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

Optimal. Leaf size=182 \[ -\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3} \]

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Rubi [A]  time = 0.22, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {852, 1635, 795, 671, 641, 217, 203} \begin {gather*} -\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(d - e*x)^4)/(e^3*Sqrt[d^2 - e^2*x^2])) - (95*d^3*Sqrt[d^2 - e^2*x^2])/(8*e^3) - (95*d^2*(d - e*x)*Sqrt[d
^2 - e^2*x^2])/(24*e^3) - (19*d*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])/(12*e^3) - ((d - e*x)^3*Sqrt[d^2 - e^2*x^2])/
(4*e^3) - (95*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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 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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

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 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=\int \frac {x^2 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\left (\frac {4 d^2}{e^2}-\frac {d x}{e}\right ) (d-e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx}{d}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {(19 d) \int \frac {(d-e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx}{4 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^2\right ) \int \frac {(d-e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx}{12 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^3\right ) \int \frac {d-e x}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {\left (95 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^2}\\ &=-\frac {d (d-e x)^4}{e^3 \sqrt {d^2-e^2 x^2}}-\frac {95 d^3 \sqrt {d^2-e^2 x^2}}{8 e^3}-\frac {95 d^2 (d-e x) \sqrt {d^2-e^2 x^2}}{24 e^3}-\frac {19 d (d-e x)^2 \sqrt {d^2-e^2 x^2}}{12 e^3}-\frac {(d-e x)^3 \sqrt {d^2-e^2 x^2}}{4 e^3}-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 103, normalized size = 0.57 \begin {gather*} \sqrt {d^2-e^2 x^2} \left (-\frac {8 d^4}{e^3 (d+e x)}-\frac {32 d^3}{3 e^3}+\frac {31 d^2 x}{8 e^2}-\frac {4 d x^2}{3 e}+\frac {x^3}{4}\right )-\frac {95 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.54, size = 121, normalized size = 0.66 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-448 d^4-163 d^3 e x+61 d^2 e^2 x^2-26 d e^3 x^3+6 e^4 x^4\right )}{24 e^3 (d+e x)}-\frac {95 d^4 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{8 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-448*d^4 - 163*d^3*e*x + 61*d^2*e^2*x^2 - 26*d*e^3*x^3 + 6*e^4*x^4))/(24*e^3*(d + e*x))
- (95*d^4*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(8*e^4)

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fricas [A]  time = 0.41, size = 124, normalized size = 0.68 \begin {gather*} -\frac {448 \, d^{4} e x + 448 \, d^{5} - 570 \, {\left (d^{4} e x + d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (6 \, e^{4} x^{4} - 26 \, d e^{3} x^{3} + 61 \, d^{2} e^{2} x^{2} - 163 \, d^{3} e x - 448 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, {\left (e^{4} x + d e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(448*d^4*e*x + 448*d^5 - 570*(d^4*e*x + d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (6*e^4*x^4 - 26
*d*e^3*x^3 + 61*d^2*e^2*x^2 - 163*d^3*e*x - 448*d^4)*sqrt(-e^2*x^2 + d^2))/(e^4*x + d*e^3)

