3.200 \(\int \frac{x^3 (d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=192 \[ \frac{101 d^4 \sqrt{d^2-e^2 x^2}}{5 e^4}-\frac{19 d^3 x \sqrt{d^2-e^2 x^2}}{2 e^3}+\frac{18 d^2 x^2 \sqrt{d^2-e^2 x^2}}{5 e^2}+\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}-\frac{d x^3 \sqrt{d^2-e^2 x^2}}{e}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}+\frac{27 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^4} \]

[Out]

(d^2*(d - e*x)^4)/(e^4*Sqrt[d^2 - e^2*x^2]) + (101*d^4*Sqrt[d^2 - e^2*x^2])/(5*e^4) - (19*d^3*x*Sqrt[d^2 - e^2
*x^2])/(2*e^3) + (18*d^2*x^2*Sqrt[d^2 - e^2*x^2])/(5*e^2) - (d*x^3*Sqrt[d^2 - e^2*x^2])/e + (x^4*Sqrt[d^2 - e^
2*x^2])/5 + (27*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^4)

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Rubi [A]  time = 0.435268, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {852, 1635, 1815, 641, 217, 203} \[ \frac{101 d^4 \sqrt{d^2-e^2 x^2}}{5 e^4}-\frac{19 d^3 x \sqrt{d^2-e^2 x^2}}{2 e^3}+\frac{18 d^2 x^2 \sqrt{d^2-e^2 x^2}}{5 e^2}+\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}-\frac{d x^3 \sqrt{d^2-e^2 x^2}}{e}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}+\frac{27 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

(d^2*(d - e*x)^4)/(e^4*Sqrt[d^2 - e^2*x^2]) + (101*d^4*Sqrt[d^2 - e^2*x^2])/(5*e^4) - (19*d^3*x*Sqrt[d^2 - e^2
*x^2])/(2*e^3) + (18*d^2*x^2*Sqrt[d^2 - e^2*x^2])/(5*e^2) - (d*x^3*Sqrt[d^2 - e^2*x^2])/e + (x^4*Sqrt[d^2 - e^
2*x^2])/5 + (27*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^4)

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 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

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^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=\int \frac{x^3 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{(d-e x)^3 \left (-\frac{4 d^3}{e^3}+\frac{d^2 x}{e^2}-\frac{d x^2}{e}\right )}{\sqrt{d^2-e^2 x^2}} \, dx}{d}\\ &=\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}+\frac{\int \frac{\frac{20 d^6}{e}-65 d^5 x+80 d^4 e x^2-54 d^3 e^2 x^3+20 d^2 e^3 x^4}{\sqrt{d^2-e^2 x^2}} \, dx}{5 d e^2}\\ &=\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}-\frac{d x^3 \sqrt{d^2-e^2 x^2}}{e}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}-\frac{\int \frac{-80 d^6 e+260 d^5 e^2 x-380 d^4 e^3 x^2+216 d^3 e^4 x^3}{\sqrt{d^2-e^2 x^2}} \, dx}{20 d e^4}\\ &=\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}+\frac{18 d^2 x^2 \sqrt{d^2-e^2 x^2}}{5 e^2}-\frac{d x^3 \sqrt{d^2-e^2 x^2}}{e}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}+\frac{\int \frac{240 d^6 e^3-1212 d^5 e^4 x+1140 d^4 e^5 x^2}{\sqrt{d^2-e^2 x^2}} \, dx}{60 d e^6}\\ &=\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}-\frac{19 d^3 x \sqrt{d^2-e^2 x^2}}{2 e^3}+\frac{18 d^2 x^2 \sqrt{d^2-e^2 x^2}}{5 e^2}-\frac{d x^3 \sqrt{d^2-e^2 x^2}}{e}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}-\frac{\int \frac{-1620 d^6 e^5+2424 d^5 e^6 x}{\sqrt{d^2-e^2 x^2}} \, dx}{120 d e^8}\\ &=\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}+\frac{101 d^4 \sqrt{d^2-e^2 x^2}}{5 e^4}-\frac{19 d^3 x \sqrt{d^2-e^2 x^2}}{2 e^3}+\frac{18 d^2 x^2 \sqrt{d^2-e^2 x^2}}{5 e^2}-\frac{d x^3 \sqrt{d^2-e^2 x^2}}{e}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}+\frac{\left (27 d^5\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{2 e^3}\\ &=\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}+\frac{101 d^4 \sqrt{d^2-e^2 x^2}}{5 e^4}-\frac{19 d^3 x \sqrt{d^2-e^2 x^2}}{2 e^3}+\frac{18 d^2 x^2 \sqrt{d^2-e^2 x^2}}{5 e^2}-\frac{d x^3 \sqrt{d^2-e^2 x^2}}{e}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}+\frac{\left (27 d^5\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^3}\\ &=\frac{d^2 (d-e x)^4}{e^4 \sqrt{d^2-e^2 x^2}}+\frac{101 d^4 \sqrt{d^2-e^2 x^2}}{5 e^4}-\frac{19 d^3 x \sqrt{d^2-e^2 x^2}}{2 e^3}+\frac{18 d^2 x^2 \sqrt{d^2-e^2 x^2}}{5 e^2}-\frac{d x^3 \sqrt{d^2-e^2 x^2}}{e}+\frac{1}{5} x^4 \sqrt{d^2-e^2 x^2}+\frac{27 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e^4}\\ \end{align*}

Mathematica [A]  time = 0.157789, size = 109, normalized size = 0.57 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (-29 d^3 e^2 x^2+16 d^2 e^3 x^3+77 d^4 e x+212 d^5-8 d e^4 x^4+2 e^5 x^5\right )}{d+e x}+135 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{10 e^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(212*d^5 + 77*d^4*e*x - 29*d^3*e^2*x^2 + 16*d^2*e^3*x^3 - 8*d*e^4*x^4 + 2*e^5*x^5))/(d +
e*x) + 135*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(10*e^4)

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Maple [A]  time = 0.066, size = 285, normalized size = 1.5 \begin{align*}{\frac{36}{5\,{e}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}+9\,{\frac{dx}{{e}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{3/2}}+{\frac{27\,{d}^{3}x}{2\,{e}^{3}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{27\,{d}^{5}}{2\,{e}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{{d}^{2}}{{e}^{8}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-4}}+6\,{\frac{d}{{e}^{7}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{7/2} \left ({\frac{d}{e}}+x \right ) ^{-3}}+7\,{\frac{1}{{e}^{6}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{7/2} \left ({\frac{d}{e}}+x \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x)

[Out]

36/5/e^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)+9/e^3*d*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+27/2/e^3*d^3*(-(d
/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x+27/2/e^3*d^5/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x)
)^(1/2))+d^2/e^8/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)+6*d/e^7/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x
))^(7/2)+7/e^6/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)

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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^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68878, size = 292, normalized size = 1.52 \begin{align*} \frac{212 \, d^{5} e x + 212 \, d^{6} - 270 \,{\left (d^{5} e x + d^{6}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (2 \, e^{5} x^{5} - 8 \, d e^{4} x^{4} + 16 \, d^{2} e^{3} x^{3} - 29 \, d^{3} e^{2} x^{2} + 77 \, d^{4} e x + 212 \, d^{5}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{10 \,{\left (e^{5} x + d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/10*(212*d^5*e*x + 212*d^6 - 270*(d^5*e*x + d^6)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^5*x^5 - 8*d
*e^4*x^4 + 16*d^2*e^3*x^3 - 29*d^3*e^2*x^2 + 77*d^4*e*x + 212*d^5)*sqrt(-e^2*x^2 + d^2))/(e^5*x + d*e^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{5}{2}}}{\left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-e**2*x**2+d**2)**(5/2)/(e*x+d)**4,x)

[Out]

Integral(x**3*(-(-d + e*x)*(d + e*x))**(5/2)/(d + e*x)**4, 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^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError