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

Optimal. Leaf size=405 \[ \frac{2 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (15 \sqrt{a} e+7 \sqrt{b} c\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{105 a^{3/4} \sqrt{a+b x^4}}-\frac{4 b^{9/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}+\frac{4 b^{5/2} c x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}-\frac{3 b^2 d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}+\frac{1}{2} b^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{b \sqrt{a+b x^4} \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right )}{1680}-\frac{1}{504} \left (a+b x^4\right )^{3/2} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \]

[Out]

-(b*((224*c)/x^5 + (315*d)/x^4 + (480*e)/x^3 + (840*f)/x^2)*Sqrt[a + b*x^4])/1680 - (4*b^2*c*Sqrt[a + b*x^4])/
(15*a*x) + (4*b^(5/2)*c*x*Sqrt[a + b*x^4])/(15*a*(Sqrt[a] + Sqrt[b]*x^2)) - (((56*c)/x^9 + (63*d)/x^8 + (72*e)
/x^7 + (84*f)/x^6)*(a + b*x^4)^(3/2))/504 + (b^(3/2)*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (3*b^2*d*Ar
cTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sqrt[a]) - (4*b^(9/4)*c*(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])/(15*a^(3/4)*Sqrt[a + b*x^4]) + (2*b^(7/4)*(7*
Sqrt[b]*c + 15*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcT
an[(b^(1/4)*x)/a^(1/4)], 1/2])/(105*a^(3/4)*Sqrt[a + b*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.425, antiderivative size = 405, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 14, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {14, 1825, 1833, 1282, 1198, 220, 1196, 1252, 844, 217, 206, 266, 63, 208} \[ \frac{2 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (15 \sqrt{a} e+7 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{105 a^{3/4} \sqrt{a+b x^4}}-\frac{4 b^{9/4} c \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 a^{3/4} \sqrt{a+b x^4}}+\frac{4 b^{5/2} c x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}-\frac{3 b^2 d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}+\frac{1}{2} b^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{b \sqrt{a+b x^4} \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right )}{1680}-\frac{1}{504} \left (a+b x^4\right )^{3/2} \left (\frac{56 c}{x^9}+\frac{63 d}{x^8}+\frac{72 e}{x^7}+\frac{84 f}{x^6}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*((224*c)/x^5 + (315*d)/x^4 + (480*e)/x^3 + (840*f)/x^2)*Sqrt[a + b*x^4])/1680 - (4*b^2*c*Sqrt[a + b*x^4])/
(15*a*x) + (4*b^(5/2)*c*x*Sqrt[a + b*x^4])/(15*a*(Sqrt[a] + Sqrt[b]*x^2)) - (((56*c)/x^9 + (63*d)/x^8 + (72*e)
/x^7 + (84*f)/x^6)*(a + b*x^4)^(3/2))/504 + (b^(3/2)*f*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (3*b^2*d*Ar
cTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(16*Sqrt[a]) - (4*b^(9/4)*c*(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])/(15*a^(3/4)*Sqrt[a + b*x^4]) + (2*b^(7/4)*(7*
Sqrt[b]*c + 15*Sqrt[a]*e)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcT
an[(b^(1/4)*x)/a^(1/4)], 1/2])/(105*a^(3/4)*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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

