3.2.89 \(\int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx\) [189]

3.2.89.1 Optimal result

Integrand size = 27, antiderivative size = 160 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {19 d^2 (d-e x)^3}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {6 d (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {(20 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^5}-\frac {19 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5} \]

output

-1/5*d^3*(-e*x+d)^4/e^5/(-e^2*x^2+d^2)^(5/2)+19/15*d^2*(-e*x+d)^3/e^5/(-e^ 
2*x^2+d^2)^(3/2)-19/2*d^2*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^5-6*d*(-e*x+d 
)^2/e^5/(-e^2*x^2+d^2)^(1/2)-1/2*(-e*x+20*d)*(-e^2*x^2+d^2)^(1/2)/e^5
 

3.2.89.2 Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-448 d^4-1059 d^3 e x-713 d^2 e^2 x^2-75 d e^3 x^3+15 e^4 x^4\right )}{30 e^5 (d+e x)^3}+\frac {19 d^2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^5} \]

input

Integrate[(x^4*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]
 

output

(Sqrt[d^2 - e^2*x^2]*(-448*d^4 - 1059*d^3*e*x - 713*d^2*e^2*x^2 - 75*d*e^3 
*x^3 + 15*e^4*x^4))/(30*e^5*(d + e*x)^3) + (19*d^2*ArcTan[(e*x)/(Sqrt[d^2] 
 - Sqrt[d^2 - e^2*x^2])])/e^5
 

3.2.89.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {570, 529, 2166, 27, 2166, 27, 676, 224, 216}

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.

\(\displaystyle \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx\)

\(\Big \downarrow \) 570

\(\displaystyle \int \frac {x^4 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{7/2}}dx\)

\(\Big \downarrow \) 529

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

\(\Big \downarrow \) 2166

\(\displaystyle -\frac {-\frac {\int \frac {15 (d-e x)^2 \left (\frac {3 d^4}{e^4}-\frac {2 x d^3}{e^3}+\frac {x^2 d^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d}-\frac {19 d^3 (d-e x)^3}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {5 \int \frac {(d-e x)^2 \left (\frac {3 d^4}{e^4}-\frac {2 x d^3}{e^3}+\frac {x^2 d^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{d}-\frac {19 d^3 (d-e x)^3}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2166

\(\displaystyle -\frac {-\frac {5 \left (-\frac {\int \frac {d^3 (d-e x) (9 d-e x)}{e^4 \sqrt {d^2-e^2 x^2}}dx}{d}-\frac {6 d^3 (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}\right )}{d}-\frac {19 d^3 (d-e x)^3}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {5 \left (-\frac {d^2 \int \frac {(d-e x) (9 d-e x)}{\sqrt {d^2-e^2 x^2}}dx}{e^4}-\frac {6 d^3 (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}\right )}{d}-\frac {19 d^3 (d-e x)^3}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 676

\(\displaystyle -\frac {-\frac {5 \left (-\frac {d^2 \left (\frac {19}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {10 d \sqrt {d^2-e^2 x^2}}{e}-\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )}{e^4}-\frac {6 d^3 (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}\right )}{d}-\frac {19 d^3 (d-e x)^3}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle -\frac {-\frac {5 \left (-\frac {d^2 \left (\frac {19}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {10 d \sqrt {d^2-e^2 x^2}}{e}-\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )}{e^4}-\frac {6 d^3 (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}\right )}{d}-\frac {19 d^3 (d-e x)^3}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {-\frac {5 \left (-\frac {d^2 \left (\frac {19 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {10 d \sqrt {d^2-e^2 x^2}}{e}-\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )}{e^4}-\frac {6 d^3 (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}\right )}{d}-\frac {19 d^3 (d-e x)^3}{3 e^5 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d}-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}\)

input

Int[(x^4*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]
 

output

-1/5*(d^3*(d - e*x)^4)/(e^5*(d^2 - e^2*x^2)^(5/2)) - ((-19*d^3*(d - e*x)^3 
)/(3*e^5*(d^2 - e^2*x^2)^(3/2)) - (5*((-6*d^3*(d - e*x)^2)/(e^5*Sqrt[d^2 - 
 e^2*x^2]) - (d^2*((10*d*Sqrt[d^2 - e^2*x^2])/e - (x*Sqrt[d^2 - e^2*x^2])/ 
2 + (19*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)))/e^4))/d)/(5*d)
 

