\(\int \frac {1}{(d+e x)^2 \sqrt [3]{a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [336]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 37, antiderivative size = 84 \[ \int \frac {1}{(d+e x)^2 \sqrt [3]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},1,-\frac {1}{3},\frac {c d (d+e x)}{c d^2-a e^2}\right )}{4 \left (c d^2-a e^2\right ) (d+e x)^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(d+e x)^2 \sqrt [3]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {3 c^2 d^2 ((a e+c d x) (d+e x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {7}{3},\frac {5}{3},\frac {e (a e+c d x)}{-c d^2+a e^2}\right )}{2 \left (c d^2-a e^2\right )^3 \left (\frac {c d (d+e x)}{c d^2-a e^2}\right )^{2/3}} \] Input:

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

Output:

(3*c^2*d^2*((a*e + c*d*x)*(d + e*x))^(2/3)*Hypergeometric2F1[2/3, 7/3, 5/3 
, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(2*(c*d^2 - a*e^2)^3*((c*d*(d + e 
*x))/(c*d^2 - a*e^2))^(2/3))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.40, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {1138, 80, 79}

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.

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

\(\Big \downarrow \) 1138

\(\displaystyle \frac {\sqrt [3]{\frac {e x}{d}+1} \sqrt [3]{a e+c d x} \int \frac {1}{\sqrt [3]{a e+c d x} \left (\frac {e x}{d}+1\right )^{7/3}}dx}{d^2 \sqrt [3]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {c^2 d^2 \sqrt [3]{a e+c d x} \sqrt [3]{\frac {c d (d+e x)}{c d^2-a e^2}} \int \frac {1}{\sqrt [3]{a e+c d x} \left (\frac {c d^2}{c d^2-a e^2}+\frac {c e x d}{c d^2-a e^2}\right )^{7/3}}dx}{\left (c d^2-a e^2\right )^2 \sqrt [3]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {3 c d (a e+c d x) \sqrt [3]{\frac {c d (d+e x)}{c d^2-a e^2}} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {7}{3},\frac {5}{3},-\frac {e (a e+c d x)}{c d^2-a e^2}\right )}{2 \left (c d^2-a e^2\right )^2 \sqrt [3]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

Input:

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

Output:

(3*c*d*(a*e + c*d*x)*((c*d*(d + e*x))/(c*d^2 - a*e^2))^(1/3)*Hypergeometri 
c2F1[2/3, 7/3, 5/3, -((e*(a*e + c*d*x))/(c*d^2 - a*e^2))])/(2*(c*d^2 - a*e 
^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(1/3))
 

Defintions of rubi rules used

rule 79
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 1138
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d^m*((a + b*x + c*x^2)^FracPart[p]/((1 + e*(x/d))^FracPart[p] 
*(a/d + (c*x)/e)^FracPart[p]))   Int[(1 + e*(x/d))^(m + p)*(a/d + (c/e)*x)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
&& (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (IntegerQ[3*p] || Integer 
Q[4*p]))
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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