\(\int \frac {x^3}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\) [140]

Optimal result

Integrand size = 27, antiderivative size = 91 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 d-3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x}{5 d^2 e^3 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {864, 833, 792, 197} \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 d-3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x}{5 d^2 e^3 \sqrt {d^2-e^2 x^2}} \]

[In]

Int[x^3/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(x^2*(d - e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (2*d - 3*e*x)/(15*e^4*(d^2 - e^2*x^2)^(3/2)) - x/(5*d^2*e^3*Sq
rt[d^2 - e^2*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 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 )^{7/2}} \, dx \\ & = \frac {x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {x \left (2 d^3-3 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2} \\ & = \frac {x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 d-3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{5 e^3} \\ & = \frac {x^2 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 d-3 e x}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {x}{5 d^2 e^3 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-2 d^4-2 d^3 e x+3 d^2 e^2 x^2+3 d e^3 x^3+3 e^4 x^4\right )}{15 d^2 e^4 (d-e x)^2 (d+e x)^3} \]

[In]

Integrate[x^3/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.77

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (-3 e^{4} x^{4}-3 d \,e^{3} x^{3}-3 d^{2} e^{2} x^{2}+2 d^{3} e x +2 d^{4}\right )}{15 d^{2} e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\)	\(70\)
trager	\(-\frac {\left (-3 e^{4} x^{4}-3 d \,e^{3} x^{3}-3 d^{2} e^{2} x^{2}+2 d^{3} e x +2 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{2} e^{4} \left (e x +d \right )^{3} \left (-e x +d \right )^{2}}\)	\(79\)
default	\(\frac {\frac {x}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2}}}{e}+\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{3}}-\frac {d}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{3} \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{6 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{4}}\)	\(311\)

[In]

int(x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (80) = 160\).

Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.88 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, e^{5} x^{5} + 2 \, d e^{4} x^{4} - 4 \, d^{2} e^{3} x^{3} - 4 \, d^{3} e^{2} x^{2} + 2 \, d^{4} e x + 2 \, d^{5} - {\left (3 \, e^{4} x^{4} + 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} - 2 \, d^{3} e x - 2 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{2} e^{9} x^{5} + d^{3} e^{8} x^{4} - 2 \, d^{4} e^{7} x^{3} - 2 \, d^{5} e^{6} x^{2} + d^{6} e^{5} x + d^{7} e^{4}\right )}} \]

[In]

integrate(x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(2*e^5*x^5 + 2*d*e^4*x^4 - 4*d^2*e^3*x^3 - 4*d^3*e^2*x^2 + 2*d^4*e*x + 2*d^5 - (3*e^4*x^4 + 3*d*e^3*x^3
+ 3*d^2*e^2*x^2 - 2*d^3*e*x - 2*d^4)*sqrt(-e^2*x^2 + d^2))/(d^2*e^9*x^5 + d^3*e^8*x^4 - 2*d^4*e^7*x^3 - 2*d^5*
e^6*x^2 + d^6*e^5*x + d^7*e^4)

Sympy [F]

\[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \]

[In]

integrate(x**3/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(x**3/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.21 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {d^{2}}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{4}\right )}} + \frac {2 \, x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} - \frac {d}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{3}} \]

[In]

integrate(x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int { \frac {x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )}} \,d x } \]

[In]

integrate(x^3/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

integrate(x^3/((-e^2*x^2 + d^2)^(5/2)*(e*x + d)), x)

Mupad [B] (verification not implemented)

Time = 12.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.86 \[ \int \frac {x^3}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (-2\,d^4-2\,d^3\,e\,x+3\,d^2\,e^2\,x^2+3\,d\,e^3\,x^3+3\,e^4\,x^4\right )}{15\,d^2\,e^4\,{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^2} \]

[In]

int(x^3/((d^2 - e^2*x^2)^(5/2)*(d + e*x)),x)

[Out]

((d^2 - e^2*x^2)^(1/2)*(3*e^4*x^4 - 2*d^4 + 3*d*e^3*x^3 + 3*d^2*e^2*x^2 - 2*d^3*e*x))/(15*d^2*e^4*(d + e*x)^3*
(d - e*x)^2)