3.2532 \(\int \frac{1}{(d+e x) \sqrt [4]{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=733 \[ \frac{\sqrt [4]{4 a c-b^2} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt [4]{4 a c-b^2} \sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}}{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a e^2-b d e+c d^2}}\right )}{\sqrt [4]{c} \sqrt{e} \sqrt [4]{a+b x+c x^2} \sqrt [4]{a e^2-b d e+c d^2}}-\frac{\sqrt [4]{4 a c-b^2} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt [4]{4 a c-b^2} \sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}}{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a e^2-b d e+c d^2}}\right )}{\sqrt [4]{c} \sqrt{e} \sqrt [4]{a+b x+c x^2} \sqrt [4]{a e^2-b d e+c d^2}}-\frac{\sqrt{4 a c-b^2} \sqrt{\frac{(b+2 c x)^2}{b^2-4 a c}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \Pi \left (-\frac{\sqrt{4 a c-b^2} e}{2 \sqrt{c} \sqrt{c d^2-b e d+a e^2}};\left .\sin ^{-1}\left (\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )\right |-1\right )}{\sqrt{2} \sqrt{c} e (b+2 c x) \sqrt [4]{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}+\frac{\sqrt{4 a c-b^2} \sqrt{\frac{(b+2 c x)^2}{b^2-4 a c}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \Pi \left (\frac{\sqrt{4 a c-b^2} e}{2 \sqrt{c} \sqrt{c d^2-b e d+a e^2}};\left .\sin ^{-1}\left (\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )\right |-1\right )}{\sqrt{2} \sqrt{c} e (b+2 c x) \sqrt [4]{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}} \]

[Out]

((-b^2 + 4*a*c)^(1/4)*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^(1/4)*ArcTan[((-b^2 + 4*a*c)^(1/4)*Sqrt[e]*(1 -
(b + 2*c*x)^2/(b^2 - 4*a*c))^(1/4))/(Sqrt[2]*c^(1/4)*(c*d^2 - b*d*e + a*e^2)^(1/4))])/(c^(1/4)*Sqrt[e]*(c*d^2
- b*d*e + a*e^2)^(1/4)*(a + b*x + c*x^2)^(1/4)) - ((-b^2 + 4*a*c)^(1/4)*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c
)))^(1/4)*ArcTanh[((-b^2 + 4*a*c)^(1/4)*Sqrt[e]*(1 - (b + 2*c*x)^2/(b^2 - 4*a*c))^(1/4))/(Sqrt[2]*c^(1/4)*(c*d
^2 - b*d*e + a*e^2)^(1/4))])/(c^(1/4)*Sqrt[e]*(c*d^2 - b*d*e + a*e^2)^(1/4)*(a + b*x + c*x^2)^(1/4)) - (Sqrt[-
b^2 + 4*a*c]*(2*c*d - b*e)*Sqrt[(b + 2*c*x)^2/(b^2 - 4*a*c)]*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^(1/4)*El
lipticPi[-(Sqrt[-b^2 + 4*a*c]*e)/(2*Sqrt[c]*Sqrt[c*d^2 - b*d*e + a*e^2]), ArcSin[(1 - (b + 2*c*x)^2/(b^2 - 4*a
*c))^(1/4)], -1])/(Sqrt[2]*Sqrt[c]*e*Sqrt[c*d^2 - b*d*e + a*e^2]*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4)) + (Sqrt[
-b^2 + 4*a*c]*(2*c*d - b*e)*Sqrt[(b + 2*c*x)^2/(b^2 - 4*a*c)]*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^(1/4)*E
llipticPi[(Sqrt[-b^2 + 4*a*c]*e)/(2*Sqrt[c]*Sqrt[c*d^2 - b*d*e + a*e^2]), ArcSin[(1 - (b + 2*c*x)^2/(b^2 - 4*a
*c))^(1/4)], -1])/(Sqrt[2]*Sqrt[c]*e*Sqrt[c*d^2 - b*d*e + a*e^2]*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))

