3.144 \(\int \frac {1}{x (d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=119 \[ \frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}+\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {857, 823, 12, 266, 63, 208} \[ \frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(5*d - 4*e*x)/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + 1/(5*d^2*(d + e*x)*(d^2 - e^2*x^2)^(3/2)) + (15*d - 8*e*x)/(15*
d^6*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^6

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

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

Rubi steps

\begin {align*} \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-5 d e^2+4 e^3 x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2}\\ &=\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-15 d^3 e^4+8 d^2 e^5 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^4}\\ &=\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {15 d^5 e^6}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^6}\\ &=\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^5}\\ &=\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^5}\\ &=\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{d^5 e^2}\\ &=\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 106, normalized size = 0.89 \[ \frac {-15 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (23 d^4+8 d^3 e x-27 d^2 e^2 x^2-7 d e^3 x^3+8 e^4 x^4\right )}{(d-e x)^2 (d+e x)^3}+15 \log (x)}{15 d^6} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(23*d^4 + 8*d^3*e*x - 27*d^2*e^2*x^2 - 7*d*e^3*x^3 + 8*e^4*x^4))/((d - e*x)^2*(d + e*x)^
3) + 15*Log[x] - 15*Log[d + Sqrt[d^2 - e^2*x^2]])/(15*d^6)

________________________________________________________________________________________

fricas [B]  time = 0.88, size = 237, normalized size = 1.99 \[ \frac {23 \, e^{5} x^{5} + 23 \, d e^{4} x^{4} - 46 \, d^{2} e^{3} x^{3} - 46 \, d^{3} e^{2} x^{2} + 23 \, d^{4} e x + 23 \, d^{5} + 15 \, {\left (e^{5} x^{5} + d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} - 2 \, d^{3} e^{2} x^{2} + d^{4} e x + d^{5}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (8 \, e^{4} x^{4} - 7 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} + 8 \, d^{3} e x + 23 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{6} e^{5} x^{5} + d^{7} e^{4} x^{4} - 2 \, d^{8} e^{3} x^{3} - 2 \, d^{9} e^{2} x^{2} + d^{10} e x + d^{11}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

1/15*(23*e^5*x^5 + 23*d*e^4*x^4 - 46*d^2*e^3*x^3 - 46*d^3*e^2*x^2 + 23*d^4*e*x + 23*d^5 + 15*(e^5*x^5 + d*e^4*
x^4 - 2*d^2*e^3*x^3 - 2*d^3*e^2*x^2 + d^4*e*x + d^5)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (8*e^4*x^4 - 7*d*e^3
*x^3 - 27*d^2*e^2*x^2 + 8*d^3*e*x + 23*d^4)*sqrt(-e^2*x^2 + d^2))/(d^6*e^5*x^5 + d^7*e^4*x^4 - 2*d^8*e^3*x^3 -
2*d^9*e^2*x^2 + d^10*e*x + d^11)

________________________________________________________________________________________

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(1/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

________________________________________________________________________________________

maple [A]  time = 0.02, size = 196, normalized size = 1.65 \[ -\frac {4 e x}{15 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{4}}+\frac {1}{5 \left (x +\frac {d}{e}\right ) \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{2} e}+\frac {1}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{5}}-\frac {8 e x}{15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{6}}+\frac {1}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x\,{\left (d^2-e^2\,x^2\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(1/(x*(-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

________________________________________________________________________________________