\(\int \frac {x^2 (d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\) [201]

Optimal result

Integrand size = 27, antiderivative size = 182 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \]

[Out]

-95/8*d^4*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^3-d*(-e*x+d)^4/e^3/(-e^2*x^2+d^2)^(1/2)-95/8*d^3*(-e^2*x^2+d^2)^(
1/2)/e^3-95/24*d^2*(-e*x+d)*(-e^2*x^2+d^2)^(1/2)/e^3-19/12*d*(-e*x+d)^2*(-e^2*x^2+d^2)^(1/2)/e^3-1/4*(-e*x+d)^
3*(-e^2*x^2+d^2)^(1/2)/e^3

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {866, 1649, 809, 685, 655, 223, 209} \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {95 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3} \]

[In]

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

[Out]

-((d*(d - e*x)^4)/(e^3*Sqrt[d^2 - e^2*x^2])) - (95*d^3*Sqrt[d^2 - e^2*x^2])/(8*e^3) - (95*d^2*(d - e*x)*Sqrt[d
^2 - e^2*x^2])/(24*e^3) - (19*d*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])/(12*e^3) - ((d - e*x)^3*Sqrt[d^2 - e^2*x^2])/
(4*e^3) - (95*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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 655

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

Rule 685

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

Rule 809

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[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

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 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[(-d)*f*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2
*a*e*(p + 1))), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)
*Q + f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p
+ 1/2, 0] && GtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx \\ & = -\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\left (\frac {4 d^2}{e^2}-\frac {d x}{e}\right ) (d-e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx}{d} \\ & = -\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {(19 d) \int \frac {(d-e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx}{4 e^2} \\ & = -\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^2\right ) \int \frac {(d-e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx}{12 e^2} \\ & = -\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^3\right ) \int \frac {d-e x}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2} \\ & = -\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2} \\ & = -\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^4\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^2} \\ & = -\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.60 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (-448 d^4-163 d^3 e x+61 d^2 e^2 x^2-26 d e^3 x^3+6 e^4 x^4\right )}{d+e x}+570 d^4 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{24 e^3} \]

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(-448*d^4 - 163*d^3*e*x + 61*d^2*e^2*x^2 - 26*d*e^3*x^3 + 6*e^4*x^4))/(d + e*x) + 570*d^
4*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(24*e^3)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {\left (-6 e^{3} x^{3}+32 d \,e^{2} x^{2}-93 d^{2} e x +256 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e^{3}}-\frac {95 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 e^{2} \sqrt {e^{2}}}-\frac {8 d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{4} \left (x +\frac {d}{e}\right )}\)	\(131\)
default	\(\text {Expression too large to display}\)	\(890\)

[In]

int(x^2*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-1/24*(-6*e^3*x^3+32*d*e^2*x^2-93*d^2*e*x+256*d^3)/e^3*(-e^2*x^2+d^2)^(1/2)-95/8*d^4/e^2/(e^2)^(1/2)*arctan((e
^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-8*d^4/e^4/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.68 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {448 \, d^{4} e x + 448 \, d^{5} - 570 \, {\left (d^{4} e x + d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (6 \, e^{4} x^{4} - 26 \, d e^{3} x^{3} + 61 \, d^{2} e^{2} x^{2} - 163 \, d^{3} e x - 448 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, {\left (e^{4} x + d e^{3}\right )}} \]

[In]

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

[Out]

-1/24*(448*d^4*e*x + 448*d^5 - 570*(d^4*e*x + d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (6*e^4*x^4 - 26
*d*e^3*x^3 + 61*d^2*e^2*x^2 - 163*d^3*e*x - 448*d^4)*sqrt(-e^2*x^2 + d^2))/(e^4*x + d*e^3)

Sympy [F]

\[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {x^{2} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]

[In]

integrate(x**2*(-e**2*x**2+d**2)**(5/2)/(e*x+d)**4,x)

[Out]

Integral(x**2*(-(-d + e*x)*(d + e*x))**(5/2)/(d + e*x)**4, x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.99 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{2 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{e^{4} x + d e^{3}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{3 \, {\left (e^{4} x + d e^{3}\right )}} - \frac {5 i \, d^{4} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{3}} - \frac {25 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{4 \, {\left (e^{4} x + d e^{3}\right )}} + \frac {5 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{2} x}{8 \, e^{2}} + \frac {5 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3}}{4 \, e^{3}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{3}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{12 \, e^{3}} \]

[In]

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

[Out]

1/2*(-e^2*x^2 + d^2)^(5/2)*d^2/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3) + 5/2*(-e^2*x^2 + d^2)^(3/2)*d^
3/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) - 15*sqrt(-e^2*x^2 + d^2)*d^4/(e^4*x + d*e^3) - 2/3*(-e^2*x^2 + d^2)^(5/2)*d
/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) - 5/3*(-e^2*x^2 + d^2)^(3/2)*d^2/(e^4*x + d*e^3) - 5/8*I*d^4*arcsin(e*x/d + 2
)/e^3 - 25/2*d^4*arcsin(e*x/d)/e^3 + 1/4*(-e^2*x^2 + d^2)^(5/2)/(e^4*x + d*e^3) + 5/8*sqrt(e^2*x^2 + 4*d*e*x +
3*d^2)*d^2*x/e^2 + 5/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^3/e^3 - 5*sqrt(-e^2*x^2 + d^2)*d^3/e^3 + 5/12*(-e^2*
x^2 + d^2)^(3/2)*d/e^3

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.64 \[ \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {95 \, d^{4} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, e^{2} {\left | e \right |}} + \frac {1}{24} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, x - \frac {16 \, d}{e}\right )} x + \frac {93 \, d^{2}}{e^{2}}\right )} x - \frac {256 \, d^{3}}{e^{3}}\right )} + \frac {16 \, d^{4}}{e^{2} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

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

[Out]

-95/8*d^4*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^2*abs(e)) + 1/24*sqrt(-e^2*x^2 + d^2)*((2*(3*x - 16*d/e)*x + 93*d^2/e
^2)*x - 256*d^3/e^3) + 16*d^4/(e^2*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs(e))

Mupad [F(-1)]

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

[In]

int((x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x)

[Out]

int((x^2*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4, x)