\(\int \frac {x^3}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx\) [120]

Optimal result

Integrand size = 27, antiderivative size = 91 \[ \int \frac {x^3}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {3 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 91, 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, 794, 223, 209} \[ \int \frac {x^3}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {3 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}+\frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4} \]

[In]

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

[Out]

(x^2*(d - e*x))/(e^2*Sqrt[d^2 - e^2*x^2]) + ((4*d - 3*e*x)*Sqrt[d^2 - e^2*x^2])/(2*e^4) + (3*d^2*ArcTan[(e*x)/
Sqrt[d^2 - e^2*x^2]])/(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 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 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 )^{3/2}} \, dx \\ & = \frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {x \left (2 d^3-3 d^2 e x\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{d^2 e^2} \\ & = \frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {\left (3 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^3} \\ & = \frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {\left (3 d^2\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3} \\ & = \frac {x^2 (d-e x)}{e^2 \sqrt {d^2-e^2 x^2}}+\frac {(4 d-3 e x) \sqrt {d^2-e^2 x^2}}{2 e^4}+\frac {3 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98 \[ \int \frac {x^3}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (4 d^2+d e x-e^2 x^2\right )}{2 e^4 (d+e x)}-\frac {3 d^2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.19

method	result	size
risch	\(\frac {\left (-e x +2 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{4}}+\frac {3 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{3} \sqrt {e^{2}}}+\frac {d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{5} \left (x +\frac {d}{e}\right )}\)	\(108\)
default	\(\frac {-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}}{e}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{3} \sqrt {e^{2}}}+\frac {d \sqrt {-e^{2} x^{2}+d^{2}}}{e^{4}}+\frac {d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{5} \left (x +\frac {d}{e}\right )}\)	\(159\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11 \[ \int \frac {x^3}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {4 \, d^{2} e x + 4 \, d^{3} - 6 \, {\left (d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (e^{2} x^{2} - d e x - 4 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (e^{5} x + d e^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int \frac {x^3}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=\frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{5} x + d e^{4}} + \frac {3 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{4}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} x}{2 \, e^{3}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d}{e^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05 \[ \int \frac {x^3}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {x}{e^{3}} - \frac {2 \, d}{e^{4}}\right )} + \frac {3 \, d^{2} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, e^{3} {\left | e \right |}} - \frac {2 \, d^{2}}{e^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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