3.2.80 \(\int \frac {x^4}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=146 \[ \frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}} \]

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Rubi [A]  time = 0.37, antiderivative size = 146, 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, 641, 217, 203} \begin {gather*} -\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-(d^3*(d - e*x)^3)/(5*e^5*(d^2 - e^2*x^2)^(5/2)) + (6*d^2*(d - e*x)^2)/(5*e^5*(d^2 - e^2*x^2)^(3/2)) - (24*d*(
d - e*x))/(5*e^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/e^5 - (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^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 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 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}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {x^4 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x)^2 \left (\frac {3 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)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {(d-e x) \left (\frac {27 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)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {\frac {45 d^4}{e^4}-\frac {15 d^3 x}{e^3}}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^4}\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {(3 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^4}\\ &=-\frac {d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {24 d (d-e x)}{5 e^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e^5}-\frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^5}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 85, normalized size = 0.58 \begin {gather*} -\frac {15 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (24 d^3+57 d^2 e x+39 d e^2 x^2+5 e^3 x^3\right )}{(d+e x)^3}}{5 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

-1/5*((Sqrt[d^2 - e^2*x^2]*(24*d^3 + 57*d^2*e*x + 39*d*e^2*x^2 + 5*e^3*x^3))/(d + e*x)^3 + 15*d*ArcTan[(e*x)/S
qrt[d^2 - e^2*x^2]])/e^5

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IntegrateAlgebraic [A]  time = 0.72, size = 106, normalized size = 0.73 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-24 d^3-57 d^2 e x-39 d e^2 x^2-5 e^3 x^3\right )}{5 e^5 (d+e x)^3}-\frac {3 d \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-24*d^3 - 57*d^2*e*x - 39*d*e^2*x^2 - 5*e^3*x^3))/(5*e^5*(d + e*x)^3) - (3*d*Sqrt[-e^2]*
Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^6

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fricas [A]  time = 0.41, size = 174, normalized size = 1.19 \begin {gather*} -\frac {24 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} + 72 \, d^{3} e x + 24 \, d^{4} - 30 \, {\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 (5 \, e^{3} x^{3} + 39 \, d e^{2} x^{2} + 57 \, d^{2} e x + 24 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\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*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/5*(24*d*e^3*x^3 + 72*d^2*e^2*x^2 + 72*d^3*e*x + 24*d^4 - 30*(d*e^3*x^3 + 3*d^2*e^2*x^2 + 3*d^3*e*x + d^4)*a
rctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (5*e^3*x^3 + 39*d*e^2*x^2 + 57*d^2*e*x + 24*d^3)*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*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-d*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^3+14*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)+6*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)^2-3*d*(-1/2
*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^5+7*d*exp(1)^4*exp(2)^3-4*d*(-1/2*(-2*d*exp(1)
-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^5-4*d*exp(2)^5+13/2*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*e
xp(1))*exp(2)^5/x/exp(2)-11*d*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^2/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+ex
p(2))^2/(-exp(1)^11+2*exp(1)^7*exp(2)^2-exp(1)*exp(2)^5)+1/2*(30*d*exp(1)^4*exp(2)^3-12*d*exp(2)^5-24*d*exp(1)
^8*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(-ex
p(1)^4+exp(2)^2)/(exp(1)^13-2*exp(1)^9*exp(2)^2+exp(1)^5*exp(2)^4)-3*d*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)^
5-4*exp(1)^4*1/4/exp(1)^9*sqrt(-exp(2)*x^2+d^2)

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maple [A]  time = 0.02, size = 187, normalized size = 1.28 \begin {gather*} -\frac {3 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{4}}-\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3}}{5 \left (x +\frac {d}{e}\right )^{3} e^{8}}+\frac {6 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2}}{5 \left (x +\frac {d}{e}\right )^{2} e^{7}}-\frac {24 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d}{5 \left (x +\frac {d}{e}\right ) e^{6}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-(-e^2*x^2+d^2)^(1/2)/e^5-3/(e^2)^(1/2)*d/e^4*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/5*d^3/e^8/(x+d/e)^3
*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)+6/5*d^2/e^7/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-24/5/e^6*d/(x+d
/e)*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)

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maxima [A]  time = 0.98, size = 160, normalized size = 1.10 \begin {gather*} -\frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{5 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} + \frac {6 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{5 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} - \frac {24 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{5 \, {\left (e^{6} x + d e^{5}\right )}} - \frac {3 \, d \arcsin \left (\frac {e x}{d}\right )}{e^{5}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-1/5*sqrt(-e^2*x^2 + d^2)*d^3/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5) + 6/5*sqrt(-e^2*x^2 + d^2)*d^2/(
e^7*x^2 + 2*d*e^6*x + d^2*e^5) - 24/5*sqrt(-e^2*x^2 + d^2)*d/(e^6*x + d*e^5) - 3*d*arcsin(e*x/d)/e^5 - sqrt(-e
^2*x^2 + d^2)/e^5

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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 )}^3} \,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)^3),x)

[Out]

int(x^4/((d^2 - e^2*x^2)^(1/2)*(d + e*x)^3), 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 )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)

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