\(\int \frac {x^3}{(d+e x)^2 (d^2-e^2 x^2)^{3/2}} \, dx\) [172]

Optimal result

Integrand size = 27, antiderivative size = 99 \[ \int \frac {x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d-2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {866, 1649, 651} \[ \int \frac {x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d-2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

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

Rule 651

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((-a)*e + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

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)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx \\ & = \frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x) \left (-\frac {2 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)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {-\frac {6 d^3}{e^3}+\frac {15 d^2 x}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2} \\ & = \frac {d^2 (d-e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d-e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d-2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int \frac {x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (2 d^3+4 d^2 e x+d e^2 x^2-2 e^3 x^3\right )}{5 d e^4 (d-e x) (d+e x)^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.66

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (-2 e^{3} x^{3}+d \,e^{2} x^{2}+4 d^{2} e x +2 d^{3}\right )}{5 \left (e x +d \right ) d \,e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\)	\(65\)
trager	\(\frac {\left (-2 e^{3} x^{3}+d \,e^{2} x^{2}+4 d^{2} e x +2 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{5 d \,e^{4} \left (e x +d \right )^{3} \left (-e x +d \right )}\)	\(67\)
default	\(\frac {1}{\sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}-\frac {2 x}{d \,e^{3} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {d^{3} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right )^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}+\frac {3 e \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{5}}+\frac {3 d^{2} \left (-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{e^{4}}\)	\(309\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \frac {x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {2 \, e^{4} x^{4} + 4 \, d e^{3} x^{3} - 4 \, d^{3} e x - 2 \, d^{4} + {\left (2 \, e^{3} x^{3} - d e^{2} x^{2} - 4 \, d^{2} e x - 2 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d e^{8} x^{4} + 2 \, d^{2} e^{7} x^{3} - 2 \, d^{4} e^{5} x - d^{5} e^{4}\right )}} \]

[In]

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

[Out]

1/5*(2*e^4*x^4 + 4*d*e^3*x^3 - 4*d^3*e*x - 2*d^4 + (2*e^3*x^3 - d*e^2*x^2 - 4*d^2*e*x - 2*d^3)*sqrt(-e^2*x^2 +
d^2))/(d*e^8*x^4 + 2*d^2*e^7*x^3 - 2*d^4*e^5*x - d^5*e^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.59 \[ \int \frac {x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {d^{2}}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{6} x^{2} + 2 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{5} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{4}\right )}} - \frac {4 \, d}{5 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} e^{5} x + \sqrt {-e^{2} x^{2} + d^{2}} d e^{4}\right )}} - \frac {2 \, x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{3}} + \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} e^{4}} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.93 \[ \int \frac {x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {\frac {16 i \, \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d e^{3}} - \frac {5}{d e^{3} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )} - \frac {d^{4} e^{12} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4} - 5 \, d^{4} e^{12} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4} + 15 \, d^{4} e^{12} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{4} \mathrm {sgn}\left (e\right )^{4}}{d^{5} e^{15} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{5} \mathrm {sgn}\left (e\right )^{5}}}{40 \, {\left | e \right |}} \]

[In]

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

[Out]

-1/40*(16*I*sgn(1/(e*x + d))*sgn(e)/(d*e^3) - 5/(d*e^3*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e)) - (d^4
*e^12*(2*d/(e*x + d) - 1)^(5/2)*sgn(1/(e*x + d))^4*sgn(e)^4 - 5*d^4*e^12*(2*d/(e*x + d) - 1)^(3/2)*sgn(1/(e*x
+ d))^4*sgn(e)^4 + 15*d^4*e^12*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))^4*sgn(e)^4)/(d^5*e^15*sgn(1/(e*x + d))
^5*sgn(e)^5))/abs(e)

Mupad [B] (verification not implemented)

Time = 11.48 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.67 \[ \int \frac {x^3}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^3+4\,d^2\,e\,x+d\,e^2\,x^2-2\,e^3\,x^3\right )}{5\,d\,e^4\,{\left (d+e\,x\right )}^3\,\left (d-e\,x\right )} \]

[In]

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

[Out]

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