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

Optimal. Leaf size=128 \[ \frac{x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x^2 (4 d-5 e x)}{3 e^4 \sqrt{d^2-e^2 x^2}}-\frac{(16 d-15 e x) \sqrt{d^2-e^2 x^2}}{6 e^6}-\frac{5 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^6} \]

[Out]

(x^4*(d - e*x))/(3*e^2*(d^2 - e^2*x^2)^(3/2)) - (x^2*(4*d - 5*e*x))/(3*e^4*Sqrt[d^2 - e^2*x^2]) - ((16*d - 15*
e*x)*Sqrt[d^2 - e^2*x^2])/(6*e^6) - (5*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^6)

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Rubi [A]  time = 0.105259, antiderivative size = 128, normalized size of antiderivative = 1., 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, 780, 217, 203} \[ \frac{x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x^2 (4 d-5 e x)}{3 e^4 \sqrt{d^2-e^2 x^2}}-\frac{(16 d-15 e x) \sqrt{d^2-e^2 x^2}}{6 e^6}-\frac{5 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(d - e*x))/(3*e^2*(d^2 - e^2*x^2)^(3/2)) - (x^2*(4*d - 5*e*x))/(3*e^4*Sqrt[d^2 - e^2*x^2]) - ((16*d - 15*
e*x)*Sqrt[d^2 - e^2*x^2])/(6*e^6) - (5*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^6)

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]))

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{x^5}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx &=\int \frac{x^5 (d-e x)}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx\\ &=\frac{x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\int \frac{x^3 \left (4 d^3-5 d^2 e x\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{3 d^2 e^2}\\ &=\frac{x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x^2 (4 d-5 e x)}{3 e^4 \sqrt{d^2-e^2 x^2}}+\frac{\int \frac{x \left (8 d^5-15 d^4 e x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{3 d^4 e^4}\\ &=\frac{x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x^2 (4 d-5 e x)}{3 e^4 \sqrt{d^2-e^2 x^2}}-\frac{(16 d-15 e x) \sqrt{d^2-e^2 x^2}}{6 e^6}-\frac{\left (5 d^2\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{2 e^5}\\ &=\frac{x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x^2 (4 d-5 e x)}{3 e^4 \sqrt{d^2-e^2 x^2}}-\frac{(16 d-15 e x) \sqrt{d^2-e^2 x^2}}{6 e^6}-\frac{\left (5 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{x^4 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x^2 (4 d-5 e x)}{3 e^4 \sqrt{d^2-e^2 x^2}}-\frac{(16 d-15 e x) \sqrt{d^2-e^2 x^2}}{6 e^6}-\frac{5 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^6}\\ \end{align*}

Mathematica [A]  time = 0.185987, size = 106, normalized size = 0.83 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (-23 d^2 e^2 x^2+d^3 e x+16 d^4-3 d e^3 x^3+3 e^4 x^4\right )}{(e x-d) (d+e x)^2}-15 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{6 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(16*d^4 + d^3*e*x - 23*d^2*e^2*x^2 - 3*d*e^3*x^3 + 3*e^4*x^4))/((-d + e*x)*(d + e*x)^2)
- 15*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(6*e^6)

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Maple [A]  time = 0.066, size = 208, normalized size = 1.6 \begin{align*} -{\frac{{x}^{3}}{2\,{e}^{3}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+{\frac{7\,{d}^{2}x}{2\,{e}^{5}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{5\,{d}^{2}}{2\,{e}^{5}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{d{x}^{2}}{{e}^{4}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-3\,{\frac{{d}^{3}}{{e}^{6}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}+{\frac{{d}^{4}}{3\,{e}^{7}} \left ({\frac{d}{e}}+x \right ) ^{-1}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}}-{\frac{2\,{d}^{2}x}{3\,{e}^{5}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66397, size = 381, normalized size = 2.98 \begin{align*} -\frac{16 \, d^{2} e^{3} x^{3} + 16 \, d^{3} e^{2} x^{2} - 16 \, d^{4} e x - 16 \, d^{5} - 30 \,{\left (d^{2} e^{3} x^{3} + 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 (3 \, e^{4} x^{4} - 3 \, d e^{3} x^{3} - 23 \, d^{2} e^{2} x^{2} + d^{3} e x + 16 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \,{\left (e^{9} x^{3} + d e^{8} x^{2} - d^{2} e^{7} x - d^{3} e^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(16*d^2*e^3*x^3 + 16*d^3*e^2*x^2 - 16*d^4*e*x - 16*d^5 - 30*(d^2*e^3*x^3 + d^3*e^2*x^2 - d^4*e*x - d^5)*a
rctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (3*e^4*x^4 - 3*d*e^3*x^3 - 23*d^2*e^2*x^2 + d^3*e*x + 16*d^4)*sqrt(
-e^2*x^2 + d^2))/(e^9*x^3 + d*e^8*x^2 - d^2*e^7*x - d^3*e^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{3}{2}} \left (d + e x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(e*x+d)/(-e**2*x**2+d**2)**(3/2),x)

[Out]

Integral(x**5/((-(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[undef, undef, undef, undef, undef, 1]