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 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 ) \left (a+b x^4\right )^{3/2}}{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 ) \left (a+b x^4\right )^{3/2}-(6 b) \int \frac{\left (-\frac{c}{9}-\frac{d x}{8}-\frac{e x^2}{7}-\frac{f x^3}{6}\right ) \sqrt{a+b x^4}}{x^6} \, dx\\ &=-\frac{b \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right ) \sqrt{a+b x^4}}{1680}-\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 ) \left (a+b x^4\right )^{3/2}+\left (12 b^2\right ) \int \frac{\frac{c}{45}+\frac{d x}{32}+\frac{e x^2}{21}+\frac{f x^3}{12}}{x^2 \sqrt{a+b x^4}} \, dx\\ &=-\frac{b \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right ) \sqrt{a+b x^4}}{1680}-\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 ) \left (a+b x^4\right )^{3/2}+\left (12 b^2\right ) \int \left (\frac{\frac{c}{45}+\frac{e x^2}{21}}{x^2 \sqrt{a+b x^4}}+\frac{\frac{d}{32}+\frac{f x^2}{12}}{x \sqrt{a+b x^4}}\right ) \, dx\\ &=-\frac{b \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right ) \sqrt{a+b x^4}}{1680}-\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 ) \left (a+b x^4\right )^{3/2}+\left (12 b^2\right ) \int \frac{\frac{c}{45}+\frac{e x^2}{21}}{x^2 \sqrt{a+b x^4}} \, dx+\left (12 b^2\right ) \int \frac{\frac{d}{32}+\frac{f x^2}{12}}{x \sqrt{a+b x^4}} \, dx\\ &=-\frac{b \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right ) \sqrt{a+b x^4}}{1680}-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}-\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 ) \left (a+b x^4\right )^{3/2}+\left (6 b^2\right ) \operatorname{Subst}\left (\int \frac{\frac{d}{32}+\frac{f x}{12}}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )-\frac{\left (12 b^2\right ) \int \frac{-\frac{a e}{21}-\frac{1}{45} b c x^2}{\sqrt{a+b x^4}} \, dx}{a}\\ &=-\frac{b \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right ) \sqrt{a+b x^4}}{1680}-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}-\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 ) \left (a+b x^4\right )^{3/2}-\frac{\left (4 b^{5/2} c\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx}{15 \sqrt{a}}+\frac{1}{16} \left (3 b^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )+\frac{1}{105} \left (4 b^2 \left (\frac{7 \sqrt{b} c}{\sqrt{a}}+15 e\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx+\frac{1}{2} \left (b^2 f\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )\\ &=-\frac{b \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right ) \sqrt{a+b x^4}}{1680}-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}+\frac{4 b^{5/2} c x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\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 ) \left (a+b x^4\right )^{3/2}-\frac{4 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^{3/4} \sqrt{a+b x^4}}+\frac{2 b^{7/4} \left (7 \sqrt{b} c+15 \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^{3/4} \sqrt{a+b x^4}}+\frac{1}{32} \left (3 b^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^4\right )+\frac{1}{2} \left (b^2 f\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )\\ &=-\frac{b \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right ) \sqrt{a+b x^4}}{1680}-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}+\frac{4 b^{5/2} c x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\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 ) \left (a+b x^4\right )^{3/2}+\frac{1}{2} b^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{4 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^{3/4} \sqrt{a+b x^4}}+\frac{2 b^{7/4} \left (7 \sqrt{b} c+15 \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^{3/4} \sqrt{a+b x^4}}+\frac{1}{16} (3 b d) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^4}\right )\\ &=-\frac{b \left (\frac{224 c}{x^5}+\frac{315 d}{x^4}+\frac{480 e}{x^3}+\frac{840 f}{x^2}\right ) \sqrt{a+b x^4}}{1680}-\frac{4 b^2 c \sqrt{a+b x^4}}{15 a x}+\frac{4 b^{5/2} c x \sqrt{a+b x^4}}{15 a \left (\sqrt{a}+\sqrt{b} x^2\right )}-\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 ) \left (a+b x^4\right )^{3/2}+\frac{1}{2} b^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{3 b^2 d \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{4 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^{3/4} \sqrt{a+b x^4}}+\frac{2 b^{7/4} \left (7 \sqrt{b} c+15 \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^{3/4} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.314219, size = 174, normalized size = 0.43 \[ -\frac{\sqrt{a+b x^4} \left (3 x \left (7 \left (8 a^2 f x^2 \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x^4}{a}\right )+9 b^2 d x^8 \tanh ^{-1}\left (\sqrt{\frac{b x^4}{a}+1}\right )+3 a d \left (2 a+5 b x^4\right ) \sqrt{\frac{b x^4}{a}+1}\right )+48 a^2 e x \, _2F_1\left (-\frac{7}{4},-\frac{3}{2};-\frac{3}{4};-\frac{b x^4}{a}\right )\right )+112 a^2 c \, _2F_1\left (-\frac{9}{4},-\frac{3}{2};-\frac{5}{4};-\frac{b x^4}{a}\right )\right )}{1008 a x^9 \sqrt{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x^4]*(112*a^2*c*Hypergeometric2F1[-9/4, -3/2, -5/4, -((b*x^4)/a)] + 3*x*(48*a^2*e*x*Hypergeometri
c2F1[-7/4, -3/2, -3/4, -((b*x^4)/a)] + 7*(3*a*d*(2*a + 5*b*x^4)*Sqrt[1 + (b*x^4)/a] + 9*b^2*d*x^8*ArcTanh[Sqrt
[1 + (b*x^4)/a]] + 8*a^2*f*x^2*Hypergeometric2F1[-3/2, -3/2, -1/2, -((b*x^4)/a)]))))/(1008*a*x^9*Sqrt[1 + (b*x
^4)/a])