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Rubi [A]  time = 1.55996, antiderivative size = 733, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {749, 748, 746, 399, 490, 1213, 537, 444, 63, 298, 205, 208} \[ \frac{\sqrt [4]{4 a c-b^2} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt [4]{4 a c-b^2} \sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}}{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a e^2-b d e+c d^2}}\right )}{\sqrt [4]{c} \sqrt{e} \sqrt [4]{a+b x+c x^2} \sqrt [4]{a e^2-b d e+c d^2}}-\frac{\sqrt [4]{4 a c-b^2} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt [4]{4 a c-b^2} \sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}}{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a e^2-b d e+c d^2}}\right )}{\sqrt [4]{c} \sqrt{e} \sqrt [4]{a+b x+c x^2} \sqrt [4]{a e^2-b d e+c d^2}}-\frac{\sqrt{4 a c-b^2} \sqrt{\frac{(b+2 c x)^2}{b^2-4 a c}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \Pi \left (-\frac{\sqrt{4 a c-b^2} e}{2 \sqrt{c} \sqrt{c d^2-b e d+a e^2}};\left .\sin ^{-1}\left (\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )\right |-1\right )}{\sqrt{2} \sqrt{c} e (b+2 c x) \sqrt [4]{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}+\frac{\sqrt{4 a c-b^2} \sqrt{\frac{(b+2 c x)^2}{b^2-4 a c}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \Pi \left (\frac{\sqrt{4 a c-b^2} e}{2 \sqrt{c} \sqrt{c d^2-b e d+a e^2}};\left .\sin ^{-1}\left (\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )\right |-1\right )}{\sqrt{2} \sqrt{c} e (b+2 c x) \sqrt [4]{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(a + b*x + c*x^2)^(1/4)),x]

[Out]

((-b^2 + 4*a*c)^(1/4)*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^(1/4)*ArcTan[((-b^2 + 4*a*c)^(1/4)*Sqrt[e]*(1 -
(b + 2*c*x)^2/(b^2 - 4*a*c))^(1/4))/(Sqrt[2]*c^(1/4)*(c*d^2 - b*d*e + a*e^2)^(1/4))])/(c^(1/4)*Sqrt[e]*(c*d^2
- b*d*e + a*e^2)^(1/4)*(a + b*x + c*x^2)^(1/4)) - ((-b^2 + 4*a*c)^(1/4)*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c
)))^(1/4)*ArcTanh[((-b^2 + 4*a*c)^(1/4)*Sqrt[e]*(1 - (b + 2*c*x)^2/(b^2 - 4*a*c))^(1/4))/(Sqrt[2]*c^(1/4)*(c*d
^2 - b*d*e + a*e^2)^(1/4))])/(c^(1/4)*Sqrt[e]*(c*d^2 - b*d*e + a*e^2)^(1/4)*(a + b*x + c*x^2)^(1/4)) - (Sqrt[-
b^2 + 4*a*c]*(2*c*d - b*e)*Sqrt[(b + 2*c*x)^2/(b^2 - 4*a*c)]*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^(1/4)*El
lipticPi[-(Sqrt[-b^2 + 4*a*c]*e)/(2*Sqrt[c]*Sqrt[c*d^2 - b*d*e + a*e^2]), ArcSin[(1 - (b + 2*c*x)^2/(b^2 - 4*a
*c))^(1/4)], -1])/(Sqrt[2]*Sqrt[c]*e*Sqrt[c*d^2 - b*d*e + a*e^2]*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4)) + (Sqrt[
-b^2 + 4*a*c]*(2*c*d - b*e)*Sqrt[(b + 2*c*x)^2/(b^2 - 4*a*c)]*(-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c)))^(1/4)*E
llipticPi[(Sqrt[-b^2 + 4*a*c]*e)/(2*Sqrt[c]*Sqrt[c*d^2 - b*d*e + a*e^2]), ArcSin[(1 - (b + 2*c*x)^2/(b^2 - 4*a
*c))^(1/4)], -1])/(Sqrt[2]*Sqrt[c]*e*Sqrt[c*d^2 - b*d*e + a*e^2]*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))

