3.521 \(\int \frac{(c+d x+e x^2+f x^3) (a+b x^4)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=392 \[ \frac{2 \sqrt [4]{a} b^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} f+9 \sqrt{b} d\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{15 \sqrt{a+b x^4}}+\frac{1}{2} b^{3/2} c \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{12 b^{3/2} d x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{12 \sqrt [4]{a} b^{5/4} d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{60} \left (a+b x^4\right )^{3/2} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right )-\frac{b \sqrt{a+b x^4} \left (2 c-3 e x^2\right )}{4 x^2}-\frac{2 b \sqrt{a+b x^4} \left (9 d-5 f x^2\right )}{15 x}-\frac{3}{4} \sqrt{a} b e \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right ) \]

[Out]

(12*b^(3/2)*d*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) - (b*(2*c - 3*e*x^2)*Sqrt[a + b*x^4])/(4*x^2) - (
2*b*(9*d - 5*f*x^2)*Sqrt[a + b*x^4])/(15*x) - (((10*c)/x^6 + (12*d)/x^5 + (15*e)/x^4 + (20*f)/x^3)*(a + b*x^4)
^(3/2))/60 + (b^(3/2)*c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (3*Sqrt[a]*b*e*ArcTanh[Sqrt[a + b*x^4]/Sqr
t[a]])/4 - (12*a^(1/4)*b^(5/4)*d*(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^(1/4)*b^(3/4)*(9*Sqrt[b]*d + 5*Sqrt[a]*f)*(Sq
rt[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 = 0.342193, antiderivative size = 392, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 15, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {14, 1825, 1833, 1252, 813, 844, 217, 206, 266, 63, 208, 1272, 1198, 220, 1196} \[ \frac{1}{2} b^{3/2} c \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )+\frac{2 \sqrt [4]{a} b^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \left (5 \sqrt{a} f+9 \sqrt{b} d\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{15 \sqrt{a+b x^4}}+\frac{12 b^{3/2} d x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{12 \sqrt [4]{a} b^{5/4} d \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 \sqrt{a+b x^4}}-\frac{1}{60} \left (a+b x^4\right )^{3/2} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right )-\frac{b \sqrt{a+b x^4} \left (2 c-3 e x^2\right )}{4 x^2}-\frac{2 b \sqrt{a+b x^4} \left (9 d-5 f x^2\right )}{15 x}-\frac{3}{4} \sqrt{a} b e \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^7,x]

[Out]

(12*b^(3/2)*d*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) - (b*(2*c - 3*e*x^2)*Sqrt[a + b*x^4])/(4*x^2) - (
2*b*(9*d - 5*f*x^2)*Sqrt[a + b*x^4])/(15*x) - (((10*c)/x^6 + (12*d)/x^5 + (15*e)/x^4 + (20*f)/x^3)*(a + b*x^4)
^(3/2))/60 + (b^(3/2)*c*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/2 - (3*Sqrt[a]*b*e*ArcTanh[Sqrt[a + b*x^4]/Sqr
t[a]])/4 - (12*a^(1/4)*b^(5/4)*d*(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^(1/4)*b^(3/4)*(9*Sqrt[b]*d + 5*Sqrt[a]*f)*(Sq
rt[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 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 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 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]

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]