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-288*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-
x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^12*exp(2)^2-252*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x
/exp(2))^4*exp(1)^10*exp(2)^3-54*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^8*ex
p(2)^4+24*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^12*exp(2)^2+64*d^4*(-1/2*(-
2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^10*exp(2)^3+8*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x
^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^14*exp(2)+96*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp
(2))^4*exp(1)^8*exp(2)^4+30*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^6*exp(2)^
5-612*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^10*exp(2)^3+216*d^4*(-1/2*(-2*d
*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^8*exp(2)^4+540*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^4*exp(1)^6*exp(2)^5+144*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp
(2))^5*exp(1)^4*exp(2)^6+66*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*exp(2)^
4+438*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^5+339*d^4*(-1/2*(-2*d*
exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^4*exp(2)^6+75*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x/exp(2))^5*exp(2)^8+1188*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp
(1)^6*exp(2)^5+972*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^4*exp(2)^6+2*d^4*e
xp(1)^8*exp(2)^4+252*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(2)^8+780*d^4*(-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)^6-72*d^4*exp(1)^6*exp(2)^5+540*d^4*(-1
/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^8+13*d^4*exp(1)^4*exp(2)^6+504*d^4*(-1/2*(-2
*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^8+252*d^4*exp(2)^8-465/2*d^4*(-2*d*exp(1)-2*sqrt(d
^2-x^2*exp(2))*exp(1))*exp(2)^8/x/exp(2)-414*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*exp(2)^6
/x/exp(2)-24*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)+189*d^4*(-2*d*exp(1)-2
*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^4/x/exp(2)-6*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp
(1)^10*exp(2)^3/x/exp(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+exp(2))^3/(3*exp(1)^13+9*exp(1)^9*exp(2)^2+3*exp(1)^7*exp(2)^3+9*exp(1)^11*exp
(2))+1/2*(48*d^4*exp(1)^10*exp(2)^2-104*d^4*exp(1)^8*exp(2)^3-280*d^4*exp(1)^6*exp(2)^4+22*d^4*exp(1)^4*exp(2)
^5+440*d^4*exp(2)^7)*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)^15-3*exp(1)^11*exp(2)^2-exp(1)^9*exp(2)^3-3*exp(1)^13*exp(2))-95/8*d^4*sig
n(d)*asin(x*exp(2)/d/exp(1))/exp(1)/exp(2)+2*(((12*exp(1)^8*1/96/exp(1)^8*x-64*exp(1)^7*d*1/96/exp(1)^8)*x+186
*exp(1)^6*d^2*1/96/exp(1)^8)*x-512*exp(1)^5*d^3*1/96/exp(1)^8)*sqrt(d^2-x^2*exp(2))

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maple [A]  time = 0.01, size = 288, normalized size = 1.58 \begin {gather*} -\frac {95 d^{4} \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}}\, e^{2}}-\frac {95 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2} x}{8 e^{2}}-\frac {95 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} x}{12 e^{2}}-\frac {19 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{3 d \,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}} d}{\left (x +\frac {d}{e}\right )^{4} e^{7}}-\frac {19 \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 \,e^{5}}-\frac {5 \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} e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d/e^7/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-5/e^6/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-19/3/
d/e^5/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-19/3/d/e^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)-95/12/e^2*(
2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x-95/8*d^2/e^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-95/8*d^4/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 [C]  time = 1.02, size = 363, normalized size = 1.99 \begin {gather*} \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{2 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{e^{4} x + d e^{3}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{3 \, {\left (e^{4} x + d e^{3}\right )}} - \frac {5 i \, d^{4} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{3}} - \frac {25 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{4 \, {\left (e^{4} x + d e^{3}\right )}} + \frac {5 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{2} x}{8 \, e^{2}} + \frac {5 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3}}{4 \, e^{3}} - \frac {5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{3}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{12 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-e^2*x^2 + d^2)^(5/2)*d^2/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3) + 5/2*(-e^2*x^2 + d^2)^(3/2)*d^
3/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) - 15*sqrt(-e^2*x^2 + d^2)*d^4/(e^4*x + d*e^3) - 2/3*(-e^2*x^2 + d^2)^(5/2)*d
/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) - 5/3*(-e^2*x^2 + d^2)^(3/2)*d^2/(e^4*x + d*e^3) - 5/8*I*d^4*arcsin(e*x/d + 2
)/e^3 - 25/2*d^4*arcsin(e*x/d)/e^3 + 1/4*(-e^2*x^2 + d^2)^(5/2)/(e^4*x + d*e^3) + 5/8*sqrt(e^2*x^2 + 4*d*e*x +
3*d^2)*d^2*x/e^2 + 5/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^3/e^3 - 5*sqrt(-e^2*x^2 + d^2)*d^3/e^3 + 5/12*(-e^2*
x^2 + d^2)^(3/2)*d/e^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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