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

Optimal. Leaf size=408 \[ \frac{2 a^{3/4} \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (9 \sqrt{a} e+5 \sqrt{b} c\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15 \sqrt{a+b x^4}}-\frac{12 a^{5/4} \sqrt [4]{b} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{\left (a+b x^4\right )^{3/2} \left (5 c-3 e x^2\right )}{15 x^3}-\frac{2 \sqrt{a+b x^4} \left (9 a e-5 b c x^2\right )}{15 x}-\frac{\left (a+b x^4\right )^{3/2} \left (3 d-f x^2\right )}{6 x^2}+\frac{1}{4} \sqrt{a+b x^4} \left (2 a f+3 b d x^2\right )+\frac{3}{4} a \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 a \sqrt{b} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

[Out]

(12*a*Sqrt[b]*e*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) - (2*(9*a*e - 5*b*c*x^2)*Sqrt[a + b*x^4])/(15*x
) + ((2*a*f + 3*b*d*x^2)*Sqrt[a + b*x^4])/4 - ((5*c - 3*e*x^2)*(a + b*x^4)^(3/2))/(15*x^3) - ((3*d - f*x^2)*(a
+ b*x^4)^(3/2))/(6*x^2) + (3*a*Sqrt[b]*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/4 - (a^(3/2)*f*ArcTanh[Sqrt[
a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*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])/(5*Sqrt[a + b*x^4]) + (2*a^(3/4)*b^(1/4)*(5*Sqrt[b]*c + 9*
Sqrt[a]*e)*(Sqrt[a] + Sqrt[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])/(15*Sqrt[a + b*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.335922, antiderivative size = 408, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 14, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {1833, 1272, 1198, 220, 1196, 1252, 813, 815, 844, 217, 206, 266, 63, 208} \[ \frac{2 a^{3/4} \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (9 \sqrt{a} e+5 \sqrt{b} c\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 \sqrt{a+b x^4}}-\frac{12 a^{5/4} \sqrt [4]{b} e \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{2} a^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{\left (a+b x^4\right )^{3/2} \left (5 c-3 e x^2\right )}{15 x^3}-\frac{2 \sqrt{a+b x^4} \left (9 a e-5 b c x^2\right )}{15 x}-\frac{\left (a+b x^4\right )^{3/2} \left (3 d-f x^2\right )}{6 x^2}+\frac{1}{4} \sqrt{a+b x^4} \left (2 a f+3 b d x^2\right )+\frac{3}{4} a \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 a \sqrt{b} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*Sqrt[b]*e*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) - (2*(9*a*e - 5*b*c*x^2)*Sqrt[a + b*x^4])/(15*x
) + ((2*a*f + 3*b*d*x^2)*Sqrt[a + b*x^4])/4 - ((5*c - 3*e*x^2)*(a + b*x^4)^(3/2))/(15*x^3) - ((3*d - f*x^2)*(a
+ b*x^4)^(3/2))/(6*x^2) + (3*a*Sqrt[b]*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/4 - (a^(3/2)*f*ArcTanh[Sqrt[
a + b*x^4]/Sqrt[a]])/2 - (12*a^(5/4)*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])/(5*Sqrt[a + b*x^4]) + (2*a^(3/4)*b^(1/4)*(5*Sqrt[b]*c + 9*
Sqrt[a]*e)*(Sqrt[a] + Sqrt[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])/(15*Sqrt[a + b*x^4])

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 1272

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((f*x)^(m + 1)*(a
+ c*x^4)^p*(d*(m + 4*p + 3) + e*(m + 1)*x^2))/(f*(m + 1)*(m + 4*p + 3)), x] + Dist[(4*p)/(f^2*(m + 1)*(m + 4*p
+ 3)), Int[(f*x)^(m + 2)*(a + c*x^4)^(p - 1)*(a*e*(m + 1) - c*d*(m + 4*p + 3)*x^2), x], x] /; FreeQ[{a, c, d,
e, f}, x] && GtQ[p, 0] && LtQ[m, -1] && m + 4*p + 3 != 0 && 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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

