3.206 \(\int \frac{(d^2-e^2 x^2)^{5/2}}{x^3 (d+e x)^4} \, dx\)

Optimal. Leaf size=110 \[ \frac{8 e^2 (d-e x)}{d \sqrt{d^2-e^2 x^2}}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d x}-\frac{\sqrt{d^2-e^2 x^2}}{2 x^2}-\frac{15 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.215146, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{8 e^2 (d-e x)}{d \sqrt{d^2-e^2 x^2}}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d x}-\frac{\sqrt{d^2-e^2 x^2}}{2 x^2}-\frac{15 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(x^3*(d + e*x)^4),x]

[Out]

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

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{\left (d^2-e^2 x^2\right )^{5/2}}{x^3 (d+e x)^4} \, dx &=\int \frac{(d-e x)^4}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac{8 e^2 (d-e x)}{d \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-d^4+4 d^3 e x-7 d^2 e^2 x^2}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{d^2}\\ &=\frac{8 e^2 (d-e x)}{d \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 x^2}+\frac{\int \frac{-8 d^5 e+15 d^4 e^2 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{2 d^4}\\ &=\frac{8 e^2 (d-e x)}{d \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 x^2}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d x}+\frac{1}{2} \left (15 e^2\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{8 e^2 (d-e x)}{d \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 x^2}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d x}+\frac{1}{4} \left (15 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )\\ &=\frac{8 e^2 (d-e x)}{d \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 x^2}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d x}-\frac{15}{2} \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 )\\ &=\frac{8 e^2 (d-e x)}{d \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{2 x^2}+\frac{4 e \sqrt{d^2-e^2 x^2}}{d x}-\frac{15 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.247161, size = 85, normalized size = 0.77 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (-d^2+7 d e x+24 e^2 x^2\right )}{x^2 (d+e x)}-15 e^2 \log \left (\sqrt{d^2-e^2 x^2}+d\right )+15 e^2 \log (x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(x^3*(d + e*x)^4),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(-d^2 + 7*d*e*x + 24*e^2*x^2))/(x^2*(d + e*x)) + 15*e^2*Log[x] - 15*e^2*Log[d + Sqrt[d^2
- e^2*x^2]])/(2*d)

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Maple [B]  time = 0.084, size = 504, normalized size = 4.6 \begin{align*} -{\frac{1}{2\,{d}^{6}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-2\,{\frac{1}{{d}^{6}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{7/2} \left ({\frac{d}{e}}+x \right ) ^{-2}}-4\,{\frac{{e}^{2}}{{d}^{6}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{5/2}}+{\frac{3\,{e}^{2}}{2\,{d}^{6}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{e}^{2}}{2\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{15\,{e}^{2}}{2\,{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{15\,{e}^{2}}{2}\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}}}}}-5\,{\frac{{e}^{3}x}{{d}^{5}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{3/2}}-{\frac{15\,{e}^{3}x}{2\,{d}^{3}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{15\,{e}^{3}}{2\,d}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{1}{{e}^{2}{d}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-4}}+5\,{\frac{{e}^{3}x \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{3/2}}{{d}^{5}}}+{\frac{15\,{e}^{3}x}{2\,{d}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{15\,{e}^{3}}{2\,d}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+4\,{\frac{e \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{7/2}}{{d}^{7}x}}+4\,{\frac{{e}^{3}x \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}{{d}^{7}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^4,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{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)^(5/2)/x^3/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate((-e^2*x^2 + d^2)^(5/2)/((e*x + d)^4*x^3), x)

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Fricas [A]  time = 1.62552, size = 225, normalized size = 2.05 \begin{align*} \frac{16 \, e^{3} x^{3} + 16 \, d e^{2} x^{2} + 15 \,{\left (e^{3} x^{3} + d e^{2} x^{2}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) +{\left (24 \, e^{2} x^{2} + 7 \, d e x - d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{2 \,{\left (d e x^{3} + d^{2} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/2*(16*e^3*x^3 + 16*d*e^2*x^2 + 15*(e^3*x^3 + d*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (24*e^2*x^2 + 7
*d*e*x - d^2)*sqrt(-e^2*x^2 + d^2))/(d*e*x^3 + d^2*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{5}{2}}}{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)**(5/2)/x**3/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(x**3*(d + e*x)**4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/x^3/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError