3.6.18 \(\int \frac {(c+d x+e x^2+f x^3) (a+b x^4)^{3/2}}{x^4} \, dx\) [518]

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

[Out]

-1/15*(-3*e*x^2+5*c)*(b*x^4+a)^(3/2)/x^3-1/6*(-f*x^2+3*d)*(b*x^4+a)^(3/2)/x^2-1/2*a^(3/2)*f*arctanh((b*x^4+a)^
(1/2)/a^(1/2))+3/4*a*d*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))*b^(1/2)-2/15*(-5*b*c*x^2+9*a*e)*(b*x^4+a)^(1/2)/x+
1/4*(3*b*d*x^2+2*a*f)*(b*x^4+a)^(1/2)+12/5*a*e*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)*e*(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*arctan(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*e*a^(1/2)+5*c*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 = 0.23, antiderivative size = 408, normalized size of antiderivative = 1.00, 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 = {1847, 1286, 1212, 226, 1210, 1266, 827, 829, 858, 223, 212, 272, 65, 214} \begin {gather*} \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 \text {ArcTan}\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 \text {ArcTan}\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 {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 (5 c-3 e x^2\right )}{15 x^3}-\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 )} \end {gather*}

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 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 223

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 272

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 827

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 829

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 858

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

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 1266

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 1286

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 1847

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)/c^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, {
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^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} \text {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} \text {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 {\text {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) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )+\frac {1}{2} \left (a^2 f\right ) \text {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) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.55, size = 327, normalized size = 0.80 \begin {gather*} \frac {\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} \left (\left (a+b x^4\right ) \left (-10 a \left (2 c+x \left (3 d+6 e x-4 f x^2\right )\right )+b x^4 (20 c+x (15 d+2 x (6 e+5 f x)))\right )+45 a \sqrt {b} d x^3 \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )-30 a^{3/2} f x^3 \sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )\right )+144 a^{3/2} \sqrt {b} e x^3 \sqrt {1+\frac {b x^4}{a}} E\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right )\right |-1\right )-16 a \sqrt {b} \left (5 i \sqrt {b} c+9 \sqrt {a} e\right ) x^3 \sqrt {1+\frac {b x^4}{a}} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x\right )\right |-1\right )}{60 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} x^3 \sqrt {a+b x^4}} \end {gather*}

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]

(Sqrt[(I*Sqrt[b])/Sqrt[a]]*((a + b*x^4)*(-10*a*(2*c + x*(3*d + 6*e*x - 4*f*x^2)) + b*x^4*(20*c + x*(15*d + 2*x
*(6*e + 5*f*x)))) + 45*a*Sqrt[b]*d*x^3*Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]] - 30*a^(3/2)*f*x
^3*Sqrt[a + b*x^4]*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]]) + 144*a^(3/2)*Sqrt[b]*e*x^3*Sqrt[1 + (b*x^4)/a]*EllipticE
[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1] - 16*a*Sqrt[b]*((5*I)*Sqrt[b]*c + 9*Sqrt[a]*e)*x^3*Sqrt[1 + (b*x^
4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/(60*Sqrt[(I*Sqrt[b])/Sqrt[a]]*x^3*Sqrt[a + b*x^4]
)

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Maple [C] Result contains complex when optimal does not.
time = 0.41, size = 349, normalized size = 0.86

method	result	size
elliptic	\(-\frac {a c \sqrt {b \,x^{4}+a}}{3 x^{3}}-\frac {a d \sqrt {b \,x^{4}+a}}{2 x^{2}}-\frac {a e \sqrt {b \,x^{4}+a}}{x}+\frac {b f \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {b e \,x^{3} \sqrt {b \,x^{4}+a}}{5}+\frac {b d \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {b c x \sqrt {b \,x^{4}+a}}{3}+\frac {2 a f \sqrt {b \,x^{4}+a}}{3}+\frac {4 a b c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 a \sqrt {b}\, d \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{4}+\frac {12 i a^{\frac {3}{2}} \sqrt {b}\, e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a^{\frac {3}{2}} f \arctanh \left (\frac {\sqrt {a}}{\sqrt {b \,x^{4}+a}}\right )}{2}\)	\(343\)
default	\(d \left (\frac {b \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {3 a \sqrt {b}\, \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4}-\frac {a \sqrt {b \,x^{4}+a}}{2 x^{2}}\right )+c \left (-\frac {a \sqrt {b \,x^{4}+a}}{3 x^{3}}+\frac {b x \sqrt {b \,x^{4}+a}}{3}+\frac {4 a b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+f \left (\frac {b \,x^{4} \sqrt {b \,x^{4}+a}}{6}+\frac {2 a \sqrt {b \,x^{4}+a}}{3}-\frac {a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}\right )+e \left (-\frac {a \sqrt {b \,x^{4}+a}}{x}+\frac {\sqrt {b \,x^{4}+a}\, b \,x^{3}}{5}+\frac {12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\)	\(349\)
risch	\(-\frac {a \sqrt {b \,x^{4}+a}\, \left (6 e \,x^{2}+3 d x +2 c \right )}{6 x^{3}}-\frac {f \sqrt {b \,x^{4}+a}\, \left (-b \,x^{4}+2 a \right )}{6}+\frac {b e \,x^{3} \sqrt {b \,x^{4}+a}}{5}-\frac {3 i \sqrt {b}\, e \,a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {3 i \sqrt {b}\, e \,a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {b d \,x^{2} \sqrt {b \,x^{4}+a}}{4}+\frac {3 a \sqrt {b}\, d \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{4}+\frac {b c x \sqrt {b \,x^{4}+a}}{3}+\frac {4 a b c \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{3 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+a f \sqrt {b \,x^{4}+a}+\frac {3 i a^{\frac {3}{2}} \sqrt {b}\, e \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}-\frac {a^{\frac {3}{2}} f \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{2}\)	\(491\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 + x^2*e + d*x + c)/x^4, x)

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Fricas [F]
time = 0.28, size = 59, normalized size = 0.14 \begin {gather*} {\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 {gather*}

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)

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Sympy [A]
time = 5.20, size = 381, normalized size = 0.93 \begin {gather*} \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 {gather*}

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))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 + x^2*e + d*x + c)/x^4, x)

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

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^4,x)

[Out]

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

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