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

Optimal. Leaf size=170 \[ -\frac {31 e^2 \sqrt {d^2-e^2 x^2}}{8 d^2 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}+\frac {4 e \sqrt {d^2-e^2 x^2}}{3 d x^3}+\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {95 e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d^3}+\frac {32 e^3 \sqrt {d^2-e^2 x^2}}{3 d^3 x} \]

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Rubi [A]  time = 0.39, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 9, 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^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}+\frac {32 e^3 \sqrt {d^2-e^2 x^2}}{3 d^3 x}-\frac {31 e^2 \sqrt {d^2-e^2 x^2}}{8 d^2 x^2}+\frac {4 e \sqrt {d^2-e^2 x^2}}{3 d x^3}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}-\frac {95 e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*e^4*(d - e*x))/(d^3*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(4*x^4) + (4*e*Sqrt[d^2 - e^2*x^2])/(3*d*x^3
) - (31*e^2*Sqrt[d^2 - e^2*x^2])/(8*d^2*x^2) + (32*e^3*Sqrt[d^2 - e^2*x^2])/(3*d^3*x) - (95*e^4*ArcTanh[Sqrt[d
^2 - e^2*x^2]/d])/(8*d^3)

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^5 (d+e x)^4} \, dx &=\int \frac {(d-e x)^4}{x^5 \left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-d^4+4 d^3 e x-7 d^2 e^2 x^2+8 d e^3 x^3-8 e^4 x^4}{x^5 \sqrt {d^2-e^2 x^2}} \, dx}{d^2}\\ &=\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}+\frac {\int \frac {-16 d^5 e+31 d^4 e^2 x-32 d^3 e^3 x^2+32 d^2 e^4 x^3}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{4 d^4}\\ &=\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}+\frac {4 e \sqrt {d^2-e^2 x^2}}{3 d x^3}-\frac {\int \frac {-93 d^6 e^2+128 d^5 e^3 x-96 d^4 e^4 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{12 d^6}\\ &=\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}+\frac {4 e \sqrt {d^2-e^2 x^2}}{3 d x^3}-\frac {31 e^2 \sqrt {d^2-e^2 x^2}}{8 d^2 x^2}+\frac {\int \frac {-256 d^7 e^3+285 d^6 e^4 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{24 d^8}\\ &=\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}+\frac {4 e \sqrt {d^2-e^2 x^2}}{3 d x^3}-\frac {31 e^2 \sqrt {d^2-e^2 x^2}}{8 d^2 x^2}+\frac {32 e^3 \sqrt {d^2-e^2 x^2}}{3 d^3 x}+\frac {\left (95 e^4\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{8 d^2}\\ &=\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}+\frac {4 e \sqrt {d^2-e^2 x^2}}{3 d x^3}-\frac {31 e^2 \sqrt {d^2-e^2 x^2}}{8 d^2 x^2}+\frac {32 e^3 \sqrt {d^2-e^2 x^2}}{3 d^3 x}+\frac {\left (95 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{16 d^2}\\ &=\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}+\frac {4 e \sqrt {d^2-e^2 x^2}}{3 d x^3}-\frac {31 e^2 \sqrt {d^2-e^2 x^2}}{8 d^2 x^2}+\frac {32 e^3 \sqrt {d^2-e^2 x^2}}{3 d^3 x}-\frac {\left (95 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 )}{8 d^2}\\ &=\frac {8 e^4 (d-e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{4 x^4}+\frac {4 e \sqrt {d^2-e^2 x^2}}{3 d x^3}-\frac {31 e^2 \sqrt {d^2-e^2 x^2}}{8 d^2 x^2}+\frac {32 e^3 \sqrt {d^2-e^2 x^2}}{3 d^3 x}-\frac {95 e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 107, normalized size = 0.63 \begin {gather*} \frac {-285 e^4 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-6 d^4+26 d^3 e x-61 d^2 e^2 x^2+163 d e^3 x^3+448 e^4 x^4\right )}{x^4 (d+e x)}+285 e^4 \log (x)}{24 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(-6*d^4 + 26*d^3*e*x - 61*d^2*e^2*x^2 + 163*d*e^3*x^3 + 448*e^4*x^4))/(x^4*(d + e*x)) +
285*e^4*Log[x] - 285*e^4*Log[d + Sqrt[d^2 - e^2*x^2]])/(24*d^3)

