\(\int \frac {x^3}{(d+e x) (d^2-e^2 x^2)^{3/2}} \, dx\) [129]

Optimal result

Integrand size = 27, antiderivative size = 89 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 d-3 e x}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

1/3*x^2*(-e*x+d)/e^2/(-e^2*x^2+d^2)^(3/2)-arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4+1/3*(3*e*x-2*d)/e^4/(-e^2*x^2+d
^2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {864, 833, 792, 223, 209} \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}+\frac {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 d-3 e x}{3 e^4 \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

(x^2*(d - e*x))/(3*e^2*(d^2 - e^2*x^2)^(3/2)) - (2*d - 3*e*x)/(3*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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 864

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3 (d-e x)}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx \\ & = \frac {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {x \left (2 d^3-3 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{3 d^2 e^2} \\ & = \frac {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 d-3 e x}{3 e^4 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^3} \\ & = \frac {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 d-3 e x}{3 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 {x^2 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 d-3 e x}{3 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 = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (-2 d^2+d e x+4 e^2 x^2\right )}{(d-e x) (d+e x)^2}+6 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{3 e^4} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(203\) vs. \(2(79)=158\).

Time = 0.38 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.29

method	result	size
default	\(\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e}+\frac {x}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}-\frac {d}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {d^{3} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{4}}\)	\(204\)

[In]

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

[Out]

1/e*(x/e^2/(-e^2*x^2+d^2)^(1/2)-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))+1/(-e^2*x^2+d^2)
^(1/2)/e^3*x-d/e^4/(-e^2*x^2+d^2)^(1/2)-d^3/e^4*(-1/3/d/e/(x+d/e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/3/e/d
^3*(-2*(x+d/e)*e^2+2*d*e)/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.76 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, e^{3} x^{3} + 2 \, d e^{2} x^{2} - 2 \, d^{2} e x - 2 \, d^{3} - 6 \, {\left (e^{3} x^{3} + d e^{2} x^{2} - d^{2} e x - d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (4 \, e^{2} x^{2} + d e x - 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (e^{7} x^{3} + d e^{6} x^{2} - d^{2} e^{5} x - d^{3} e^{4}\right )}} \]

[In]

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

[Out]

-1/3*(2*e^3*x^3 + 2*d*e^2*x^2 - 2*d^2*e*x - 2*d^3 - 6*(e^3*x^3 + d*e^2*x^2 - d^2*e*x - d^3)*arctan(-(d - sqrt(
-e^2*x^2 + d^2))/(e*x)) + (4*e^2*x^2 + d*e*x - 2*d^2)*sqrt(-e^2*x^2 + d^2))/(e^7*x^3 + d*e^6*x^2 - d^2*e^5*x -
d^3*e^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {d^{2}}{3 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{5} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{4}\right )}} + \frac {4 \, x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{3}} - \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{4}} - \frac {d}{\sqrt {-e^{2} x^{2} + d^{2}} e^{4}} \]

[In]

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

[Out]

1/3*d^2/(sqrt(-e^2*x^2 + d^2)*e^5*x + sqrt(-e^2*x^2 + d^2)*d*e^4) + 4/3*x/(sqrt(-e^2*x^2 + d^2)*e^3) - arcsin(
e*x/d)/e^4 - d/(sqrt(-e^2*x^2 + d^2)*e^4)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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