3.2486 \(\int \frac{(a+b x+c x^2)^{4/3}}{d+e x} \, dx\)

Optimal. Leaf size=180 \[ \frac{3 \left (a+b x+c x^2\right )^{4/3} F_1\left (-\frac{8}{3};-\frac{4}{3},-\frac{4}{3};-\frac{5}{3};\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{\sqrt [3]{2} e \left (\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3}} \]

[Out]

(3*(a + b*x + c*x^2)^(4/3)*AppellF1[-8/3, -4/3, -4/3, -5/3, (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x)
), (2*d - ((b + Sqrt[b^2 - 4*a*c])*e)/c)/(2*(d + e*x))])/(2^(1/3)*e*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d
 + e*x)))^(4/3)*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(4/3))

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Rubi [A]  time = 0.198309, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {758, 133} \[ \frac{3 \left (a+b x+c x^2\right )^{4/3} F_1\left (-\frac{8}{3};-\frac{4}{3},-\frac{4}{3};-\frac{5}{3};\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{\sqrt [3]{2} e \left (\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x + c*x^2)^(4/3)*AppellF1[-8/3, -4/3, -4/3, -5/3, (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x)
), (2*d - ((b + Sqrt[b^2 - 4*a*c])*e)/c)/(2*(d + e*x))])/(2^(1/3)*e*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d
 + e*x)))^(4/3)*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(4/3))

Rule 758

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
 2]}, -Dist[((1/(d + e*x))^(2*p)*(a + b*x + c*x^2)^p)/(e*((e*(b - q + 2*c*x))/(2*c*(d + e*x)))^p*((e*(b + q +
2*c*x))/(2*c*(d + e*x)))^p), Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - (e*(b - q))/(2*c))*x, x]^p*Simp[1 - (d
 - (e*(b + q))/(2*c))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, 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] &&  !IntegerQ[p] && ILtQ[m, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
 x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{4/3}}{d+e x} \, dx &=-\frac{\left (4\ 2^{2/3} \left (\frac{1}{d+e x}\right )^{8/3} \left (a+b x+c x^2\right )^{4/3}\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{1}{2} \left (2 d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{c}\right ) x\right )^{4/3} \left (1-\frac{1}{2} \left (2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}\right ) x\right )^{4/3}}{x^{11/3}} \, dx,x,\frac{1}{d+e x}\right )}{e \left (\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3}}\\ &=\frac{3 \left (a+b x+c x^2\right )^{4/3} F_1\left (-\frac{8}{3};-\frac{4}{3},-\frac{4}{3};-\frac{5}{3};\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{\sqrt [3]{2} e \left (\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3}}\\ \end{align*}

Mathematica [F]  time = 1.30788, size = 0, normalized size = 0. \[ \int \frac{\left (a+b x+c x^2\right )^{4/3}}{d+e x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 1.302, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ex+d} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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