3.2.90 \(\int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx\)
Optimal. Leaf size=148 \[ \frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}+\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]
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Rubi [A] time = 0.25, antiderivative size = 148, 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 =
{1639, 1637, 659, 651, 663, 217, 203} \begin {gather*} \frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}+\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(x^3*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]
[Out]
(8*d*Sqrt[d^2 - e^2*x^2])/(e^4*(d + e*x)) + (d^2*(d^2 - e^2*x^2)^(3/2))/(5*e^4*(d + e*x)^4) - (14*d*(d^2 - e^2
*x^2)^(3/2))/(15*e^4*(d + e*x)^3) - (d^2 - e^2*x^2)^(3/2)/(e^4*(d + e*x)^2) + (4*d*ArcTan[(e*x)/Sqrt[d^2 - e^2
*x^2]])/e^4
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]
Rule 1639
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && !IGtQ[m, 0]
Rubi steps
\begin {align*} \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {\int \frac {\sqrt {d^2-e^2 x^2} \left (2 d^3 e^2+5 d^2 e^3 x+4 d e^4 x^2\right )}{(d+e x)^4} \, dx}{e^5}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {\int \left (\frac {d^3 e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4}-\frac {3 d^2 e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^3}+\frac {4 d e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^2}\right ) \, dx}{e^5}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {(4 d) \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx}{e^3}+\frac {\left (3 d^2\right ) \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{e^3}-\frac {d^3 \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx}{e^3}\\ &=\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {(4 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^3}-\frac {d^2 \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{5 e^3}\\ &=\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {(4 d) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}\\ &=\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 85, normalized size = 0.57 \begin {gather*} \frac {60 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (94 d^3+222 d^2 e x+149 d e^2 x^2+15 e^3 x^3\right )}{(d+e x)^3}}{15 e^4} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(x^3*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]
[Out]
((Sqrt[d^2 - e^2*x^2]*(94*d^3 + 222*d^2*e*x + 149*d*e^2*x^2 + 15*e^3*x^3))/(d + e*x)^3 + 60*d*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]])/(15*e^4)
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IntegrateAlgebraic [A] time = 0.65, size = 106, normalized size = 0.72 \begin {gather*} \frac {4 d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^5}+\frac {\sqrt {d^2-e^2 x^2} \left (94 d^3+222 d^2 e x+149 d e^2 x^2+15 e^3 x^3\right )}{15 e^4 (d+e x)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(x^3*Sqrt[d^2 - e^2*x^2])/(d + e*x)^4,x]
[Out]
(Sqrt[d^2 - e^2*x^2]*(94*d^3 + 222*d^2*e*x + 149*d*e^2*x^2 + 15*e^3*x^3))/(15*e^4*(d + e*x)^3) + (4*d*Sqrt[-e^
2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^5
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fricas [A] time = 0.42, size = 174, normalized size = 1.18 \begin {gather*} \frac {94 \, d e^{3} x^{3} + 282 \, d^{2} e^{2} x^{2} + 282 \, d^{3} e x + 94 \, d^{4} - 120 \, {\left (d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (15 \, e^{3} x^{3} + 149 \, d e^{2} x^{2} + 222 \, d^{2} e x + 94 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="fricas")
[Out]
1/15*(94*d*e^3*x^3 + 282*d^2*e^2*x^2 + 282*d^3*e*x + 94*d^4 - 120*(d*e^3*x^3 + 3*d^2*e^2*x^2 + 3*d^3*e*x + d^4
)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (15*e^3*x^3 + 149*d*e^2*x^2 + 222*d^2*e*x + 94*d^3)*sqrt(-e^2*x^
2 + d^2))/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^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(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")
[Out]
Exception raised: NotImplementedError >> Unable to parse Giac output: (-30*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))/x/exp(2))^4*exp(1)^12*exp(2)^2-6*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2)
)^5*exp(1)^10*exp(2)^3+144*d*(-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+
132*d*(-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+30*d*(-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)^4-102*d*(-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+62*d*(-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+87*d*(-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+21*d*(-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+300*d*(-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+144*d*(-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)^4-24*d*(-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)
^5-12*d*(-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+264*d*(-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+120*d*(-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)^5-24*d*(-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)^6-12*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(2)^8-84*d*(-1/2*(-2*d*ex
p(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^6*exp(2)^5-108*d*(-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)^6-11*d*exp(1)^8*exp(2)^4-36*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2)
)*exp(1))/x/exp(2))^4*exp(2)^8-96*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)^6+36*d*exp(1)^6*exp(2)^5-108*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^8+32*d
*exp(1)^4*exp(2)^6-72*d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^8-36*d*exp(2)^8-4
4*d*(-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)+48*d*(-2*d*exp(1)-2*sqrt(d^
2-x^2*exp(2))*exp(1))*exp(2)^8/x/exp(2)+48*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*exp(2)^6/x/e
xp(2)-171/2*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)-93*d*(-2*d*exp(1)-2*sqrt(
d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^4/x/exp(2)+30*d*(-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)^14-6*exp(1)^10*exp(2)^2-6*exp(1)^8*exp(2)^3+3*exp(1)^6*exp(2)^4+3
*exp(1)^4*exp(2)^5+3*exp(1)^12*exp(2))+1/2*(-36*d*exp(1)^10*exp(2)^2-30*d*exp(1)^8*exp(2)^3+40*d*exp(1)^6*exp(
2)^4+40*d*exp(1)^4*exp(2)^5-32*d*exp(2)^7+4*d*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)^16+2*exp(1)^12*exp(2)^2+2*exp(1
)^10*exp(2)^3-exp(1)^8*exp(2)^4-exp(1)^6*exp(2)^5-exp(1)^14*exp(2))+4*d*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)
^4+2*exp(1)^3*1/2/exp(1)^7*sqrt(d^2-x^2*exp(2))
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maple [A] time = 0.01, size = 212, normalized size = 1.43 \begin {gather*} \frac {4 d \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^{3}}+\frac {4 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}{e^{4}}+\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^{2}}{5 \left (x +\frac {d}{e}\right )^{4} e^{8}}-\frac {14 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d}{15 \left (x +\frac {d}{e}\right )^{3} e^{7}}+\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}}}{\left (x +\frac {d}{e}\right )^{2} e^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x)
[Out]
1/5*d^2/e^8/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)-14/15*d/e^7/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^
(3/2)+4/e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)+4/e^3*d/(e^2)^(1/2)*arctan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^
2*e^2)^(1/2)*x)+3/e^6/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\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^3*(d^2 - e^2*x^2)^(1/2))/(d + e*x)^4,x)
[Out]
int((x^3*(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^{3} \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**3*(-e**2*x**2+d**2)**(1/2)/(e*x+d)**4,x)
[Out]
Integral(x**3*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x)**4, x)
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