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

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

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Rubi [A]  time = 0.22, antiderivative size = 110, 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, 1805, 1807, 807, 266, 63, 208} \begin {gather*} \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} \end {gather*}

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 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 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^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*}

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Mathematica [A]  time = 0.22, size = 85, normalized size = 0.77 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.62, size = 140, normalized size = 1.27 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-d^2+7 d e x+24 e^2 x^2\right )}{2 d x^2 (d+e x)}-\frac {15 e^2 \log \left (\sqrt {d^2-e^2 x^2}+d-\sqrt {-e^2} x\right )}{2 d}+\frac {15 e^2 \log \left (-d \sqrt {d^2-e^2 x^2}+d^2+d \sqrt {-e^2} x\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.41, size = 112, normalized size = 1.02 \begin {gather*} \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 {gather*}

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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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^3/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/16*(-2*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2
-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^11-16*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*
exp(2)^10/x/exp(2))/d^2/exp(1)^6/exp(2)^9+1/24*((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*
(-3216*exp(1)^14*exp(2)^2-7776*exp(1)^12*exp(2)^3-6300*exp(1)^10*exp(2)^4-2868*exp(1)^8*exp(2)^5-2571*exp(1)^6
*exp(2)^6-225*exp(1)^4*exp(2)^7+36*exp(2)^9)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*(-3
456*exp(1)^14*exp(2)^2-4688*exp(1)^12*exp(2)^3-6336*exp(1)^10*exp(2)^4-4638*exp(1)^8*exp(2)^5-90*exp(1)^6*exp(
2)^6-378*exp(1)^4*exp(2)^7-126*exp(2)^9-1504*exp(1)^16*exp(2))+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1
))/x/exp(2))^6*(-1680*exp(1)^14*exp(2)^2-3744*exp(1)^12*exp(2)^3-2376*exp(1)^10*exp(2)^4-864*exp(1)^8*exp(2)^5
-1293*exp(1)^6*exp(2)^6-135*exp(1)^4*exp(2)^7+12*exp(2)^9)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x
/exp(2))^7*(-480*exp(1)^12*exp(2)^3-1008*exp(1)^10*exp(2)^4-408*exp(1)^8*exp(2)^5+144*exp(1)^6*exp(2)^6-144*ex
p(1)^4*exp(2)^7-48*exp(2)^9)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*(-2328*exp(1)^12*ex
p(2)^3-5832*exp(1)^10*exp(2)^4-4188*exp(1)^8*exp(2)^5-300*exp(1)^6*exp(2)^6-324*exp(1)^4*exp(2)^7-108*exp(2)^9
)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*(-628*exp(1)^10*exp(2)^4-1620*exp(1)^8*exp(2)^
5-1211*exp(1)^6*exp(2)^6-81*exp(1)^4*exp(2)^7+36*exp(2)^9)+3*exp(1)^6*exp(2)^6+9*exp(1)^4*exp(2)^7+12*exp(2)^9
-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*(-30*exp(1)^8*exp(2)^5-90*exp(1)^6*exp(2)^6-90*exp(1)^4*exp(2
)^7-30*exp(2)^9)/x/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/((-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+ex
p(2))^3/exp(1)^3/d/exp(1)+1/2*(5*exp(2)^3-20*exp(1)^4*exp(2))*ln(1/2*abs(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*ex
p(1))/abs(x)/exp(2))/exp(1)^3/d/exp(1)+1/2*(-108*exp(1)^7*exp(2)^2-66*exp(1)^5*exp(2)^3+40*exp(1)^3*exp(2)^4-4
0*exp(1)^9*exp(2)+48*exp(1)*exp(2)^5)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-e
xp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/(-d*exp(1)^7-3*d*exp(1)^5*exp(2)-4*d*exp(1)*exp(2)^3)

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maple [B]  time = 0.01, size = 504, normalized size = 4.58 \begin {gather*} -\frac {15 e^{3} \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 {15 e^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, d}-\frac {15 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}+\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3} x}{2 d^{3}}-\frac {15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{3} x}{2 d^{3}}+\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{2}}{2 d^{2}}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3} x}{d^{5}}-\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^{3} x}{d^{5}}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}{2 d^{4}}+\frac {4 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3} x}{d^{7}}+\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}{2 d^{6}}-\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^{2}}{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}}}{\left (x +\frac {d}{e}\right )^{4} d^{4} e^{2}}-\frac {2 \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 )^{2} d^{6}}+\frac {4 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{d^{7} x}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{2 d^{6} x^{2}} \end {gather*}

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]

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

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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^{3}}\,{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^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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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^3\,{\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^3*(d + e*x)^4),x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(x^3*(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^{3} \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**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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