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

Optimal. Leaf size=146 \[ -\frac{d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{24 d (d-e x)}{5 e^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{e^5}-\frac{3 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5} \]

[Out]

-(d^3*(d - e*x)^3)/(5*e^5*(d^2 - e^2*x^2)^(5/2)) + (6*d^2*(d - e*x)^2)/(5*e^5*(d^2 - e^2*x^2)^(3/2)) - (24*d*(
d - e*x))/(5*e^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/e^5 - (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^5

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Rubi [A]  time = 0.365028, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {852, 1635, 641, 217, 203} \[ -\frac{d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{24 d (d-e x)}{5 e^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{e^5}-\frac{3 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^3*(d - e*x)^3)/(5*e^5*(d^2 - e^2*x^2)^(5/2)) + (6*d^2*(d - e*x)^2)/(5*e^5*(d^2 - e^2*x^2)^(3/2)) - (24*d*(
d - e*x))/(5*e^5*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/e^5 - (3*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^5

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 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)^3 \sqrt{d^2-e^2 x^2}} \, dx &=\int \frac{x^4 (d-e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac{d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{(d-e x)^2 \left (\frac{3 d^4}{e^4}-\frac{5 d^3 x}{e^3}+\frac{5 d^2 x^2}{e^2}-\frac{5 d x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=-\frac{d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{(d-e x) \left (\frac{27 d^4}{e^4}-\frac{30 d^3 x}{e^3}+\frac{15 d^2 x^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=-\frac{d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{24 d (d-e x)}{5 e^5 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{\frac{45 d^4}{e^4}-\frac{15 d^3 x}{e^3}}{\sqrt{d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=-\frac{d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{24 d (d-e x)}{5 e^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{e^5}-\frac{(3 d) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{e^4}\\ &=-\frac{d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{24 d (d-e x)}{5 e^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{e^5}-\frac{(3 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{d^3 (d-e x)^3}{5 e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{6 d^2 (d-e x)^2}{5 e^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{24 d (d-e x)}{5 e^5 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{e^5}-\frac{3 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5}\\ \end{align*}

Mathematica [A]  time = 0.159434, size = 85, normalized size = 0.58 \[ -\frac{\frac{\sqrt{d^2-e^2 x^2} \left (57 d^2 e x+24 d^3+39 d e^2 x^2+5 e^3 x^3\right )}{(d+e x)^3}+15 d \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{5 e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[d^2 - e^2*x^2]*(24*d^3 + 57*d^2*e*x + 39*d*e^2*x^2 + 5*e^3*x^3))/(d + e*x)^3 + 15*d*ArcTan[(e*x)/Sqrt[
d^2 - e^2*x^2]])/(5*e^5)

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Maple [A]  time = 0.061, size = 187, normalized size = 1.3 \begin{align*} -{\frac{1}{{e}^{5}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-3\,{\frac{d}{{e}^{4}\sqrt{{e}^{2}}}\arctan \left ({\frac{\sqrt{{e}^{2}}x}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}} \right ) }-{\frac{24\,d}{5\,{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}}+{\frac{6\,{d}^{2}}{5\,{e}^{7}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-2}}-{\frac{{d}^{3}}{5\,{e}^{8}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) } \left ({\frac{d}{e}}+x \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-(-e^2*x^2+d^2)^(1/2)/e^5-3*d/e^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-24/5/e^6*d/(d/e+x)*(-
(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)+6/5/e^7*d^2/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)-1/5*d^3/e^8/(d/e
+x)^3*(-(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)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70304, size = 369, normalized size = 2.53 \begin{align*} -\frac{24 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} + 72 \, d^{3} e x + 24 \, d^{4} - 30 \,{\left (d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x + d^{4}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (5 \, e^{3} x^{3} + 39 \, d e^{2} x^{2} + 57 \, d^{2} e x + 24 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{5 \,{\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, 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)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/5*(24*d*e^3*x^3 + 72*d^2*e^2*x^2 + 72*d^3*e*x + 24*d^4 - 30*(d*e^3*x^3 + 3*d^2*e^2*x^2 + 3*d^3*e*x + d^4)*a
rctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (5*e^3*x^3 + 39*d*e^2*x^2 + 57*d^2*e*x + 24*d^3)*sqrt(-e^2*x^2 + d^
2))/(e^8*x^3 + 3*d*e^7*x^2 + 3*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}}{\sqrt{- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), 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)^3/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError