\(\int \frac {x (d+e x)}{(d^2-e^2 x^2)^{7/2}} \, dx\) [25]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \frac {x (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d+e x}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{15 d^4 e \sqrt {d^2-e^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {792, 198, 197} \[ \int \frac {x (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {d+e x}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 x}{15 d^4 e \sqrt {d^2-e^2 x^2}} \]

[In]

Int[(x*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d + e*x)/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - x/(15*d^2*e*(d^2 - e^2*x^2)^(3/2)) - (2*x)/(15*d^4*e*Sqrt[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 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {d+e x}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 e} \\ & = \frac {d+e x}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2 e} \\ & = \frac {d+e x}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{15 d^4 e \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {x (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (3 d^4-3 d^3 e x+3 d^2 e^2 x^2+2 d e^3 x^3-2 e^4 x^4\right )}{15 d^4 e^2 (d-e x)^3 (d+e x)^2} \]

[In]

Integrate[(x*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (e x +d \right )^{2} \left (-2 e^{4} x^{4}+2 d \,e^{3} x^{3}+3 d^{2} e^{2} x^{2}-3 d^{3} e x +3 d^{4}\right )}{15 d^{4} e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(77\)
trager	\(\frac {\left (-2 e^{4} x^{4}+2 d \,e^{3} x^{3}+3 d^{2} e^{2} x^{2}-3 d^{3} e x +3 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{4} \left (-e x +d \right )^{3} \left (e x +d \right )^{2} e^{2}}\)	\(79\)
default	\(e \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )+\frac {d}{5 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\)	\(120\)

[In]

int(x*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(-e*x+d)*(e*x+d)^2*(-2*e^4*x^4+2*d*e^3*x^3+3*d^2*e^2*x^2-3*d^3*e*x+3*d^4)/d^4/e^2/(-e^2*x^2+d^2)^(7/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (71) = 142\).

Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.07 \[ \int \frac {x (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {3 \, e^{5} x^{5} - 3 \, d e^{4} x^{4} - 6 \, d^{2} e^{3} x^{3} + 6 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x - 3 \, d^{5} + {\left (2 \, e^{4} x^{4} - 2 \, d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x - 3 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{4} e^{7} x^{5} - d^{5} e^{6} x^{4} - 2 \, d^{6} e^{5} x^{3} + 2 \, d^{7} e^{4} x^{2} + d^{8} e^{3} x - d^{9} e^{2}\right )}} \]

[In]

integrate(x*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 6.60 (sec) , antiderivative size = 432, normalized size of antiderivative = 5.20 \[ \int \frac {x (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=d \left (\begin {cases} \frac {1}{5 d^{4} e^{2} \sqrt {d^{2} - e^{2} x^{2}} - 10 d^{2} e^{4} x^{2} \sqrt {d^{2} - e^{2} x^{2}} + 5 e^{6} x^{4} \sqrt {d^{2} - e^{2} x^{2}}} & \text {for}\: e \neq 0 \\\frac {x^{2}}{2 \left (d^{2}\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} - \frac {5 i d^{2} x^{3}}{15 d^{9} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} - 30 d^{7} e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} + 15 d^{5} e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {2 i e^{2} x^{5}}{15 d^{9} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} - 30 d^{7} e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} + 15 d^{5} e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {5 d^{2} x^{3}}{15 d^{9} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} - 30 d^{7} e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} + 15 d^{5} e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {2 e^{2} x^{5}}{15 d^{9} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} - 30 d^{7} e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} + 15 d^{5} e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate(x*(e*x+d)/(-e**2*x**2+d**2)**(7/2),x)

[Out]

d*Piecewise((1/(5*d**4*e**2*sqrt(d**2 - e**2*x**2) - 10*d**2*e**4*x**2*sqrt(d**2 - e**2*x**2) + 5*e**6*x**4*sq
rt(d**2 - e**2*x**2)), Ne(e, 0)), (x**2/(2*(d**2)**(7/2)), True)) + e*Piecewise((-5*I*d**2*x**3/(15*d**9*sqrt(
-1 + e**2*x**2/d**2) - 30*d**7*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 15*d**5*e**4*x**4*sqrt(-1 + e**2*x**2/d**
2)) + 2*I*e**2*x**5/(15*d**9*sqrt(-1 + e**2*x**2/d**2) - 30*d**7*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 15*d**5
*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (5*d**2*x**3/(15*d**9*sqrt(1 - e**2*x**2/d**2
) - 30*d**7*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**5*e**4*x**4*sqrt(1 - e**2*x**2/d**2)) - 2*e**2*x**5/(15
*d**9*sqrt(1 - e**2*x**2/d**2) - 30*d**7*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**5*e**4*x**4*sqrt(1 - e**2*
x**2/d**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05 \[ \int \frac {x (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e} - \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} e} \]

[In]

integrate(x*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )} x}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(x*(e*x+d)/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((e*x + d)*x/(-e^2*x^2 + d^2)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 11.53 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \frac {x (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (3\,d^4-3\,d^3\,e\,x+3\,d^2\,e^2\,x^2+2\,d\,e^3\,x^3-2\,e^4\,x^4\right )}{15\,d^4\,e^2\,{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^3} \]

[In]

int((x*(d + e*x))/(d^2 - e^2*x^2)^(7/2),x)

[Out]

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