Integrand size = 22, antiderivative size = 187 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\frac {6\ 2^{2/3} \left (a+b x+c x^2\right )^{4/3} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {4}{3},-\frac {4}{3},\frac {1}{3},\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} (d+e x)^2} \] Output:
6*2^(2/3)*(c*x^2+b*x+a)^(4/3)*AppellF1(-2/3,-4/3,-4/3,1/3,(2*d-(b+(-4*a*c+ b^2)^(1/2))*e/c)/(2*e*x+2*d),1/2*(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)/c/(e*x+d ))/e/(e*(b-(-4*a*c+b^2)^(1/2)+2*c*x)/c/(e*x+d))^(4/3)/(e*(2*c*x+(-4*a*c+b^ 2)^(1/2)+b)/c/(e*x+d))^(4/3)/(e*x+d)^2
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx \] Input:
Integrate[(a + b*x + c*x^2)^(4/3)/(d + e*x)^3,x]
Output:
Integrate[(a + b*x + c*x^2)^(4/3)/(d + e*x)^3, x]
Time = 0.31 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1178, 27, 150}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 1178 |
\(\displaystyle -\frac {4\ 2^{2/3} \left (\frac {1}{d+e x}\right )^{8/3} \left (a+b x+c x^2\right )^{4/3} \int \frac {\left (2-\frac {2 d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{c}}{d+e x}\right )^{4/3} \left (2-\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{d+e x}\right )^{4/3}}{4\ 2^{2/3} \left (\frac {1}{d+e x}\right )^{5/3}}d\frac {1}{d+e x}}{e \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\left (\frac {1}{d+e x}\right )^{8/3} \left (a+b x+c x^2\right )^{4/3} \int \frac {\left (2-\frac {2 d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{c}}{d+e x}\right )^{4/3} \left (2-\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{d+e x}\right )^{4/3}}{\left (\frac {1}{d+e x}\right )^{5/3}}d\frac {1}{d+e x}}{e \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {6\ 2^{2/3} \left (a+b x+c x^2\right )^{4/3} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {4}{3},-\frac {4}{3},\frac {1}{3},\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (d+e x)^2 \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3}}\) |
Input:
Int[(a + b*x + c*x^2)^(4/3)/(d + e*x)^3,x]
Output:
(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)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> 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] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(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] && ILtQ[m, 0]
\[\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (e x +d \right )^{3}}d x\]
Input:
int((c*x^2+b*x+a)^(4/3)/(e*x+d)^3,x)
Output:
int((c*x^2+b*x+a)^(4/3)/(e*x+d)^3,x)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\text {Timed out} \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(e*x+d)^3,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {4}{3}}}{\left (d + e x\right )^{3}}\, dx \] Input:
integrate((c*x**2+b*x+a)**(4/3)/(e*x+d)**3,x)
Output:
Integral((a + b*x + c*x**2)**(4/3)/(d + e*x)**3, x)
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (e x + d\right )}^{3}} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(e*x+d)^3,x, algorithm="maxima")
Output:
integrate((c*x^2 + b*x + a)^(4/3)/(e*x + d)^3, x)
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (e x + d\right )}^{3}} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(e*x+d)^3,x, algorithm="giac")
Output:
integrate((c*x^2 + b*x + a)^(4/3)/(e*x + d)^3, x)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (d+e\,x\right )}^3} \,d x \] Input:
int((a + b*x + c*x^2)^(4/3)/(d + e*x)^3,x)
Output:
int((a + b*x + c*x^2)^(4/3)/(d + e*x)^3, x)
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(d+e x)^3} \, dx=\text {too large to display} \] Input:
int((c*x^2+b*x+a)^(4/3)/(e*x+d)^3,x)
Output:
(15*(a + b*x + c*x**2)**(1/3)*a*b*e**2 - 54*(a + b*x + c*x**2)**(1/3)*a*c* d*e - 36*(a + b*x + c*x**2)**(1/3)*b**2*d*e - 45*(a + b*x + c*x**2)**(1/3) *b**2*e**2*x + 96*(a + b*x + c*x**2)**(1/3)*b*c*d**2 + 138*(a + b*x + c*x* *2)**(1/3)*b*c*d*e*x + 15*(a + b*x + c*x**2)**(1/3)*b*c*e**2*x**2 - 48*(a + b*x + c*x**2)**(1/3)*c**2*d**2*x - 6*(a + b*x + c*x**2)**(1/3)*c**2*d*e* x**2 + 200*int((a + b*x + c*x**2)**(1/3)/(5*a*b*d**3*e + 15*a*b*d**2*e**2* x + 15*a*b*d*e**3*x**2 + 5*a*b*e**4*x**3 - 2*a*c*d**4 - 6*a*c*d**3*e*x - 6 *a*c*d**2*e**2*x**2 - 2*a*c*d*e**3*x**3 + 5*b**2*d**3*e*x + 15*b**2*d**2*e **2*x**2 + 15*b**2*d*e**3*x**3 + 5*b**2*e**4*x**4 - 2*b*c*d**4*x - b*c*d** 3*e*x**2 + 9*b*c*d**2*e**2*x**3 + 13*b*c*d*e**3*x**4 + 5*b*c*e**4*x**5 - 2 *c**2*d**4*x**2 - 6*c**2*d**3*e*x**3 - 6*c**2*d**2*e**2*x**4 - 2*c**2*d*e* *3*x**5),x)*a**2*b**2*d**2*e**4 + 400*int((a + b*x + c*x**2)**(1/3)/(5*a*b *d**3*e + 15*a*b*d**2*e**2*x + 15*a*b*d*e**3*x**2 + 5*a*b*e**4*x**3 - 2*a* c*d**4 - 6*a*c*d**3*e*x - 6*a*c*d**2*e**2*x**2 - 2*a*c*d*e**3*x**3 + 5*b** 2*d**3*e*x + 15*b**2*d**2*e**2*x**2 + 15*b**2*d*e**3*x**3 + 5*b**2*e**4*x* *4 - 2*b*c*d**4*x - b*c*d**3*e*x**2 + 9*b*c*d**2*e**2*x**3 + 13*b*c*d*e**3 *x**4 + 5*b*c*e**4*x**5 - 2*c**2*d**4*x**2 - 6*c**2*d**3*e*x**3 - 6*c**2*d **2*e**2*x**4 - 2*c**2*d*e**3*x**5),x)*a**2*b**2*d*e**5*x + 200*int((a + b *x + c*x**2)**(1/3)/(5*a*b*d**3*e + 15*a*b*d**2*e**2*x + 15*a*b*d*e**3*x** 2 + 5*a*b*e**4*x**3 - 2*a*c*d**4 - 6*a*c*d**3*e*x - 6*a*c*d**2*e**2*x**...