3.1.51 \(\int \frac {(d+e x)^2}{x^2 (d^2-e^2 x^2)^{7/2}} \, dx\) [51]

Optimal. Leaf size=145 \[ \frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (30 d+41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {2 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6} \]

[Out]

2/5*e*(e*x+d)/d^2/(-e^2*x^2+d^2)^(5/2)+1/15*e*(13*e*x+10*d)/d^4/(-e^2*x^2+d^2)^(3/2)-2*e*arctanh((-e^2*x^2+d^2
)^(1/2)/d)/d^6+1/15*e*(41*e*x+30*d)/d^6/(-e^2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/d^6/x

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Rubi [A]
time = 0.17, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1819, 821, 272, 65, 214} \begin {gather*} \frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {e (30 d+41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {2 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}+\frac {e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(d + e*x))/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (e*(10*d + 13*e*x))/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (e*(30*d
+ 41*e*x))/(15*d^6*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(d^6*x) - (2*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d
^6

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2-10 d e x-8 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^2+30 d e x+26 e^2 x^2}{x^2 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (30 d+41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^2-30 d e x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (30 d+41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {(2 e) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^5}\\ &=\frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (30 d+41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^6 x}+\frac {e \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{d^5}\\ &=\frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (30 d+41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {2 \text {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{d^5 e}\\ &=\frac {2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {e (30 d+41 e x)}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^6 x}-\frac {2 e \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}\\ \end {align*}

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Mathematica [A]
time = 0.46, size = 123, normalized size = 0.85 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (15 d^4-76 d^3 e x+32 d^2 e^2 x^2+82 d e^3 x^3-56 e^4 x^4\right )}{x (-d+e x)^3 (d+e x)}+60 e \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(15*d^4 - 76*d^3*e*x + 32*d^2*e^2*x^2 + 82*d*e^3*x^3 - 56*e^4*x^4))/(x*(-d + e*x)^3*(d +
e*x)) + 60*e*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(15*d^6)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(287\) vs. \(2(129)=258\).
time = 0.09, size = 288, normalized size = 1.99

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{6} x}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{8 d^{6} \left (x +\frac {d}{e}\right )}-\frac {313 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{120 d^{6} \left (x -\frac {d}{e}\right )}-\frac {2 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{5} \sqrt {d^{2}}}+\frac {29 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{60 d^{5} e \left (x -\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{10 d^{4} e^{2} \left (x -\frac {d}{e}\right )^{3}}\)	\(250\)
default	\(e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+d^{2} \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )+2 e d \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )\)	\(288\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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/d^2/x/(-e^2*x^2+d^2)^(5/2)+6*e^2/d^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))))+2*e*d*(1/5/d^2/(-e^2*x^2+d^2)^(5/2)+1/d^2*(1/3/d^2/(-e^2*x^2+d^2)^(3/2
)+1/d^2*(1/d^2/(-e^2*x^2+d^2)^(1/2)-1/d^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x))))

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Maxima [A]
time = 0.27, size = 180, normalized size = 1.24 \begin {gather*} \frac {7 \, x e^{2}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {2 \, e}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d} - \frac {1}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} x} + \frac {28 \, x e^{2}}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {2 \, e}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} - \frac {2 \, e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{6}} + \frac {56 \, x e^{2}}{15 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{6}} + \frac {2 \, e}{\sqrt {-x^{2} e^{2} + d^{2}} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/5*x*e^2/((-x^2*e^2 + d^2)^(5/2)*d^2) + 2/5*e/((-x^2*e^2 + d^2)^(5/2)*d) - 1/((-x^2*e^2 + d^2)^(5/2)*x) + 28/
15*x*e^2/((-x^2*e^2 + d^2)^(3/2)*d^4) + 2/3*e/((-x^2*e^2 + d^2)^(3/2)*d^3) - 2*e*log(2*d^2/abs(x) + 2*sqrt(-x^
2*e^2 + d^2)*d/abs(x))/d^6 + 56/15*x*e^2/(sqrt(-x^2*e^2 + d^2)*d^6) + 2*e/(sqrt(-x^2*e^2 + d^2)*d^5)

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Fricas [A]
time = 2.08, size = 186, normalized size = 1.28 \begin {gather*} \frac {46 \, x^{5} e^{5} - 92 \, d x^{4} e^{4} + 92 \, d^{3} x^{2} e^{2} - 46 \, d^{4} x e + 30 \, {\left (x^{5} e^{5} - 2 \, d x^{4} e^{4} + 2 \, d^{3} x^{2} e^{2} - d^{4} x e\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) - {\left (56 \, x^{4} e^{4} - 82 \, d x^{3} e^{3} - 32 \, d^{2} x^{2} e^{2} + 76 \, d^{3} x e - 15 \, d^{4}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d^{6} x^{5} e^{4} - 2 \, d^{7} x^{4} e^{3} + 2 \, d^{9} x^{2} e - d^{10} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(46*x^5*e^5 - 92*d*x^4*e^4 + 92*d^3*x^2*e^2 - 46*d^4*x*e + 30*(x^5*e^5 - 2*d*x^4*e^4 + 2*d^3*x^2*e^2 - d^
4*x*e)*log(-(d - sqrt(-x^2*e^2 + d^2))/x) - (56*x^4*e^4 - 82*d*x^3*e^3 - 32*d^2*x^2*e^2 + 76*d^3*x*e - 15*d^4)
*sqrt(-x^2*e^2 + d^2))/(d^6*x^5*e^4 - 2*d^7*x^4*e^3 + 2*d^9*x^2*e - d^10*x)

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^2}{x^2\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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