}{2}}}-\frac {b c x}{2 \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, a^{2}}-\frac {5 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, b c \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{6 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, a^{2}}-\frac {f \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{2 a^{\frac {3}{2}}}+\frac {f}{2 \sqrt {b \,x^{4}+a}\, a}-\frac {\sqrt {b \,x^{4}+a}\, e}{a^{2} x}-\frac {\left (2 b \,x^{4}+a \right ) d}{2 \sqrt {b \,x^{4}+a}\, a^{2} x^{2}}-\frac {\sqrt {b \,x^{4}+a}\, c}{3 a^{2} x^{3}} \]

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)^(3/2),x)

[Out]

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

________________________________________________________________________________________

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

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)^(3/2)),x)

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 33.73, size = 321, normalized size = 0.83 \[ d \left (- \frac {1}{2 a \sqrt {b} x^{4} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {\sqrt {b}}{a^{2} \sqrt {\frac {a}{b x^{4}} + 1}}\right ) + f \left (\frac {2 a^{3} \sqrt {1 + \frac {b x^{4}}{a}}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} + \frac {a^{3} \log {\left (\frac {b x^{4}}{a} \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{4}}{a}} + 1 \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} + \frac {a^{2} b x^{4} \log {\left (\frac {b x^{4}}{a} \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}} - \frac {2 a^{2} b x^{4} \log {\left (\sqrt {1 + \frac {b x^{4}}{a}} + 1 \right )}}{4 a^{\frac {9}{2}} + 4 a^{\frac {7}{2}} b x^{4}}\right ) + \frac {c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x^{3} \Gamma \left (\frac {1}{4}\right )} + \frac {e \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x \Gamma \left (\frac {3}{4}\right )} \]

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)**(3/2),x)

[Out]

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

________________________________________________________________________________________