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

Optimal. Leaf size=113 \[ \frac{x^3 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x (3 d-4 e x)}{3 e^4 \sqrt{d^2-e^2 x^2}}+\frac{8 \sqrt{d^2-e^2 x^2}}{3 e^5}+\frac{d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5} \]

[Out]

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

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Rubi [A]  time = 0.0938843, antiderivative size = 113, 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, 641, 217, 203} \[ \frac{x^3 (d-e x)}{3 e^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{x (3 d-4 e x)}{3 e^4 \sqrt{d^2-e^2 x^2}}+\frac{8 \sqrt{d^2-e^2 x^2}}{3 e^5}+\frac{d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.150518, size = 93, normalized size = 0.82 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (5 d^2 e x+8 d^3-7 d e^2 x^2-3 e^3 x^3\right )}{(d-e x) (d+e x)^2}+3 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{3 e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(8*d^3 + 5*d^2*e*x - 7*d*e^2*x^2 - 3*e^3*x^3))/((d - e*x)*(d + e*x)^2) + 3*d*ArcTan[(e*x
)/Sqrt[d^2 - e^2*x^2]])/(3*e^5)

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Maple [A]  time = 0.062, size = 179, normalized size = 1.6 \begin{align*} -{\frac{{x}^{2}}{{e}^{3}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+3\,{\frac{{d}^{2}}{{e}^{5}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}-2\,{\frac{dx}{{e}^{4}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}+{\frac{d}{{e}^{4}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{{d}^{3}}{3\,{e}^{6}} \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\,dx}{3\,{e}^{4}}{\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^4/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x)

[Out]

-1/e^3*x^2/(-e^2*x^2+d^2)^(1/2)+3*d^2/e^5/(-e^2*x^2+d^2)^(1/2)-2*d/e^4*x/(-e^2*x^2+d^2)^(1/2)+d/e^4/(e^2)^(1/2
)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/3*d^3/e^6/(d/e+x)/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)+2/3*d/e^
4/(-(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^4/(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 = 1.71407, size = 346, normalized size = 3.06 \begin{align*} \frac{8 \, d e^{3} x^{3} + 8 \, d^{2} e^{2} x^{2} - 8 \, d^{3} e x - 8 \, d^{4} - 6 \,{\left (d e^{3} x^{3} + d^{2} e^{2} x^{2} - d^{3} e x - d^{4}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (3 \, e^{3} x^{3} + 7 \, d e^{2} x^{2} - 5 \, d^{2} e x - 8 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{3 \,{\left (e^{8} x^{3} + d e^{7} x^{2} - d^{2} e^{6} x - d^{3} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\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**4/(e*x+d)/(-e**2*x**2+d**2)**(3/2),x)

[Out]

Integral(x**4/((-(-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}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[undef, undef, undef, undef, 1]