3.51 \(\int \frac{(d+e x)^2}{x^2 (d^2-e^2 x^2)^{7/2}} \, dx\)

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

[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

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Rubi [A]  time = 0.275118, antiderivative size = 145, normalized size of antiderivative = 1., 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 = {1805, 807, 266, 63, 208} \[ \frac{2 e (d+e x)}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\sqrt{d^2-e^2 x^2}}{d^6 x}+\frac{e (30 d+41 e x)}{15 d^6 \sqrt{d^2-e^2 x^2}}+\frac{e (10 d+13 e x)}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{2 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^6} \]

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 1805

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 807

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 266

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 63

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 208

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

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 \operatorname{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 \operatorname{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*}

Mathematica [C]  time = 0.0542935, size = 90, normalized size = 0.62 \[ \frac{6 d^5 e x \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};1-\frac{e^2 x^2}{d^2}\right )+105 d^4 e^2 x^2-140 d^2 e^4 x^4-15 d^6+56 e^6 x^6}{15 d^6 x \left (d^2-e^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*d^6 + 105*d^4*e^2*x^2 - 140*d^2*e^4*x^4 + 56*e^6*x^6 + 6*d^5*e*x*Hypergeometric2F1[-5/2, 1, -3/2, 1 - (e^
2*x^2)/d^2])/(15*d^6*x*(d^2 - e^2*x^2)^(5/2))

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Maple [A]  time = 0.064, size = 193, normalized size = 1.3 \begin{align*}{\frac{7\,{e}^{2}x}{5\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{28\,{e}^{2}x}{15\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{56\,{e}^{2}x}{15\,{d}^{6}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+{\frac{2\,e}{5\,d} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{2\,e}{3\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{e}{{d}^{5}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}-2\,{\frac{e}{{d}^{5}\sqrt{{d}^{2}}}\ln \left ({\frac{2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{x}} \right ) }-{\frac{1}{x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.95366, size = 401, normalized size = 2.77 \begin{align*} \frac{46 \, e^{5} x^{5} - 92 \, d e^{4} x^{4} + 92 \, d^{3} e^{2} x^{2} - 46 \, d^{4} e x + 30 \,{\left (e^{5} x^{5} - 2 \, d e^{4} x^{4} + 2 \, d^{3} e^{2} x^{2} - d^{4} e x\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (56 \, e^{4} x^{4} - 82 \, d e^{3} x^{3} - 32 \, d^{2} e^{2} x^{2} + 76 \, d^{3} e x - 15 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{6} e^{4} x^{5} - 2 \, d^{7} e^{3} x^{4} + 2 \, d^{9} e x^{2} - d^{10} x\right )}} \end{align*}

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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 [A]  time = 1.16364, size = 254, normalized size = 1.75 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left ({\left (x{\left (\frac{41 \, x e^{6}}{d^{6}} + \frac{30 \, e^{5}}{d^{5}}\right )} - \frac{95 \, e^{4}}{d^{4}}\right )} x - \frac{70 \, e^{3}}{d^{3}}\right )} x + \frac{60 \, e^{2}}{d^{2}}\right )} x + \frac{46 \, e}{d}\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac{2 \, e \log \left (\frac{{\left | -2 \, d e - 2 \, \sqrt{-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \,{\left | x \right |}}\right )}{d^{6}} + \frac{x e^{3}}{2 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{6}} - \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-1\right )}}{2 \, d^{6} x} \end{align*}

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]

-1/15*sqrt(-x^2*e^2 + d^2)*((((x*(41*x*e^6/d^6 + 30*e^5/d^5) - 95*e^4/d^4)*x - 70*e^3/d^3)*x + 60*e^2/d^2)*x +
46*e/d)/(x^2*e^2 - d^2)^3 - 2*e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^6 + 1/2*x*e^3
/((d*e + sqrt(-x^2*e^2 + d^2)*e)*d^6) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-1)/(d^6*x)