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^4} \, dx &=\int \left (\frac{\left (c+e x^2\right ) \left (a+b x^4\right )^{3/2}}{x^4}+\frac{\left (d+f x^2\right ) \left (a+b x^4\right )^{3/2}}{x^3}\right ) \, dx\\ &=\int \frac{\left (c+e x^2\right ) \left (a+b x^4\right )^{3/2}}{x^4} \, dx+\int \frac{\left (d+f x^2\right ) \left (a+b x^4\right )^{3/2}}{x^3} \, dx\\ &=-\frac{\left (5 c-3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{15 x^3}-\frac{2}{5} \int \frac{\left (-3 a e-5 b c x^2\right ) \sqrt{a+b x^4}}{x^2} \, dx+\frac{1}{2} \operatorname{Subst}\left (\int \frac{(d+f x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{2 \left (9 a e-5 b c x^2\right ) \sqrt{a+b x^4}}{15 x}-\frac{\left (5 c-3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{15 x^3}-\frac{\left (3 d-f x^2\right ) \left (a+b x^4\right )^{3/2}}{6 x^2}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{(-2 a f-6 b d x) \sqrt{a+b x^2}}{x} \, dx,x,x^2\right )+\frac{4}{15} \int \frac{5 a b c+9 a b e x^2}{\sqrt{a+b x^4}} \, dx\\ &=-\frac{2 \left (9 a e-5 b c x^2\right ) \sqrt{a+b x^4}}{15 x}+\frac{1}{4} \left (2 a f+3 b d x^2\right ) \sqrt{a+b x^4}-\frac{\left (5 c-3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{15 x^3}-\frac{\left (3 d-f x^2\right ) \left (a+b x^4\right )^{3/2}}{6 x^2}-\frac{\operatorname{Subst}\left (\int \frac{-4 a^2 b f-6 a b^2 d x}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )}{8 b}-\frac{1}{5} \left (12 a^{3/2} \sqrt{b} e\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx+\frac{1}{15} \left (4 a \sqrt{b} \left (5 \sqrt{b} c+9 \sqrt{a} e\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx\\ &=\frac{12 a \sqrt{b} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{2 \left (9 a e-5 b c x^2\right ) \sqrt{a+b x^4}}{15 x}+\frac{1}{4} \left (2 a f+3 b d x^2\right ) \sqrt{a+b x^4}-\frac{\left (5 c-3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{15 x^3}-\frac{\left (3 d-f x^2\right ) \left (a+b x^4\right )^{3/2}}{6 x^2}-\frac{12 a^{5/4} \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 )}{5 \sqrt{a+b x^4}}+\frac{2 a^{3/4} \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 )}{15 \sqrt{a+b x^4}}+\frac{1}{4} (3 a b d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )+\frac{1}{2} \left (a^2 f\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac{12 a \sqrt{b} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{2 \left (9 a e-5 b c x^2\right ) \sqrt{a+b x^4}}{15 x}+\frac{1}{4} \left (2 a f+3 b d x^2\right ) \sqrt{a+b x^4}-\frac{\left (5 c-3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{15 x^3}-\frac{\left (3 d-f x^2\right ) \left (a+b x^4\right )^{3/2}}{6 x^2}-\frac{12 a^{5/4} \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 )}{5 \sqrt{a+b x^4}}+\frac{2 a^{3/4} \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 )}{15 \sqrt{a+b x^4}}+\frac{1}{4} (3 a b d) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )+\frac{1}{4} \left (a^2 f\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^4\right )\\ &=\frac{12 a \sqrt{b} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{2 \left (9 a e-5 b c x^2\right ) \sqrt{a+b x^4}}{15 x}+\frac{1}{4} \left (2 a f+3 b d x^2\right ) \sqrt{a+b x^4}-\frac{\left (5 c-3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{15 x^3}-\frac{\left (3 d-f x^2\right ) \left (a+b x^4\right )^{3/2}}{6 x^2}+\frac{3}{4} a \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{12 a^{5/4} \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 )}{5 \sqrt{a+b x^4}}+\frac{2 a^{3/4} \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 )}{15 \sqrt{a+b x^4}}+\frac{\left (a^2 f\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^4}\right )}{2 b}\\ &=\frac{12 a \sqrt{b} e x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{2 \left (9 a e-5 b c x^2\right ) \sqrt{a+b x^4}}{15 x}+\frac{1}{4} \left (2 a f+3 b d x^2\right ) \sqrt{a+b x^4}-\frac{\left (5 c-3 e x^2\right ) \left (a+b x^4\right )^{3/2}}{15 x^3}-\frac{\left (3 d-f x^2\right ) \left (a+b x^4\right )^{3/2}}{6 x^2}+\frac{3}{4} a \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{1}{2} a^{3/2} f \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{12 a^{5/4} \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 )}{5 \sqrt{a+b x^4}}+\frac{2 a^{3/4} \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 )}{15 \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.357646, size = 194, normalized size = 0.48 \[ \frac{x^2 \left (f x \sqrt{\frac{b x^4}{a}+1} \left (\sqrt{a+b x^4} \left (4 a+b x^4\right )-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )\right )-6 a e \sqrt{a+b x^4} \, _2F_1\left (-\frac{3}{2},-\frac{1}{4};\frac{3}{4};-\frac{b x^4}{a}\right )\right )-2 a c \sqrt{a+b x^4} \, _2F_1\left (-\frac{3}{2},-\frac{3}{4};\frac{1}{4};-\frac{b x^4}{a}\right )-3 a d x \sqrt{a+b x^4} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x^4}{a}\right )}{6 x^3 \sqrt{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*c*Sqrt[a + b*x^4]*Hypergeometric2F1[-3/2, -3/4, 1/4, -((b*x^4)/a)] - 3*a*d*x*Sqrt[a + b*x^4]*Hypergeomet
ric2F1[-3/2, -1/2, 1/2, -((b*x^4)/a)] + x^2*(f*x*Sqrt[1 + (b*x^4)/a]*(Sqrt[a + b*x^4]*(4*a + b*x^4) - 3*a^(3/2
)*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]]) - 6*a*e*Sqrt[a + b*x^4]*Hypergeometric2F1[-3/2, -1/4, 3/4, -((b*x^4)/a)]))
/(6*x^3*Sqrt[1 + (b*x^4)/a])

