\(\int \frac {x}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\) [142]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 85 \[ \int \frac {x}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 x}{15 d^4 e \sqrt {d^2-e^2 x^2}} \]

[Out]

1/15*x/d^2/e/(-e^2*x^2+d^2)^(3/2)+1/5/e^2/(e*x+d)/(-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 = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {807, 198, 197} \[ \int \frac {x}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 x}{15 d^4 e \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

x/(15*d^2*e*(d^2 - e^2*x^2)^(3/2)) + 1/(5*e^2*(d + e*x)*(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 807

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 e} \\ & = \frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 e^2 (d+e x) \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 {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 e^2 (d+e x) \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 = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int \frac {x}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (3 d^4+3 d^3 e x+3 d^2 e^2 x^2-2 d e^3 x^3-2 e^4 x^4\right )}{15 d^4 e^2 (d-e x)^2 (d+e x)^3} \]

[In]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82

method result size
gosper \(\frac {\left (-e x +d \right ) \left (-2 e^{4} x^{4}-2 d \,e^{3} x^{3}+3 d^{2} e^{2} x^{2}+3 d^{3} e x +3 d^{4}\right )}{15 d^{4} e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\) \(70\)
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 \(\frac {\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{e}-\frac {d \left (-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{6 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}\right )}{e^{2}}\) \(212\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (73) = 146\).

Time = 0.31 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.01 \[ \int \frac {x}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {3 \, e^{5} x^{5} + 3 \, d e^{4} x^{4} - 6 \, d^{2} e^{3} x^{3} - 6 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x + 3 \, d^{5} - {\left (2 \, e^{4} x^{4} + 2 \, d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} - 3 \, d^{3} e x - 3 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{4} e^{7} x^{5} + d^{5} e^{6} x^{4} - 2 \, d^{6} e^{5} x^{3} - 2 \, d^{7} e^{4} x^{2} + d^{8} e^{3} x + d^{9} e^{2}\right )}} \]

[In]

integrate(x/(e*x+d)/(-e^2*x^2+d^2)^(5/2),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 [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \frac {x}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {1}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{2}\right )}} + \frac {x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e} + \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} e} \]

[In]

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

[Out]

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

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.79 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {x}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (3\,d^4+3\,d^3\,e\,x+3\,d^2\,e^2\,x^2-2\,d\,e^3\,x^3-2\,e^4\,x^4\right )}{15\,d^4\,e^2\,{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^2} \]

[In]

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