\(\int (d+e x)^2 \sqrt [4]{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [337]

Optimal result

Integrand size = 37, antiderivative size = 350 \[ \int (d+e x)^2 \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {3 \left (c d^2-a e^2\right )^3 \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{28 c^3 d^3 e}+\frac {9 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{35 c^2 d^2}+\frac {3 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{14 c^3 d^3 (d+e x)}+\frac {2 (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{7 c d}+\frac {3 \left (c d^2-a e^2\right )^{7/2} (a e+c d x)^{3/2} \left (\frac {c d (d+e x)}{e (a e+c d x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {c d^2-a e^2}}{\sqrt {e} \sqrt {a e+c d x}}\right ),2\right )}{28 c^4 d^4 \sqrt {e} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \] Output:

3/28*(-a*e^2+c*d^2)^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/4)/c^3/d^3/e+9/ 
35*(-a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/c^2/d^2+3/14*(-a 
*e^2+c*d^2)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/c^3/d^3/(e*x+d)+2/7* 
(e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/c/d+3/28*(-a*e^2+c*d^2)^(7 
/2)*(c*d*x+a*e)^(3/2)*(c*d*(e*x+d)/e/(c*d*x+a*e))^(3/4)*InverseJacobiAM(1/ 
2*arctan((-a*e^2+c*d^2)^(1/2)/e^(1/2)/(c*d*x+a*e)^(1/2)),2^(1/2))/c^4/d^4/ 
e^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.31 \[ \int (d+e x)^2 \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {4 \left (c d^2-a e^2\right )^2 (a e+c d x) \sqrt [4]{(a e+c d x) (d+e x)} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},\frac {5}{4},\frac {9}{4},\frac {e (a e+c d x)}{-c d^2+a e^2}\right )}{5 c^3 d^3 \sqrt [4]{\frac {c d (d+e x)}{c d^2-a e^2}}} \] Input:

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

Output:

(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)*((a*e + c*d*x)*(d + e*x))^(1/4)*Hyperge 
ometric2F1[-9/4, 5/4, 9/4, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(5*c^3*d 
^3*((c*d*(d + e*x))/(c*d^2 - a*e^2))^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.98 (sec) , antiderivative size = 581, normalized size of antiderivative = 1.66, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1166, 27, 1160, 1087, 1094, 761}

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.

\(\displaystyle \int (d+e x)^2 \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \, dx\)

\(\Big \downarrow \) 1166

\(\displaystyle \frac {2 \int \frac {9}{4} e \left (c d^2-a e^2\right ) (d+e x) \sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}dx}{7 c d e}+\frac {2 (d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {9 \left (c d^2-a e^2\right ) \int (d+e x) \sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}dx}{14 c d}+\frac {2 (d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {9 \left (c d^2-a e^2\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}dx}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{5 c d}\right )}{14 c d}+\frac {2 (d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {9 \left (c d^2-a e^2\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}-\frac {\left (c d^2-a e^2\right )^2 \int \frac {1}{\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/4}}dx}{12 c d e}\right )}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{5 c d}\right )}{14 c d}+\frac {2 (d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d}\)

\(\Big \downarrow \) 1094

\(\displaystyle \frac {9 \left (c d^2-a e^2\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}-\frac {\left (c d^2-a e^2\right )^2 \sqrt {\left (a e^2+c d^2+2 c d e x\right )^2} \int \frac {1}{\sqrt {\left (c d^2-a e^2\right )^2+4 c d e \left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )}}d\sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{3 c d e \left (a e^2+c d^2+2 c d e x\right )}\right )}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{5 c d}\right )}{14 c d}+\frac {2 (d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {9 \left (c d^2-a e^2\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}-\frac {\left (c d^2-a e^2\right )^{5/2} \sqrt {\left (a e^2+c d^2+2 c d e x\right )^2} \left (\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d^2-a e^2}+1\right ) \sqrt {\frac {4 c d e \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )+\left (c d^2-a e^2\right )^2}{\left (c d^2-a e^2\right )^2 \left (\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d^2-a e^2}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt [4]{e} \sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{\sqrt {c d^2-a e^2}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} d^{5/4} e^{5/4} \left (a e^2+c d^2+2 c d e x\right ) \sqrt {4 c d e \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )+\left (c d^2-a e^2\right )^2}}\right )}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{5 c d}\right )}{14 c d}+\frac {2 (d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d}\)

