3.2.91 \(\int \frac {x^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=115 \[ \frac {3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}-\frac {2 \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

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Rubi [A]  time = 0.15, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1637, 659, 651, 663, 217, 203} \begin {gather*} \frac {3 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^3}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{5 e^3 (d+e x)^4}-\frac {2 \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d^2 - e^2*x^2])/(e^3*(d + e*x)) - (d*(d^2 - e^2*x^2)^(3/2))/(5*e^3*(d + e*x)^4) + (3*(d^2 - e^2*x^2)^
(3/2))/(5*e^3*(d + e*x)^3) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/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 651

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

Rule 659

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

Rule 663

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

Rule 1637

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p,
(d + e*x)^m*Pq, x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq
, x] + 2*p + 1, 0] && ILtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 73, normalized size = 0.63 \begin {gather*} -\frac {\frac {\sqrt {d^2-e^2 x^2} \left (8 d^2+19 d e x+13 e^2 x^2\right )}{(d+e x)^3}+5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{5 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*((Sqrt[d^2 - e^2*x^2]*(8*d^2 + 19*d*e*x + 13*e^2*x^2))/(d + e*x)^3 + 5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])
/e^3

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IntegrateAlgebraic [A]  time = 0.69, size = 94, normalized size = 0.82 \begin {gather*} \frac {\left (-8 d^2-19 d e x-13 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{5 e^3 (d+e x)^3}-\frac {\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*d^2 - 19*d*e*x - 13*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(5*e^3*(d + e*x)^3) - (Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) +
Sqrt[d^2 - e^2*x^2]])/e^4

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fricas [A]  time = 0.41, size = 157, normalized size = 1.37 \begin {gather*} -\frac {8 \, e^{3} x^{3} + 24 \, d e^{2} x^{2} + 24 \, d^{2} e x + 8 \, d^{3} - 10 \, {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (13 \, e^{2} x^{2} + 19 \, d e x + 8 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} 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)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/5*(8*e^3*x^3 + 24*d*e^2*x^2 + 24*d^2*e*x + 8*d^3 - 10*(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*arctan(-(d
- sqrt(-e^2*x^2 + d^2))/(e*x)) + (13*e^2*x^2 + 19*d*e*x + 8*d^2)*sqrt(-e^2*x^2 + d^2))/(e^6*x^3 + 3*d*e^5*x^2
+ 3*d^2*e^4*x + d^3*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)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-96*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*e
xp(2))*exp(1))/x/exp(2))^3*exp(1)^12*exp(2)^2-84*(-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)^3-18*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^8*exp(2)^4+24*(-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)^2-32*(-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*(-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)-24*(-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)^4-6*(-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)^5-204*(-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)^3-120*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/ex
p(2))^3*exp(1)^8*exp(2)^4-12*(-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)^5-1
02*(-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)^4-42*(-1/2*(-2*d*exp(1)-2*sqr
t(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^5+3*(-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+3*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(2)^8+12*(-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+36*(-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*exp(1)^8*exp(2)^4+12*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2)
)*exp(1))/x/exp(2))^4*exp(2)^8+36*(-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
)^6-24*exp(1)^6*exp(2)^5+36*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^8-11*exp(1)^4
*exp(2)^6+24*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^8+12*exp(2)^8-33/2*(-2*d*exp
(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^8/x/exp(2)-18*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*e
xp(2)^6/x/exp(2)+30*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)+63*(-2*d*exp(1)-2*s
qrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^4/x/exp(2)-6*(-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)^13-6*exp(1)^9*exp(2)^2-6*exp(1)^7*exp(2)^3+3*exp(1)^5*exp(2)^4+3
*exp(1)^11*exp(2)+3*exp(1)*exp(2)^6)+1/2*(16*exp(1)^10*exp(2)^2+8*exp(1)^8*exp(2)^3-8*exp(1)^6*exp(2)^4-10*exp
(1)^4*exp(2)^5+8*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+2*exp(1)^11*exp(2)^2+2*exp(1)^9*exp(2)^3-exp(1)^7*exp(2)^4-exp(1)
^5*exp(2)^5-exp(1)^13*exp(2))-sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)/exp(2)

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maple [B]  time = 0.01, size = 214, normalized size = 1.86 \begin {gather*} -\frac {\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 )}{\sqrt {e^{2}}\, e^{2}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{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 {3}{2}} d}{5 \left (x +\frac {d}{e}\right )^{4} e^{7}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{\left (x +\frac {d}{e}\right )^{2} d \,e^{5}}+\frac {3 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}}}{5 \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)^(1/2)/(e*x+d)^4,x)

[Out]

-1/5*d/e^7/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)+3/5/e^6/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)
-1/e^5/d/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)-1/e^3/d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-1/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 {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{{\left (e x + d\right )}^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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

[Out]

int((x^2*(d^2 - e^2*x^2)^(1/2))/(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} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{\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)**(1/2)/(e*x+d)**4,x)

[Out]

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

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