3.1.52 \(\int \frac {(d+e x)^2}{x^3 (d^2-e^2 x^2)^{7/2}} \, dx\) [52]

Optimal. Leaf size=182 \[ \frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7} \]

[Out]

2/5*e^2*(e*x+d)/d^3/(-e^2*x^2+d^2)^(5/2)+1/5*e^2*(6*e*x+5*d)/d^5/(-e^2*x^2+d^2)^(3/2)-9/2*e^2*arctanh((-e^2*x^
2+d^2)^(1/2)/d)/d^7+2/5*e^2*(11*e*x+10*d)/d^7/(-e^2*x^2+d^2)^(1/2)-1/2*(-e^2*x^2+d^2)^(1/2)/d^6/x^2-2*e*(-e^2*
x^2+d^2)^(1/2)/d^7/x

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Rubi [A]
time = 0.23, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1819, 1821, 821, 272, 65, 214} \begin {gather*} \frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*(d + e*x))/(5*d^3*(d^2 - e^2*x^2)^(5/2)) + (e^2*(5*d + 6*e*x))/(5*d^5*(d^2 - e^2*x^2)^(3/2)) + (2*e^2*(
10*d + 11*e*x))/(5*d^7*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(2*d^6*x^2) - (2*e*Sqrt[d^2 - e^2*x^2])/(d^7
*x) - (9*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*d^7)

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]

Rule 1821

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2-10 d e x-10 e^2 x^2-\frac {8 e^3 x^3}{d}}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^2+30 d e x+45 e^2 x^2+\frac {36 e^3 x^3}{d}}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^2-30 d e x-60 e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {\int \frac {60 d^3 e+135 d^2 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}+\frac {\left (9 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^6}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}+\frac {\left (9 e^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^6}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 \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 )}{2 d^6}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7}\\ \end {align*}

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Mathematica [A]
time = 0.48, size = 136, normalized size = 0.75 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (5 d^5+10 d^4 e x-94 d^3 e^2 x^2+58 d^2 e^3 x^3+83 d e^4 x^4-64 e^5 x^5\right )}{x^2 (-d+e x)^3 (d+e x)}+90 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{10 d^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(5*d^5 + 10*d^4*e*x - 94*d^3*e^2*x^2 + 58*d^2*e^3*x^3 + 83*d*e^4*x^4 - 64*e^5*x^5))/(x^2
*(-d + e*x)^3*(d + e*x)) + 90*e^2*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(10*d^7)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(360\) vs. \(2(160)=320\).
time = 0.08, size = 361, normalized size = 1.98

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (4 e x +d \right )}{2 d^{7} x^{2}}+\frac {e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{8 d^{7} \left (x +\frac {d}{e}\right )}-\frac {181 e \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{40 d^{7} \left (x -\frac {d}{e}\right )}-\frac {9 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{6} \sqrt {d^{2}}}+\frac {13 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{20 d^{6} \left (x -\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d \left (x -\frac {d}{e}\right ) e}}{10 e \,d^{5} \left (x -\frac {d}{e}\right )^{3}}\)	\(257\)
default	\(d^{2} \left (-\frac {1}{2 d^{2} x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {7 e^{2} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )}{2 d^{2}}\right )+2 d e \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )+e^{2} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )\)	\(361\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*(-1/2/d^2/x^2/(-e^2*x^2+d^2)^(5/2)+7/2*e^2/d^2*(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))))
)+2*d*e*(-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))))+e^2*(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 = 203, normalized size = 1.12 \begin {gather*} \frac {12 \, x e^{3}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {9 \, e^{2}}{10 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} - \frac {2 \, e}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d x} - \frac {1}{2 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} x^{2}} + \frac {16 \, x e^{3}}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {3 \, e^{2}}{2 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} - \frac {9 \, e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d^{7}} + \frac {32 \, x e^{3}}{5 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{7}} + \frac {9 \, e^{2}}{2 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12/5*x*e^3/((-x^2*e^2 + d^2)^(5/2)*d^3) + 9/10*e^2/((-x^2*e^2 + d^2)^(5/2)*d^2) - 2*e/((-x^2*e^2 + d^2)^(5/2)*
d*x) - 1/2/((-x^2*e^2 + d^2)^(5/2)*x^2) + 16/5*x*e^3/((-x^2*e^2 + d^2)^(3/2)*d^5) + 3/2*e^2/((-x^2*e^2 + d^2)^
(3/2)*d^4) - 9/2*e^2*log(2*d^2/abs(x) + 2*sqrt(-x^2*e^2 + d^2)*d/abs(x))/d^7 + 32/5*x*e^3/(sqrt(-x^2*e^2 + d^2
)*d^7) + 9/2*e^2/(sqrt(-x^2*e^2 + d^2)*d^6)

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Fricas [A]
time = 1.67, size = 202, normalized size = 1.11 \begin {gather*} \frac {54 \, x^{6} e^{6} - 108 \, d x^{5} e^{5} + 108 \, d^{3} x^{3} e^{3} - 54 \, d^{4} x^{2} e^{2} + 45 \, {\left (x^{6} e^{6} - 2 \, d x^{5} e^{5} + 2 \, d^{3} x^{3} e^{3} - d^{4} x^{2} e^{2}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) - {\left (64 \, x^{5} e^{5} - 83 \, d x^{4} e^{4} - 58 \, d^{2} x^{3} e^{3} + 94 \, d^{3} x^{2} e^{2} - 10 \, d^{4} x e - 5 \, d^{5}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{10 \, {\left (d^{7} x^{6} e^{4} - 2 \, d^{8} x^{5} e^{3} + 2 \, d^{10} x^{3} e - d^{11} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(54*x^6*e^6 - 108*d*x^5*e^5 + 108*d^3*x^3*e^3 - 54*d^4*x^2*e^2 + 45*(x^6*e^6 - 2*d*x^5*e^5 + 2*d^3*x^3*e^
3 - d^4*x^2*e^2)*log(-(d - sqrt(-x^2*e^2 + d^2))/x) - (64*x^5*e^5 - 83*d*x^4*e^4 - 58*d^2*x^3*e^3 + 94*d^3*x^2
*e^2 - 10*d^4*x*e - 5*d^5)*sqrt(-x^2*e^2 + d^2))/(d^7*x^6*e^4 - 2*d^8*x^5*e^3 + 2*d^10*x^3*e - d^11*x^2)

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

[Out]

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

[Out]

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

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