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

Optimal. Leaf size=172 \[ -\frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}-\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{3 d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{128 e^4} \]

[Out]

(-3*d^6*x*Sqrt[d^2 - e^2*x^2])/(128*e^3) - (d^4*x*(d^2 - e^2*x^2)^(3/2))/(64*e^3) - (d*x^2*(d^2 - e^2*x^2)^(5/
2))/(7*e^2) + (x^3*(d^2 - e^2*x^2)^(5/2))/(8*e) - (d^2*(32*d - 35*e*x)*(d^2 - e^2*x^2)^(5/2))/(560*e^4) - (3*d
^8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(128*e^4)

________________________________________________________________________________________

Rubi [A]  time = 0.123488, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {850, 833, 780, 195, 217, 203} \[ -\frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}-\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{3 d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{128 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d^6*x*Sqrt[d^2 - e^2*x^2])/(128*e^3) - (d^4*x*(d^2 - e^2*x^2)^(3/2))/(64*e^3) - (d*x^2*(d^2 - e^2*x^2)^(5/
2))/(7*e^2) + (x^3*(d^2 - e^2*x^2)^(5/2))/(8*e) - (d^2*(32*d - 35*e*x)*(d^2 - e^2*x^2)^(5/2))/(560*e^4) - (3*d
^8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(128*e^4)

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 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 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^3 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx &=\int x^3 (d-e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{\int x^2 \left (3 d^2 e-8 d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{8 e^2}\\ &=-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}+\frac{\int x \left (16 d^3 e^2-21 d^2 e^3 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{56 e^4}\\ &=-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{d^4 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{16 e^3}\\ &=-\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{\left (3 d^6\right ) \int \sqrt{d^2-e^2 x^2} \, dx}{64 e^3}\\ &=-\frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}-\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{\left (3 d^8\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{128 e^3}\\ &=-\frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}-\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{\left (3 d^8\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{128 e^3}\\ &=-\frac{3 d^6 x \sqrt{d^2-e^2 x^2}}{128 e^3}-\frac{d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac{d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac{x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac{d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac{3 d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{128 e^4}\\ \end{align*}

Mathematica [A]  time = 0.139402, size = 124, normalized size = 0.72 \[ \frac{\sqrt{d^2-e^2 x^2} \left (-128 d^5 e^2 x^2+70 d^4 e^3 x^3+1024 d^3 e^4 x^4-840 d^2 e^5 x^5+105 d^6 e x-256 d^7-640 d e^6 x^6+560 e^7 x^7\right )-105 d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{4480 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-256*d^7 + 105*d^6*e*x - 128*d^5*e^2*x^2 + 70*d^4*e^3*x^3 + 1024*d^3*e^4*x^4 - 840*d^2*e
^5*x^5 - 640*d*e^6*x^6 + 560*e^7*x^7) - 105*d^8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(4480*e^4)

________________________________________________________________________________________

Maple [B]  time = 0.059, size = 305, normalized size = 1.8 \begin{align*} -{\frac{x}{8\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+{\frac{3\,{d}^{2}x}{16\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{15\,{d}^{4}x}{64\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{45\,{d}^{6}x}{128\,{e}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{45\,{d}^{8}}{128\,{e}^{3}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{d}{7\,{e}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{{d}^{3}}{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}}}}-{\frac{{d}^{4}x}{4\,{e}^{3}} \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}^{6}x}{8\,{e}^{3}}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{3\,{d}^{8}}{8\,{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}}}}} \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),x)

[Out]

-1/8/e^3*x*(-e^2*x^2+d^2)^(7/2)+3/16*d^2/e^3*x*(-e^2*x^2+d^2)^(5/2)+15/64*d^4*x*(-e^2*x^2+d^2)^(3/2)/e^3+45/12
8*d^6*x*(-e^2*x^2+d^2)^(1/2)/e^3+45/128*d^8/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+1/7*d/e
^4*(-e^2*x^2+d^2)^(7/2)-1/5*d^3/e^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)-1/4*d^4/e^3*(-(d/e+x)^2*e^2+2*d*e*(d/
e+x))^(3/2)*x-3/8*d^6/e^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x-3/8*d^8/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/
(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))

________________________________________________________________________________________

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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.65833, size = 288, normalized size = 1.67 \begin{align*} \frac{210 \, d^{8} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (560 \, e^{7} x^{7} - 640 \, d e^{6} x^{6} - 840 \, d^{2} e^{5} x^{5} + 1024 \, d^{3} e^{4} x^{4} + 70 \, d^{4} e^{3} x^{3} - 128 \, d^{5} e^{2} x^{2} + 105 \, d^{6} e x - 256 \, d^{7}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{4480 \, e^{4}} \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),x, algorithm="fricas")

[Out]

1/4480*(210*d^8*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (560*e^7*x^7 - 640*d*e^6*x^6 - 840*d^2*e^5*x^5 + 1
024*d^3*e^4*x^4 + 70*d^4*e^3*x^3 - 128*d^5*e^2*x^2 + 105*d^6*e*x - 256*d^7)*sqrt(-e^2*x^2 + d^2))/e^4

________________________________________________________________________________________

Sympy [A]  time = 20.617, size = 779, normalized size = 4.53 \begin{align*} \text{result too large to display} \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),x)

[Out]

d**3*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*s
qrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) - d**2*e*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)) - d*e*
*2*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))
+ e**3*Piecewise((-5*I*d**8*acosh(e*x/d)/(128*e**7) + 5*I*d**7*x/(128*e**6*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*
*5*x**3/(384*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**5/(192*e**2*sqrt(-1 + e**2*x**2/d**2)) - 7*I*d*x**7/(
48*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**9/(8*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (5
*d**8*asin(e*x/d)/(128*e**7) - 5*d**7*x/(128*e**6*sqrt(1 - e**2*x**2/d**2)) + 5*d**5*x**3/(384*e**4*sqrt(1 - e
**2*x**2/d**2)) + d**3*x**5/(192*e**2*sqrt(1 - e**2*x**2/d**2)) + 7*d*x**7/(48*sqrt(1 - e**2*x**2/d**2)) - e**
2*x**9/(8*d*sqrt(1 - e**2*x**2/d**2)), True))

________________________________________________________________________________________

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),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError