3.1.100 \(\int \frac {\sqrt {d^2-e^2 x^2}}{x^4 (d+e x)} \, dx\)

Optimal. Leaf size=114 \[ \frac {e \sqrt {d^2-e^2 x^2}}{2 d^2 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{3 d x^3}-\frac {2 e^2 \sqrt {d^2-e^2 x^2}}{3 d^3 x}+\frac {e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^3} \]

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Rubi [A]  time = 0.11, antiderivative size = 114, 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 = {850, 835, 807, 266, 63, 208} \begin {gather*} -\frac {2 e^2 \sqrt {d^2-e^2 x^2}}{3 d^3 x}+\frac {e \sqrt {d^2-e^2 x^2}}{2 d^2 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{3 d x^3}+\frac {e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d^2 - e^2*x^2]/(x^4*(d + e*x)),x]

[Out]

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

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

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Mathematica [A]  time = 0.11, size = 84, normalized size = 0.74 \begin {gather*} \frac {\left (-2 d^2+3 d e x-4 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}+3 e^3 x^3 \log \left (\sqrt {d^2-e^2 x^2}+d\right )-3 e^3 x^3 \log (x)}{6 d^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d^2 - e^2*x^2]/(x^4*(d + e*x)),x]

[Out]

((-2*d^2 + 3*d*e*x - 4*e^2*x^2)*Sqrt[d^2 - e^2*x^2] - 3*e^3*x^3*Log[x] + 3*e^3*x^3*Log[d + Sqrt[d^2 - e^2*x^2]
])/(6*d^3*x^3)

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IntegrateAlgebraic [A]  time = 0.38, size = 137, normalized size = 1.20 \begin {gather*} \frac {\left (-2 d^2+3 d e x-4 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{6 d^3 x^3}+\frac {e^3 \log \left (\sqrt {d^2-e^2 x^2}+d-\sqrt {-e^2} x\right )}{2 d^3}-\frac {e^3 \log \left (d^4+d^3 \sqrt {-e^2} x-d^3 \sqrt {d^2-e^2 x^2}\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[d^2 - e^2*x^2]/(x^4*(d + e*x)),x]

[Out]

((-2*d^2 + 3*d*e*x - 4*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(6*d^3*x^3) + (e^3*Log[d - Sqrt[-e^2]*x + Sqrt[d^2 - e^2*
x^2]])/(2*d^3) - (e^3*Log[d^4 + d^3*Sqrt[-e^2]*x - d^3*Sqrt[d^2 - e^2*x^2]])/(2*d^3)

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fricas [A]  time = 0.40, size = 75, normalized size = 0.66 \begin {gather*} -\frac {3 \, e^{3} x^{3} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (4 \, e^{2} x^{2} - 3 \, d e x + 2 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, d^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^4/(e*x+d),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^4/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/24*((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*
exp(2))*exp(1))/x/exp(2))^2*(12*exp(1)^4*exp(2)^2-3*exp(2)^4)+exp(2)^4+3/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))
*exp(1))*exp(2)^4/x/exp(2))/d^3/(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3/exp(1)^5+1/512*(
64*d^6*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^10*exp(2)^7-64/3*d^6*(-1/2*(-2*d*e
xp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^8*exp(2)^8+96*d^6*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*
exp(1))*exp(1)^8*exp(2)^8/x/exp(2)-128*d^6*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^10*exp(2)^7/x/ex
p(2)+128*d^6*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^12*exp(2)^6/x/exp(2))/d^9/exp(1)^15/exp(2)^3+1
/2*(4*exp(2)^3-4*exp(1)^4*exp(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^3/sqrt(-exp(1)^4+exp(2)^2)/exp(1)+1/2*(-exp(2)^3+2*exp(1)^4*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^3/exp(1)/exp(2)

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maple [B]  time = 0.01, size = 280, normalized size = 2.46 \begin {gather*} \frac {e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}\, d^{2}}+\frac {e^{4} \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^{3}}-\frac {e^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, d^{3}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{4} x}{d^{5}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{3}}{2 d^{4}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{3}}{d^{4}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}{d^{5} x}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}{2 d^{4} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3 d^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(1/2)/x^4/(e*x+d),x)

[Out]

-1/d^5*e^2/x*(-e^2*x^2+d^2)^(3/2)-1/d^5*e^4*x*(-e^2*x^2+d^2)^(1/2)-1/d^3*e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-
e^2*x^2+d^2)^(1/2)*x)+1/2*e/d^4/x^2*(-e^2*x^2+d^2)^(3/2)-1/2*e^3/d^4*(-e^2*x^2+d^2)^(1/2)+1/2*e^3/d^2/(d^2)^(1
/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/3/d^3/x^3*(-e^2*x^2+d^2)^(3/2)+1/d^4*e^3*(2*(x+d/e)*d*e
-(x+d/e)^2*e^2)^(1/2)+1/d^3*e^4/(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 \begin {gather*} \int \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^4/(e*x+d),x, algorithm="maxima")

[Out]

integrate(sqrt(-e^2*x^2 + d^2)/((e*x + d)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d^2-e^2\,x^2}}{x^4\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^(1/2)/(x^4*(d + e*x)),x)

[Out]

int((d^2 - e^2*x^2)^(1/2)/(x^4*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{x^{4} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(1/2)/x**4/(e*x+d),x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))/(x**4*(d + e*x)), x)

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