Input:

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

Output:

(2*(d + e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/4))/(7*c*d) + (9*( 
c*d^2 - a*e^2)*((2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/4))/(5*c*d) 
+ ((d^2 - (a*e^2)/c)*(((c*d^2 + a*e^2 + 2*c*d*e*x)*(a*d*e + (c*d^2 + a*e^2 
)*x + c*d*e*x^2)^(1/4))/(3*c*d*e) - ((c*d^2 - a*e^2)^(5/2)*Sqrt[(c*d^2 + a 
*e^2 + 2*c*d*e*x)^2]*(1 + (2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + 
 a*e^2)*x + c*d*e*x^2])/(c*d^2 - a*e^2))*Sqrt[((c*d^2 - a*e^2)^2 + 4*c*d*e 
*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2))/((c*d^2 - a*e^2)^2*(1 + (2*Sqrt[ 
c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(c*d^2 - a 
*e^2))^2)]*EllipticF[2*ArcTan[(Sqrt[2]*c^(1/4)*d^(1/4)*e^(1/4)*(a*d*e + (c 
*d^2 + a*e^2)*x + c*d*e*x^2)^(1/4))/Sqrt[c*d^2 - a*e^2]], 1/2])/(6*Sqrt[2] 
*c^(5/4)*d^(5/4)*e^(5/4)*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[(c*d^2 - a*e^2)^ 
2 + 4*c*d*e*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)])))/(2*d)))/(14*c*d)
 

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 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1094

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[4*(Sqrt[(b 
+ 2*c*x)^2]/(b + 2*c*x))   Subst[Int[x^(4*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4 
*c*x^4], x], x, (a + b*x + c*x^2)^(1/4)], x] /; FreeQ[{a, b, c}, x] && Inte 
gerQ[4*p]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1166

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[1/(c*(m + 2*p + 1))   Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m 
+ 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* 
(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration 
alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat 
icQ[a, b, c, d, e, m, p, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(1/4)*(e^2*x^2 + 2*d*e*x 
+ d^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (d+e x)^2 \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(48*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**3*d*e**5 - 12*(a* 
d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**3*e**6*x - 160*(a*d*e + 
a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*c*d**3*e**3 + 28*(a*d*e + a* 
e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*c*d**2*e**4*x + 8*(a*d*e + a*e 
**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*c*d*e**5*x**2 + 272*(a*d*e + a* 
e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**2*d**5*e + 252*(a*d*e + a*e**2 
*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**2*d**4*e**2*x + 240*(a*d*e + a*e** 
2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**2*d**3*e**3*x**2 + 80*(a*d*e + a* 
e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**2*d**2*e**4*x**3 + 212*(a*d*e 
+ a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*c**3*d**6*x + 232*(a*d*e + a*e* 
*2*x + c*d**2*x + c*d*e*x**2)**(1/4)*c**3*d**5*e*x**2 + 80*(a*d*e + a*e**2 
*x + c*d**2*x + c*d*e*x**2)**(1/4)*c**3*d**4*e**2*x**3 + 15*int(((a*d*e + 
a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*x)/(a**2*d*e**3 + a**2*e**4*x + a 
*c*d**3*e + 2*a*c*d**2*e**2*x + a*c*d*e**3*x**2 + c**2*d**4*x + c**2*d**3* 
e*x**2),x)*a**5*e**10 - 45*int(((a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2) 
**(1/4)*x)/(a**2*d*e**3 + a**2*e**4*x + a*c*d**3*e + 2*a*c*d**2*e**2*x + a 
*c*d*e**3*x**2 + c**2*d**4*x + c**2*d**3*e*x**2),x)*a**4*c*d**2*e**8 + 30* 
int(((a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*x)/(a**2*d*e**3 + a 
**2*e**4*x + a*c*d**3*e + 2*a*c*d**2*e**2*x + a*c*d*e**3*x**2 + c**2*d**4* 
x + c**2*d**3*e*x**2),x)*a**3*c**2*d**4*e**6 + 30*int(((a*d*e + a*e**2*...