3.504 \(\int \frac{(c+d x+e x^2+f x^3) \sqrt{a+b x^4}}{x^5} \, dx\)

Optimal. Leaf size=329 \[ \frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{a} f+\sqrt{b} d\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{3 \sqrt [4]{a} \sqrt{a+b x^4}}-\frac{1}{12} \sqrt{a+b x^4} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right )-\frac{b c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{1}{2} \sqrt{b} e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{2 \sqrt{b} f x \sqrt{a+b x^4}}{\sqrt{a}+\sqrt{b} x^2}-\frac{2 \sqrt [4]{a} \sqrt [4]{b} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{\sqrt{a+b x^4}} \]

[Out]

-(((3*c)/x^4 + (4*d)/x^3 + (6*e)/x^2 + (12*f)/x)*Sqrt[a + b*x^4])/12 + (2*Sqrt[b]*f*x*Sqrt[a + b*x^4])/(Sqrt[a
] + Sqrt[b]*x^2) + (Sqrt[b]*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (b*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]
])/(4*Sqrt[a]) - (2*a^(1/4)*b^(1/4)*f*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elli
pticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/Sqrt[a + b*x^4] + (b^(1/4)*(Sqrt[b]*d + 3*Sqrt[a]*f)*(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])/(3*a^(1/4
)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.280873, antiderivative size = 329, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 13, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.433, Rules used = {14, 1825, 1832, 266, 63, 208, 1885, 275, 217, 206, 1198, 220, 1196} \[ -\frac{1}{12} \sqrt{a+b x^4} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right )-\frac{b c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (3 \sqrt{a} f+\sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{3 \sqrt [4]{a} \sqrt{a+b x^4}}+\frac{1}{2} \sqrt{b} e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{2 \sqrt{b} f x \sqrt{a+b x^4}}{\sqrt{a}+\sqrt{b} x^2}-\frac{2 \sqrt [4]{a} \sqrt [4]{b} f \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{\sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((3*c)/x^4 + (4*d)/x^3 + (6*e)/x^2 + (12*f)/x)*Sqrt[a + b*x^4])/12 + (2*Sqrt[b]*f*x*Sqrt[a + b*x^4])/(Sqrt[a
] + Sqrt[b]*x^2) + (Sqrt[b]*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (b*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]
])/(4*Sqrt[a]) - (2*a^(1/4)*b^(1/4)*f*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elli
pticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/Sqrt[a + b*x^4] + (b^(1/4)*(Sqrt[b]*d + 3*Sqrt[a]*f)*(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])/(3*a^(1/4
)*Sqrt[a + b*x^4])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1825

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Module[{u = IntHide[x^m*Pq, x]}, Simp[u*(a +
b*x^n)^p, x] - Dist[b*n*p, Int[x^(m + n)*(a + b*x^n)^(p - 1)*ExpandToSum[u/x^(m + 1), x], x], x]] /; FreeQ[{a
, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && GtQ[p, 0] && LtQ[m + Expon[Pq, x] + 1, 0]

Rule 1832

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[Coeff[Pq, x, 0], Int[1/(x*Sqrt[a + b*x^n]), x
], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] &
& IGtQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]

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 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 1885

