\(\int \frac {d+e x}{\sqrt [4]{a+b x+c x^2}} \, dx\) [2530]

Optimal result

Integrand size = 20, antiderivative size = 469 \[ \int \frac {d+e x}{\sqrt [4]{a+b x+c x^2}} \, dx=\frac {2 e \left (a+b x+c x^2\right )^{3/4}}{3 c}+\frac {(2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{c^{3/2} \sqrt {b^2-4 a c} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {\left (b^2-4 a c\right )^{3/4} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} c^{7/4} (b+2 c x)}+\frac {\left (b^2-4 a c\right )^{3/4} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{2 \sqrt {2} c^{7/4} (b+2 c x)} \]

[Out]

2/3*e*(c*x^2+b*x+a)^(3/4)/c+(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c^(3/2)/(-4*a*c+b^2)^(1/2)/(1+2*c^(1/2)
*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))-1/2*(-4*a*c+b^2)^(3/4)*(-b*e+2*c*d)*(cos(2*arctan(c^(1/4)*(c*x^2+b*x+
a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(
1/4)))*EllipticE(sin(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4))),1/2*2^(1/2))*(1+2*c^(1/
2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))*((2*c*x+b)^2/(-4*a*c+b^2)/(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+
b^2)^(1/2))^2)^(1/2)/c^(7/4)/(2*c*x+b)*2^(1/2)+1/4*(-4*a*c+b^2)^(3/4)*(-b*e+2*c*d)*(cos(2*arctan(c^(1/4)*(c*x^
2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+
b^2)^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4))),1/2*2^(1/2))*(1+2
*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))*((2*c*x+b)^2/(-4*a*c+b^2)/(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-
4*a*c+b^2)^(1/2))^2)^(1/2)/c^(7/4)/(2*c*x+b)*2^(1/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {654, 637, 311, 226, 1210} \[ \int \frac {d+e x}{\sqrt [4]{a+b x+c x^2}} \, dx=\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) (2 c d-b e) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{2 \sqrt {2} c^{7/4} (b+2 c x)}-\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) (2 c d-b e) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} c^{7/4} (b+2 c x)}+\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2} (2 c d-b e)}{c^{3/2} \sqrt {b^2-4 a c} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}+\frac {2 e \left (a+b x+c x^2\right )^{3/4}}{3 c} \]

[In]

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

[Out]

(2*e*(a + b*x + c*x^2)^(3/4))/(3*c) + ((2*c*d - b*e)*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(c^(3/2)*Sqrt[b^2 -
4*a*c]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])) - ((b^2 - 4*a*c)^(3/4)*(2*c*d - b*e)*Sqrt[(b
+ 2*c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[
a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*EllipticE[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c
)^(1/4)], 1/2])/(Sqrt[2]*c^(7/4)*(b + 2*c*x)) + ((b^2 - 4*a*c)^(3/4)*(2*c*d - b*e)*Sqrt[(b + 2*c*x)^2/((b^2 -
4*a*c)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sq
rt[b^2 - 4*a*c])*EllipticF[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(2*S
qrt[2]*c^(7/4)*(b + 2*c*x))

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 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 637

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[d*(Sqrt[(b + 2*c*x)
^2]/(b + 2*c*x)), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rule 654

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 e \left (a+b x+c x^2\right )^{3/4}}{3 c}+\frac {(2 c d-b e) \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx}{2 c} \\ & = \frac {2 e \left (a+b x+c x^2\right )^{3/4}}{3 c}+\frac {\left (2 (2 c d-b e) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c (b+2 c x)} \\ & = \frac {2 e \left (a+b x+c x^2\right )^{3/4}}{3 c}+\frac {\left (\sqrt {b^2-4 a c} (2 c d-b e) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c^{3/2} (b+2 c x)}-\frac {\left (\sqrt {b^2-4 a c} (2 c d-b e) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1-\frac {2 \sqrt {c} x^2}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c^{3/2} (b+2 c x)} \\ & = \frac {2 e \left (a+b x+c x^2\right )^{3/4}}{3 c}+\frac {(2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{c^{3/2} \sqrt {b^2-4 a c} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {\left (b^2-4 a c\right )^{3/4} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} c^{7/4} (b+2 c x)}+\frac {\left (b^2-4 a c\right )^{3/4} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{2 \sqrt {2} c^{7/4} (b+2 c x)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.24 \[ \int \frac {d+e x}{\sqrt [4]{a+b x+c x^2}} \, dx=\frac {8 c e (a+x (b+c x))+3 \sqrt {2} (2 c d-b e) (b+2 c x) \sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{12 c^2 \sqrt [4]{a+x (b+c x)}} \]

[In]

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

[Out]

(8*c*e*(a + x*(b + c*x)) + 3*Sqrt[2]*(2*c*d - b*e)*(b + 2*c*x)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/4)*Hy
pergeometric2F1[1/4, 1/2, 3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(12*c^2*(a + x*(b + c*x))^(1/4))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x}{\sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {d+e\,x}{{\left (c\,x^2+b\,x+a\right )}^{1/4}} \,d x \]

[In]

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

[Out]

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