\(\int (d+e x)^2 \sqrt [4]{a+b x+c x^2} \, dx\) [2511]

Optimal result

Integrand size = 22, antiderivative size = 319 \[ \int (d+e x)^2 \sqrt [4]{a+b x+c x^2} \, dx=\frac {\left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^3}+\frac {9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {\left (b^2-4 a c\right )^{5/4} \left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{168 \sqrt {2} c^{13/4} (b+2 c x)} \]

[Out]

1/84*(28*c^2*d^2+9*b^2*e^2-4*c*e*(2*a*e+7*b*d))*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c^3+9/35*e*(-b*e+2*c*d)*(c*x^2+b
*x+a)^(5/4)/c^2+2/7*e*(e*x+d)*(c*x^2+b*x+a)^(5/4)/c-1/336*(-4*a*c+b^2)^(5/4)*(28*c^2*d^2+9*b^2*e^2-4*c*e*(2*a*
e+7*b*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^(13/4)/(2*c*x+b)*2^(1/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {756, 654, 626, 637, 226} \[ \int (d+e x)^2 \sqrt [4]{a+b x+c x^2} \, dx=-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+7 b d)+9 b^2 e^2+28 c^2 d^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{168 \sqrt {2} c^{13/4} (b+2 c x)}+\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+7 b d)+9 b^2 e^2+28 c^2 d^2\right )}{84 c^3}+\frac {9 e \left (a+b x+c x^2\right )^{5/4} (2 c d-b e)}{35 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c} \]

[In]

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

[Out]

((28*c^2*d^2 + 9*b^2*e^2 - 4*c*e*(7*b*d + 2*a*e))*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(84*c^3) + (9*e*(2*c*d
- b*e)*(a + b*x + c*x^2)^(5/4))/(35*c^2) + (2*e*(d + e*x)*(a + b*x + c*x^2)^(5/4))/(7*c) - ((b^2 - 4*a*c)^(5/4
)*(28*c^2*d^2 + 9*b^2*e^2 - 4*c*e*(7*b*d + 2*a*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])*EllipticF[2*Arc
Tan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(168*Sqrt[2]*c^(13/4)*(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 626

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

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 756

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac {2 \int \left (\frac {1}{4} \left (14 c d^2-4 e \left (\frac {5 b d}{4}+a e\right )\right )+\frac {9}{4} e (2 c d-b e) x\right ) \sqrt [4]{a+b x+c x^2} \, dx}{7 c} \\ & = \frac {9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac {\left (-\frac {9}{4} b e (2 c d-b e)+\frac {1}{2} c \left (14 c d^2-4 e \left (\frac {5 b d}{4}+a e\right )\right )\right ) \int \sqrt [4]{a+b x+c x^2} \, dx}{7 c^2} \\ & = \frac {\left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^3}+\frac {9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {\left (\left (b^2-4 a c\right ) \left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right )\right ) \int \frac {1}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{336 c^3} \\ & = \frac {\left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^3}+\frac {9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {\left (\left (b^2-4 a c\right ) \left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) \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 )}{84 c^3 (b+2 c x)} \\ & = \frac {\left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^3}+\frac {9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {\left (b^2-4 a c\right )^{5/4} \left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) \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 )}{168 \sqrt {2} c^{13/4} (b+2 c x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 8.90 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.60 \[ \int (d+e x)^2 \sqrt [4]{a+b x+c x^2} \, dx=\frac {-54 e (-2 c d+b e) (a+x (b+c x))^2+60 c e (d+e x) (a+x (b+c x))^2+\frac {5 \left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) \left (2 c (b+2 c x) (a+x (b+c x))-\sqrt {2} \left (b^2-4 a c\right )^{3/2} \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )\right )}{4 c^2}}{210 c^2 (a+x (b+c x))^{3/4}} \]

[In]

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

[Out]

(-54*e*(-2*c*d + b*e)*(a + x*(b + c*x))^2 + 60*c*e*(d + e*x)*(a + x*(b + c*x))^2 + (5*(28*c^2*d^2 + 9*b^2*e^2
- 4*c*e*(7*b*d + 2*a*e))*(2*c*(b + 2*c*x)*(a + x*(b + c*x)) - Sqrt[2]*(b^2 - 4*a*c)^(3/2)*((c*(a + x*(b + c*x)
))/(-b^2 + 4*a*c))^(3/4)*EllipticF[ArcSin[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]/2, 2]))/(4*c^2))/(210*c^2*(a + x*(b +
c*x))^(3/4))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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