3.2.89.3.1 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 216

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

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 529

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem 
ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ 
(2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b* 
x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; 
 FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* 
c^2 + a*d^2, 0]
 

rule 570

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, -1] &&  !(IGtQ[m, 0] && ILtQ[m + n, 0] &&  !GtQ[p, 1])
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 0]
 

3.2.89.4 Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.24

method	result	size
risch	\(-\frac {\left (-e x +8 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{5}}-\frac {19 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{4} \sqrt {e^{2}}}-\frac {2 d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{8} \left (x +\frac {d}{e}\right )^{3}}+\frac {41 d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 e^{7} \left (x +\frac {d}{e}\right )^{2}}-\frac {199 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 e^{6} \left (x +\frac {d}{e}\right )}\)	\(199\)
default	\(\frac {\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}}{e^{4}}+\frac {d^{4} \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}\right )}{e^{8}}+\frac {6 d^{2} \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}\right )}{e^{6}}-\frac {4 d \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{5}}+\frac {4 d^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{8} \left (x +\frac {d}{e}\right )^{3}}\)	\(408\)

input

int(x^4*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)
 

output

-1/2*(-e*x+8*d)/e^5*(-e^2*x^2+d^2)^(1/2)-19/2/e^4*d^2/(e^2)^(1/2)*arctan(( 
e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-2/5/e^8*d^4/(x+d/e)^3*(-(x+d/e)^2*e^2+2 
*d*e*(x+d/e))^(1/2)+41/15/e^7*d^3/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e)) 
^(1/2)-199/15/e^6*d^2/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)
 

3.2.89.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.19 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=-\frac {448 \, d^{2} e^{3} x^{3} + 1344 \, d^{3} e^{2} x^{2} + 1344 \, d^{4} e x + 448 \, d^{5} - 570 \, {\left (d^{2} e^{3} x^{3} + 3 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x + d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (15 \, e^{4} x^{4} - 75 \, d e^{3} x^{3} - 713 \, d^{2} e^{2} x^{2} - 1059 \, d^{3} e x - 448 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \]

input

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

output

-1/30*(448*d^2*e^3*x^3 + 1344*d^3*e^2*x^2 + 1344*d^4*e*x + 448*d^5 - 570*( 
d^2*e^3*x^3 + 3*d^3*e^2*x^2 + 3*d^4*e*x + d^5)*arctan(-(d - sqrt(-e^2*x^2 
+ d^2))/(e*x)) - (15*e^4*x^4 - 75*d*e^3*x^3 - 713*d^2*e^2*x^2 - 1059*d^3*e 
*x - 448*d^4)*sqrt(-e^2*x^2 + d^2))/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + 
 d^3*e^5)
 

3.2.89.6 Sympy [F]

\[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {x^{4} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{4}}\, dx \]

input

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

output

Integral(x**4*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x)**4, x)
 

3.2.89.7 Maxima [F(-1)]

Timed out. \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.2.89.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.46 \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {x}{e^{4}} - \frac {8 \, d}{e^{5}}\right )} - \frac {19 \, d^{2} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e^{4} {\left | e \right |}} + \frac {2 \, {\left (164 \, d^{2} + \frac {685 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{2}}{e^{2} x} + \frac {1025 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{2}}{e^{4} x^{2}} + \frac {615 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d^{2}}{e^{6} x^{3}} + \frac {135 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} d^{2}}{e^{8} x^{4}}\right )}}{15 \, e^{4} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

input

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

output

1/2*sqrt(-e^2*x^2 + d^2)*(x/e^4 - 8*d/e^5) - 19/2*d^2*arcsin(e*x/d)*sgn(d) 
*sgn(e)/(e^4*abs(e)) + 2/15*(164*d^2 + 685*(d*e + sqrt(-e^2*x^2 + d^2)*abs 
(e))*d^2/(e^2*x) + 1025*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d^2/(e^4*x^2 
) + 615*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*d^2/(e^6*x^3) + 135*(d*e + s 
qrt(-e^2*x^2 + d^2)*abs(e))^4*d^2/(e^8*x^4))/(e^4*((d*e + sqrt(-e^2*x^2 + 
d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))
 

3.2.89.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {x^4\,\sqrt {d^2-e^2\,x^2}}{{\left (d+e\,x\right )}^4} \,d x \]

input

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

output

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