\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/4}}{(d+e x)^3} \, dx\) [357]

Optimal result

Integrand size = 37, antiderivative size = 214 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx=\frac {10 c d \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 e^2}-\frac {4 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{3 e (d+e x)^2}+\frac {10 \sqrt {c d^2-a e^2} (a e+c d x)^{3/2} \left (\frac {c d (d+e x)}{e (a e+c d x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {c d^2-a e^2}}{\sqrt {e} \sqrt {a e+c d x}}\right ),2\right )}{3 e^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \] Output:

10/3*c*d*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/4)/e^2-4/3*(a*d*e+(a*e^2+c*d 
^2)*x+c*d*e*x^2)^(5/4)/e/(e*x+d)^2+10/3*(-a*e^2+c*d^2)^(1/2)*(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))/e^(3/2)/(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 = 108, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx=\frac {4 c d (a e+c d x)^2 \sqrt [4]{(a e+c d x) (d+e x)} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {9}{4},\frac {13}{4},\frac {e (a e+c d x)}{-c d^2+a e^2}\right )}{9 \left (c d^2-a e^2\right )^2 \sqrt [4]{\frac {c d (d+e x)}{c d^2-a e^2}}} \] Input:

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/4)/(d + e*x)^3,x]
 

Output:

(4*c*d*(a*e + c*d*x)^2*((a*e + c*d*x)*(d + e*x))^(1/4)*Hypergeometric2F1[7 
/4, 9/4, 13/4, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(9*(c*d^2 - a*e^2)^2 
*((c*d*(d + e*x))/(c*d^2 - a*e^2))^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.69 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.31, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {1138, 57, 60, 73, 768, 858, 807, 229}

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.

\(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx\)

\(\Big \downarrow \) 1138

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 768

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 229

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

Input:

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/4)/(d + e*x)^3,x]
 

Output:

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

Defintions of rubi rules used

rule 57

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

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 768

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 
/4)/(a + b*x^4)^(3/4))   Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ 
[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

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]

\[\int \frac {{\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e \right )}^{\frac {5}{4}}}{\left (e x +d \right )^{3}}d x\]

Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/4)/(e*x+d)^3,x)
 

Output:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/4)/(e*x+d)^3,x)
 

Fricas [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}}}{{\left (e x + d\right )}^{3}} \,d x } \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/(e*x+d)^3,x, algorithm=" 
fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx=\int \frac {\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {5}{4}}}{\left (d + e x\right )^{3}}\, dx \] Input:

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/4)/(e*x+d)**3,x)
 

Output:

Integral(((d + e*x)*(a*e + c*d*x))**(5/4)/(d + e*x)**3, x)
 

Maxima [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}}}{{\left (e x + d\right )}^{3}} \,d x } \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/(e*x+d)^3,x, algorithm=" 
maxima")
 

Output:

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/4)/(e*x + d)^3, x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx=\int { \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}}}{{\left (e x + d\right )}^{3}} \,d x } \] Input:

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/(e*x+d)^3,x, algorithm=" 
giac")
 

Output:

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/4)/(e*x + d)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx=\int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/4}}{{\left (d+e\,x\right )}^3} \,d x \] Input:

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/4)/(d + e*x)^3,x)
 

Output:

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/4)/(d + e*x)^3, x)
 

Reduce [F]

\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{(d+e x)^3} \, dx =\text {Too large to display} \] Input:

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/(e*x+d)^3,x)
 

Output:

( - 8*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*e**3 + 16*(a* 
d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c*d**2*e + 12*(a*d*e + a* 
e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c*d*e**2*x - 4*(a*d*e + a*e**2*x 
+ c*d**2*x + c*d*e*x**2)**(1/4)*c**2*d**3*x + 15*int(((a*d*e + a*e**2*x + 
c*d**2*x + c*d*e*x**2)**(1/4)*x)/(3*a**2*d**2*e**3 + 6*a**2*d*e**4*x + 3*a 
**2*e**5*x**2 - a*c*d**4*e + a*c*d**3*e**2*x + 5*a*c*d**2*e**3*x**2 + 3*a* 
c*d*e**4*x**3 - c**2*d**5*x - 2*c**2*d**4*e*x**2 - c**2*d**3*e**2*x**3),x) 
*a**3*c*d**2*e**6 + 15*int(((a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1 
/4)*x)/(3*a**2*d**2*e**3 + 6*a**2*d*e**4*x + 3*a**2*e**5*x**2 - a*c*d**4*e 
 + a*c*d**3*e**2*x + 5*a*c*d**2*e**3*x**2 + 3*a*c*d*e**4*x**3 - c**2*d**5* 
x - 2*c**2*d**4*e*x**2 - c**2*d**3*e**2*x**3),x)*a**3*c*d*e**7*x - 35*int( 
((a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*x)/(3*a**2*d**2*e**3 + 
6*a**2*d*e**4*x + 3*a**2*e**5*x**2 - a*c*d**4*e + a*c*d**3*e**2*x + 5*a*c* 
d**2*e**3*x**2 + 3*a*c*d*e**4*x**3 - c**2*d**5*x - 2*c**2*d**4*e*x**2 - c* 
*2*d**3*e**2*x**3),x)*a**2*c**2*d**4*e**4 - 35*int(((a*d*e + a*e**2*x + c* 
d**2*x + c*d*e*x**2)**(1/4)*x)/(3*a**2*d**2*e**3 + 6*a**2*d*e**4*x + 3*a** 
2*e**5*x**2 - a*c*d**4*e + a*c*d**3*e**2*x + 5*a*c*d**2*e**3*x**2 + 3*a*c* 
d*e**4*x**3 - c**2*d**5*x - 2*c**2*d**4*e*x**2 - c**2*d**3*e**2*x**3),x)*a 
**2*c**2*d**3*e**5*x + 25*int(((a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)* 
*(1/4)*x)/(3*a**2*d**2*e**3 + 6*a**2*d*e**4*x + 3*a**2*e**5*x**2 - a*c*...