________________________________________________________________________________________

Maple [C]  time = 0.014, size = 408, normalized size = 1. \begin{align*}{\frac{bf{x}^{4}}{6}\sqrt{b{x}^{4}+a}}+{\frac{2\,af}{3}\sqrt{b{x}^{4}+a}}-{\frac{f}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ) }-{\frac{ac}{3\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{bcx}{3}\sqrt{b{x}^{4}+a}}+{\frac{4\,abc}{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{bd{x}^{2}}{4}\sqrt{b{x}^{4}+a}}+{\frac{3\,ad}{4}\sqrt{b}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{ad}{2\,{x}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{ae}{x}\sqrt{b{x}^{4}+a}}+{\frac{be{x}^{3}}{5}\sqrt{b{x}^{4}+a}}+{{\frac{12\,i}{5}}e{a}^{{\frac{3}{2}}}\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}}}}-{{\frac{12\,i}{5}}e{a}^{{\frac{3}{2}}}\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)^(3/2)/x^4,x)

[Out]

1/6*f*b*x^4*(b*x^4+a)^(1/2)+2/3*f*a*(b*x^4+a)^(1/2)-1/2*f*a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^2)-1/3*
c*a*(b*x^4+a)^(1/2)/x^3+1/3*c*b*x*(b*x^4+a)^(1/2)+4/3*c*a*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*d*b*x^2*(
b*x^4+a)^(1/2)+3/4*d*a*b^(1/2)*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))-1/2*d*a/x^2*(b*x^4+a)^(1/2)-e*a*(b*x^4+a)^(1/2)
/x+1/5*e*b*x^3*(b*x^4+a)^(1/2)+12/5*I*e*a^(3/2)*b^(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)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-12/5*I*e*a^(3/2)*
b^(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)

________________________________________________________________________________________

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^{4}}\,{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^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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^{4}}, 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^4,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^4, x)

________________________________________________________________________________________

Sympy [A]  time = 9.12796, size = 381, normalized size = 0.93 \begin{align*} \frac{a^{\frac{3}{2}} c \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{a^{\frac{3}{2}} d}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{a^{\frac{3}{2}} e \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{a^{\frac{3}{2}} f \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{2} + \frac{\sqrt{a} b c x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{\sqrt{a} b d x^{2} \sqrt{1 + \frac{b x^{4}}{a}}}{4} - \frac{\sqrt{a} b d x^{2}}{2 \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b e x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + \frac{a^{2} f}{2 \sqrt{b} x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} + \frac{3 a \sqrt{b} d \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4} + \frac{a \sqrt{b} f x^{2}}{2 \sqrt{\frac{a}{b x^{4}} + 1}} + b f \left (\begin{cases} \frac{\sqrt{a} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\left (a + b x^{4}\right )^{\frac{3}{2}}}{6 b} & \text{otherwise} \end{cases}\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**4,x)

[Out]

a**(3/2)*c*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**3*gamma(1/4)) - a**(3/2)*d/
(2*x**2*sqrt(1 + b*x**4/a)) + a**(3/2)*e*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*
x*gamma(3/4)) - a**(3/2)*f*asinh(sqrt(a)/(sqrt(b)*x**2))/2 + sqrt(a)*b*c*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,
), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + sqrt(a)*b*d*x**2*sqrt(1 + b*x**4/a)/4 - sqrt(a)*b*d*x**2/(2*sqrt
(1 + b*x**4/a)) + sqrt(a)*b*e*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/
4)) + a**2*f/(2*sqrt(b)*x**2*sqrt(a/(b*x**4) + 1)) + 3*a*sqrt(b)*d*asinh(sqrt(b)*x**2/sqrt(a))/4 + a*sqrt(b)*f
*x**2/(2*sqrt(a/(b*x**4) + 1)) + b*f*Piecewise((sqrt(a)*x**4/4, Eq(b, 0)), ((a + b*x**4)**(3/2)/(6*b), True))

________________________________________________________________________________________

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^{4}}\,{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^4,x, algorithm="giac")

[Out]

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