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IntegrateAlgebraic [A]  time = 0.94, size = 122, normalized size = 0.72 \begin {gather*} \frac {95 e^4 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^3}+\frac {\sqrt {d^2-e^2 x^2} \left (-6 d^4+26 d^3 e x-61 d^2 e^2 x^2+163 d e^3 x^3+448 e^4 x^4\right )}{24 d^3 x^4 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-6*d^4 + 26*d^3*e*x - 61*d^2*e^2*x^2 + 163*d*e^3*x^3 + 448*e^4*x^4))/(24*d^3*x^4*(d + e*
x)) + (95*e^4*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])/(4*d^3)

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fricas [A]  time = 0.40, size = 136, normalized size = 0.80 \begin {gather*} \frac {192 \, e^{5} x^{5} + 192 \, d e^{4} x^{4} + 285 \, {\left (e^{5} x^{5} + d e^{4} x^{4}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (448 \, e^{4} x^{4} + 163 \, d e^{3} x^{3} - 61 \, d^{2} e^{2} x^{2} + 26 \, d^{3} e x - 6 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, {\left (d^{3} e x^{5} + d^{4} x^{4}\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)^4,x, algorithm="fricas")

[Out]

1/24*(192*e^5*x^5 + 192*d*e^4*x^4 + 285*(e^5*x^5 + d*e^4*x^4)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (448*e^4*x^
4 + 163*d*e^3*x^3 - 61*d^2*e^2*x^2 + 26*d^3*e*x - 6*d^4)*sqrt(-e^2*x^2 + d^2))/(d^3*e*x^5 + d^4*x^4)

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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^5/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/65536*(-81920*d^9*(-1/2*(-2*d*exp(1)-2
*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^22*exp(2)^19+32768/3*d^9*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^3*exp(1)^20*exp(2)^20-1024*d^9*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp
(2))^4*exp(1)^18*exp(2)^21+24576*d^9*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^20*e
xp(2)^20-8192*d^9*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^18*exp(2)^21-49152*d^9*
(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^20*exp(2)^20/x/exp(2)+196608*d^9*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))*exp(1)^22*exp(2)^19/x/exp(2)-327680*d^9*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^2
4*exp(2)^18/x/exp(2))/d^12/exp(1)^24/exp(2)^16+1/192*((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(
2))^6*(-67968*exp(1)^18*exp(2)^2-126720*exp(1)^16*exp(2)^3-53184*exp(1)^14*exp(2)^4-21408*exp(1)^12*exp(2)^5-2
3472*exp(1)^10*exp(2)^6+19800*exp(1)^8*exp(2)^7+3699*exp(1)^6*exp(2)^8-3063*exp(1)^4*exp(2)^9+84*exp(2)^11)+(-
1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^7*(-46080*exp(1)^18*exp(2)^2-62080*exp(1)^16*exp(2)^
3-101376*exp(1)^14*exp(2)^4-33888*exp(1)^12*exp(2)^5+32688*exp(1)^10*exp(2)^6-3488*exp(1)^8*exp(2)^7-960*exp(1
)^6*exp(2)^8+768*exp(1)^4*exp(2)^9-752*exp(2)^11-27392*exp(1)^20*exp(2))+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp
(2))*exp(1))/x/exp(2))^8*(-29568*exp(1)^18*exp(2)^2-48384*exp(1)^16*exp(2)^3-13632*exp(1)^14*exp(2)^4-13824*ex
p(1)^12*exp(2)^5-16848*exp(1)^10*exp(2)^6+10440*exp(1)^8*exp(2)^7+1872*exp(1)^6*exp(2)^8-1632*exp(1)^4*exp(2)^
9+24*exp(2)^11)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^9*(-8064*exp(1)^16*exp(2)^3-12672*
exp(1)^14*exp(2)^4+192*exp(1)^12*exp(2)^5+2304*exp(1)^10*exp(2)^6-3360*exp(1)^8*exp(2)^7+672*exp(1)^6*exp(2)^8
+288*exp(1)^4*exp(2)^9-288*exp(2)^11)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*(-54144*ex
p(1)^16*exp(2)^3-106560*exp(1)^14*exp(2)^4-29184*exp(1)^12*exp(2)^5+29280*exp(1)^10*exp(2)^6-750*exp(1)^8*exp(
2)^7-714*exp(1)^6*exp(2)^8+630*exp(1)^4*exp(2)^9-654*exp(2)^11)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(
1))/x/exp(2))^4*(-15744*exp(1)^14*exp(2)^4-32160*exp(1)^12*exp(2)^5-9100*exp(1)^10*exp(2)^6+11588*exp(1)^8*exp
(2)^7+1893*exp(1)^6*exp(2)^8-1473*exp(1)^4*exp(2)^9+108*exp(2)^11)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*e
xp(1))/x/exp(2))^3*(-840*exp(1)^12*exp(2)^5-1800*exp(1)^10*exp(2)^6-564*exp(1)^8*exp(2)^7+708*exp(1)^6*exp(2)^
8+108*exp(1)^4*exp(2)^9-204*exp(2)^11)+(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*(84*exp(1
)^10*exp(2)^6+180*exp(1)^8*exp(2)^7+69*exp(1)^6*exp(2)^8-33*exp(1)^4*exp(2)^9+60*exp(2)^11)+3*exp(1)^6*exp(2)^
8+9*exp(1)^4*exp(2)^9+12*exp(2)^11-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*(-14*exp(1)^8*exp(2)^7-42*e
xp(1)^6*exp(2)^8-42*exp(1)^4*exp(2)^9-14*exp(2)^11)/x/exp(2))/d^3/(2*exp(2))^3/(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x
^2*exp(2))*exp(1))/x/exp(2))^4/((-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/exp(1)^6+1/8*(240*exp(1)^6*exp(2)^2-64*exp(1)^4*exp(2)^3+9*exp(
2)^5-280*exp(1)^8*exp(2))*ln(1/2*abs(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/abs(x)/exp(2))/d^3/exp(1)^5/ex
p(1)+1/2*(-300*exp(1)^9*exp(2)^2-22*exp(1)^7*exp(2)^3+280*exp(1)^5*exp(2)^4+104*exp(1)^3*exp(2)^5-140*exp(1)^1
1*exp(2)-48*exp(1)*exp(2)^6)*atan((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+e
xp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/(-d^3*exp(1)^7-3*d^3*exp(1)^5*exp(2)-4*d^3*exp(1)*exp(2)^3)

