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

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

[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])

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Rubi [A]  time = 0.382575, antiderivative size = 344, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 12, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {1829, 1833, 1835, 1584, 1198, 220, 1196, 21, 266, 50, 63, 208} \[ \frac{x \left (a e+a f x-b c x^2-b d x^3\right )}{2 a^2 \sqrt{a+b x^4}}+\frac{\left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (\sqrt{a} e+3 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 a^{7/4} \sqrt [4]{b} \sqrt{a+b x^4}}-\frac{c \sqrt{a+b x^4}}{a^2 x}+\frac{3 \sqrt{b} c x \sqrt{a+b x^4}}{2 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{3 \sqrt [4]{b} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 a^{7/4} \sqrt{a+b x^4}}+\frac{d \sqrt{a+b x^4}}{2 a^2}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x + e*x^2 + f*x^3)/(x^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 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]]

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 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 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 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 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 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 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]

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 \operatorname{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 \operatorname{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 \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 (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*}

Mathematica [C]  time = 0.114255, size = 123, normalized size = 0.36 \[ \frac{-2 c \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{1}{4},\frac{3}{2};\frac{3}{4};-\frac{b x^4}{a}\right )+d x \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b x^4}{a}+1\right )+x^2 \left (e \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^4}{a}\right )+e+f x\right )}{2 a x \sqrt{a+b x^4}} \]

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]

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

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Maple [C]  time = 0.01, size = 355, normalized size = 1. \begin{align*}{\frac{f{x}^{2}}{2\,a}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{ex}{2\,a}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}+{\frac{e}{2\,a}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{d}{2\,a}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{d}{2}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{bc{x}^{3}}{2\,{a}^{2}}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}-{\frac{c}{{a}^{2}x}\sqrt{b{x}^{4}+a}}+{{\frac{3\,i}{2}}c\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{3\,i}{2}}c\sqrt{b}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \end{align*}

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)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f x^{3} + e x^{2} + d x + c}{{\left (b x^{4} + a\right )}^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{b^{2} x^{10} + 2 \, a b x^{6} + a^{2} x^{2}}, x\right ) \end{align*}

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]

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/(b^2*x^10 + 2*a*b*x^6 + a^2*x^2), x)

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Sympy [C]  time = 34.1444, size = 291, normalized size = 0.85 \begin{align*} 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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{f x^{3} + e x^{2} + d x + c}{{\left (b x^{4} + a\right )}^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

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