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

Optimal. Leaf size=140 \[ \frac{d^5 x \sqrt{d^2-e^2 x^2}}{16 e^2}+\frac{d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac{d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac{d^7 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e^3} \]

[Out]

(d^5*x*Sqrt[d^2 - e^2*x^2])/(16*e^2) + (d^3*x*(d^2 - e^2*x^2)^(3/2))/(24*e^2) + (d*(6*d - 5*e*x)*(d^2 - e^2*x^
2)^(5/2))/(30*e^3) - (d^2 - e^2*x^2)^(7/2)/(7*e^3) + (d^7*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e^3)

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Rubi [A]  time = 0.150291, antiderivative size = 140, 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 = {1639, 12, 785, 780, 195, 217, 203} \[ \frac{d^5 x \sqrt{d^2-e^2 x^2}}{16 e^2}+\frac{d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac{d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac{d^7 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^5*x*Sqrt[d^2 - e^2*x^2])/(16*e^2) + (d^3*x*(d^2 - e^2*x^2)^(3/2))/(24*e^2) + (d*(6*d - 5*e*x)*(d^2 - e^2*x^
2)^(5/2))/(30*e^3) - (d^2 - e^2*x^2)^(7/2)/(7*e^3) + (d^7*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e^3)

Rule 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 785

Int[(x_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^m*e^m, Int[(x*(a + c*x^2)^(m
+ p))/(a*e + c*d*x)^m, x], x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[
m, 0] && EqQ[m, -1] &&  !ILtQ[p - 1/2, 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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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} \, dx &=-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac{\int \frac{7 d e^3 x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx}{7 e^4}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac{d \int \frac{x \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx}{e}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}-\frac{\int x \left (d^2 e-d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{e^2}\\ &=\frac{d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac{d^3 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{6 e^2}\\ &=\frac{d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac{d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac{d^5 \int \sqrt{d^2-e^2 x^2} \, dx}{8 e^2}\\ &=\frac{d^5 x \sqrt{d^2-e^2 x^2}}{16 e^2}+\frac{d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac{d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac{d^7 \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{16 e^2}\\ &=\frac{d^5 x \sqrt{d^2-e^2 x^2}}{16 e^2}+\frac{d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac{d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac{d^7 \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e^2}\\ &=\frac{d^5 x \sqrt{d^2-e^2 x^2}}{16 e^2}+\frac{d^3 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e^2}+\frac{d (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^3}-\frac{\left (d^2-e^2 x^2\right )^{7/2}}{7 e^3}+\frac{d^7 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e^3}\\ \end{align*}

Mathematica [A]  time = 0.108896, size = 113, normalized size = 0.81 \[ \frac{\sqrt{d^2-e^2 x^2} \left (48 d^4 e^2 x^2+490 d^3 e^3 x^3-384 d^2 e^4 x^4-105 d^5 e x+96 d^6-280 d e^5 x^5+240 e^6 x^6\right )+105 d^7 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{1680 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(96*d^6 - 105*d^5*e*x + 48*d^4*e^2*x^2 + 490*d^3*e^3*x^3 - 384*d^2*e^4*x^4 - 280*d*e^5*x^
5 + 240*e^6*x^6) + 105*d^7*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(1680*e^3)

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Maple [B]  time = 0.058, size = 282, normalized size = 2. \begin{align*} -{\frac{1}{7\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{dx}{6\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{d}^{3}x}{24\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{d}^{5}x}{16\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{5\,{d}^{7}}{16\,{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{{d}^{2}}{5\,{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{{d}^{3}x}{4\,{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{3\,{d}^{5}x}{8\,{e}^{2}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{3\,{d}^{7}}{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),x)

[Out]

-1/7*(-e^2*x^2+d^2)^(7/2)/e^3-1/6*d*x*(-e^2*x^2+d^2)^(5/2)/e^2-5/24*d^3*x*(-e^2*x^2+d^2)^(3/2)/e^2-5/16*d^5*x*
(-e^2*x^2+d^2)^(1/2)/e^2-5/16*d^7/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+1/5*d^2/e^3*(-(d/
e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)+1/4*d^3/e^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+3/8*d^5/e^2*(-(d/e+x)^2*e^2
+2*d*e*(d/e+x))^(1/2)*x+3/8*d^7/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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58616, size = 262, normalized size = 1.87 \begin{align*} -\frac{210 \, d^{7} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (240 \, e^{6} x^{6} - 280 \, d e^{5} x^{5} - 384 \, d^{2} e^{4} x^{4} + 490 \, d^{3} e^{3} x^{3} + 48 \, d^{4} e^{2} x^{2} - 105 \, d^{5} e x + 96 \, d^{6}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{1680 \, e^{3}} \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),x, algorithm="fricas")

[Out]

-1/1680*(210*d^7*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (240*e^6*x^6 - 280*d*e^5*x^5 - 384*d^2*e^4*x^4 +
490*d^3*e^3*x^3 + 48*d^4*e^2*x^2 - 105*d^5*e*x + 96*d^6)*sqrt(-e^2*x^2 + d^2))/e^3

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Sympy [C]  time = 13.9106, size = 656, normalized size = 4.69 \begin{align*} \text{result too large to display} \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),x)

[Out]

d**3*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sq
rt(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**4*a
sin(e*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*
x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) - d**2*e*Piecewise((-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**
2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*sqrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True))
- d*e**2*Piecewise((-I*d**6*acosh(e*x/d)/(16*e**5) + I*d**5*x/(16*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x*
*3/(48*e**2*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*x**5/(24*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1
+ e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**6*asin(e*x/d)/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2
*x**2/d**2)) + d**3*x**3/(48*e**2*sqrt(1 - e**2*x**2/d**2)) + 5*d*x**5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x*
*7/(6*d*sqrt(1 - e**2*x**2/d**2)), True)) + e**3*Piecewise((-8*d**6*sqrt(d**2 - e**2*x**2)/(105*e**6) - 4*d**4
*x**2*sqrt(d**2 - e**2*x**2)/(105*e**4) - d**2*x**4*sqrt(d**2 - e**2*x**2)/(35*e**2) + x**6*sqrt(d**2 - e**2*x
**2)/7, Ne(e, 0)), (x**6*sqrt(d**2)/6, True))

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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),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError