3.1.86 \(\int \frac {x^2 (d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\) [86]

Optimal. Leaf size=93 \[ \frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 (d+e x)}{15 d e^3 \sqrt {d^2-e^2 x^2}} \]

[Out]

1/5*d*(e*x+d)^3/e^3/(-e^2*x^2+d^2)^(5/2)-8/15*(e*x+d)^2/e^3/(-e^2*x^2+d^2)^(3/2)+7/15*(e*x+d)/d/e^3/(-e^2*x^2+
d^2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1649, 803, 651} \begin {gather*} \frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 (d+e x)}{15 d e^3 \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(d + e*x)^3)/(5*e^3*(d^2 - e^2*x^2)^(5/2)) - (8*(d + e*x)^2)/(15*e^3*(d^2 - e^2*x^2)^(3/2)) + (7*(d + e*x))
/(15*d*e^3*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 803

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*(p + 1))), x] - Dist[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 0]

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^2 (d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {\left (\frac {3 d^2}{e^2}+\frac {5 d x}{e}\right ) (d+e x)^2}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 \int \frac {d+e x}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 e^2}\\ &=\frac {d (d+e x)^3}{5 e^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8 (d+e x)^2}{15 e^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 (d+e x)}{15 d e^3 \sqrt {d^2-e^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 53, normalized size = 0.57 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (2 d^2-6 d e x+7 e^2 x^2\right )}{15 d e^3 (d-e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(2*d^2 - 6*d*e*x + 7*e^2*x^2))/(15*d*e^3*(d - e*x)^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(364\) vs. \(2(81)=162\).
time = 0.06, size = 365, normalized size = 3.92

method	result	size
trager	\(\frac {\left (7 e^{2} x^{2}-6 d e x +2 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d \,e^{3} \left (-e x +d \right )^{3}}\)	\(50\)
gosper	\(\frac {\left (-e x +d \right ) \left (e x +d \right )^{4} \left (7 e^{2} x^{2}-6 d e x +2 d^{2}\right )}{15 d \,e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(55\)
default	\(e^{3} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+3 e^{2} d \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+3 e \,d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+d^{3} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )\)	\(365\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^3*(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
)))+3*e^2*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)))))+3*e*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))+d^3*(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))))

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Maxima [A]
time = 0.29, size = 143, normalized size = 1.54 \begin {gather*} \frac {x^{4} e}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {d^{2} x^{2} e^{\left (-1\right )}}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {7 \, d^{3} x e^{\left (-2\right )}}{10 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {2 \, d^{4} e^{\left (-3\right )}}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {3 \, d x^{3}}{2 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {7 \, d x e^{\left (-2\right )}}{30 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {7 \, x e^{\left (-2\right )}}{15 \, \sqrt {-x^{2} e^{2} + d^{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4*e/(-x^2*e^2 + d^2)^(5/2) - 1/3*d^2*x^2*e^(-1)/(-x^2*e^2 + d^2)^(5/2) - 7/10*d^3*x*e^(-2)/(-x^2*e^2 + d^2)^
(5/2) + 2/15*d^4*e^(-3)/(-x^2*e^2 + d^2)^(5/2) + 3/2*d*x^3/(-x^2*e^2 + d^2)^(5/2) + 7/30*d*x*e^(-2)/(-x^2*e^2
+ d^2)^(3/2) + 7/15*x*e^(-2)/(sqrt(-x^2*e^2 + d^2)*d)

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Fricas [A]
time = 3.63, size = 100, normalized size = 1.08 \begin {gather*} \frac {2 \, x^{3} e^{3} - 6 \, d x^{2} e^{2} + 6 \, d^{2} x e - 2 \, d^{3} - {\left (7 \, x^{2} e^{2} - 6 \, d x e + 2 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d x^{3} e^{6} - 3 \, d^{2} x^{2} e^{5} + 3 \, d^{3} x e^{4} - d^{4} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*x^3*e^3 - 6*d*x^2*e^2 + 6*d^2*x*e - 2*d^3 - (7*x^2*e^2 - 6*d*x*e + 2*d^2)*sqrt(-x^2*e^2 + d^2))/(d*x^3
*e^6 - 3*d^2*x^2*e^5 + 3*d^3*x*e^4 - d^4*e^3)

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

[Out]

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

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Giac [A]
time = 0.69, size = 98, normalized size = 1.05 \begin {gather*} -\frac {4 \, {\left (\frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - \frac {10 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{\left (-4\right )}}{x^{2}} - 1\right )} e^{\left (-3\right )}}{15 \, d {\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^2*(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.69, size = 49, normalized size = 0.53 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2-6\,d\,e\,x+7\,e^2\,x^2\right )}{15\,d\,e^3\,{\left (d-e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^2 - e^2*x^2)^(1/2)*(2*d^2 + 7*e^2*x^2 - 6*d*e*x))/(15*d*e^3*(d - e*x)^3)

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