3.2.79 \(\int \frac {x^5}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=177 \[ \frac {127 d^2 (d-e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {3 d \sqrt {d^2-e^2 x^2}}{e^6}+\frac {13 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^6}-\frac {x \sqrt {d^2-e^2 x^2}}{2 e^5}+\frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.44, antiderivative size = 177, 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 = {852, 1635, 1815, 641, 217, 203} \begin {gather*} \frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {127 d^2 (d-e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {3 d \sqrt {d^2-e^2 x^2}}{e^6}-\frac {x \sqrt {d^2-e^2 x^2}}{2 e^5}+\frac {13 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(d^4*(d - e*x)^3)/(5*e^6*(d^2 - e^2*x^2)^(5/2)) - (23*d^3*(d - e*x)^2)/(15*e^6*(d^2 - e^2*x^2)^(3/2)) + (127*d
^2*(d - e*x))/(15*e^6*Sqrt[d^2 - e^2*x^2]) + (3*d*Sqrt[d^2 - e^2*x^2])/e^6 - (x*Sqrt[d^2 - e^2*x^2])/(2*e^5) +
(13*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^6)

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]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^5}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx &=\int \frac {x^5 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d-e x)^2 \left (-\frac {3 d^5}{e^5}+\frac {5 d^4 x}{e^4}-\frac {5 d^3 x^2}{e^3}+\frac {5 d^2 x^3}{e^2}-\frac {5 d x^4}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {(d-e x) \left (-\frac {37 d^5}{e^5}+\frac {45 d^4 x}{e^4}-\frac {30 d^3 x^2}{e^3}+\frac {15 d^2 x^3}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {127 d^2 (d-e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-\frac {90 d^5}{e^5}+\frac {45 d^4 x}{e^4}-\frac {15 d^3 x^2}{e^3}}{\sqrt {d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=\frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {127 d^2 (d-e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}-\frac {x \sqrt {d^2-e^2 x^2}}{2 e^5}+\frac {\int \frac {\frac {195 d^5}{e^3}-\frac {90 d^4 x}{e^2}}{\sqrt {d^2-e^2 x^2}} \, dx}{30 d^3 e^2}\\ &=\frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {127 d^2 (d-e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {3 d \sqrt {d^2-e^2 x^2}}{e^6}-\frac {x \sqrt {d^2-e^2 x^2}}{2 e^5}+\frac {\left (13 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^5}\\ &=\frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {127 d^2 (d-e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {3 d \sqrt {d^2-e^2 x^2}}{e^6}-\frac {x \sqrt {d^2-e^2 x^2}}{2 e^5}+\frac {\left (13 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^5}\\ &=\frac {d^4 (d-e x)^3}{5 e^6 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {23 d^3 (d-e x)^2}{15 e^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {127 d^2 (d-e x)}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {3 d \sqrt {d^2-e^2 x^2}}{e^6}-\frac {x \sqrt {d^2-e^2 x^2}}{2 e^5}+\frac {13 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^6}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 98, normalized size = 0.55 \begin {gather*} \frac {195 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 (304 d^4+717 d^3 e x+479 d^2 e^2 x^2+45 d e^3 x^3-15 e^4 x^4\right )}{(d+e x)^3}}{30 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(304*d^4 + 717*d^3*e*x + 479*d^2*e^2*x^2 + 45*d*e^3*x^3 - 15*e^4*x^4))/(d + e*x)^3 + 195
*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(30*e^6)

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IntegrateAlgebraic [A]  time = 0.69, size = 121, normalized size = 0.68 \begin {gather*} \frac {13 d^2 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{2 e^7}+\frac {\sqrt {d^2-e^2 x^2} \left (304 d^4+717 d^3 e x+479 d^2 e^2 x^2+45 d e^3 x^3-15 e^4 x^4\right )}{30 e^6 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/((d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(304*d^4 + 717*d^3*e*x + 479*d^2*e^2*x^2 + 45*d*e^3*x^3 - 15*e^4*x^4))/(30*e^6*(d + e*x)^
3) + (13*d^2*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*e^7)

