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

Optimal. Leaf size=182 \[ -\frac{95 d^3 \sqrt{d^2-e^2 x^2}}{8 e^3}-\frac{95 d^2 (d-e x) \sqrt{d^2-e^2 x^2}}{24 e^3}-\frac{19 d (d-e x)^2 \sqrt{d^2-e^2 x^2}}{12 e^3}-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{(d-e x)^3 \sqrt{d^2-e^2 x^2}}{4 e^3}-\frac{95 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3} \]

[Out]

-((d*(d - e*x)^4)/(e^3*Sqrt[d^2 - e^2*x^2])) - (95*d^3*Sqrt[d^2 - e^2*x^2])/(8*e^3) - (95*d^2*(d - e*x)*Sqrt[d
^2 - e^2*x^2])/(24*e^3) - (19*d*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])/(12*e^3) - ((d - e*x)^3*Sqrt[d^2 - e^2*x^2])/
(4*e^3) - (95*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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Rubi [A]  time = 0.219433, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {852, 1635, 795, 671, 641, 217, 203} \[ -\frac{95 d^3 \sqrt{d^2-e^2 x^2}}{8 e^3}-\frac{95 d^2 (d-e x) \sqrt{d^2-e^2 x^2}}{24 e^3}-\frac{19 d (d-e x)^2 \sqrt{d^2-e^2 x^2}}{12 e^3}-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{(d-e x)^3 \sqrt{d^2-e^2 x^2}}{4 e^3}-\frac{95 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(d - e*x)^4)/(e^3*Sqrt[d^2 - e^2*x^2])) - (95*d^3*Sqrt[d^2 - e^2*x^2])/(8*e^3) - (95*d^2*(d - e*x)*Sqrt[d
^2 - e^2*x^2])/(24*e^3) - (19*d*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])/(12*e^3) - ((d - e*x)^3*Sqrt[d^2 - e^2*x^2])/
(4*e^3) - (95*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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 795

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[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

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^2 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=\int \frac{x^2 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{\left (\frac{4 d^2}{e^2}-\frac{d x}{e}\right ) (d-e x)^3}{\sqrt{d^2-e^2 x^2}} \, dx}{d}\\ &=-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{(d-e x)^3 \sqrt{d^2-e^2 x^2}}{4 e^3}-\frac{(19 d) \int \frac{(d-e x)^3}{\sqrt{d^2-e^2 x^2}} \, dx}{4 e^2}\\ &=-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{19 d (d-e x)^2 \sqrt{d^2-e^2 x^2}}{12 e^3}-\frac{(d-e x)^3 \sqrt{d^2-e^2 x^2}}{4 e^3}-\frac{\left (95 d^2\right ) \int \frac{(d-e x)^2}{\sqrt{d^2-e^2 x^2}} \, dx}{12 e^2}\\ &=-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{95 d^2 (d-e x) \sqrt{d^2-e^2 x^2}}{24 e^3}-\frac{19 d (d-e x)^2 \sqrt{d^2-e^2 x^2}}{12 e^3}-\frac{(d-e x)^3 \sqrt{d^2-e^2 x^2}}{4 e^3}-\frac{\left (95 d^3\right ) \int \frac{d-e x}{\sqrt{d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{95 d^3 \sqrt{d^2-e^2 x^2}}{8 e^3}-\frac{95 d^2 (d-e x) \sqrt{d^2-e^2 x^2}}{24 e^3}-\frac{19 d (d-e x)^2 \sqrt{d^2-e^2 x^2}}{12 e^3}-\frac{(d-e x)^3 \sqrt{d^2-e^2 x^2}}{4 e^3}-\frac{\left (95 d^4\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{95 d^3 \sqrt{d^2-e^2 x^2}}{8 e^3}-\frac{95 d^2 (d-e x) \sqrt{d^2-e^2 x^2}}{24 e^3}-\frac{19 d (d-e x)^2 \sqrt{d^2-e^2 x^2}}{12 e^3}-\frac{(d-e x)^3 \sqrt{d^2-e^2 x^2}}{4 e^3}-\frac{\left (95 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^2}\\ &=-\frac{d (d-e x)^4}{e^3 \sqrt{d^2-e^2 x^2}}-\frac{95 d^3 \sqrt{d^2-e^2 x^2}}{8 e^3}-\frac{95 d^2 (d-e x) \sqrt{d^2-e^2 x^2}}{24 e^3}-\frac{19 d (d-e x)^2 \sqrt{d^2-e^2 x^2}}{12 e^3}-\frac{(d-e x)^3 \sqrt{d^2-e^2 x^2}}{4 e^3}-\frac{95 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3}\\ \end{align*}

Mathematica [A]  time = 0.135274, size = 103, normalized size = 0.57 \[ \sqrt{d^2-e^2 x^2} \left (-\frac{8 d^4}{e^3 (d+e x)}+\frac{31 d^2 x}{8 e^2}-\frac{32 d^3}{3 e^3}-\frac{4 d x^2}{3 e}+\frac{x^3}{4}\right )-\frac{95 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[d^2 - e^2*x^2]*((-32*d^3)/(3*e^3) + (31*d^2*x)/(8*e^2) - (4*d*x^2)/(3*e) + x^3/4 - (8*d^4)/(e^3*(d + e*x)
)) - (95*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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Maple [A]  time = 0.065, size = 288, normalized size = 1.6 \begin{align*} -{\frac{d}{{e}^{7}} \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}}-5\,{\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 ) ^{-3}}-{\frac{19}{3\,d{e}^{5}} \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 ) ^{-2}}-{\frac{19}{3\,d{e}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{95\,x}{12\,{e}^{2}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{95\,{d}^{2}x}{8\,{e}^{2}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{95\,{d}^{4}}{8\,{e}^{2}}\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}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d/e^7/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-5/e^6/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-19/
3/d/e^5/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-19/3/d/e^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)-95/12/e
^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x-95/8*d^2/e^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x-95/8*d^4/e^2/(e^
2)^(1/2)*arctan((e^2)^(1/2)*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^2*(-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 = 1.59704, size = 273, normalized size = 1.5 \begin{align*} -\frac{448 \, d^{4} e x + 448 \, d^{5} - 570 \,{\left (d^{4} e x + d^{5}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (6 \, e^{4} x^{4} - 26 \, d e^{3} x^{3} + 61 \, d^{2} e^{2} x^{2} - 163 \, d^{3} e x - 448 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{24 \,{\left (e^{4} x + d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(448*d^4*e*x + 448*d^5 - 570*(d^4*e*x + d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (6*e^4*x^4 - 26
*d*e^3*x^3 + 61*d^2*e^2*x^2 - 163*d^3*e*x - 448*d^4)*sqrt(-e^2*x^2 + d^2))/(e^4*x + d*e^3)

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

[Out]

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

[Out]

Exception raised: NotImplementedError