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

Optimal. Leaf size=94 \[ -\frac {8 e (d-e x)}{\sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{x}-e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+4 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 94, normalized size of antiderivative = 1.00, 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, 1807, 844, 217, 203, 266, 63, 208} \begin {gather*} -\frac {8 e (d-e x)}{\sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{x}-e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+4 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 84, normalized size = 0.89 \begin {gather*} \sqrt {d^2-e^2 x^2} \left (-\frac {8 e}{d+e x}-\frac {1}{x}\right )+4 e \log \left (\sqrt {d^2-e^2 x^2}+d\right )-e \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-4 e \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.54, size = 117, normalized size = 1.24 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} (-d-9 e x)}{x (d+e x)}-\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )-8 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-d - 9*e*x)*Sqrt[d^2 - e^2*x^2])/(x*(d + e*x)) - 8*e*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d] - Sqr
t[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]]

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fricas [A]  time = 0.42, size = 127, normalized size = 1.35 \begin {gather*} -\frac {8 \, e^{2} x^{2} + 8 \, d e x - 2 \, {\left (e^{2} x^{2} + d e x\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 4 \, {\left (e^{2} x^{2} + d e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + \sqrt {-e^{2} x^{2} + d^{2}} {\left (9 \, e x + d\right )}}{e x^{2} + d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(8*e^2*x^2 + 8*d*e*x - 2*(e^2*x^2 + d*e*x)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 4*(e^2*x^2 + d*e*x)*lo
g(-(d - sqrt(-e^2*x^2 + d^2))/x) + sqrt(-e^2*x^2 + d^2)*(9*e*x + d))/(e*x^2 + d*x)

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*
exp(1))*exp(2)^3/exp(1)^4/x/exp(1)/exp(2)+1/2*(-48*exp(1)^10*exp(2)^2-40*exp(1)^8*exp(2)^3-8*exp(1)^6*exp(2)^4
+2*exp(1)^4*exp(2)^5-16*exp(2)^7-16*exp(1)^12*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))/sqrt(-exp(1)^4+exp(2)^2)/(exp(1)^11+3*exp(1)^7*exp(2)^2+exp(1)^5*exp(2)^3+3
*exp(1)^9*exp(2))+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))/exp(1)-1/3*x*(
(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*(408*exp(1)^12*exp(2)^2+1152*exp(1)^10*exp(2)^3+
1392*exp(1)^8*exp(2)^4+780*exp(1)^6*exp(2)^5+516*exp(1)^4*exp(2)^6+132*exp(2)^8)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2
-x^2*exp(2))*exp(1))/x/exp(2))^4*(576*exp(1)^12*exp(2)^2+932*exp(1)^10*exp(2)^3+1116*exp(1)^8*exp(2)^4+1041*ex
p(1)^6*exp(2)^5+279*exp(1)^4*exp(2)^6+108*exp(2)^8+208*exp(1)^14*exp(2))+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp
(2))*exp(1))/x/exp(2))^5*(240*exp(1)^12*exp(2)^2+648*exp(1)^10*exp(2)^3+642*exp(1)^8*exp(2)^4+270*exp(1)^6*exp
(2)^5+228*exp(1)^4*exp(2)^6+66*exp(2)^8)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^6*(72*exp
(1)^10*exp(2)^3+180*exp(1)^8*exp(2)^4+135*exp(1)^6*exp(2)^5+9*exp(1)^4*exp(2)^6+18*exp(2)^8)+(-1/2*(-2*d*exp(1
)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*(276*exp(1)^10*exp(2)^3+792*exp(1)^8*exp(2)^4+861*exp(1)^6*exp(2)
^5+279*exp(1)^4*exp(2)^6+102*exp(2)^8)+3*exp(1)^6*exp(2)^5+9*exp(1)^4*exp(2)^6+12*exp(2)^8-1/2*(-2*d*exp(1)-2*
sqrt(d^2-x^2*exp(2))*exp(1))*(70*exp(1)^8*exp(2)^4+198*exp(1)^6*exp(2)^5+200*exp(1)^4*exp(2)^6+66*exp(2)^8)/x/
exp(2))*exp(2)/(2*exp(2))^3/((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)
-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))^3/(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/exp(1)/exp(2)-sign(d)*
asin(x*exp(2)/d/exp(1))*exp(2)/exp(1)

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maple [B]  time = 0.01, size = 515, normalized size = 5.48 \begin {gather*} \frac {4 d e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+\frac {7 e^{2} \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 )}{8 \sqrt {e^{2}}}-\frac {15 e^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{2} x}{8 d^{2}}+\frac {7 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{2} x}{8 d^{2}}-\frac {4 \sqrt {-e^{2} x^{2}+d^{2}}\, e}{d}-\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2} x}{4 d^{4}}+\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{2} x}{12 d^{4}}-\frac {4 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}{3 d^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2} x}{d^{6}}-\frac {4 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{5 d^{5}}+\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} e}{15 d^{5}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{4} d^{3} e^{3}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{3} d^{4} e^{2}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{3 \left (x +\frac {d}{e}\right )^{2} d^{5} e}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{d^{6} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/8/(e^2)^(1/2)*e^2*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/d^6*e^2*x*(-e^2*x^2+d^2)^(5/2)-5/4/d^4*e^2*
x*(-e^2*x^2+d^2)^(3/2)-15/8/d^2*e^2*x*(-e^2*x^2+d^2)^(1/2)-1/d^3/e^3/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(
7/2)-1/d^4/e^2/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-1/3/d^5/e/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)
^(7/2)+7/12/d^4*e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+7/8/d^2*e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+4/
(d^2)^(1/2)*d*e*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-4/d*e*(-e^2*x^2+d^2)^(1/2)-4/5/d^5*e*(-e^2*x^
2+d^2)^(5/2)-4/3/d^3*e*(-e^2*x^2+d^2)^(3/2)-1/d^6/x*(-e^2*x^2+d^2)^(7/2)+7/15/d^5*e*(2*(x+d/e)*d*e-(x+d/e)^2*e
^2)^(5/2)+7/8*e^2/(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 {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{4} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2\,{\left (d+e\,x\right )}^4} \,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^2*(d + e*x)^4),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{x^{2} \left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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