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fricas [A]  time = 0.43, size = 190, normalized size = 1.07 \begin {gather*} \frac {304 \, d^{2} e^{3} x^{3} + 912 \, d^{3} e^{2} x^{2} + 912 \, d^{4} e x + 304 \, d^{5} - 390 \, {\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} - 45 \, d e^{3} x^{3} - 479 \, d^{2} e^{2} x^{2} - 717 \, d^{3} e x - 304 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

1/30*(304*d^2*e^3*x^3 + 912*d^3*e^2*x^2 + 912*d^4*e*x + 304*d^5 - 390*(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 - 45*d*e^3*x^3 - 479*d^2*e^2*x^2 - 717*d^3*e*x
- 304*d^4)*sqrt(-e^2*x^2 + d^2))/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6)

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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^5/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (3*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)^3-18*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)-8*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)^2+5*
d^2*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^5-9*d^2*exp(1)^4*exp(2)^3+6*d^2*(-1/2
*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^5+6*d^2*exp(2)^5-19/2*d^2*(-2*d*exp(1)-2*sqrt(
d^2-x^2*exp(2))*exp(1))*exp(2)^5/x/exp(2)+14*d^2*(-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*e
xp(2))*exp(1))/x+exp(2))^2/(-exp(1)^12+2*exp(1)^8*exp(2)^2-exp(1)^4*exp(2)^4)+1/2*(-58*d^2*exp(1)^4*exp(2)^3+2
4*d^2*exp(2)^5+40*d^2*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(-exp(1)^4+exp(2)^2)/(exp(1)^14-2*exp(1)^10*exp(2)^2+exp(1)^6*exp(2)^4)+13/2*d^2*sign(
d)*asin(x*exp(2)/d/exp(1))/exp(1)^6+2*(-2*exp(1)^11*1/8/exp(1)^16*x+12*exp(1)^10*d*1/8/exp(1)^16)*sqrt(-exp(2)
*x^2+d^2)

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maple [A]  time = 0.02, size = 212, normalized size = 1.20 \begin {gather*} \frac {13 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{5}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, x}{2 e^{5}}+\frac {\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{4}}{5 \left (x +\frac {d}{e}\right )^{3} e^{9}}-\frac {23 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3}}{15 \left (x +\frac {d}{e}\right )^{2} e^{8}}+\frac {127 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{2}}{15 \left (x +\frac {d}{e}\right ) e^{7}}+\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, d}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/2*x*(-e^2*x^2+d^2)^(1/2)/e^5+13/2/(e^2)^(1/2)*d^2/e^5*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+3*d*(-e^2*
x^2+d^2)^(1/2)/e^6+1/5*d^4/e^9/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)-23/15*d^3/e^8/(x+d/e)^2*(2*(x+d/e
)*d*e-(x+d/e)^2*e^2)^(1/2)+127/15/e^7*d^2/(x+d/e)*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)

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maxima [A]  time = 1.00, size = 185, normalized size = 1.05 \begin {gather*} \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{5 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} - \frac {23 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{15 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} + \frac {127 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{15 \, {\left (e^{7} x + d e^{6}\right )}} + \frac {13 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{6}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} x}{2 \, e^{5}} + \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(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^4/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6) - 23/15*sqrt(-e^2*x^2 + d^2)*d^3/
(e^8*x^2 + 2*d*e^7*x + d^2*e^6) + 127/15*sqrt(-e^2*x^2 + d^2)*d^2/(e^7*x + d*e^6) + 13/2*d^2*arcsin(e*x/d)/e^6
- 1/2*sqrt(-e^2*x^2 + d^2)*x/e^5 + 3*sqrt(-e^2*x^2 + d^2)*d/e^6

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5}{\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^5/((d^2 - e^2*x^2)^(1/2)*(d + e*x)^3),x)

[Out]

int(x^5/((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^{5}}{\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**5/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(x**5/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)

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