3.2529 \(\int \frac{(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=573 \[ \frac{\left (b^2-4 a c\right )^{3/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{20 \sqrt{2} c^{11/4} (b+2 c x)}+\frac{(b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right )}{10 c^{5/2} \sqrt{b^2-4 a c} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )}-\frac{\left (b^2-4 a c\right )^{3/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{10 \sqrt{2} c^{11/4} (b+2 c x)}+\frac{7 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{15 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c} \]

[Out]

(7*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(3/4))/(15*c^2) + (2*e*(d + e*x)*(a + b*x + c*x^2)^(3/4))/(5*c) + ((20*c^
2*d^2 + 7*b^2*e^2 - 4*c*e*(5*b*d + 2*a*e))*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(10*c^(5/2)*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)*(20*c^2*d^2 + 7*b^2*e^2 - 4*
c*e*(5*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])*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])/(10*Sqrt[2]*c^(11/4)*(b + 2*c*x)) + ((b^2 - 4*a*c)^(3/4)*(20*c^2*d
^2 + 7*b^2*e^2 - 4*c*e*(5*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*ArcTan[(Sqrt[2
]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(20*Sqrt[2]*c^(11/4)*(b + 2*c*x))

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Rubi [A]  time = 0.725086, antiderivative size = 573, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {742, 640, 623, 305, 220, 1196} \[ \frac{(b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right )}{10 c^{5/2} \sqrt{b^2-4 a c} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )}+\frac{\left (b^2-4 a c\right )^{3/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{20 \sqrt{2} c^{11/4} (b+2 c x)}-\frac{\left (b^2-4 a c\right )^{3/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{10 \sqrt{2} c^{11/4} (b+2 c x)}+\frac{7 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{15 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(3/4))/(15*c^2) + (2*e*(d + e*x)*(a + b*x + c*x^2)^(3/4))/(5*c) + ((20*c^
2*d^2 + 7*b^2*e^2 - 4*c*e*(5*b*d + 2*a*e))*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(10*c^(5/2)*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)*(20*c^2*d^2 + 7*b^2*e^2 - 4*
c*e*(5*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])*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])/(10*Sqrt[2]*c^(11/4)*(b + 2*c*x)) + ((b^2 - 4*a*c)^(3/4)*(20*c^2*d
^2 + 7*b^2*e^2 - 4*c*e*(5*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*ArcTan[(Sqrt[2
]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(20*Sqrt[2]*c^(11/4)*(b + 2*c*x))

Rule 742

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]

Rule 640

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 623

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 305

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 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 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]

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx &=\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac{2 \int \frac{\frac{1}{4} \left (10 c d^2-4 e \left (\frac{3 b d}{4}+a e\right )\right )+\frac{7}{4} e (2 c d-b e) x}{\sqrt [4]{a+b x+c x^2}} \, dx}{5 c}\\ &=\frac{7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac{\left (-\frac{7}{4} b e (2 c d-b e)+\frac{1}{2} c \left (10 c d^2-4 e \left (\frac{3 b d}{4}+a e\right )\right )\right ) \int \frac{1}{\sqrt [4]{a+b x+c x^2}} \, dx}{5 c^2}\\ &=\frac{7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac{\left (4 \left (-\frac{7}{4} b e (2 c d-b e)+\frac{1}{2} c \left (10 c d^2-4 e \left (\frac{3 b d}{4}+a e\right )\right )\right ) \sqrt{(b+2 c x)^2}\right ) \operatorname{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 )}{5 c^2 (b+2 c x)}\\ &=\frac{7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac{\left (2 \sqrt{b^2-4 a c} \left (-\frac{7}{4} b e (2 c d-b e)+\frac{1}{2} c \left (10 c d^2-4 e \left (\frac{3 b d}{4}+a e\right )\right )\right ) \sqrt{(b+2 c x)^2}\right ) \operatorname{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 )}{5 c^{5/2} (b+2 c x)}-\frac{\left (2 \sqrt{b^2-4 a c} \left (-\frac{7}{4} b e (2 c d-b e)+\frac{1}{2} c \left (10 c d^2-4 e \left (\frac{3 b d}{4}+a e\right )\right )\right ) \sqrt{(b+2 c x)^2}\right ) \operatorname{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 )}{5 c^{5/2} (b+2 c x)}\\ &=\frac{7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac{\left (20 c^2 d^2+7 b^2 e^2-4 c e (5 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{10 c^{5/2} \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} \left (20 c^2 d^2+7 b^2 e^2-4 c e (5 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 ) 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 )}{10 \sqrt{2} c^{11/4} (b+2 c x)}+\frac{\left (b^2-4 a c\right )^{3/4} \left (20 c^2 d^2+7 b^2 e^2-4 c e (5 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 )}{20 \sqrt{2} c^{11/4} (b+2 c x)}\\ \end{align*}

Mathematica [C]  time = 0.206822, size = 165, normalized size = 0.29 \[ \frac{\frac{3 (b+2 c x) \sqrt [4]{\frac{c (a+x (b+c x))}{4 a c-b^2}} \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right ) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{2 \sqrt{2} c^2}-\frac{14 e (a+x (b+c x)) (b e-2 c d)}{c}+12 e (d+e x) (a+x (b+c x))}{30 c \sqrt [4]{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-14*e*(-2*c*d + b*e)*(a + x*(b + c*x)))/c + 12*e*(d + e*x)*(a + x*(b + c*x)) + (3*(20*c^2*d^2 + 7*b^2*e^2 -
4*c*e*(5*b*d + 2*a*e))*(b + 2*c*x)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/
2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(2*Sqrt[2]*c^2))/(30*c*(a + x*(b + c*x))^(1/4))

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Maple [F]  time = 0.954, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{2}{\frac{1}{\sqrt [4]{c{x}^{2}+bx+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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