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Maple [C]  time = 0.018, size = 437, normalized size = 1.1 \begin{align*} -{\frac{3\,{b}^{2}d}{16}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{ad}{8\,{x}^{8}}\sqrt{b{x}^{4}+a}}-{\frac{5\,bd}{16\,{x}^{4}}\sqrt{b{x}^{4}+a}}+{\frac{f}{2}{b}^{{\frac{3}{2}}}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{af}{6\,{x}^{6}}\sqrt{b{x}^{4}+a}}-{\frac{2\,fb}{3\,{x}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{ae}{7\,{x}^{7}}\sqrt{b{x}^{4}+a}}-{\frac{3\,be}{7\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{4\,{b}^{2}e}{7}\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{ac}{9\,{x}^{9}}\sqrt{b{x}^{4}+a}}-{\frac{11\,bc}{45\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{4\,{b}^{2}c}{15\,ax}\sqrt{b{x}^{4}+a}}+{{\frac{4\,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 ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{{\frac{4\,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 ){\frac{1}{\sqrt{a}}}{\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)^(3/2)/x^10,x)

[Out]

-3/16*d*b^2/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2)-1/8*d*a/x^8*(b*x^4+a)^(1/2)-5/16*d*b/x^4*(b*x^4+a)
^(1/2)+1/2*f*b^(3/2)*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))-1/6*f*a/x^6*(b*x^4+a)^(1/2)-2/3*f*b/x^2*(b*x^4+a)^(1/2)-1
/7*e*a*(b*x^4+a)^(1/2)/x^7-3/7*e*b*(b*x^4+a)^(1/2)/x^3+4/7*e*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*a*
(b*x^4+a)^(1/2)/x^9-11/45*c*b*(b*x^4+a)^(1/2)/x^5-4/15*b^2*c*(b*x^4+a)^(1/2)/a/x+4/15*I*c/a^(1/2)*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)*Elliptic
F(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-4/15*I*c/a^(1/2)*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{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\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)^(3/2)/x^10,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/2)*(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{{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt{b x^{4} + a}}{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)^(3/2)/x^10,x, algorithm="fricas")

[Out]

integral((b*f*x^7 + b*e*x^6 + b*d*x^5 + b*c*x^4 + a*f*x^3 + a*e*x^2 + a*d*x + a*c)*sqrt(b*x^4 + a)/x^10, x)

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Sympy [C]  time = 13.5714, size = 449, normalized size = 1.11 \begin{align*} \frac{a^{\frac{3}{2}} 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{a^{\frac{3}{2}} 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{\sqrt{a} b c \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} + \frac{\sqrt{a} b e \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} b f}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} - \frac{a^{2} d}{8 \sqrt{b} x^{10} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{3 a \sqrt{b} d}{16 x^{6} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{a \sqrt{b} f \sqrt{\frac{a}{b x^{4}} + 1}}{6 x^{4}} - \frac{b^{\frac{3}{2}} d \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} - \frac{b^{\frac{3}{2}} d}{16 x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{3}{2}} f \sqrt{\frac{a}{b x^{4}} + 1}}{6} + \frac{b^{\frac{3}{2}} f \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2} - \frac{3 b^{2} d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{16 \sqrt{a}} - \frac{b^{2} f 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)**(3/2)/x**10,x)

[Out]

a**(3/2)*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)) + a**(3/2)*
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)) + sqrt(a)*b*c*gamma(
-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**5*gamma(-1/4)) + sqrt(a)*b*e*gamma(-3/4)*hy
per((-3/4, -1/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**3*gamma(1/4)) - sqrt(a)*b*f/(2*x**2*sqrt(1 + b*x**4/
a)) - a**2*d/(8*sqrt(b)*x**10*sqrt(a/(b*x**4) + 1)) - 3*a*sqrt(b)*d/(16*x**6*sqrt(a/(b*x**4) + 1)) - a*sqrt(b)
*f*sqrt(a/(b*x**4) + 1)/(6*x**4) - b**(3/2)*d*sqrt(a/(b*x**4) + 1)/(4*x**2) - b**(3/2)*d/(16*x**2*sqrt(a/(b*x*
*4) + 1)) - b**(3/2)*f*sqrt(a/(b*x**4) + 1)/6 + b**(3/2)*f*asinh(sqrt(b)*x**2/sqrt(a))/2 - 3*b**2*d*asinh(sqrt
(a)/(sqrt(b)*x**2))/(16*sqrt(a)) - b**2*f*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{{\left (b x^{4} + a\right )}^{\frac{3}{2}}{\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)^(3/2)/x^10,x, algorithm="giac")

[Out]

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