3.2.38 \(\int \frac {x^5}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=122 \[ \frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {8 d-15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}-\frac {x^2 (4 d-5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.10, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {850, 819, 778, 217, 203} \begin {gather*} \frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d-5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d-15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(d - e*x))/(5*e^2*(d^2 - e^2*x^2)^(5/2)) - (x^2*(4*d - 5*e*x))/(15*e^4*(d^2 - e^2*x^2)^(3/2)) + (8*d - 15
*e*x)/(15*e^6*Sqrt[d^2 - e^2*x^2]) + ArcTan[(e*x)/Sqrt[d^2 - e^2*x^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 778

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

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 850

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

Rubi steps

\begin {align*} \int \frac {x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\int \frac {x^5 (d-e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=\frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {x^3 \left (4 d^3-5 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2}\\ &=\frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d-5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {x \left (8 d^5-15 d^4 e x\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4 e^4}\\ &=\frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d-5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d-15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^5}\\ &=\frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d-5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d-15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}+\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^5}\\ &=\frac {x^4 (d-e x)}{5 e^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {x^2 (4 d-5 e x)}{15 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d-15 e x}{15 e^6 \sqrt {d^2-e^2 x^2}}+\frac {\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^6}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 103, normalized size = 0.84 \begin {gather*} \frac {15 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (8 d^4-7 d^3 e x-27 d^2 e^2 x^2+8 d e^3 x^3+23 e^4 x^4\right )}{(d-e x)^2 (d+e x)^3}}{15 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(8*d^4 - 7*d^3*e*x - 27*d^2*e^2*x^2 + 8*d*e^3*x^3 + 23*e^4*x^4))/((d - e*x)^2*(d + e*x)^
3) + 15*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(15*e^6)

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IntegrateAlgebraic [A]  time = 0.53, size = 124, normalized size = 1.02 \begin {gather*} \frac {\sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{e^7}+\frac {\sqrt {d^2-e^2 x^2} \left (8 d^4-7 d^3 e x-27 d^2 e^2 x^2+8 d e^3 x^3+23 e^4 x^4\right )}{15 e^6 (e x-d)^2 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(8*d^4 - 7*d^3*e*x - 27*d^2*e^2*x^2 + 8*d*e^3*x^3 + 23*e^4*x^4))/(15*e^6*(-d + e*x)^2*(d
+ e*x)^3) + (Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^7

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fricas [B]  time = 0.43, size = 241, normalized size = 1.98 \begin {gather*} \frac {8 \, e^{5} x^{5} + 8 \, d e^{4} x^{4} - 16 \, d^{2} e^{3} x^{3} - 16 \, d^{3} e^{2} x^{2} + 8 \, d^{4} e x + 8 \, d^{5} - 30 \, {\left (e^{5} x^{5} + d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} - 2 \, d^{3} e^{2} x^{2} + d^{4} e x + d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (23 \, e^{4} x^{4} + 8 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} - 7 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{11} x^{5} + d e^{10} x^{4} - 2 \, d^{2} e^{9} x^{3} - 2 \, d^{3} e^{8} x^{2} + d^{4} e^{7} x + d^{5} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(8*e^5*x^5 + 8*d*e^4*x^4 - 16*d^2*e^3*x^3 - 16*d^3*e^2*x^2 + 8*d^4*e*x + 8*d^5 - 30*(e^5*x^5 + d*e^4*x^4
- 2*d^2*e^3*x^3 - 2*d^3*e^2*x^2 + d^4*e*x + d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (23*e^4*x^4 + 8*d
*e^3*x^3 - 27*d^2*e^2*x^2 - 7*d^3*e*x + 8*d^4)*sqrt(-e^2*x^2 + d^2))/(e^11*x^5 + d*e^10*x^4 - 2*d^2*e^9*x^3 -
2*d^3*e^8*x^2 + d^4*e^7*x + d^5*e^6)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [B]  time = 0.01, size = 259, normalized size = 2.12 \begin {gather*} \frac {x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}-\frac {d \,x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{4}}+\frac {2 d^{2} x}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5}}-\frac {4 d^{2} x}{15 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{5}}+\frac {d^{4}}{5 \left (x +\frac {d}{e}\right ) \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} e^{7}}+\frac {d^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{6}}-\frac {2 x}{3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{5}}-\frac {8 x}{15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, e^{5}}+\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}\, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.06, size = 234, normalized size = 1.92 \begin {gather*} \frac {d^{4}}{5 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{7} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{6}\right )}} + \frac {x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} - \frac {8 \, d x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}} - \frac {4 \, d^{2} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5}} - \frac {x^{2}}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{4}} + \frac {2 \, d^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{6}} - \frac {8 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}} + \frac {\arcsin \left (\frac {e x}{d}\right )}{e^{6}} - \frac {4 \, d}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{6}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{3 \, d e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*d^4/((-e^2*x^2 + d^2)^(3/2)*e^7*x + (-e^2*x^2 + d^2)^(3/2)*d*e^6) + x^3/((-e^2*x^2 + d^2)^(3/2)*e^3) - 8/3
*d*x^2/((-e^2*x^2 + d^2)^(3/2)*e^4) - 4/15*d^2*x/((-e^2*x^2 + d^2)^(3/2)*e^5) - 1/3*x^2/(sqrt(-e^2*x^2 + d^2)*
d*e^4) + 2*d^3/((-e^2*x^2 + d^2)^(3/2)*e^6) - 8/15*x/(sqrt(-e^2*x^2 + d^2)*e^5) + arcsin(e*x/d)/e^6 - 4/3*d/(s
qrt(-e^2*x^2 + d^2)*e^6) - 1/3*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}{{\left (d^2-e^2\,x^2\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(x**5/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

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