\(\int \frac {1}{(d+e x)^2 (a+b x+c x^2)^{4/3}} \, dx\) [695]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx=-\frac {3 \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} \operatorname {AppellF1}\left (\frac {11}{3},\frac {4}{3},\frac {4}{3},\frac {14}{3},\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{44\ 2^{2/3} e (d+e x) \left (a+b x+c x^2\right )^{4/3}} \] Output:

-3/88*(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)*AppellF1(11/3,4/3,4/3,14/3,(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) 
)*2^(1/3)/e/(e*x+d)/(c*x^2+b*x+a)^(4/3)
 

Mathematica [A] (warning: unable to verify)

Time = 11.61 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx=-\frac {3 \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{4/3} \operatorname {AppellF1}\left (\frac {11}{3},\frac {4}{3},\frac {4}{3},\frac {14}{3},\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 c d-b e+\sqrt {b^2-4 a c} e}{2 c d+2 c e x}\right )}{44\ 2^{2/3} e (d+e x) (a+x (b+c x))^{4/3}} \] Input:

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

Output:

(-3*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(4/3)*((e*(b + Sqr 
t[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(4/3)*AppellF1[11/3, 4/3, 4/3, 14/ 
3, (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x)), (2*c*d - b*e + Sqr 
t[b^2 - 4*a*c]*e)/(2*c*d + 2*c*e*x)])/(44*2^(2/3)*e*(d + e*x)*(a + x*(b + 
c*x))^(4/3))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 189, 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.

\(\displaystyle \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx\)

\(\Big \downarrow \) 1178

\(\displaystyle -\frac {\left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \int \frac {4\ 2^{2/3} \left (\frac {1}{d+e x}\right )^{8/3}}{\left (2-\frac {2 d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{c}}{d+e x}\right )^{4/3} \left (2-\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{d+e x}\right )^{4/3}}d\frac {1}{d+e x}}{4\ 2^{2/3} e \left (\frac {1}{d+e x}\right )^{8/3} \left (a+b x+c x^2\right )^{4/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \int \frac {\left (\frac {1}{d+e x}\right )^{8/3}}{\left (2-\frac {2 d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{c}}{d+e x}\right )^{4/3} \left (2-\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{d+e x}\right )^{4/3}}d\frac {1}{d+e x}}{e \left (\frac {1}{d+e x}\right )^{8/3} \left (a+b x+c x^2\right )^{4/3}}\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {3 \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{4/3} \operatorname {AppellF1}\left (\frac {11}{3},\frac {4}{3},\frac {4}{3},\frac {14}{3},\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{44\ 2^{2/3} e (d+e x) \left (a+b x+c x^2\right )^{4/3}}\)

Input:

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

Output:

(-3*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(4/3)*((e*(b + Sqr 
t[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^(4/3)*AppellF1[11/3, 4/3, 4/3, 14/ 
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))])/(44*2^(2/3)*e*(d + e*x)*(a + b*x + c*x 
^2)^(4/3))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 150

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])
 

rule 1178

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]
 

Maple [F]

\[\int \frac {1}{\left (e x +d \right )^{2} \left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}d x\]

Input:

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

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x+d)^2/(c*x^2+b*x+a)^(4/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx=\int \frac {1}{\left (d + e x\right )^{2} \left (a + b x + c x^{2}\right )^{\frac {4}{3}}}\, dx \] Input:

integrate(1/(e*x+d)**2/(c*x**2+b*x+a)**(4/3),x)
 

Output:

Integral(1/((d + e*x)**2*(a + b*x + c*x**2)**(4/3)), x)
 

Maxima [F]

\[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}} {\left (e x + d\right )}^{2}} \,d x } \] Input:

integrate(1/(e*x+d)^2/(c*x^2+b*x+a)^(4/3),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}} {\left (e x + d\right )}^{2}} \,d x } \] Input:

integrate(1/(e*x+d)^2/(c*x^2+b*x+a)^(4/3),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx=\int \frac {1}{{\left (d+e\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{4/3}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{4/3}} \, dx=\int \frac {1}{\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} a \,d^{2}+2 \left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} a d e x +\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} a \,e^{2} x^{2}+\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} b \,d^{2} x +2 \left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} b d e \,x^{2}+\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} b \,e^{2} x^{3}+\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} c \,d^{2} x^{2}+2 \left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} c d e \,x^{3}+\left (c \,x^{2}+b x +a \right )^{\frac {1}{3}} c \,e^{2} x^{4}}d x \] Input:

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

Output:

int(1/((a + b*x + c*x**2)**(1/3)*a*d**2 + 2*(a + b*x + c*x**2)**(1/3)*a*d* 
e*x + (a + b*x + c*x**2)**(1/3)*a*e**2*x**2 + (a + b*x + c*x**2)**(1/3)*b* 
d**2*x + 2*(a + b*x + c*x**2)**(1/3)*b*d*e*x**2 + (a + b*x + c*x**2)**(1/3 
)*b*e**2*x**3 + (a + b*x + c*x**2)**(1/3)*c*d**2*x**2 + 2*(a + b*x + c*x** 
2)**(1/3)*c*d*e*x**3 + (a + b*x + c*x**2)**(1/3)*c*e**2*x**4),x)