Rule 749

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

Rule 748

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

Rule 746

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

Rule 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 205

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

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{1}{(d+e x) \sqrt [4]{a+b x+c x^2}} \, dx &=\frac{\sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac{1}{(d+e x) \sqrt [4]{-\frac{a c}{b^2-4 a c}-\frac{b c x}{b^2-4 a c}-\frac{c^2 x^2}{b^2-4 a c}}} \, dx}{\sqrt [4]{a+b x+c x^2}}\\ &=\frac{\left (\sqrt{2} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{c (2 c d-b e)}{b^2-4 a c}+e x\right ) \sqrt [4]{1-\frac{\left (b^2-4 a c\right ) x^2}{c^2}}} \, dx,x,-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )}{\sqrt [4]{a+b x+c x^2}}\\ &=-\frac{\left (\sqrt{2} e \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt [4]{1-\frac{\left (b^2-4 a c\right ) x^2}{c^2}} \left (\frac{c^2 (2 c d-b e)^2}{\left (b^2-4 a c\right )^2}-e^2 x^2\right )} \, dx,x,-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )}{\sqrt [4]{a+b x+c x^2}}-\frac{\left (\sqrt{2} c (2 c d-b e) \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{1-\frac{\left (b^2-4 a c\right ) x^2}{c^2}} \left (\frac{c^2 (2 c d-b e)^2}{\left (b^2-4 a c\right )^2}-e^2 x^2\right )} \, dx,x,-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\\ &=-\frac{\left (e \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{1-\frac{\left (b^2-4 a c\right ) x}{c^2}} \left (\frac{c^2 (2 c d-b e)^2}{\left (b^2-4 a c\right )^2}-e^2 x\right )} \, dx,x,\left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2\right )}{\sqrt{2} \sqrt [4]{a+b x+c x^2}}-\frac{\left (2 \sqrt{2} c (2 c d-b e) \sqrt{\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^4} \left (e^2-\frac{(2 c d-b e)^2}{b^2-4 a c}-e^2 x^4\right )} \, dx,x,\sqrt [4]{1-\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}}\right )}{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right ) \sqrt [4]{a+b x+c x^2}}\\ &=\frac{\left (2 \sqrt{2} c^2 e \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{x^2}{-\frac{c^2 e^2}{b^2-4 a c}+\frac{c^2 (2 c d-b e)^2}{\left (b^2-4 a c\right )^2}+\frac{c^2 e^2 x^4}{b^2-4 a c}} \, dx,x,\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}-\frac{\left (\sqrt{2} c \sqrt{-b^2+4 a c} (2 c d-b e) \sqrt{\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (2 \sqrt{c} \sqrt{c d^2-b d e+a e^2}-\sqrt{-b^2+4 a c} e x^2\right ) \sqrt{1-x^4}} \, dx,x,\sqrt [4]{1-\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}}\right )}{\left (b^2-4 a c\right ) e \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right ) \sqrt [4]{a+b x+c x^2}}+\frac{\left (\sqrt{2} c \sqrt{-b^2+4 a c} (2 c d-b e) \sqrt{\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (2 \sqrt{c} \sqrt{c d^2-b d e+a e^2}+\sqrt{-b^2+4 a c} e x^2\right ) \sqrt{1-x^4}} \, dx,x,\sqrt [4]{1-\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}}\right )}{\left (b^2-4 a c\right ) e \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right ) \sqrt [4]{a+b x+c x^2}}\\ &=\frac{\left (\sqrt{2} \left (-b^2+4 a c\right )^{3/2} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{2 \sqrt{c} \sqrt{c d^2-b d e+a e^2}-\sqrt{-b^2+4 a c} e x^2} \, dx,x,\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}-\frac{\left (\sqrt{2} \left (-b^2+4 a c\right )^{3/2} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{2 \sqrt{c} \sqrt{c d^2-b d e+a e^2}+\sqrt{-b^2+4 a c} e x^2} \, dx,x,\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}-\frac{\left (\sqrt{2} c \sqrt{-b^2+4 a c} (2 c d-b e) \sqrt{\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (2 \sqrt{c} \sqrt{c d^2-b d e+a e^2}-\sqrt{-b^2+4 a c} e x^2\right )} \, dx,x,\sqrt [4]{1-\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}}\right )}{\left (b^2-4 a c\right ) e \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right ) \sqrt [4]{a+b x+c x^2}}+\frac{\left (\sqrt{2} c \sqrt{-b^2+4 a c} (2 c d-b e) \sqrt{\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (2 \sqrt{c} \sqrt{c d^2-b d e+a e^2}+\sqrt{-b^2+4 a c} e x^2\right )} \, dx,x,\sqrt [4]{1-\frac{\left (b^2-4 a c\right ) \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right )^2}{c^2}}\right )}{\left (b^2-4 a c\right ) e \left (-\frac{b c}{b^2-4 a c}-\frac{2 c^2 x}{b^2-4 a c}\right ) \sqrt [4]{a+b x+c x^2}}\\ &=\frac{\sqrt [4]{-b^2+4 a c} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt [4]{-b^2+4 a c} \sqrt{e} \sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}}{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c d^2-b d e+a e^2}}\right )}{\sqrt [4]{c} \sqrt{e} \sqrt [4]{c d^2-b d e+a e^2} \sqrt [4]{a+b x+c x^2}}-\frac{\sqrt [4]{-b^2+4 a c} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \tanh ^{-1}\left (\frac{\sqrt [4]{-b^2+4 a c} \sqrt{e} \sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}}{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c d^2-b d e+a e^2}}\right )}{\sqrt [4]{c} \sqrt{e} \sqrt [4]{c d^2-b d e+a e^2} \sqrt [4]{a+b x+c x^2}}+\frac{\left (b^2-4 a c\right ) (2 c d-b e) \sqrt{\frac{(b+2 c x)^2}{b^2-4 a c}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \Pi \left (-\frac{\sqrt{-b^2+4 a c} e}{2 \sqrt{c} \sqrt{c d^2-b d e+a e^2}};\left .\sin ^{-1}\left (\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )\right |-1\right )}{\sqrt{2} \sqrt{c} \sqrt{-b^2+4 a c} e \sqrt{c d^2-b d e+a e^2} (b+2 c x) \sqrt [4]{a+b x+c x^2}}-\frac{\left (b^2-4 a c\right ) (2 c d-b e) \sqrt{\frac{(b+2 c x)^2}{b^2-4 a c}} \sqrt [4]{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \Pi \left (\frac{\sqrt{-b^2+4 a c} e}{2 \sqrt{c} \sqrt{c d^2-b d e+a e^2}};\left .\sin ^{-1}\left (\sqrt [4]{1-\frac{(b+2 c x)^2}{b^2-4 a c}}\right )\right |-1\right )}{\sqrt{2} \sqrt{c} \sqrt{-b^2+4 a c} e \sqrt{c d^2-b d e+a e^2} (b+2 c x) \sqrt [4]{a+b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.326815, size = 178, normalized size = 0.24 \[ -\frac{\sqrt{2} \sqrt [4]{\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}} \sqrt [4]{\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}} F_1\left (\frac{1}{2};\frac{1}{4},\frac{1}{4};\frac{3}{2};\frac{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 c d-b e+\sqrt{b^2-4 a c} e}{2 c d+2 c e x}\right )}{e \sqrt [4]{a+x (b+c x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((d + e*x)*(a + b*x + c*x^2)^(1/4)),x]

[Out]

-((Sqrt[2]*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(1/4)*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(
d + e*x)))^(1/4)*AppellF1[1/2, 1/4, 1/4, 3/2, (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x)), (2*c*d - b*
e + Sqrt[b^2 - 4*a*c]*e)/(2*c*d + 2*c*e*x)])/(e*(a + x*(b + c*x))^(1/4)))

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Maple [F]  time = 1.245, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ex+d}{\frac{1}{\sqrt [4]{c{x}^{2}+bx+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(c*x^2+b*x+a)^(1/4),x)

[Out]

int(1/(e*x+d)/(c*x^2+b*x+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}{\left (e x + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+b*x+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x + a)^(1/4)*(e*x + d)), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+b*x+a)^(1/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x\right ) \sqrt [4]{a + b x + c x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x**2+b*x+a)**(1/4),x)

[Out]

Integral(1/((d + e*x)*(a + b*x + c*x**2)**(1/4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}{\left (e x + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+b*x+a)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((c*x^2 + b*x + a)^(1/4)*(e*x + d)), x)