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

Optimal. Leaf size=108 \[ -\frac {5 e^2 \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}+\frac {5 e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d} \]

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Rubi [A]  time = 0.15, antiderivative size = 108, normalized size of antiderivative = 1.00, 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, 1807, 807, 266, 47, 63, 208} \begin {gather*} -\frac {5 e^2 \sqrt {d^2-e^2 x^2}}{8 x^2}+\frac {2 e \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+\frac {5 e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*e^2*Sqrt[d^2 - e^2*x^2])/(8*x^2) - (d^2 - e^2*x^2)^(3/2)/(4*x^4) + (2*e*(d^2 - e^2*x^2)^(3/2))/(3*d*x^3) +
(5*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(8*d)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 95, normalized size = 0.88 \begin {gather*} -\frac {-15 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^3-16 d^2 e x+9 d e^2 x^2+16 e^3 x^3\right )+15 e^4 x^4 \log (x)}{24 d x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[d^2 - e^2*x^2]*(6*d^3 - 16*d^2*e*x + 9*d*e^2*x^2 + 16*e^3*x^3) + 15*e^4*x^4*Log[x] - 15*e^4*x^4*Lo
g[d + Sqrt[d^2 - e^2*x^2]])/(d*x^4)

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IntegrateAlgebraic [A]  time = 0.72, size = 144, normalized size = 1.33 \begin {gather*} \frac {5 e^4 \log \left (\sqrt {d^2-e^2 x^2}+d-\sqrt {-e^2} x\right )}{8 d}-\frac {5 e^4 \log \left (-d \sqrt {d^2-e^2 x^2}+d^2+d \sqrt {-e^2} x\right )}{8 d}+\frac {\sqrt {d^2-e^2 x^2} \left (-6 d^3+16 d^2 e x-9 d e^2 x^2-16 e^3 x^3\right )}{24 d x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-6*d^3 + 16*d^2*e*x - 9*d*e^2*x^2 - 16*e^3*x^3))/(24*d*x^4) + (5*e^4*Log[d - Sqrt[-e^2]*
x + Sqrt[d^2 - e^2*x^2]])/(8*d) - (5*e^4*Log[d^2 + d*Sqrt[-e^2]*x - d*Sqrt[d^2 - e^2*x^2]])/(8*d)

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fricas [A]  time = 0.39, size = 86, normalized size = 0.80 \begin {gather*} -\frac {15 \, 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} - 16 \, d^{2} e x + 6 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, d x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(15*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 - 16*d^2*e*x + 6*d^3)*sqrt(-e
^2*x^2 + d^2))/(d*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueEvaluat
ion time: 0.69Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

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maple [B]  time = 0.02, size = 513, normalized size = 4.75 \begin {gather*} -\frac {5 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 )}{2 \sqrt {e^{2}}\, d}+\frac {5 e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, d}+\frac {5 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}}}-\frac {5 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{5} x}{2 d^{3}}+\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{5} x}{2 d^{3}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}{8 d^{2}}-\frac {5 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{5} x}{3 d^{5}}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5} x}{3 d^{5}}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}{24 d^{4}}+\frac {4 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5} x}{3 d^{7}}-\frac {4 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e^{4}}{3 d^{6}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}{8 d^{6}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}} e^{2}}{3 \left (x +\frac {d}{e}\right )^{2} d^{6}}+\frac {4 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{3}}{3 d^{7} x}-\frac {9 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{8 d^{6} x^{2}}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{3 d^{5} x^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{4 d^{4} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/d^6*e^2/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-5/3/d^5*e^5*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x-5
/2/d^3*e^5*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-5/2/d*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)+2/3/d^5*e/x^3*(-e^2*x^2+d^2)^(7/2)+4/3/d^7*e^5*x*(-e^2*x^2+d^2)^(5/2)+5/3/d^5*e^5*x*(-e^2*x
^2+d^2)^(3/2)+5/2/d^3*e^5*x*(-e^2*x^2+d^2)^(1/2)+5/2/d*e^5/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)
*x)+4/3/d^7*e^3/x*(-e^2*x^2+d^2)^(7/2)-9/8/d^6*e^2/x^2*(-e^2*x^2+d^2)^(7/2)-4/3/d^6*e^4*(2*(x+d/e)*d*e-(x+d/e)
^2*e^2)^(5/2)-5/8/d^2*e^4*(-e^2*x^2+d^2)^(1/2)+5/8*e^4/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2
))/x)-1/4/d^4/x^4*(-e^2*x^2+d^2)^(7/2)-1/8/d^6*e^4*(-e^2*x^2+d^2)^(5/2)-5/24/d^4*e^4*(-e^2*x^2+d^2)^(3/2)

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maxima [A]  time = 1.01, size = 130, normalized size = 1.20 \begin {gather*} \frac {5 \, e^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{8 \, d} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}}{8 \, d^{2}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{8 \, d^{2} x^{2}} + \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{3 \, d x^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*e^4*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d - 5/8*sqrt(-e^2*x^2 + d^2)*e^4/d^2 - 5/8*(-e^2*x
^2 + d^2)^(3/2)*e^2/(d^2*x^2) + 2/3*(-e^2*x^2 + d^2)^(3/2)*e/(d*x^3) - 1/4*(-e^2*x^2 + d^2)^(3/2)/x^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 12.36, size = 422, normalized size = 3.91 \begin {gather*} d^{2} \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {d^{2}}{2 e x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e}{2 x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{2 x} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*Piecewise((-d**2/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**3/(
8*d**2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(4*e*
x**5*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e/(8*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**3/(8*d**2*x*sqrt(-d**2/(
e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True)) - 2*d*e*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x
**2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**2), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)
/(3*x**2) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**2), True)) + e**2*Piecewise((-d**2/(2*e*x**3*sqrt(d**2/(e
**2*x**2) - 1)) + e/(2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**2*acosh(d/(e*x))/(2*d), Abs(d**2/(e**2*x**2)) > 1),
(-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(2*x) - I*e**2*asin(d/(e*x))/(2*d), True))

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