3.1.42 \(\int \sqrt [4]{b x+c x^2} \, dx\) [42]

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

[Out]

1/3*(2*c*x+b)*(c*x^2+b*x)^(1/4)/c-1/6*b^3*(-c*(c*x^2+b*x)/b^2)^(3/4)*(cos(1/2*arcsin(1+2*c*x/b))^2)^(1/2)/cos(
1/2*arcsin(1+2*c*x/b))*EllipticF(sin(1/2*arcsin(1+2*c*x/b)),2^(1/2))/c^2/(c*x^2+b*x)^(3/4)*2^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 = {626, 636, 633, 238} \begin {gather*} \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} \text {ArcSin}\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}} \end {gather*}

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 238

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2]))*EllipticF[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[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 633

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 636

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

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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 45, normalized size = 0.50 \begin {gather*} \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]{1+\frac {c x}{b}}} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \left (c \,x^{2}+b x \right )^{\frac {1}{4}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [4]{b x + c x^{2}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 0.17, size = 36, normalized size = 0.40 \begin {gather*} \frac {4\,x\,{\left (c\,x^2+b\,x\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {5}{4};\ \frac {9}{4};\ -\frac {c\,x}{b}\right )}{5\,{\left (\frac {c\,x}{b}+1\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/4),x)

[Out]

(4*x*(b*x + c*x^2)^(1/4)*hypergeom([-1/4, 5/4], 9/4, -(c*x)/b))/(5*((c*x)/b + 1)^(1/4))

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