3.509 \(\int \frac{(c+d x+e x^2+f x^3) \sqrt{a+b x^4}}{x^{10}} \, dx\)

Optimal. Leaf size=425 \[ -\frac{b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} e+7 \sqrt{b} c\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{105 a^{7/4} \sqrt{a+b x^4}}-\frac{2 b^{5/2} c x \sqrt{a+b x^4}}{15 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{2 b^2 c \sqrt{a+b x^4}}{15 a^2 x}+\frac{2 b^{9/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{7/4} \sqrt{a+b x^4}}+\frac{b^2 d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 a^{3/2}}-\frac{1}{504} \sqrt{a+b x^4} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right )-\frac{2 b c \sqrt{a+b x^4}}{45 a x^5}-\frac{b d \sqrt{a+b x^4}}{16 a x^4}-\frac{2 b e \sqrt{a+b x^4}}{21 a x^3}-\frac{b f \sqrt{a+b x^4}}{6 a x^2} \]

[Out]

-(((56*c)/x^9 + (63*d)/x^8 + (72*e)/x^7 + (84*f)/x^6)*Sqrt[a + b*x^4])/504 - (2*b*c*Sqrt[a + b*x^4])/(45*a*x^5
) - (b*d*Sqrt[a + b*x^4])/(16*a*x^4) - (2*b*e*Sqrt[a + b*x^4])/(21*a*x^3) - (b*f*Sqrt[a + b*x^4])/(6*a*x^2) +
(2*b^2*c*Sqrt[a + b*x^4])/(15*a^2*x) - (2*b^(5/2)*c*x*Sqrt[a + b*x^4])/(15*a^2*(Sqrt[a] + Sqrt[b]*x^2)) + (b^2
*d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*a^(3/2)) + (2*b^(9/4)*c*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqr
t[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*a^(7/4)*Sqrt[a + b*x^4]) - (b^(7/4)*
(7*Sqrt[b]*c + 5*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*Ar
cTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(105*a^(7/4)*Sqrt[a + b*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.433128, antiderivative size = 425, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 13, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.433, Rules used = {14, 1825, 1833, 1282, 1198, 220, 1196, 1252, 835, 807, 266, 63, 208} \[ -\frac{b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} e+7 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 a^{7/4} \sqrt{a+b x^4}}-\frac{2 b^{5/2} c x \sqrt{a+b x^4}}{15 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{2 b^2 c \sqrt{a+b x^4}}{15 a^2 x}+\frac{2 b^{9/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{7/4} \sqrt{a+b x^4}}+\frac{b^2 d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 a^{3/2}}-\frac{1}{504} \sqrt{a+b x^4} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right )-\frac{2 b c \sqrt{a+b x^4}}{45 a x^5}-\frac{b d \sqrt{a+b x^4}}{16 a x^4}-\frac{2 b e \sqrt{a+b x^4}}{21 a x^3}-\frac{b f \sqrt{a+b x^4}}{6 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^10,x]

[Out]

-(((56*c)/x^9 + (63*d)/x^8 + (72*e)/x^7 + (84*f)/x^6)*Sqrt[a + b*x^4])/504 - (2*b*c*Sqrt[a + b*x^4])/(45*a*x^5
) - (b*d*Sqrt[a + b*x^4])/(16*a*x^4) - (2*b*e*Sqrt[a + b*x^4])/(21*a*x^3) - (b*f*Sqrt[a + b*x^4])/(6*a*x^2) +
(2*b^2*c*Sqrt[a + b*x^4])/(15*a^2*x) - (2*b^(5/2)*c*x*Sqrt[a + b*x^4])/(15*a^2*(Sqrt[a] + Sqrt[b]*x^2)) + (b^2
*d*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*a^(3/2)) + (2*b^(9/4)*c*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqr
t[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*a^(7/4)*Sqrt[a + b*x^4]) - (b^(7/4)*
(7*Sqrt[b]*c + 5*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*Ar
cTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(105*a^(7/4)*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 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 1282

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 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 1252

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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

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 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]