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[P
q, 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
, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (c+d x+e x^2+f x^3\right ) \sqrt{a+b x^4}}{x^5} \, dx &=-\frac{1}{12} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right ) \sqrt{a+b x^4}-(2 b) \int \frac{-\frac{c}{4}-\frac{d x}{3}-\frac{e x^2}{2}-f x^3}{x \sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{12} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right ) \sqrt{a+b x^4}-(2 b) \int \frac{-\frac{d}{3}-\frac{e x}{2}-f x^2}{\sqrt{a+b x^4}} \, dx+\frac{1}{2} (b c) \int \frac{1}{x \sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{12} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right ) \sqrt{a+b x^4}-(2 b) \int \left (-\frac{e x}{2 \sqrt{a+b x^4}}+\frac{-\frac{d}{3}-f x^2}{\sqrt{a+b x^4}}\right ) \, dx+\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^4\right )\\ &=-\frac{1}{12} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right ) \sqrt{a+b x^4}-(2 b) \int \frac{-\frac{d}{3}-f x^2}{\sqrt{a+b x^4}} \, dx+\frac{1}{4} c \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^4}\right )+(b e) \int \frac{x}{\sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{12} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right ) \sqrt{a+b x^4}-\frac{b c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{1}{2} (b e) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )-\left (2 \sqrt{a} \sqrt{b} f\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx+\frac{1}{3} \left (2 b \left (d+\frac{3 \sqrt{a} f}{\sqrt{b}}\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{12} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right ) \sqrt{a+b x^4}+\frac{2 \sqrt{b} f x \sqrt{a+b x^4}}{\sqrt{a}+\sqrt{b} x^2}-\frac{b c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{2 \sqrt [4]{a} \sqrt [4]{b} f \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]{b} \left (\sqrt{b} d+3 \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]{a} \sqrt{a+b x^4}}+\frac{1}{2} (b e) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )\\ &=-\frac{1}{12} \left (\frac{3 c}{x^4}+\frac{4 d}{x^3}+\frac{6 e}{x^2}+\frac{12 f}{x}\right ) \sqrt{a+b x^4}+\frac{2 \sqrt{b} f x \sqrt{a+b x^4}}{\sqrt{a}+\sqrt{b} x^2}+\frac{1}{2} \sqrt{b} e \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{b c \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{2 \sqrt [4]{a} \sqrt [4]{b} f \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]{b} \left (\sqrt{b} d+3 \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]{a} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.233866, size = 175, normalized size = 0.53 \[ -\frac{\sqrt{\frac{b x^4}{a}+1} \left (3 a c \sqrt{\frac{b x^4}{a}+1}+3 b c x^4 \tanh ^{-1}\left (\sqrt{\frac{b x^4}{a}+1}\right )+4 a d x \, _2F_1\left (-\frac{3}{4},-\frac{1}{2};\frac{1}{4};-\frac{b x^4}{a}\right )+6 a e x^2 \sqrt{\frac{b x^4}{a}+1}-6 \sqrt{a} \sqrt{b} e x^4 \sinh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )+12 a f x^3 \, _2F_1\left (-\frac{1}{2},-\frac{1}{4};\frac{3}{4};-\frac{b x^4}{a}\right )\right )}{12 x^4 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 + (b*x^4)/a]*(3*a*c*Sqrt[1 + (b*x^4)/a] + 6*a*e*x^2*Sqrt[1 + (b*x^4)/a] - 6*Sqrt[a]*Sqrt[b]*e*x^4*Arc
Sinh[(Sqrt[b]*x^2)/Sqrt[a]] + 3*b*c*x^4*ArcTanh[Sqrt[1 + (b*x^4)/a]] + 4*a*d*x*Hypergeometric2F1[-3/4, -1/2, 1
/4, -((b*x^4)/a)] + 12*a*f*x^3*Hypergeometric2F1[-1/2, -1/4, 3/4, -((b*x^4)/a)]))/(12*x^4*Sqrt[a + b*x^4])

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Maple [C]  time = 0.016, size = 385, normalized size = 1.2 \begin{align*} -{\frac{d}{3\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{2\,bd}{3}\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{c}{4\,a{x}^{4}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}-{\frac{bc}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{bc}{4\,a}\sqrt{b{x}^{4}+a}}-{\frac{e}{2\,a{x}^{2}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{be{x}^{2}}{2\,a}\sqrt{b{x}^{4}+a}}+{\frac{e}{2}\sqrt{b}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{f}{x}\sqrt{b{x}^{4}+a}}+{2\,if\sqrt{a}\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 ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{2\,if\sqrt{a}\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 ){\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)*(b*x^4+a)^(1/2)/x^5,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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 )}}{x^{5}}, 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)*(b*x^4+a)^(1/2)/x^5,x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^5, x)

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Sympy [C]  time = 5.73538, size = 211, normalized size = 0.64 \begin{align*} \frac{\sqrt{a} d \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 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{\sqrt{a} e}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} f \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{b} c \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} + \frac{\sqrt{b} e \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2} - \frac{b c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4 \sqrt{a}} - \frac{b e x^{2}}{2 \sqrt{a} \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)*(b*x**4+a)**(1/2)/x**5,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x^{4} + a}{\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{5}}\,{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)*(b*x^4+a)^(1/2)/x^5,x, algorithm="giac")

[Out]

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^5, x)