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

Optimal. Leaf size=143 \[ -\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}+\frac {e \sqrt {d^2-e^2 x^2}}{3 d^2 x^3}-\frac {3 e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d^4}+\frac {2 e^3 \sqrt {d^2-e^2 x^2}}{3 d^4 x}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{8 d^3 x^2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {850, 835, 807, 266, 63, 208} \[ \frac {2 e^3 \sqrt {d^2-e^2 x^2}}{3 d^4 x}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{8 d^3 x^2}+\frac {e \sqrt {d^2-e^2 x^2}}{3 d^2 x^3}-\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}-\frac {3 e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 807

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

Rule 835

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

Rule 850

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*} \int \frac {\sqrt {d^2-e^2 x^2}}{x^5 (d+e x)} \, dx &=\int \frac {d-e x}{x^5 \sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}-\frac {\int \frac {4 d^2 e-3 d e^2 x}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{4 d^2}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}+\frac {e \sqrt {d^2-e^2 x^2}}{3 d^2 x^3}+\frac {\int \frac {9 d^3 e^2-8 d^2 e^3 x}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{12 d^4}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}+\frac {e \sqrt {d^2-e^2 x^2}}{3 d^2 x^3}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{8 d^3 x^2}-\frac {\int \frac {16 d^4 e^3-9 d^3 e^4 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{24 d^6}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}+\frac {e \sqrt {d^2-e^2 x^2}}{3 d^2 x^3}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{8 d^3 x^2}+\frac {2 e^3 \sqrt {d^2-e^2 x^2}}{3 d^4 x}+\frac {\left (3 e^4\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{8 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}+\frac {e \sqrt {d^2-e^2 x^2}}{3 d^2 x^3}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{8 d^3 x^2}+\frac {2 e^3 \sqrt {d^2-e^2 x^2}}{3 d^4 x}+\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{16 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}+\frac {e \sqrt {d^2-e^2 x^2}}{3 d^2 x^3}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{8 d^3 x^2}+\frac {2 e^3 \sqrt {d^2-e^2 x^2}}{3 d^4 x}-\frac {\left (3 e^2\right ) \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 )}{8 d^3}\\ &=-\frac {\sqrt {d^2-e^2 x^2}}{4 d x^4}+\frac {e \sqrt {d^2-e^2 x^2}}{3 d^2 x^3}-\frac {3 e^2 \sqrt {d^2-e^2 x^2}}{8 d^3 x^2}+\frac {2 e^3 \sqrt {d^2-e^2 x^2}}{3 d^4 x}-\frac {3 e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d^4}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 95, normalized size = 0.66 \[ \frac {-9 e^4 x^4 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\sqrt {d^2-e^2 x^2} \left (-6 d^3+8 d^2 e x-9 d e^2 x^2+16 e^3 x^3\right )+9 e^4 x^4 \log (x)}{24 d^4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-6*d^3 + 8*d^2*e*x - 9*d*e^2*x^2 + 16*e^3*x^3) + 9*e^4*x^4*Log[x] - 9*e^4*x^4*Log[d + Sq
rt[d^2 - e^2*x^2]])/(24*d^4*x^4)

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fricas [A]  time = 0.89, size = 86, normalized size = 0.60 \[ \frac {9 \, e^{4} x^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (16 \, e^{3} x^{3} - 9 \, d e^{2} x^{2} + 8 \, d^{2} e x - 6 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, d^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*e^4*x^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (16*e^3*x^3 - 9*d*e^2*x^2 + 8*d^2*e*x - 6*d^3)*sqrt(-e^2*
x^2 + d^2))/(d^4*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/192*((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^3*(-96*exp(1)^6*exp(2)^2+96*exp(1)^4*exp(2)^3-72*exp(2)^5)+24*exp(1)^4*exp(2)^3*(-1
/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2+3*exp(2)^5+4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*ex
p(1))*exp(2)^5/x/exp(2))/d^4/(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4/exp(1)^6+1/65536*(-
8192*d^12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^22*exp(2)^7+8192/3*d^12*(-1/2*(
-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^20*exp(2)^8-1024*d^12*(-1/2*(-2*d*exp(1)-2*sqrt(
d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^18*exp(2)^9+8192*d^12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp
(1))/x/exp(2))^2*exp(1)^20*exp(2)^8-8192*d^12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*ex
p(1)^18*exp(2)^9-12288*d^12*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^20*exp(2)^8/x/exp(2)+16384*d^12
*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^22*exp(2)^7/x/exp(2)-16384*d^12*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))*exp(1)^24*exp(2)^6/x/exp(2))/d^16/exp(1)^24/exp(2)^4+1/2*(-4*exp(1)^3*exp(2)^2+4*exp(1)^5*ex
p(2))*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/d^4/sqrt(-exp
(1)^4+exp(2)^2)/exp(1)+1/8*(8*exp(1)^6*exp(2)^2-4*exp(1)^4*exp(2)^3+exp(2)^5-8*exp(1)^8*exp(2))*ln(1/2*abs(-2*
d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/d^4/exp(1)^5/exp(1)

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maple [B]  time = 0.01, size = 304, normalized size = 2.13 \[ -\frac {3 e^{4} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}\, d^{3}}-\frac {e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{\sqrt {e^{2}}\, d^{4}}+\frac {e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, d^{4}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{5} x}{d^{6}}+\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}{8 d^{5}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{4}}{d^{5}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}{d^{6} x}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}{8 d^{5} x^{2}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}{3 d^{4} x^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4 d^{3} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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