Rubi steps

\begin{align*} \int \frac{\left (c+d x+e x^2+f x^3\right ) \sqrt{a+b x^4}}{x^{10}} \, dx &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-(2 b) \int \frac{-\frac{c}{9}-\frac{d x}{8}-\frac{e x^2}{7}-\frac{f x^3}{6}}{x^6 \sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-(2 b) \int \left (\frac{-\frac{c}{9}-\frac{e x^2}{7}}{x^6 \sqrt{a+b x^4}}+\frac{-\frac{d}{8}-\frac{f x^2}{6}}{x^5 \sqrt{a+b x^4}}\right ) \, dx\\ &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-(2 b) \int \frac{-\frac{c}{9}-\frac{e x^2}{7}}{x^6 \sqrt{a+b x^4}} \, dx-(2 b) \int \frac{-\frac{d}{8}-\frac{f x^2}{6}}{x^5 \sqrt{a+b x^4}} \, dx\\ &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{45 a x^5}-b \operatorname{Subst}\left (\int \frac{-\frac{d}{8}-\frac{f x}{6}}{x^3 \sqrt{a+b x^2}} \, dx,x,x^2\right )+\frac{(2 b) \int \frac{\frac{5 a e}{7}-\frac{1}{3} b c x^2}{x^4 \sqrt{a+b x^4}} \, dx}{5 a}\\ &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{45 a x^5}-\frac{b d \sqrt{a+b x^4}}{16 a x^4}-\frac{2 b e \sqrt{a+b x^4}}{21 a x^3}-\frac{(2 b) \int \frac{a b c+\frac{5}{7} a b e x^2}{x^2 \sqrt{a+b x^4}} \, dx}{15 a^2}+\frac{b \operatorname{Subst}\left (\int \frac{\frac{a f}{3}-\frac{b d x}{8}}{x^2 \sqrt{a+b x^2}} \, dx,x,x^2\right )}{2 a}\\ &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{45 a x^5}-\frac{b d \sqrt{a+b x^4}}{16 a x^4}-\frac{2 b e \sqrt{a+b x^4}}{21 a x^3}-\frac{b f \sqrt{a+b x^4}}{6 a x^2}+\frac{2 b^2 c \sqrt{a+b x^4}}{15 a^2 x}+\frac{(2 b) \int \frac{-\frac{5}{7} a^2 b e-a b^2 c x^2}{\sqrt{a+b x^4}} \, dx}{15 a^3}-\frac{\left (b^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )}{16 a}\\ &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{45 a x^5}-\frac{b d \sqrt{a+b x^4}}{16 a x^4}-\frac{2 b e \sqrt{a+b x^4}}{21 a x^3}-\frac{b f \sqrt{a+b x^4}}{6 a x^2}+\frac{2 b^2 c \sqrt{a+b x^4}}{15 a^2 x}+\frac{\left (2 b^{5/2} c\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx}{15 a^{3/2}}-\frac{\left (b^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^4\right )}{32 a}-\frac{\left (2 b^2 \left (7 \sqrt{b} c+5 \sqrt{a} e\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx}{105 a^{3/2}}\\ &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{45 a x^5}-\frac{b d \sqrt{a+b x^4}}{16 a x^4}-\frac{2 b e \sqrt{a+b x^4}}{21 a x^3}-\frac{b f \sqrt{a+b x^4}}{6 a x^2}+\frac{2 b^2 c \sqrt{a+b x^4}}{15 a^2 x}-\frac{2 b^{5/2} c x \sqrt{a+b x^4}}{15 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{2 b^{9/4} 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 )}{15 a^{7/4} \sqrt{a+b x^4}}-\frac{b^{7/4} \left (7 \sqrt{b} c+5 \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 )}{105 a^{7/4} \sqrt{a+b x^4}}-\frac{(b d) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^4}\right )}{16 a}\\ &=-\frac{1}{504} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \sqrt{a+b x^4}-\frac{2 b c \sqrt{a+b x^4}}{45 a x^5}-\frac{b d \sqrt{a+b x^4}}{16 a x^4}-\frac{2 b e \sqrt{a+b x^4}}{21 a x^3}-\frac{b f \sqrt{a+b x^4}}{6 a x^2}+\frac{2 b^2 c \sqrt{a+b x^4}}{15 a^2 x}-\frac{2 b^{5/2} c x \sqrt{a+b x^4}}{15 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{b^2 d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 a^{3/2}}+\frac{2 b^{9/4} 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 )}{15 a^{7/4} \sqrt{a+b x^4}}-\frac{b^{7/4} \left (7 \sqrt{b} c+5 \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 )}{105 a^{7/4} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.16672, size = 148, normalized size = 0.35 \[ -\frac{\sqrt{a+b x^4} \left (3 x^2 \left (7 x \left (a+b x^4\right ) \sqrt{\frac{b x^4}{a}+1} \left (a^2 f+b^2 d x^6 \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{b x^4}{a}+1\right )\right )+6 a^3 e \, _2F_1\left (-\frac{7}{4},-\frac{1}{2};-\frac{3}{4};-\frac{b x^4}{a}\right )\right )+14 a^3 c \, _2F_1\left (-\frac{9}{4},-\frac{1}{2};-\frac{5}{4};-\frac{b x^4}{a}\right )\right )}{126 a^3 x^9 \sqrt{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((c + d*x + e*x^2 + f*x^3)*Sqrt[a + b*x^4])/x^10,x]

