3.2.90 \(\int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx\) [190]

Optimal. Leaf size=148 \[ \frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]

[Out]

1/5*d^2*(-e^2*x^2+d^2)^(3/2)/e^4/(e*x+d)^4-14/15*d*(-e^2*x^2+d^2)^(3/2)/e^4/(e*x+d)^3-(-e^2*x^2+d^2)^(3/2)/e^4
/(e*x+d)^2+4*d*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4+8*d*(-e^2*x^2+d^2)^(1/2)/e^4/(e*x+d)

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Rubi [A]
time = 0.16, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1653, 1651, 673, 665, 677, 223, 209} \begin {gather*} \frac {4 d \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}+\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*d*Sqrt[d^2 - e^2*x^2])/(e^4*(d + e*x)) + (d^2*(d^2 - e^2*x^2)^(3/2))/(5*e^4*(d + e*x)^4) - (14*d*(d^2 - e^2
*x^2)^(3/2))/(15*e^4*(d + e*x)^3) - (d^2 - e^2*x^2)^(3/2)/(e^4*(d + e*x)^2) + (4*d*ArcTan[(e*x)/Sqrt[d^2 - e^2
*x^2]])/e^4

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 665

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

Rule 673

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a + c*x^2)^(p +
1)/(2*c*d*(m + p + 1))), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^
p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p +
2], 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 1651

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p,
(d + e*x)^m*Pq, x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq
, x] + 2*p + 1, 0] && ILtQ[m, 0]

Rule 1653

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]

Rubi steps

\begin {align*} \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {\int \frac {\sqrt {d^2-e^2 x^2} \left (2 d^3 e^2+5 d^2 e^3 x+4 d e^4 x^2\right )}{(d+e x)^4} \, dx}{e^5}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {\int \left (\frac {d^3 e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4}-\frac {3 d^2 e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^3}+\frac {4 d e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^2}\right ) \, dx}{e^5}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {(4 d) \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx}{e^3}+\frac {\left (3 d^2\right ) \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{e^3}-\frac {d^3 \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx}{e^3}\\ &=\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {(4 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^3}-\frac {d^2 \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{5 e^3}\\ &=\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {(4 d) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}\\ &=\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 106, normalized size = 0.72 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (94 d^3+222 d^2 e x+149 d e^2 x^2+15 e^3 x^3\right )}{15 e^4 (d+e x)^3}+\frac {4 d \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(94*d^3 + 222*d^2*e*x + 149*d*e^2*x^2 + 15*e^3*x^3))/(15*e^4*(d + e*x)^3) + (4*d*Sqrt[-e^
2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/e^5

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(134)=268\).
time = 0.08, size = 349, normalized size = 2.36

method	result	size
risch	\(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{4}}+\frac {4 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{3} \sqrt {e^{2}}}+\frac {104 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 e^{5} \left (x +\frac {d}{e}\right )}-\frac {31 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 e^{6} \left (x +\frac {d}{e}\right )^{2}}+\frac {2 d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{7} \left (x +\frac {d}{e}\right )^{3}}\)	\(186\)
default	\(\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}}{e^{4}}-\frac {d \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{e^{7} \left (x +\frac {d}{e}\right )^{3}}-\frac {d^{3} \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}\right )}{e^{7}}-\frac {3 d \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}\right )}{e^{5}}\)	\(349\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

1/e^4*((-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+d*e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e)
)^(1/2)))-1/e^7*d/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)-d^3/e^7*(-1/5/d/e/(x+d/e)^4*(-(x+d/e)^2*e^2+2
*d*e*(x+d/e))^(3/2)-1/15/d^2/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2))-3*d/e^5*(-1/d/e/(x+d/e)^2*(-(x+d/
e)^2*e^2+2*d*e*(x+d/e))^(3/2)-e/d*((-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+d*e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(
-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))))

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 1.94, size = 164, normalized size = 1.11 \begin {gather*} \frac {94 \, d x^{3} e^{3} + 282 \, d^{2} x^{2} e^{2} + 282 \, d^{3} x e + 94 \, d^{4} - 120 \, {\left (d x^{3} e^{3} + 3 \, d^{2} x^{2} e^{2} + 3 \, 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 (15 \, x^{3} e^{3} + 149 \, d x^{2} e^{2} + 222 \, d^{2} x e + 94 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (x^{3} e^{7} + 3 \, d x^{2} e^{6} + 3 \, d^{2} x e^{5} + d^{3} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/15*(94*d*x^3*e^3 + 282*d^2*x^2*e^2 + 282*d^3*x*e + 94*d^4 - 120*(d*x^3*e^3 + 3*d^2*x^2*e^2 + 3*d^3*x*e + d^4
)*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (15*x^3*e^3 + 149*d*x^2*e^2 + 222*d^2*x*e + 94*d^3)*sqrt(-x^2
*e^2 + d^2))/(x^3*e^7 + 3*d*x^2*e^6 + 3*d^2*x*e^5 + d^3*e^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-e**2*x**2+d**2)**(1/2)/(e*x+d)**4,x)

[Out]

Integral(x**3*sqrt(-(-d + e*x)*(d + e*x))/(d + e*x)**4, x)

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Giac [A]
time = 1.51, size = 193, normalized size = 1.30 \begin {gather*} 4 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-4\right )} \mathrm {sgn}\left (d\right ) + \sqrt {-x^{2} e^{2} + d^{2}} e^{\left (-4\right )} - \frac {2 \, {\left (\frac {335 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d e^{\left (-2\right )}}{x} + \frac {505 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{\left (-4\right )}}{x^{2}} + \frac {285 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d e^{\left (-6\right )}}{x^{3}} + \frac {60 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d e^{\left (-8\right )}}{x^{4}} + 79 \, d\right )} e^{\left (-4\right )}}{15 \, {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(1/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

4*d*arcsin(x*e/d)*e^(-4)*sgn(d) + sqrt(-x^2*e^2 + d^2)*e^(-4) - 2/15*(335*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^(
-2)/x + 505*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d*e^(-4)/x^2 + 285*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d*e^(-6)/x^3
+ 60*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d*e^(-8)/x^4 + 79*d)*e^(-4)/((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1
)^5

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\sqrt {d^2-e^2\,x^2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d^2 - e^2*x^2)^(1/2))/(d + e*x)^4,x)

[Out]

int((x^3*(d^2 - e^2*x^2)^(1/2))/(d + e*x)^4, x)

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