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maple [B]  time = 0.02, size = 600, normalized size = 3.53 \begin {gather*} -\frac {95 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}}\, d^{2}}-\frac {55 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^{3}}+\frac {55 e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, d^{3}}+\frac {55 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{5} x}{2 d^{5}}-\frac {55 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{5} x}{2 d^{5}}+\frac {95 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}{8 d^{4}}+\frac {55 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5} x}{3 d^{7}}-\frac {55 \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^{7}}+\frac {95 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}{24 d^{6}}+\frac {44 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5} x}{3 d^{9}}+\frac {19 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}{8 d^{8}}-\frac {44 \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^{8}}+\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^{6}}-\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}} e}{\left (x +\frac {d}{e}\right )^{3} d^{7}}-\frac {23 \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^{8}}+\frac {44 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{3}}{3 d^{9} x}-\frac {37 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{8 d^{8} x^{2}}+\frac {4 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{3 d^{7} x^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{4 d^{6} 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)^4,x)

[Out]

4/3/d^7*e/x^3*(-e^2*x^2+d^2)^(7/2)-37/8/d^8*e^2/x^2*(-e^2*x^2+d^2)^(7/2)+44/3/d^9*e^3/x*(-e^2*x^2+d^2)^(7/2)+4
4/3/d^9*e^5*x*(-e^2*x^2+d^2)^(5/2)+55/3/d^7*e^5*x*(-e^2*x^2+d^2)^(3/2)+55/2/d^5*e^5*x*(-e^2*x^2+d^2)^(1/2)+55/
2/d^3*e^5/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-2/d^7*e/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^
2)^(7/2)-23/3/d^8*e^2/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-55/3/d^7*e^5*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)
^(3/2)*x-55/2/d^5*e^5*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-55/2/d^3*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)-95/8/(d^2)^(1/2)/d^2*e^4*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/4
/d^6/x^4*(-e^2*x^2+d^2)^(7/2)+19/8/d^8*e^4*(-e^2*x^2+d^2)^(5/2)+95/24/d^6*e^4*(-e^2*x^2+d^2)^(3/2)+95/8/d^4*e^
4*(-e^2*x^2+d^2)^(1/2)+1/d^6/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-44/3/d^8*e^4*(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^{5}}\,{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^5/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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

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