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^7} \, dx &=-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}-(6 b) \int \frac{\left (-\frac{c}{6}-\frac{d x}{5}-\frac{e x^2}{4}-\frac{f x^3}{3}\right ) \sqrt{a+b x^4}}{x^3} \, dx\\ &=-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}-(6 b) \int \left (\frac{\left (-\frac{c}{6}-\frac{e x^2}{4}\right ) \sqrt{a+b x^4}}{x^3}+\frac{\left (-\frac{d}{5}-\frac{f x^2}{3}\right ) \sqrt{a+b x^4}}{x^2}\right ) \, dx\\ &=-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}-(6 b) \int \frac{\left (-\frac{c}{6}-\frac{e x^2}{4}\right ) \sqrt{a+b x^4}}{x^3} \, dx-(6 b) \int \frac{\left (-\frac{d}{5}-\frac{f x^2}{3}\right ) \sqrt{a+b x^4}}{x^2} \, dx\\ &=-\frac{2 b \left (9 d-5 f x^2\right ) \sqrt{a+b x^4}}{15 x}-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}-(3 b) \operatorname{Subst}\left (\int \frac{\left (-\frac{c}{6}-\frac{e x}{4}\right ) \sqrt{a+b x^2}}{x^2} \, dx,x,x^2\right )+(4 b) \int \frac{\frac{a f}{3}+\frac{3}{5} b d x^2}{\sqrt{a+b x^4}} \, dx\\ &=-\frac{b \left (2 c-3 e x^2\right ) \sqrt{a+b x^4}}{4 x^2}-\frac{2 b \left (9 d-5 f x^2\right ) \sqrt{a+b x^4}}{15 x}-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}+\frac{1}{2} (3 b) \operatorname{Subst}\left (\int \frac{\frac{a e}{2}+\frac{b c x}{3}}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )-\frac{1}{5} \left (12 \sqrt{a} b^{3/2} d\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx+\frac{1}{15} \left (4 \sqrt{a} b \left (9 \sqrt{b} d+5 \sqrt{a} f\right )\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx\\ &=\frac{12 b^{3/2} d x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{b \left (2 c-3 e x^2\right ) \sqrt{a+b x^4}}{4 x^2}-\frac{2 b \left (9 d-5 f x^2\right ) \sqrt{a+b x^4}}{15 x}-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}-\frac{12 \sqrt [4]{a} b^{5/4} d \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 \sqrt [4]{a} b^{3/4} \left (9 \sqrt{b} d+5 \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 )}{15 \sqrt{a+b x^4}}+\frac{1}{2} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )+\frac{1}{4} (3 a b e) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac{12 b^{3/2} d x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{b \left (2 c-3 e x^2\right ) \sqrt{a+b x^4}}{4 x^2}-\frac{2 b \left (9 d-5 f x^2\right ) \sqrt{a+b x^4}}{15 x}-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}-\frac{12 \sqrt [4]{a} b^{5/4} d \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 \sqrt [4]{a} b^{3/4} \left (9 \sqrt{b} d+5 \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 )}{15 \sqrt{a+b x^4}}+\frac{1}{2} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )+\frac{1}{8} (3 a b e) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^4\right )\\ &=\frac{12 b^{3/2} d x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{b \left (2 c-3 e x^2\right ) \sqrt{a+b x^4}}{4 x^2}-\frac{2 b \left (9 d-5 f x^2\right ) \sqrt{a+b x^4}}{15 x}-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}+\frac{1}{2} b^{3/2} c \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{12 \sqrt [4]{a} b^{5/4} d \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 \sqrt [4]{a} b^{3/4} \left (9 \sqrt{b} d+5 \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 )}{15 \sqrt{a+b x^4}}+\frac{1}{4} (3 a e) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^4}\right )\\ &=\frac{12 b^{3/2} d x \sqrt{a+b x^4}}{5 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{b \left (2 c-3 e x^2\right ) \sqrt{a+b x^4}}{4 x^2}-\frac{2 b \left (9 d-5 f x^2\right ) \sqrt{a+b x^4}}{15 x}-\frac{1}{60} \left (\frac{10 c}{x^6}+\frac{12 d}{x^5}+\frac{15 e}{x^4}+\frac{20 f}{x^3}\right ) \left (a+b x^4\right )^{3/2}+\frac{1}{2} b^{3/2} c \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )-\frac{3}{4} \sqrt{a} b e \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )-\frac{12 \sqrt [4]{a} b^{5/4} d \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 \sqrt [4]{a} b^{3/4} \left (9 \sqrt{b} d+5 \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 )}{15 \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.201954, size = 163, normalized size = 0.42 \[ \frac{\sqrt{a+b x^4} \left (-5 a^3 c \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x^4}{a}\right )-6 a^3 d x \, _2F_1\left (-\frac{3}{2},-\frac{5}{4};-\frac{1}{4};-\frac{b x^4}{a}\right )-10 a^3 f x^3 \, _2F_1\left (-\frac{3}{2},-\frac{3}{4};\frac{1}{4};-\frac{b x^4}{a}\right )+3 b e x^6 \left (a+b x^4\right )^2 \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{b x^4}{a}+1\right )\right )}{30 a^2 x^6 \sqrt{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^(3/2))/x^7,x]

[Out]

(Sqrt[a + b*x^4]*(-5*a^3*c*Hypergeometric2F1[-3/2, -3/2, -1/2, -((b*x^4)/a)] - 6*a^3*d*x*Hypergeometric2F1[-3/
2, -5/4, -1/4, -((b*x^4)/a)] - 10*a^3*f*x^3*Hypergeometric2F1[-3/2, -3/4, 1/4, -((b*x^4)/a)] + 3*b*e*x^6*(a +
b*x^4)^2*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[2, 5/2, 7/2, 1 + (b*x^4)/a]))/(30*a^2*x^6*Sqrt[1 + (b*x^4)/a])

