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

Optimal. Leaf size=118 \[ -\frac{d (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{6 e^5}-\frac{4 x^2 \sqrt{d^2-e^2 x^2}}{3 e^3}+\frac{x^3 (d-e x)}{e^2 \sqrt{d^2-e^2 x^2}}-\frac{3 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^5} \]

[Out]

(x^3*(d - e*x))/(e^2*Sqrt[d^2 - e^2*x^2]) - (4*x^2*Sqrt[d^2 - e^2*x^2])/(3*e^3) - (d*(16*d - 9*e*x)*Sqrt[d^2 -
e^2*x^2])/(6*e^5) - (3*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^5)

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Rubi [A]  time = 0.0961943, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {850, 819, 833, 780, 217, 203} \[ -\frac{d (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{6 e^5}-\frac{4 x^2 \sqrt{d^2-e^2 x^2}}{3 e^3}+\frac{x^3 (d-e x)}{e^2 \sqrt{d^2-e^2 x^2}}-\frac{3 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(d - e*x))/(e^2*Sqrt[d^2 - e^2*x^2]) - (4*x^2*Sqrt[d^2 - e^2*x^2])/(3*e^3) - (d*(16*d - 9*e*x)*Sqrt[d^2 -
e^2*x^2])/(6*e^5) - (3*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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^4}{(d+e x) \sqrt{d^2-e^2 x^2}} \, dx &=\int \frac{x^4 (d-e x)}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac{x^3 (d-e x)}{e^2 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{x^2 \left (3 d^3-4 d^2 e x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{d^2 e^2}\\ &=\frac{x^3 (d-e x)}{e^2 \sqrt{d^2-e^2 x^2}}-\frac{4 x^2 \sqrt{d^2-e^2 x^2}}{3 e^3}+\frac{\int \frac{x \left (8 d^4 e-9 d^3 e^2 x\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{3 d^2 e^4}\\ &=\frac{x^3 (d-e x)}{e^2 \sqrt{d^2-e^2 x^2}}-\frac{4 x^2 \sqrt{d^2-e^2 x^2}}{3 e^3}-\frac{d (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{6 e^5}-\frac{\left (3 d^3\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{2 e^4}\\ &=\frac{x^3 (d-e x)}{e^2 \sqrt{d^2-e^2 x^2}}-\frac{4 x^2 \sqrt{d^2-e^2 x^2}}{3 e^3}-\frac{d (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{6 e^5}-\frac{\left (3 d^3\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^4}\\ &=\frac{x^3 (d-e x)}{e^2 \sqrt{d^2-e^2 x^2}}-\frac{4 x^2 \sqrt{d^2-e^2 x^2}}{3 e^3}-\frac{d (16 d-9 e x) \sqrt{d^2-e^2 x^2}}{6 e^5}-\frac{3 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^5}\\ \end{align*}

Mathematica [A]  time = 0.095506, size = 91, normalized size = 0.77 \[ \frac{\sqrt{d^2-e^2 x^2} \left (-7 d^2 e x-16 d^3+d e^2 x^2-2 e^3 x^3\right )-9 d^3 (d+e x) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{6 e^5 (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-16*d^3 - 7*d^2*e*x + d*e^2*x^2 - 2*e^3*x^3) - 9*d^3*(d + e*x)*ArcTan[(e*x)/Sqrt[d^2 - e
^2*x^2]])/(6*e^5*(d + e*x))

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Maple [A]  time = 0.063, size = 147, normalized size = 1.3 \begin{align*} -{\frac{{x}^{2}}{3\,{e}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{5\,{d}^{2}}{3\,{e}^{5}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{dx}{2\,{e}^{4}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{3\,{d}^{3}}{2\,{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}}{{e}^{6}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-1}} \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)^(1/2),x)

[Out]

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

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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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69335, size = 236, normalized size = 2. \begin{align*} -\frac{16 \, d^{3} e x + 16 \, d^{4} - 18 \,{\left (d^{3} e x + d^{4}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (2 \, e^{3} x^{3} - d e^{2} x^{2} + 7 \, d^{2} e x + 16 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \,{\left (e^{6} x + d 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)^(1/2),x, algorithm="fricas")

[Out]

-1/6*(16*d^3*e*x + 16*d^4 - 18*(d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^3*x^3 - d*e^2*
x^2 + 7*d^2*e*x + 16*d^3)*sqrt(-e^2*x^2 + d^2))/(e^6*x + d*e^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{- \left (- d + e x\right ) \left (d + e x\right )} \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)**(1/2),x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError