\(\int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [181]

Optimal result

Integrand size = 27, antiderivative size = 120 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

1/5*d^2*(-e*x+d)^3/e^4/(-e^2*x^2+d^2)^(5/2)-13/15*d*(-e*x+d)^2/e^4/(-e^2*x^2+d^2)^(3/2)+arctan(e*x/(-e^2*x^2+d
^2)^(1/2))/e^4+32/15*(-e*x+d)/e^4/(-e^2*x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {866, 1649, 792, 223, 209} \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}+\frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}} \]

[In]

Int[x^3/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(d^2*(d - e*x)^3)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (13*d*(d - e*x)^2)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (32*(d -
e*x))/(15*e^4*Sqrt[d^2 - e^2*x^2]) + ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e^4

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 792

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

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^3 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx \\ & = \frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x)^2 \left (-\frac {3 d^3}{e^3}+\frac {5 d^2 x}{e^2}-\frac {5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d} \\ & = \frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\left (-\frac {17 d^3}{e^3}+\frac {15 d^2 x}{e^2}\right ) (d-e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2} \\ & = \frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^3} \\ & = \frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \\ & = \frac {d^2 (d-e x)^3}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {13 d (d-e x)^2}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {32 (d-e x)}{15 e^4 \sqrt {d^2-e^2 x^2}}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (22 d^2+51 d e x+32 e^2 x^2\right )}{15 e^4 (d+e x)^3}-\frac {2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[In]

Integrate[x^3/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(22*d^2 + 51*d*e*x + 32*e^2*x^2))/(15*e^4*(d + e*x)^3) - (2*ArcTan[(e*x)/(Sqrt[d^2] - Sqr
t[d^2 - e^2*x^2])])/e^4

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(318\) vs. \(2(106)=212\).

Time = 0.43 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.66

method	result	size
default	\(\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{3} \sqrt {e^{2}}}+\frac {3 d^{2} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{e^{5}}+\frac {3 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{5} \left (x +\frac {d}{e}\right )}-\frac {d^{3} \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}\right )}{e^{6}}\)	\(319\)

[In]

int(x^3/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.31 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {22 \, e^{3} x^{3} + 66 \, d e^{2} x^{2} + 66 \, d^{2} e x + 22 \, d^{3} - 30 \, {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (32 \, e^{2} x^{2} + 51 \, d e x + 22 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]

[In]

integrate(x^3/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

1/15*(22*e^3*x^3 + 66*d*e^2*x^2 + 66*d^2*e*x + 22*d^3 - 30*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (32*e^2*x^2 + 51*d*e*x + 22*d^2)*sqrt(-e^2*x^2 + d^2))/(e^7*x^3 + 3*d*e^6*x
^2 + 3*d^2*e^5*x + d^3*e^4)

Sympy [F]

\[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^{3}}{\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]

[In]

integrate(x**3/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(x**3/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.13 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{5 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} - \frac {13 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{15 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {32 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{5} x + d e^{4}\right )}} + \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{4}} \]

[In]

integrate(x^3/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(-e^2*x^2 + d^2)*d^2/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) - 13/15*sqrt(-e^2*x^2 + d^2)*d/(e
^6*x^2 + 2*d*e^5*x + d^2*e^4) + 32/15*sqrt(-e^2*x^2 + d^2)/(e^5*x + d*e^4) + arcsin(e*x/d)/e^4

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.54 \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{3} {\left | e \right |}} - \frac {2 \, {\left (\frac {95 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} + \frac {145 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {75 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} + 22\right )}}{15 \, e^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

integrate(x^3/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) - 2/15*(95*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 145*(d*e + s
qrt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 75*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) + 15*(d*e + sqrt(
-e^2*x^2 + d^2)*abs(e))^4/(e^8*x^4) + 22)/(e^3*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {x^3}{\sqrt {d^2-e^2\,x^2}\,{\left (d+e\,x\right )}^3} \,d x \]

[In]

int(x^3/((d^2 - e^2*x^2)^(1/2)*(d + e*x)^3),x)

[Out]

int(x^3/((d^2 - e^2*x^2)^(1/2)*(d + e*x)^3), x)