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

Optimal. Leaf size=89 \[ \frac{8 d (d-e x)}{\sqrt{d^2-e^2 x^2}}+\sqrt{d^2-e^2 x^2}+4 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

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

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Rubi [A]  time = 0.211394, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {852, 1805, 1809, 844, 217, 203, 266, 63, 208} \[ \frac{8 d (d-e x)}{\sqrt{d^2-e^2 x^2}}+\sqrt{d^2-e^2 x^2}+4 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*d*(d - e*x))/Sqrt[d^2 - e^2*x^2] + Sqrt[d^2 - e^2*x^2] + 4*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] - d*ArcTanh[
Sqrt[d^2 - e^2*x^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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 (d+e x)^4} \, dx &=\int \frac{(d-e x)^4}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac{8 d (d-e x)}{\sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-d^4-4 d^3 e x+d^2 e^2 x^2}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^2}\\ &=\frac{8 d (d-e x)}{\sqrt{d^2-e^2 x^2}}+\sqrt{d^2-e^2 x^2}+\frac{\int \frac{d^4 e^2+4 d^3 e^3 x}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^2 e^2}\\ &=\frac{8 d (d-e x)}{\sqrt{d^2-e^2 x^2}}+\sqrt{d^2-e^2 x^2}+d^2 \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx+(4 d e) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{8 d (d-e x)}{\sqrt{d^2-e^2 x^2}}+\sqrt{d^2-e^2 x^2}+\frac{1}{2} d^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )+(4 d e) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )\\ &=\frac{8 d (d-e x)}{\sqrt{d^2-e^2 x^2}}+\sqrt{d^2-e^2 x^2}+4 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\frac{d^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 )}{e^2}\\ &=\frac{8 d (d-e x)}{\sqrt{d^2-e^2 x^2}}+\sqrt{d^2-e^2 x^2}+4 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\\ \end{align*}

Mathematica [A]  time = 0.169612, size = 79, normalized size = 0.89 \[ \sqrt{d^2-e^2 x^2} \left (\frac{8 d}{d+e x}+1\right )-d \log \left (\sqrt{d^2-e^2 x^2}+d\right )+4 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+d \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.075, size = 378, normalized size = 4.3 \begin{align*}{\frac{32}{15\,{d}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}+4\,{\frac{de}{\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 ) }+\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}+{\frac{1}{5\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{1}{3\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+4\,{\frac{ex}{d}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{1}{{e}^{4}{d}^{2}} \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}}+2\,{\frac{1}{{e}^{3}{d}^{3}} \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 ) ^{-3}}+{\frac{7}{3\,{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 ) ^{-2}}-{{d}^{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}}}}}+{\frac{8\,ex}{3\,{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/15/d^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)+4*d*e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d
/e+x))^(1/2))+(-e^2*x^2+d^2)^(1/2)+1/5/d^4*(-e^2*x^2+d^2)^(5/2)+1/3/d^2*(-e^2*x^2+d^2)^(3/2)+4/d*e*(-(d/e+x)^2
*e^2+2*d*e*(d/e+x))^(1/2)*x+1/e^4/d^2/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)+2/e^3/d^3/(d/e+x)^3*(-(d/
e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)+7/3/e^2/d^4/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-d^2/(d^2)^(1/2)*ln(
(2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)+8/3/d^3*e*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x

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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}\,{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/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.65044, size = 236, normalized size = 2.65 \begin{align*} \frac{9 \, d e x + 9 \, d^{2} - 8 \,{\left (d e x + d^{2}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (d e x + d^{2}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt{-e^{2} x^{2} + d^{2}}{\left (e x + 9 \, d\right )}}{e x + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*d*e*x + 9*d^2 - 8*(d*e*x + d^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (d*e*x + d^2)*log(-(d - sqrt(-e
^2*x^2 + d^2))/x) + sqrt(-e^2*x^2 + d^2)*(e*x + 9*d))/(e*x + d)

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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 \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/(e*x+d)**4,x)

[Out]

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

[Out]

Exception raised: NotImplementedError