3.2537 \(\int \frac{d+e x}{(a+b x+c x^2)^{3/4}} \, dx\)

Optimal. Leaf size=200 \[ \frac{\sqrt [4]{b^2-4 a c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) (2 c d-b e) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{\sqrt{2} c^{5/4} (b+2 c x)}+\frac{2 e \sqrt [4]{a+b x+c x^2}}{c} \]

[Out]

(2*e*(a + b*x + c*x^2)^(1/4))/c + ((b^2 - 4*a*c)^(1/4)*(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
])*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])/(Sqrt[2]*c^(5/4)*(
b + 2*c*x))

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Rubi [A]  time = 0.142158, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {640, 623, 220} \[ \frac{\sqrt [4]{b^2-4 a c} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) (2 c d-b e) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{\sqrt{2} c^{5/4} (b+2 c x)}+\frac{2 e \sqrt [4]{a+b x+c x^2}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(a + b*x + c*x^2)^(3/4),x]

[Out]

(2*e*(a + b*x + c*x^2)^(1/4))/c + ((b^2 - 4*a*c)^(1/4)*(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
])*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])/(Sqrt[2]*c^(5/4)*(
b + 2*c*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 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]

Rubi steps

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

Mathematica [A]  time = 0.172171, size = 111, normalized size = 0.56 \[ \frac{2 c e (a+x (b+c x))-\sqrt{2} \sqrt{b^2-4 a c} \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{3/4} (b e-2 c d) \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ),2\right )}{c^2 (a+x (b+c x))^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(a + b*x + c*x^2)^(3/4),x]

[Out]

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

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Maple [F]  time = 0.999, size = 0, normalized size = 0. \begin{align*} \int{(ex+d) \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(c*x^2+b*x+a)^(3/4),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{4}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(c*x^2+b*x+a)^(3/4),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x}{\left (a + b x + c x^{2}\right )^{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(c*x**2+b*x+a)**(3/4),x)

[Out]

Integral((d + e*x)/(a + b*x + c*x**2)**(3/4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)/(c*x^2+b*x+a)^(3/4),x, algorithm="giac")

[Out]

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