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Maple [C]  time = 0.02, size = 408, normalized size = 1. \begin{align*} -{\frac{af}{3\,{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{fbx}{3}\sqrt{b{x}^{4}+a}}+{\frac{4\,abf}{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{ad}{5\,{x}^{5}}\sqrt{b{x}^{4}+a}}-{\frac{7\,bd}{5\,x}\sqrt{b{x}^{4}+a}}+{{\frac{12\,i}{5}}d{b}^{{\frac{3}{2}}}\sqrt{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{12\,i}{5}}d{b}^{{\frac{3}{2}}}\sqrt{a}\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}}}}+{\frac{be}{2}\sqrt{b{x}^{4}+a}}-{\frac{3\,be}{4}\sqrt{a}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ) }-{\frac{ae}{4\,{x}^{4}}\sqrt{b{x}^{4}+a}}+{\frac{c}{2}{b}^{{\frac{3}{2}}}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ) }-{\frac{ac}{6\,{x}^{6}}\sqrt{b{x}^{4}+a}}-{\frac{2\,bc}{3\,{x}^{2}}\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^7,x)

[Out]

-1/3*f*a*(b*x^4+a)^(1/2)/x^3+1/3*f*b*x*(b*x^4+a)^(1/2)+4/3*f*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)-1/5*d*a*
(b*x^4+a)^(1/2)/x^5-7/5*d*b*(b*x^4+a)^(1/2)/x+12/5*I*d*b^(3/2)*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)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-12
/5*I*d*b^(3/2)*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)+1/2*e*b*(b*x^4+a)^(1/2)-3/4*e*a^(1/2)*b*ln((2*a+2*
a^(1/2)*(b*x^4+a)^(1/2))/x^2)-1/4*e*a/x^4*(b*x^4+a)^(1/2)+1/2*c*b^(3/2)*ln(x^2*b^(1/2)+(b*x^4+a)^(1/2))-1/6*c*
a/x^6*(b*x^4+a)^(1/2)-2/3*c*b/x^2*(b*x^4+a)^(1/2)

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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)^(3/2)/x^7,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{{\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^{7}}, 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^7,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^7, x)

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Sympy [C]  time = 9.89324, size = 406, normalized size = 1.04 \begin{align*} \frac{a^{\frac{3}{2}} d \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{a^{\frac{3}{2}} f \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 c}{2 x^{2} \sqrt{1 + \frac{b x^{4}}{a}}} + \frac{\sqrt{a} b d \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{3 \sqrt{a} b e \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4} + \frac{\sqrt{a} b f 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{a \sqrt{b} c \sqrt{\frac{a}{b x^{4}} + 1}}{6 x^{4}} - \frac{a \sqrt{b} e \sqrt{\frac{a}{b x^{4}} + 1}}{4 x^{2}} + \frac{a \sqrt{b} e}{2 x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{\frac{3}{2}} c \sqrt{\frac{a}{b x^{4}} + 1}}{6} + \frac{b^{\frac{3}{2}} c \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2} + \frac{b^{\frac{3}{2}} e x^{2}}{2 \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{b^{2} c 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**7,x)

[Out]

a**(3/2)*d*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**5*gamma(-1/4)) + a**(3/2)*
f*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)*b*c/(2*x**2*
sqrt(1 + b*x**4/a)) + sqrt(a)*b*d*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**4*exp_polar(I*pi)/a)/(4*x*gamma
(3/4)) - 3*sqrt(a)*b*e*asinh(sqrt(a)/(sqrt(b)*x**2))/4 + sqrt(a)*b*f*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b
*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) - a*sqrt(b)*c*sqrt(a/(b*x**4) + 1)/(6*x**4) - a*sqrt(b)*e*sqrt(a/(b*x*
*4) + 1)/(4*x**2) + a*sqrt(b)*e/(2*x**2*sqrt(a/(b*x**4) + 1)) - b**(3/2)*c*sqrt(a/(b*x**4) + 1)/6 + b**(3/2)*c
*asinh(sqrt(b)*x**2/sqrt(a))/2 + b**(3/2)*e*x**2/(2*sqrt(a/(b*x**4) + 1)) - b**2*c*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^{7}}\,{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^7,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/2)*(f*x^3 + e*x^2 + d*x + c)/x^7, x)