\(\int \frac {(a+b x+c x^2)^{4/3}}{(d+e x)^3} \, dx\) [2488]

Optimal result

Integrand size = 22, antiderivative size = 187 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\frac {6\ 2^{2/3} \left (a+b x+c x^2\right )^{4/3} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {4}{3},-\frac {4}{3},\frac {1}{3},\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} (d+e x)^2} \]

[Out]

6*2^(2/3)*(c*x^2+b*x+a)^(4/3)*AppellF1(-2/3,-4/3,-4/3,1/3,1/2*(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))/e/(e*x+d)^2/(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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {772, 138} \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\frac {6\ 2^{2/3} \left (a+b x+c x^2\right )^{4/3} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {4}{3},-\frac {4}{3},\frac {1}{3},\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (d+e x)^2 \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3}} \]

[In]

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

[Out]

(6*2^(2/3)*(a + b*x + c*x^2)^(4/3)*AppellF1[-2/3, -4/3, -4/3, 1/3, (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))])/(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)*(d + e*x)^2)

Rule 138

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

Rule 772

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*} \text {integral}& = -\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 ) \text {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^{5/3}} \, 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 {6\ 2^{2/3} \left (a+b x+c x^2\right )^{4/3} F_1\left (-\frac {2}{3};-\frac {4}{3},-\frac {4}{3};\frac {1}{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 )}{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} (d+e x)^2} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (d+e\,x\right )}^3} \,d x \]

[In]

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

[Out]

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