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]

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

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Rubi [A]  time = 0.20, antiderivative size = 180, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [F]  time = 1.22, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{d+e x} \, dx \]

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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{e x + d}\,{d x} \]

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)

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maple [F]  time = 1.88, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{e x +d}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{e x + d}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{d + e x}\, dx \]

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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