3.196 \(\int \frac{\sqrt{d^2-e^2 x^2}}{x^3 (d+e x)^4} \, dx\)

Optimal. Leaf size=183 \[ \frac{e^2 (135 d-164 e x)}{15 d^5 \sqrt{d^2-e^2 x^2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d^5 x}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^4 x^2}-\frac{19 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^5} \]

[Out]

(8*e^2*(d - e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + (4*e^2*(10*d - 13*e*x))/(15*d^3*(d^2 - e^2*x^2)^(3/2)) + (e^2*
(135*d - 164*e*x))/(15*d^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^4*x^2) + (4*e*Sqrt[d^2 - e^2*x^2])/
(d^5*x) - (19*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^5)

________________________________________________________________________________________

Rubi [A]  time = 0.391562, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {852, 1805, 1807, 807, 266, 63, 208} \[ \frac{e^2 (135 d-164 e x)}{15 d^5 \sqrt{d^2-e^2 x^2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d^5 x}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^4 x^2}-\frac{19 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*e^2*(d - e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + (4*e^2*(10*d - 13*e*x))/(15*d^3*(d^2 - e^2*x^2)^(3/2)) + (e^2*
(135*d - 164*e*x))/(15*d^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^4*x^2) + (4*e*Sqrt[d^2 - e^2*x^2])/
(d^5*x) - (19*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^5)

Rule 852

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

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 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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{d^2-e^2 x^2}}{x^3 (d+e x)^4} \, dx &=\int \frac{(d-e x)^4}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{-5 d^4+20 d^3 e x-35 d^2 e^2 x^2+32 d e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{15 d^4-60 d^3 e x+120 d^2 e^2 x^2-104 d e^3 x^3}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (135 d-164 e x)}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-15 d^4+60 d^3 e x-135 d^2 e^2 x^2}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (135 d-164 e x)}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^4 x^2}+\frac{\int \frac{-120 d^5 e+285 d^4 e^2 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (135 d-164 e x)}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^4 x^2}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d^5 x}+\frac{\left (19 e^2\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{2 d^4}\\ &=\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (135 d-164 e x)}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^4 x^2}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d^5 x}+\frac{\left (19 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^4}\\ &=\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (135 d-164 e x)}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^4 x^2}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d^5 x}-\frac{19 \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^4}\\ &=\frac{8 e^2 (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac{4 e^2 (10 d-13 e x)}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{e^2 (135 d-164 e x)}{15 d^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^4 x^2}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d^5 x}-\frac{19 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^5}\\ \end{align*}

Mathematica [A]  time = 0.242391, size = 107, normalized size = 0.58 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (713 d^2 e^2 x^2+75 d^3 e x-15 d^4+1059 d e^3 x^3+448 e^4 x^4\right )}{x^2 (d+e x)^3}-285 e^2 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+285 e^2 \log (x)}{30 d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(-15*d^4 + 75*d^3*e*x + 713*d^2*e^2*x^2 + 1059*d*e^3*x^3 + 448*e^4*x^4))/(x^2*(d + e*x)^
3) + 285*e^2*Log[x] - 285*e^2*Log[d + Sqrt[d^2 - e^2*x^2]])/(30*d^5)

________________________________________________________________________________________

Maple [B]  time = 0.073, size = 389, normalized size = 2.1 \begin{align*}{\frac{19\,{e}^{2}}{2\,{d}^{6}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{19\,{e}^{2}}{2\,{d}^{4}}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}-4\,{\frac{{e}^{2}}{{d}^{6}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-4\,{\frac{{e}^{3}}{{d}^{5}\sqrt{{e}^{2}}}\arctan \left ({\sqrt{{e}^{2}}x{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ) }+{\frac{1}{5\,{d}^{4}{e}^{2}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-4}}+{\frac{16}{15\,e{d}^{5}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-3}}+6\,{\frac{1}{{d}^{6}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{3/2} \left ({\frac{d}{e}}+x \right ) ^{-2}}-{\frac{1}{2\,{d}^{6}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+4\,{\frac{e \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{3/2}}{{d}^{7}x}}+4\,{\frac{{e}^{3}x\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{{d}^{7}}}+4\,{\frac{{e}^{3}}{{d}^{5}\sqrt{{e}^{2}}}\arctan \left ({\frac{\sqrt{{e}^{2}}x}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(1/2)/x^3/(e*x+d)^4,x)

[Out]

19/2/d^6*e^2*(-e^2*x^2+d^2)^(1/2)-19/2/d^4*e^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-4/
d^6*e^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)-4/d^5*e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*
(d/e+x))^(1/2))+1/5/e^2/d^4/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)+16/15/e/d^5/(d/e+x)^3*(-(d/e+x)^2*e
^2+2*d*e*(d/e+x))^(3/2)+6/d^6/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)-1/2/d^6/x^2*(-e^2*x^2+d^2)^(3/2)+
4/d^7*e/x*(-e^2*x^2+d^2)^(3/2)+4/d^7*e^3*x*(-e^2*x^2+d^2)^(1/2)+4/d^5*e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e
^2*x^2+d^2)^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-e^{2} x^{2} + d^{2}}}{{\left (e x + d\right )}^{4} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^3/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(-e^2*x^2 + d^2)/((e*x + d)^4*x^3), x)

________________________________________________________________________________________

Fricas [A]  time = 1.65852, size = 428, normalized size = 2.34 \begin{align*} \frac{398 \, e^{5} x^{5} + 1194 \, d e^{4} x^{4} + 1194 \, d^{2} e^{3} x^{3} + 398 \, d^{3} e^{2} x^{2} + 285 \,{\left (e^{5} x^{5} + 3 \, d e^{4} x^{4} + 3 \, d^{2} e^{3} x^{3} + d^{3} e^{2} x^{2}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (448 \, e^{4} x^{4} + 1059 \, d e^{3} x^{3} + 713 \, d^{2} e^{2} x^{2} + 75 \, d^{3} e x - 15 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{30 \,{\left (d^{5} e^{3} x^{5} + 3 \, d^{6} e^{2} x^{4} + 3 \, d^{7} e x^{3} + d^{8} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^3/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/30*(398*e^5*x^5 + 1194*d*e^4*x^4 + 1194*d^2*e^3*x^3 + 398*d^3*e^2*x^2 + 285*(e^5*x^5 + 3*d*e^4*x^4 + 3*d^2*e
^3*x^3 + d^3*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (448*e^4*x^4 + 1059*d*e^3*x^3 + 713*d^2*e^2*x^2 + 7
5*d^3*e*x - 15*d^4)*sqrt(-e^2*x^2 + d^2))/(d^5*e^3*x^5 + 3*d^6*e^2*x^4 + 3*d^7*e*x^3 + d^8*x^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (- d + e x\right ) \left (d + e x\right )}}{x^{3} \left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(1/2)/x**3/(e*x+d)**4,x)

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))/(x**3*(d + e*x)**4), x)

________________________________________________________________________________________

Giac [A]  time = 1.19897, size = 1, normalized size = 0.01 \begin{align*} +\infty \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(1/2)/x^3/(e*x+d)^4,x, algorithm="giac")

[Out]

+Infinity