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

Optimal. Leaf size=160 \[ \frac {19 d^2 (d-e x)^3}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {6 d (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {(20 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^5}-\frac {19 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5}-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \]

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Rubi [A]  time = 0.42, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {852, 1635, 780, 217, 203} \begin {gather*} -\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {19 d^2 (d-e x)^3}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {6 d (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {(20 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^5}-\frac {19 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^3*(d - e*x)^4)/(5*e^5*(d^2 - e^2*x^2)^(5/2)) + (19*d^2*(d - e*x)^3)/(15*e^5*(d^2 - e^2*x^2)^(3/2)) - (6*d*
(d - e*x)^2)/(e^5*Sqrt[d^2 - e^2*x^2]) - ((20*d - e*x)*Sqrt[d^2 - e^2*x^2])/(2*e^5) - (19*d^2*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]])/(2*e^5)

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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^4 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx &=\int \frac {x^4 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x)^3 \left (\frac {4 d^4}{e^4}-\frac {5 d^3 x}{e^3}+\frac {5 d^2 x^2}{e^2}-\frac {5 d x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {19 d^2 (d-e x)^3}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {(d-e x)^2 \left (\frac {45 d^4}{e^4}-\frac {30 d^3 x}{e^3}+\frac {15 d^2 x^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {19 d^2 (d-e x)^3}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {6 d (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\left (\frac {135 d^4}{e^4}-\frac {15 d^3 x}{e^3}\right ) (d-e x)}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {19 d^2 (d-e x)^3}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {6 d (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {(20 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^5}-\frac {\left (19 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^4}\\ &=-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {19 d^2 (d-e x)^3}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {6 d (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {(20 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^5}-\frac {\left (19 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}\\ &=-\frac {d^3 (d-e x)^4}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {19 d^2 (d-e x)^3}{15 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {6 d (d-e x)^2}{e^5 \sqrt {d^2-e^2 x^2}}-\frac {(20 d-e x) \sqrt {d^2-e^2 x^2}}{2 e^5}-\frac {19 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^5}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 98, normalized size = 0.61 \begin {gather*} -\frac {285 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (448 d^4+1059 d^3 e x+713 d^2 e^2 x^2+75 d e^3 x^3-15 e^4 x^4\right )}{(d+e x)^3}}{30 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*((Sqrt[d^2 - e^2*x^2]*(448*d^4 + 1059*d^3*e*x + 713*d^2*e^2*x^2 + 75*d*e^3*x^3 - 15*e^4*x^4))/(d + e*x)^
3 + 285*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^5

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IntegrateAlgebraic [A]  time = 0.82, size = 121, normalized size = 0.76 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-448 d^4-1059 d^3 e x-713 d^2 e^2 x^2-75 d e^3 x^3+15 e^4 x^4\right )}{30 e^5 (d+e x)^3}-\frac {19 d^2 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{2 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-448*d^4 - 1059*d^3*e*x - 713*d^2*e^2*x^2 - 75*d*e^3*x^3 + 15*e^4*x^4))/(30*e^5*(d + e*x
)^3) - (19*d^2*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*e^6)

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fricas [A]  time = 0.43, size = 190, normalized size = 1.19 \begin {gather*} -\frac {448 \, d^{2} e^{3} x^{3} + 1344 \, d^{3} e^{2} x^{2} + 1344 \, d^{4} e x + 448 \, d^{5} - 570 \, {\left (d^{2} e^{3} x^{3} + 3 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x + d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (15 \, e^{4} x^{4} - 75 \, d e^{3} x^{3} - 713 \, d^{2} e^{2} x^{2} - 1059 \, d^{3} e x - 448 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(448*d^2*e^3*x^3 + 1344*d^3*e^2*x^2 + 1344*d^4*e*x + 448*d^5 - 570*(d^2*e^3*x^3 + 3*d^3*e^2*x^2 + 3*d^4*
e*x + d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (15*e^4*x^4 - 75*d*e^3*x^3 - 713*d^2*e^2*x^2 - 1059*d^3
*e*x - 448*d^4)*sqrt(-e^2*x^2 + d^2))/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (84*d^2*(-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+18*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/ex
p(2))^5*exp(1)^10*exp(2)^3-192*d^2*(-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)^2-180*d^2*(-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)^3-42*d^2*(-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+228*d^2*(-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)^2-104*d^2*(-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)^3-192*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^8*e
xp(2)^4-48*d^2*(-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)^5-396*d^2*(-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)^3-168*d^2*(-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+60*d^2*(-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+24*d^2*(-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)^6-510*d^2*(-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)^4-246*d^2*(-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+57*d^2*(-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+27*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/ex
p(2))^5*exp(2)^8+156*d^2*(-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+180*d
^2*(-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+26*d^2*exp(1)^8*exp(2)^4+66
*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(2)^8+180*d^2*(-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)^6-48*d^2*exp(1)^6*exp(2)^5+216*d^2*(-1/2*(-2*d*exp(1)-2*sq
rt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^8-65*d^2*exp(1)^4*exp(2)^6+132*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2
-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^8+66*d^2*exp(2)^8+104*d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*ex
p(1))/x/exp(2))^3*exp(1)^14*exp(2)-189/2*d^2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^8/x/exp(2)-78*
d^2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*exp(2)^6/x/exp(2)+171*d^2*(-2*d*exp(1)-2*sqrt(d^2-x^2
*exp(2))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)+123*d^2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2
)^4/x/exp(2)-69*d^2*(-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)^15-6*exp(1)^11*exp(2)^2-6*exp(1)^9*exp(2)^3+3*exp(1)^7*exp(2)^4+3*exp(1)^5*exp(2)^5+3*exp(1)^13*exp(2
))+1/2*(64*d^2*exp(1)^10*exp(2)^2+80*d^2*exp(1)^8*exp(2)^3-88*d^2*exp(1)^6*exp(2)^4-102*d^2*exp(1)^4*exp(2)^5+
76*d^2*exp(2)^7-16*d^2*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)^17+2*exp(1)^13*exp(2)^2+2*exp(1)^11*exp(2)^3-exp(1)^9*
exp(2)^4-exp(1)^7*exp(2)^5-exp(1)^15*exp(2))-19/2*d^2*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)^5+2*(2*exp(1)^9*1
/8/exp(1)^13*x-16*exp(1)^8*d*1/8/exp(1)^13)*sqrt(d^2-x^2*exp(2))

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maple [A]  time = 0.01, size = 273, normalized size = 1.71 \begin {gather*} -\frac {10 d^{2} \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^{4}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{4}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x}{2 e^{4}}-\frac {10 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d}{e^{5}}-\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^{3}}{5 \left (x +\frac {d}{e}\right )^{4} e^{9}}+\frac {19 \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}}{15 \left (x +\frac {d}{e}\right )^{3} e^{8}}-\frac {6 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d}{\left (x +\frac {d}{e}\right )^{2} e^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x)

[Out]

1/2/e^4*x*(-e^2*x^2+d^2)^(1/2)+1/2/e^4*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/5*d^3/e^9/
(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)+19/15/e^8*d^2/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)-10/e
^5*d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-10/e^4*d^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)-6/e^7*d/(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^4*(-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^4\,\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^4*(d^2 - e^2*x^2)^(1/2))/(d + e*x)^4,x)

[Out]

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

[Out]

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

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