3.1.50 \(\int \frac {(d+e x)^2}{x (d^2-e^2 x^2)^{7/2}} \, dx\) [50]

Optimal. Leaf size=117 \[ \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \]

[Out]

2/5*(e*x+d)/d/(-e^2*x^2+d^2)^(5/2)+1/15*(8*e*x+5*d)/d^3/(-e^2*x^2+d^2)^(3/2)-arctanh((-e^2*x^2+d^2)^(1/2)/d)/d
^5+1/15*(16*e*x+15*d)/d^5/(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1819, 837, 12, 272, 65, 214} \begin {gather*} \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d + e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + (5*d + 8*e*x)/(15*d^3*(d^2 - e^2*x^2)^(3/2)) + (15*d + 16*e*x)/(15
*d^5*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2-8 d e x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-15 d^4 e^2-16 d^3 e^3 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^2}\\ &=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {15 d^6 e^4}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^4}\\ &=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^4}\\ &=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^4}\\ &=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\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^4 e^2}\\ &=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}\\ \end {align*}

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Mathematica [A]
time = 0.47, size = 107, normalized size = 0.91 \begin {gather*} \frac {\frac {\sqrt {d^2-e^2 x^2} \left (26 d^3-22 d^2 e x-17 d e^2 x^2+16 e^3 x^3\right )}{(d-e x)^3 (d+e x)}+30 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d^2 - e^2*x^2]*(26*d^3 - 22*d^2*e*x - 17*d*e^2*x^2 + 16*e^3*x^3))/((d - e*x)^3*(d + e*x)) + 30*ArcTanh[
(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(15*d^5)

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Maple [A]
time = 0.07, size = 202, normalized size = 1.73

method	result	size
default	\(\frac {1}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+2 e d \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+d^{2} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/(-e^2*x^2+d^2)^(5/2)+2*e*d*(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*(1/5/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 = 150, normalized size = 1.28 \begin {gather*} \frac {2 \, x e}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d} + \frac {2}{5 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {8 \, x e}{15 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {1}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {16 \, x e}{15 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{5}} - \frac {\log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-x^{2} e^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{5}} + \frac {1}{\sqrt {-x^{2} e^{2} + d^{2}} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x*e/((-x^2*e^2 + d^2)^(5/2)*d) + 2/5/(-x^2*e^2 + d^2)^(5/2) + 8/15*x*e/((-x^2*e^2 + d^2)^(3/2)*d^3) + 1/3/
((-x^2*e^2 + d^2)^(3/2)*d^2) + 16/15*x*e/(sqrt(-x^2*e^2 + d^2)*d^5) - log(2*d^2/abs(x) + 2*sqrt(-x^2*e^2 + d^2
)*d/abs(x))/d^5 + 1/(sqrt(-x^2*e^2 + d^2)*d^4)

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Fricas [A]
time = 2.33, size = 163, normalized size = 1.39 \begin {gather*} \frac {26 \, x^{4} e^{4} - 52 \, d x^{3} e^{3} + 52 \, d^{3} x e - 26 \, d^{4} + 15 \, {\left (x^{4} e^{4} - 2 \, d x^{3} e^{3} + 2 \, d^{3} x e - d^{4}\right )} \log \left (-\frac {d - \sqrt {-x^{2} e^{2} + d^{2}}}{x}\right ) - {\left (16 \, x^{3} e^{3} - 17 \, d x^{2} e^{2} - 22 \, d^{2} x e + 26 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{15 \, {\left (d^{5} x^{4} e^{4} - 2 \, d^{6} x^{3} e^{3} + 2 \, d^{8} x e - d^{9}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(26*x^4*e^4 - 52*d*x^3*e^3 + 52*d^3*x*e - 26*d^4 + 15*(x^4*e^4 - 2*d*x^3*e^3 + 2*d^3*x*e - d^4)*log(-(d -
sqrt(-x^2*e^2 + d^2))/x) - (16*x^3*e^3 - 17*d*x^2*e^2 - 22*d^2*x*e + 26*d^3)*sqrt(-x^2*e^2 + d^2))/(d^5*x^4*e
^4 - 2*d^6*x^3*e^3 + 2*d^8*x*e - d^9)

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

[Out]

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

[Out]

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

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