[Out]

-(Sqrt[a + b*x^4]*(14*a^3*c*Hypergeometric2F1[-9/4, -1/2, -5/4, -((b*x^4)/a)] + 3*x^2*(6*a^3*e*Hypergeometric2
F1[-7/4, -1/2, -3/4, -((b*x^4)/a)] + 7*x*(a + b*x^4)*Sqrt[1 + (b*x^4)/a]*(a^2*f + b^2*d*x^6*Hypergeometric2F1[
3/2, 3, 5/2, 1 + (b*x^4)/a]))))/(126*a^3*x^9*Sqrt[1 + (b*x^4)/a])

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Maple [C]  time = 0.018, size = 429, normalized size = 1. \begin{align*} -{\frac{d}{8\,a{x}^{8}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{bd}{16\,{a}^{2}{x}^{4}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}+{\frac{{b}^{2}d}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{{b}^{2}d}{16\,{a}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{f}{6\,{x}^{6}a} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{2}}}}-{\frac{e}{7\,{x}^{7}}\sqrt{b{x}^{4}+a}}-{\frac{2\,be}{21\,a{x}^{3}}\sqrt{b{x}^{4}+a}}-{\frac{2\,{b}^{2}e}{21\,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{c}{9\,{x}^{9}}\sqrt{b{x}^{4}+a}}-{\frac{2\,bc}{45\,a{x}^{5}}\sqrt{b{x}^{4}+a}}+{\frac{2\,{b}^{2}c}{15\,{a}^{2}x}\sqrt{b{x}^{4}+a}}-{{\frac{2\,i}{15}}c{b}^{{\frac{5}{2}}}\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{2\,i}{15}}c{b}^{{\frac{5}{2}}}\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)*(b*x^4+a)^(1/2)/x^10,x)

[Out]

-1/8*d/a/x^8*(b*x^4+a)^(3/2)+1/16*d*b/a^2/x^4*(b*x^4+a)^(3/2)+1/16*d*b^2/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(
1/2))/x^2)-1/16*d*b^2/a^2*(b*x^4+a)^(1/2)-1/6*f*(b*x^4+a)^(3/2)/x^6/a-1/7*e/x^7*(b*x^4+a)^(1/2)-2/21*b*e*(b*x^
4+a)^(1/2)/a/x^3-2/21*e/a*b^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)-1/9*c/x^9*(b*x^4+a)^(1/2)-2/45*b*c*(b*x^4+a
)^(1/2)/a/x^5+2/15*b^2*c*(b*x^4+a)^(1/2)/a^2/x-2/15*I*c/a^(3/2)*b^(5/2)/(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)+2
/15*I*c/a^(3/2)*b^(5/2)/(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]  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^{10}}\,{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^10,x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^4 + a)*(f*x^3 + e*x^2 + d*x + c)/x^10, 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 )}}{x^{10}}, 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^10,x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 8.46443, size = 246, normalized size = 0.58 \begin{align*} \frac{\sqrt{a} c \Gamma \left (- \frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{9}{4}, - \frac{1}{2} \\ - \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{9} \Gamma \left (- \frac{5}{4}\right )} + \frac{\sqrt{a} e \Gamma \left (- \frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{4}, - \frac{1}{2} \\ - \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac{3}{4}\right )} - \frac{a d}{8 \sqrt{b} x^{10} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{3 \sqrt{b} d}{16 x^{6} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{\sqrt{b} f \sqrt{\frac{a}{b x^{4}} + 1}}{6 x^{4}} - \frac{b^{\frac{3}{2}} d}{16 a x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{3}{2}} f \sqrt{\frac{a}{b x^{4}} + 1}}{6 a} + \frac{b^{2} d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{16 a^{\frac{3}{2}}} \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**10,x)

[Out]

sqrt(a)*c*gamma(-9/4)*hyper((-9/4, -1/2), (-5/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**9*gamma(-5/4)) + sqrt(a)*e*
gamma(-7/4)*hyper((-7/4, -1/2), (-3/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**7*gamma(-3/4)) - a*d/(8*sqrt(b)*x**10
*sqrt(a/(b*x**4) + 1)) - 3*sqrt(b)*d/(16*x**6*sqrt(a/(b*x**4) + 1)) - sqrt(b)*f*sqrt(a/(b*x**4) + 1)/(6*x**4)
- b**(3/2)*d/(16*a*x**2*sqrt(a/(b*x**4) + 1)) - b**(3/2)*f*sqrt(a/(b*x**4) + 1)/(6*a) + b**2*d*asinh(sqrt(a)/(
sqrt(b)*x**2))/(16*a**(3/2))

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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^{10}}\,{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^10,x, algorithm="giac")

[Out]

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