3.1.46 \(\int \frac {x^3 (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\) [46]

Optimal. Leaf size=97 \[ \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d+2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1649, 651} \begin {gather*} \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d+2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

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

Rule 651

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((-a)*e + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 1649

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]

Rubi steps

\begin {align*} \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x) \left (\frac {2 d^3}{e^3}+\frac {5 d^2 x}{e^2}+\frac {5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\frac {6 d^3}{e^3}+\frac {15 d^2 x}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d+2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 70, normalized size = 0.72 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (2 d^3-4 d^2 e x+d e^2 x^2+2 e^3 x^3\right )}{5 d e^4 (d-e x)^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(260\) vs. \(2(85)=170\).
time = 0.07, size = 261, normalized size = 2.69

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (e x +d \right )^{3} \left (2 e^{3} x^{3}+d \,e^{2} x^{2}-4 d^{2} e x +2 d^{3}\right )}{5 e^{4} d \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(65\)
trager	\(\frac {\left (2 e^{3} x^{3}+d \,e^{2} x^{2}-4 d^{2} e x +2 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{5 e^{4} d \left (-e x +d \right )^{3} \left (e x +d \right )}\)	\(67\)
default	\(e^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+2 e d \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 142, normalized size = 1.46 \begin {gather*} \frac {d x^{3} e^{\left (-1\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {d^{2} x^{2} e^{\left (-2\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {3 \, d^{3} x e^{\left (-3\right )}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {2 \, d^{4} e^{\left (-4\right )}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {x^{4}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {d x e^{\left (-3\right )}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, x e^{\left (-3\right )}}{5 \, \sqrt {-x^{2} e^{2} + d^{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.06, size = 109, normalized size = 1.12 \begin {gather*} \frac {2 \, x^{4} e^{4} - 4 \, d x^{3} e^{3} + 4 \, d^{3} x e - 2 \, d^{4} - {\left (2 \, x^{3} e^{3} + d x^{2} e^{2} - 4 \, d^{2} x e + 2 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{5 \, {\left (d x^{4} e^{8} - 2 \, d^{2} x^{3} e^{7} + 2 \, d^{4} x e^{5} - d^{5} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**3*(d + e*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((x*e + d)^2*x^3/(-x^2*e^2 + d^2)^(7/2), x)

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Mupad [B]
time = 2.89, size = 66, normalized size = 0.68 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^3-4\,d^2\,e\,x+d\,e^2\,x^2+2\,e^3\,x^3\right )}{5\,d\,e^4\,\left (d+e\,x\right )\,{\left (d-e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

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

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