3.3.2 \(\int \frac {x (d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\) [202]

Optimal. Leaf size=130 \[ \frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {10 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2} \]

[Out]

20/3*(-e^2*x^2+d^2)^(3/2)/e^2+8*(-e^2*x^2+d^2)^(5/2)/e^2/(e*x+d)^2+(-e^2*x^2+d^2)^(7/2)/e^2/(e*x+d)^4+10*d^3*a
rctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^2+10*d*x*(-e^2*x^2+d^2)^(1/2)/e

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {807, 677, 679, 201, 223, 209} \begin {gather*} \frac {10 d^3 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {10 d x \sqrt {d^2-e^2 x^2}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*d*x*Sqrt[d^2 - e^2*x^2])/e + (20*(d^2 - e^2*x^2)^(3/2))/(3*e^2) + (8*(d^2 - e^2*x^2)^(5/2))/(e^2*(d + e*x)
^2) + (d^2 - e^2*x^2)^(7/2)/(e^2*(d + e*x)^4) + (10*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e^2

Rule 201

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 677

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

Rule 679

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

Rule 807

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d
+ e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(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] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {x \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {4 \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx}{e}\\ &=\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {20 \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{d+e x} \, dx}{e}\\ &=\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {(20 d) \int \sqrt {d^2-e^2 x^2} \, dx}{e}\\ &=\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {\left (10 d^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e}\\ &=\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {\left (10 d^3\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e}\\ &=\frac {10 d x \sqrt {d^2-e^2 x^2}}{e}+\frac {20 \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}+\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{e^2 (d+e x)^2}+\frac {\left (d^2-e^2 x^2\right )^{7/2}}{e^2 (d+e x)^4}+\frac {10 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.29, size = 107, normalized size = 0.82 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (47 d^3+17 d^2 e x-5 d e^2 x^2+e^3 x^3\right )}{3 e^2 (d+e x)}+\frac {10 d^3 \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(47*d^3 + 17*d^2*e*x - 5*d*e^2*x^2 + e^3*x^3))/(3*e^2*(d + e*x)) + (10*d^3*Sqrt[-e^2]*Log
[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^3

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(643\) vs. \(2(118)=236\).
time = 0.08, size = 644, normalized size = 4.95

method	result	size
risch	\(\frac {\left (e^{2} x^{2}-6 d e x +23 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}+\frac {10 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e \sqrt {e^{2}}}+\frac {8 d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{3} \left (x +\frac {d}{e}\right )}\)	\(119\)
default	\(\frac {\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{3}}+\frac {4 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}\right )}{d}}{e^{4}}-\frac {d \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{4}}-\frac {3 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{3}}+\frac {4 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}\right )}{d}\right )}{d}\right )}{e^{5}}\)	\(644\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.51, size = 214, normalized size = 1.65 \begin {gather*} 10 \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} + \frac {5}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-2\right )} - \frac {{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d}{2 \, {\left (x^{3} e^{5} + 3 \, d x^{2} e^{4} + 3 \, d^{2} x e^{3} + d^{3} e^{2}\right )}} - \frac {5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{2 \, {\left (x^{2} e^{4} + 2 \, d x e^{3} + d^{2} e^{2}\right )}} + \frac {15 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{3}}{x e^{3} + d e^{2}} + \frac {{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}}{3 \, {\left (x^{2} e^{4} + 2 \, d x e^{3} + d^{2} e^{2}\right )}} + \frac {5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d}{6 \, {\left (x e^{3} + d e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*d^3*arcsin(x*e/d)*e^(-2) + 5/2*sqrt(-x^2*e^2 + d^2)*d^2*e^(-2) - 1/2*(-x^2*e^2 + d^2)^(5/2)*d/(x^3*e^5 + 3*
d*x^2*e^4 + 3*d^2*x*e^3 + d^3*e^2) - 5/2*(-x^2*e^2 + d^2)^(3/2)*d^2/(x^2*e^4 + 2*d*x*e^3 + d^2*e^2) + 15*sqrt(
-x^2*e^2 + d^2)*d^3/(x*e^3 + d*e^2) + 1/3*(-x^2*e^2 + d^2)^(5/2)/(x^2*e^4 + 2*d*x*e^3 + d^2*e^2) + 5/6*(-x^2*e
^2 + d^2)^(3/2)*d/(x*e^3 + d*e^2)

________________________________________________________________________________________

Fricas [A]
time = 1.86, size = 107, normalized size = 0.82 \begin {gather*} \frac {47 \, d^{3} x e + 47 \, d^{4} - 60 \, {\left (d^{3} x e + d^{4}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (x^{3} e^{3} - 5 \, d x^{2} e^{2} + 17 \, d^{2} x e + 47 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (x e^{3} + d e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(47*d^3*x*e + 47*d^4 - 60*(d^3*x*e + d^4)*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (x^3*e^3 - 5*d*x^
2*e^2 + 17*d^2*x*e + 47*d^3)*sqrt(-x^2*e^2 + d^2))/(x*e^3 + d*e^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.50, size = 91, normalized size = 0.70 \begin {gather*} 10 \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-2\right )} \mathrm {sgn}\left (d\right ) - \frac {16 \, d^{3} e^{\left (-2\right )}}{\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1} + \frac {1}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (23 \, d^{2} e^{\left (-2\right )} - {\left (6 \, d e^{\left (-1\right )} - x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*d^3*arcsin(x*e/d)*e^(-2)*sgn(d) - 16*d^3*e^(-2)/((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1) + 1/3*sqrt(-x
^2*e^2 + d^2)*(23*d^2*e^(-2) - (6*d*e^(-1) - x)*x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________