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

Optimal. Leaf size=386 \[ \frac {2 a^{3/4} \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (9 \sqrt {a} f+5 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 \sqrt {a+b x^4}}-\frac {12 a^{5/4} \sqrt [4]{b} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+b x^4}}-\frac {1}{12} \left (a+b x^4\right )^{3/2} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )+\frac {3}{4} b \sqrt {a+b x^4} \left (c+e x^2\right )-\frac {3}{4} \sqrt {a} b c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {2}{15} b x \sqrt {a+b x^4} \left (5 d+9 f x^2\right )+\frac {3}{4} a \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )+\frac {12 a \sqrt {b} f x \sqrt {a+b x^4}}{5 \left (\sqrt {a}+\sqrt {b} x^2\right )} \]

[Out]

-1/12*(3*c/x^4+4*d/x^3+6*e/x^2+12*f/x)*(b*x^4+a)^(3/2)-3/4*b*c*arctanh((b*x^4+a)^(1/2)/a^(1/2))*a^(1/2)+3/4*a*
e*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))*b^(1/2)+3/4*b*(e*x^2+c)*(b*x^4+a)^(1/2)+2/15*b*x*(9*f*x^2+5*d)*(b*x^4+a
)^(1/2)+12/5*a*f*x*b^(1/2)*(b*x^4+a)^(1/2)/(a^(1/2)+x^2*b^(1/2))-12/5*a^(5/4)*b^(1/4)*f*(cos(2*arctan(b^(1/4)*
x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*
(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/(b*x^4+a)^(1/2)+2/15*a^(3/4)*b^(1/4)*(cos(2*ar
ctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),
1/2*2^(1/2))*(9*f*a^(1/2)+5*d*b^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/(b*x^4+
a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 386, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 15, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {14, 1825, 1833, 1252, 815, 844, 217, 206, 266, 63, 208, 1177, 1198, 220, 1196} \[ \frac {2 a^{3/4} \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (9 \sqrt {a} f+5 \sqrt {b} d\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 \sqrt {a+b x^4}}-\frac {12 a^{5/4} \sqrt [4]{b} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+b x^4}}-\frac {1}{12} \left (a+b x^4\right )^{3/2} \left (\frac {3 c}{x^4}+\frac {4 d}{x^3}+\frac {6 e}{x^2}+\frac {12 f}{x}\right )+\frac {3}{4} b \sqrt {a+b x^4} \left (c+e x^2\right )-\frac {3}{4} \sqrt {a} b c \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )+\frac {2}{15} b x \sqrt {a+b x^4} \left (5 d+9 f x^2\right )+\frac {3}{4} a \sqrt {b} e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )+\frac {12 a \sqrt {b} f x \sqrt {a+b x^4}}{5 \left (\sqrt {a}+\sqrt {b} x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*Sqrt[b]*f*x*Sqrt[a + b*x^4])/(5*(Sqrt[a] + Sqrt[b]*x^2)) + (3*b*(c + e*x^2)*Sqrt[a + b*x^4])/4 + (2*b*x*
(5*d + 9*f*x^2)*Sqrt[a + b*x^4])/15 - (((3*c)/x^4 + (4*d)/x^3 + (6*e)/x^2 + (12*f)/x)*(a + b*x^4)^(3/2))/12 +
(3*a*Sqrt[b]*e*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/4 - (3*Sqrt[a]*b*c*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/4
- (12*a^(5/4)*b^(1/4)*f*(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]*d + 9*Sqrt[a]*f)*(Sqrt[a] + S
qrt[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*Sqr
t[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 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 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 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 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 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 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 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 1177

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(d*(4*p + 3) + e*(4*p + 1)*x^2)*(a
+ c*x^4)^p)/((4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/((4*p + 1)*(4*p + 3)), Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4
*p + 1))*x^2, x]*(a + c*x^4)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] &
& FractionQ[p] && IntegerQ[2*p]

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 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 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 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)]

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

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Mathematica [C]  time = 0.25, size = 163, normalized size = 0.42 \[ \frac {\sqrt {a+b x^4} \left (3 x \left (-5 a^3 e \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x^4}{a}\right )-10 a^3 f x \, _2F_1\left (-\frac {3}{2},-\frac {1}{4};\frac {3}{4};-\frac {b x^4}{a}\right )+b c x^2 \left (a+b x^4\right )^2 \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {b x^4}{a}+1\right )\right )-10 a^3 d \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};\frac {1}{4};-\frac {b x^4}{a}\right )\right )}{30 a^2 x^3 \sqrt {\frac {b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b f x^{7} + b e x^{6} + b d x^{5} + b c x^{4} + a f x^{3} + a e x^{2} + a d x + a c\right )} \sqrt {b x^{4} + a}}{x^{5}}, x\right ) \]

