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

Optimal. Leaf size=85 \[ \frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 x}{15 d^4 e \sqrt {d^2-e^2 x^2}} \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {793, 192, 191} \[ \frac {2 x}{15 d^4 e \sqrt {d^2-e^2 x^2}}+\frac {x}{15 d^2 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 e^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[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 191

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 192

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 793

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

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Mathematica [A]  time = 0.05, size = 82, normalized size = 0.96 \[ \frac {\sqrt {d^2-e^2 x^2} \left (3 d^4+3 d^3 e x+3 d^2 e^2 x^2-2 d e^3 x^3-2 e^4 x^4\right )}{15 d^4 e^2 (d-e x)^2 (d+e x)^3} \]

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 0.69, size = 171, normalized size = 2.01 \[ \frac {3 \, e^{5} x^{5} + 3 \, d e^{4} x^{4} - 6 \, d^{2} e^{3} x^{3} - 6 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x + 3 \, d^{5} - {\left (2 \, e^{4} x^{4} + 2 \, d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} - 3 \, d^{3} e x - 3 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{4} e^{7} x^{5} + d^{5} e^{6} x^{4} - 2 \, d^{6} e^{5} x^{3} - 2 \, d^{7} e^{4} x^{2} + d^{8} e^{3} x + d^{9} e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [A]  time = 0.01, size = 70, normalized size = 0.82 \[ \frac {\left (-e x +d \right ) \left (-2 x^{4} e^{4}-2 x^{3} d \,e^{3}+3 d^{2} x^{2} e^{2}+3 d^{3} x e +3 d^{4}\right )}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{4} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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maxima [A]  time = 0.49, size = 90, normalized size = 1.06 \[ \frac {1}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{2}\right )}} + \frac {x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e} + \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.78, size = 78, normalized size = 0.92 \[ \frac {\sqrt {d^2-e^2\,x^2}\,\left (3\,d^4+3\,d^3\,e\,x+3\,d^2\,e^2\,x^2-2\,d\,e^3\,x^3-2\,e^4\,x^4\right )}{15\,d^4\,e^2\,{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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