\(\int \frac {x^2 (d^2-e^2 x^2)^{5/2}}{(d+e x)^2} \, dx\) [160]

Optimal result

Integrand size = 27, antiderivative size = 142 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {3 d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \]

[Out]

2/5*d*x^2*(-e^2*x^2+d^2)^(3/2)/e-1/6*x^3*(-e^2*x^2+d^2)^(3/2)+1/120*d^2*(-45*e*x+32*d)*(-e^2*x^2+d^2)^(3/2)/e^
3+3/16*d^6*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3+3/16*d^4*x*(-e^2*x^2+d^2)^(1/2)/e^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {866, 1823, 847, 794, 201, 223, 209} \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {3 d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2} \]

[In]

Int[(x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x]

[Out]

(3*d^4*x*Sqrt[d^2 - e^2*x^2])/(16*e^2) + (2*d*x^2*(d^2 - e^2*x^2)^(3/2))/(5*e) - (x^3*(d^2 - e^2*x^2)^(3/2))/6
+ (d^2*(32*d - 45*e*x)*(d^2 - e^2*x^2)^(3/2))/(120*e^3) + (3*d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e^3)

Rule 201

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

Rule 209

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

Rule 223

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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(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 847

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^
m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[(f + g*x)^n*((a + c*x^2)^(m + p)/(d - e*x)^m), x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1823

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps \begin{align*} \text {integral}& = \int x^2 (d-e x)^2 \sqrt {d^2-e^2 x^2} \, dx \\ & = -\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}-\frac {\int x^2 \left (-9 d^2 e^2+12 d e^3 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{6 e^2} \\ & = \frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {\int x \left (-24 d^3 e^3+45 d^2 e^4 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{30 e^4} \\ & = \frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 d^4\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{8 e^2} \\ & = \frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 d^6\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e^2} \\ & = \frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {\left (3 d^6\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2} \\ & = \frac {3 d^4 x \sqrt {d^2-e^2 x^2}}{16 e^2}+\frac {2 d x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac {1}{6} x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac {d^2 (32 d-45 e x) \left (d^2-e^2 x^2\right )^{3/2}}{120 e^3}+\frac {3 d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.80 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (64 d^5-45 d^4 e x+32 d^3 e^2 x^2+50 d^2 e^3 x^3-96 d e^4 x^4+40 e^5 x^5\right )-90 d^6 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{240 e^3} \]

[In]

Integrate[(x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(64*d^5 - 45*d^4*e*x + 32*d^3*e^2*x^2 + 50*d^2*e^3*x^3 - 96*d*e^4*x^4 + 40*e^5*x^5) - 90*
d^6*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(240*e^3)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {\left (40 e^{5} x^{5}-96 d \,e^{4} x^{4}+50 d^{2} e^{3} x^{3}+32 d^{3} e^{2} x^{2}-45 d^{4} e x +64 d^{5}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{240 e^{3}}+\frac {3 d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 e^{2} \sqrt {e^{2}}}\)	\(108\)
default	\(\frac {\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\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}}}\right )}{4}\right )}{6}}{e^{2}}+\frac {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 {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \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 )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}\right )}{e^{4}}-\frac {2 d \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \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 )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{e^{3}}\)	\(541\)

[In]

int(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/240*(40*e^5*x^5-96*d*e^4*x^4+50*d^2*e^3*x^3+32*d^3*e^2*x^2-45*d^4*e*x+64*d^5)/e^3*(-e^2*x^2+d^2)^(1/2)+3/16*
d^6/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.75 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {90 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (40 \, e^{5} x^{5} - 96 \, d e^{4} x^{4} + 50 \, d^{2} e^{3} x^{3} + 32 \, d^{3} e^{2} x^{2} - 45 \, d^{4} e x + 64 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{3}} \]

[In]

integrate(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="fricas")

[Out]

-1/240*(90*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (40*e^5*x^5 - 96*d*e^4*x^4 + 50*d^2*e^3*x^3 + 32*d^
3*e^2*x^2 - 45*d^4*e*x + 64*d^5)*sqrt(-e^2*x^2 + d^2))/e^3

Sympy [A] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.10 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=d^{2} \left (\begin {cases} \frac {d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2} x}{8 e^{2}} + \frac {x^{3}}{4}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{3} \sqrt {d^{2}}}{3} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{4}}{15 e^{4}} - \frac {d^{2} x^{2}}{15 e^{2}} + \frac {x^{4}}{5}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \frac {d^{6} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 e^{4}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{4} x}{16 e^{4}} - \frac {d^{2} x^{3}}{24 e^{2}} + \frac {x^{5}}{6}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{5} \sqrt {d^{2}}}{5} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate(x**2*(-e**2*x**2+d**2)**(5/2)/(e*x+d)**2,x)

