3.42 \(\int \sqrt [4]{b x+c x^2} \, dx\)

Optimal. Leaf size=90 \[ \frac{(b+2 c x) \sqrt [4]{b x+c x^2}}{3 c}-\frac{b^3 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{2 c x}{b}+1\right ),2\right )}{3 \sqrt{2} c^2 \left (b x+c x^2\right )^{3/4}} \]

[Out]

((b + 2*c*x)*(b*x + c*x^2)^(1/4))/(3*c) - (b^3*(-((c*(b*x + c*x^2))/b^2))^(3/4)*EllipticF[ArcSin[1 + (2*c*x)/b
]/2, 2])/(3*Sqrt[2]*c^2*(b*x + c*x^2)^(3/4))

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Rubi [A]  time = 0.0327646, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {612, 622, 619, 232} \[ \frac{(b+2 c x) \sqrt [4]{b x+c x^2}}{3 c}-\frac{b^3 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{2 c x}{b}+1\right )\right |2\right )}{3 \sqrt{2} c^2 \left (b x+c x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(1/4),x]

[Out]

((b + 2*c*x)*(b*x + c*x^2)^(1/4))/(3*c) - (b^3*(-((c*(b*x + c*x^2))/b^2))^(3/4)*EllipticF[ArcSin[1 + (2*c*x)/b
]/2, 2])/(3*Sqrt[2]*c^2*(b*x + c*x^2)^(3/4))

Rule 612

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 622

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

Rule 619

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

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

\begin{align*} \int \sqrt [4]{b x+c x^2} \, dx &=\frac{(b+2 c x) \sqrt [4]{b x+c x^2}}{3 c}-\frac{b^2 \int \frac{1}{\left (b x+c x^2\right )^{3/4}} \, dx}{12 c}\\ &=\frac{(b+2 c x) \sqrt [4]{b x+c x^2}}{3 c}-\frac{\left (b^2 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{3/4}\right ) \int \frac{1}{\left (-\frac{c x}{b}-\frac{c^2 x^2}{b^2}\right )^{3/4}} \, dx}{12 c \left (b x+c x^2\right )^{3/4}}\\ &=\frac{(b+2 c x) \sqrt [4]{b x+c x^2}}{3 c}+\frac{\left (b^4 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{b^2 x^2}{c^2}\right )^{3/4}} \, dx,x,-\frac{c}{b}-\frac{2 c^2 x}{b^2}\right )}{6 \sqrt{2} c^3 \left (b x+c x^2\right )^{3/4}}\\ &=\frac{(b+2 c x) \sqrt [4]{b x+c x^2}}{3 c}-\frac{b^3 \left (-\frac{c \left (b x+c x^2\right )}{b^2}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (1+\frac{2 c x}{b}\right )\right |2\right )}{3 \sqrt{2} c^2 \left (b x+c x^2\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.009837, size = 45, normalized size = 0.5 \[ \frac{4 x \sqrt [4]{x (b+c x)} \, _2F_1\left (-\frac{1}{4},\frac{5}{4};\frac{9}{4};-\frac{c x}{b}\right )}{5 \sqrt [4]{\frac{c x}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(1/4),x]

[Out]

(4*x*(x*(b + c*x))^(1/4)*Hypergeometric2F1[-1/4, 5/4, 9/4, -((c*x)/b)])/(5*(1 + (c*x)/b)^(1/4))

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Maple [F]  time = 0.401, size = 0, normalized size = 0. \begin{align*} \int \sqrt [4]{c{x}^{2}+bx}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x)^(1/4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [4]{b x + c x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x + c*x**2)**(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^(1/4), x)