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^5,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^5, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} {\left (f x^{3} + e x^{2} + d x + c\right )}}{x^{5}}\,{d x} \]

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^5,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.18, size = 409, normalized size = 1.06 \[ \frac {\sqrt {b \,x^{4}+a}\, b f \,x^{3}}{5}-\frac {12 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {3}{2}} \sqrt {b}\, f \EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {12 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a^{\frac {3}{2}} \sqrt {b}\, f \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {4 \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, a b d \EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {\sqrt {b \,x^{4}+a}\, b e \,x^{2}}{4}+\frac {3 a \sqrt {b}\, e \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4}-\frac {3 \sqrt {a}\, b c \ln \left (\frac {2 a +2 \sqrt {b \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{4}+\frac {\sqrt {b \,x^{4}+a}\, b d x}{3}+\frac {\sqrt {b \,x^{4}+a}\, b c}{2}-\frac {\sqrt {b \,x^{4}+a}\, a f}{x}-\frac {\sqrt {b \,x^{4}+a}\, a e}{2 x^{2}}-\frac {\sqrt {b \,x^{4}+a}\, a d}{3 x^{3}}-\frac {\sqrt {b \,x^{4}+a}\, a c}{4 x^{4}} \]

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^5,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, {\left (3 \, \sqrt {a} b \log \left (\frac {\sqrt {b x^{4} + a} - \sqrt {a}}{\sqrt {b x^{4} + a} + \sqrt {a}}\right ) + 4 \, \sqrt {b x^{4} + a} b - \frac {2 \, \sqrt {b x^{4} + a} a}{x^{4}}\right )} c + \int \frac {{\left (b f x^{6} + b e x^{5} + b d x^{4} + a f x^{2} + a e x + a d\right )} \sqrt {b x^{4} + a}}{x^{4}}\,{d x} \]

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^5,x, algorithm="maxima")

[Out]

1/8*(3*sqrt(a)*b*log((sqrt(b*x^4 + a) - sqrt(a))/(sqrt(b*x^4 + a) + sqrt(a))) + 4*sqrt(b*x^4 + a)*b - 2*sqrt(b
*x^4 + a)*a/x^4)*c + integrate((b*f*x^6 + b*e*x^5 + b*d*x^4 + a*f*x^2 + a*e*x + a*d)*sqrt(b*x^4 + a)/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^4+a\right )}^{3/2}\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 13.19, size = 379, normalized size = 0.98 \[ \frac {a^{\frac {3}{2}} d \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {a^{\frac {3}{2}} e}{2 x^{2} \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {a^{\frac {3}{2}} f \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {3 \sqrt {a} b c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4} + \frac {\sqrt {a} b d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {a} b e x^{2} \sqrt {1 + \frac {b x^{4}}{a}}}{4} - \frac {\sqrt {a} b e x^{2}}{2 \sqrt {1 + \frac {b x^{4}}{a}}} + \frac {\sqrt {a} b f x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {a \sqrt {b} c \sqrt {\frac {a}{b x^{4}} + 1}}{4 x^{2}} + \frac {a \sqrt {b} c}{2 x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {3 a \sqrt {b} e \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4} + \frac {b^{\frac {3}{2}} c x^{2}}{2 \sqrt {\frac {a}{b x^{4}} + 1}} \]

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**5,x)

[Out]

a**(3/2)*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)) - a**(3/2)*e/
(2*x**2*sqrt(1 + b*x**4/a)) + a**(3/2)*f*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*c*asinh(sqrt(a)/(sqrt(b)*x**2))/4 + sqrt(a)*b*d*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*e*x**2*sqrt(1 + b*x**4/a)/4 - sqrt(a)*b*e*x**2/(2*s
qrt(1 + b*x**4/a)) + sqrt(a)*b*f*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*sqrt(b)*c*sqrt(a/(b*x**4) + 1)/(4*x**2) + a*sqrt(b)*c/(2*x**2*sqrt(a/(b*x**4) + 1)) + 3*a*sqrt(b)*e
*asinh(sqrt(b)*x**2/sqrt(a))/4 + b**(3/2)*c*x**2/(2*sqrt(a/(b*x**4) + 1))

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