[Out]

d**2*Piecewise((d**4*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)
), (x*log(x)/sqrt(-e**2*x**2), True))/(8*e**2) + sqrt(d**2 - e**2*x**2)*(-d**2*x/(8*e**2) + x**3/4), Ne(e**2,
0)), (x**3*sqrt(d**2)/3, True)) - 2*d*e*Piecewise((sqrt(d**2 - e**2*x**2)*(-2*d**4/(15*e**4) - d**2*x**2/(15*e
**2) + x**4/5), Ne(e**2, 0)), (x**4*sqrt(d**2)/4, True)) + e**2*Piecewise((d**6*Piecewise((log(-2*e**2*x + 2*s
qrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)), (x*log(x)/sqrt(-e**2*x**2), True))/(16*e**4) + s
qrt(d**2 - e**2*x**2)*(-d**4*x/(16*e**4) - d**2*x**3/(24*e**2) + x**5/6), Ne(e**2, 0)), (x**5*sqrt(d**2)/5, Tr
ue))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.62 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {i \, d^{6} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{3}} + \frac {5 \, d^{6} \arcsin \left (\frac {e x}{d}\right )}{16 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{4 \, {\left (e^{4} x + d e^{3}\right )}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4} x}{8 \, e^{2}} + \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e^{2}} - \frac {\sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5}}{4 \, e^{3}} - \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x}{24 \, e^{2}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{12 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x}{6 \, e^{2}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{5 \, e^{3}} \]

[In]

integrate(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="maxima")

[Out]

1/8*I*d^6*arcsin(e*x/d + 2)/e^3 + 5/16*d^6*arcsin(e*x/d)/e^3 + 1/4*(-e^2*x^2 + d^2)^(5/2)*d^2/(e^4*x + d*e^3)
- 1/8*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4*x/e^2 + 5/16*sqrt(-e^2*x^2 + d^2)*d^4*x/e^2 - 1/4*sqrt(e^2*x^2 + 4*d
*e*x + 3*d^2)*d^5/e^3 - 7/24*(-e^2*x^2 + d^2)^(3/2)*d^2*x/e^2 + 5/12*(-e^2*x^2 + d^2)^(3/2)*d^3/e^3 + 1/6*(-e^
2*x^2 + d^2)^(5/2)*x/e^2 - 2/5*(-e^2*x^2 + d^2)^(5/2)*d/e^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (122) = 244\).

Time = 0.33 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.75 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {{\left (2880 \, d^{7} e^{7} \arctan \left (\sqrt {\frac {2 \, d}{e x + d} - 1}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + \frac {{\left (45 \, d^{7} e^{7} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {11}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 1025 \, d^{7} e^{7} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {9}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 174 \, d^{7} e^{7} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 594 \, d^{7} e^{7} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 255 \, d^{7} e^{7} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 45 \, d^{7} e^{7} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )} {\left (e x + d\right )}^{6}}{d^{6}}\right )} {\left | e \right |}}{7680 \, d e^{11}} \]

[In]

integrate(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

-1/7680*(2880*d^7*e^7*arctan(sqrt(2*d/(e*x + d) - 1))*sgn(1/(e*x + d))*sgn(e) + (45*d^7*e^7*(2*d/(e*x + d) - 1
)^(11/2)*sgn(1/(e*x + d))*sgn(e) - 1025*d^7*e^7*(2*d/(e*x + d) - 1)^(9/2)*sgn(1/(e*x + d))*sgn(e) - 174*d^7*e^
7*(2*d/(e*x + d) - 1)^(7/2)*sgn(1/(e*x + d))*sgn(e) - 594*d^7*e^7*(2*d/(e*x + d) - 1)^(5/2)*sgn(1/(e*x + d))*s
gn(e) - 255*d^7*e^7*(2*d/(e*x + d) - 1)^(3/2)*sgn(1/(e*x + d))*sgn(e) - 45*d^7*e^7*sqrt(2*d/(e*x + d) - 1)*sgn
(1/(e*x + d))*sgn(e))*(e*x + d)^6/d^6)*abs(e)/(d*e^11)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^2} \,d x \]

[In]

int((x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2,x)

[Out]

int((x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^2, x)