3.53 \(\int \frac{(d+e x)^2}{x^4 (d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=209 \[ \frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 e^2 \sqrt{d^2-e^2 x^2}}{3 d^8 x}-\frac{e \sqrt{d^2-e^2 x^2}}{d^7 x^2}-\frac{\sqrt{d^2-e^2 x^2}}{3 d^6 x^3}-\frac{7 e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^8} \]

[Out]

(2*e^3*(d + e*x))/(5*d^4*(d^2 - e^2*x^2)^(5/2)) + (e^3*(20*d + 23*e*x))/(15*d^6*(d^2 - e^2*x^2)^(3/2)) + (2*e^
3*(45*d + 53*e*x))/(15*d^8*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(3*d^6*x^3) - (e*Sqrt[d^2 - e^2*x^2])/(d
^7*x^2) - (14*e^2*Sqrt[d^2 - e^2*x^2])/(3*d^8*x) - (7*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^8

________________________________________________________________________________________

Rubi [A]  time = 0.473886, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1805, 1807, 807, 266, 63, 208} \[ \frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{14 e^2 \sqrt{d^2-e^2 x^2}}{3 d^8 x}-\frac{e \sqrt{d^2-e^2 x^2}}{d^7 x^2}-\frac{\sqrt{d^2-e^2 x^2}}{3 d^6 x^3}-\frac{7 e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^8} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(2*e^3*(d + e*x))/(5*d^4*(d^2 - e^2*x^2)^(5/2)) + (e^3*(20*d + 23*e*x))/(15*d^6*(d^2 - e^2*x^2)^(3/2)) + (2*e^
3*(45*d + 53*e*x))/(15*d^8*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(3*d^6*x^3) - (e*Sqrt[d^2 - e^2*x^2])/(d
^7*x^2) - (14*e^2*Sqrt[d^2 - e^2*x^2])/(3*d^8*x) - (7*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^8

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 1807

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])

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^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{2 e^3 (d+e x)}{5 d^4 \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{10 e^3 x^3}{d}-\frac{8 e^4 x^4}{d^2}}{x^4 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \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{60 e^3 x^3}{d}+\frac{46 e^4 x^4}{d^2}}{x^4 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{-15 d^2-30 d e x-60 e^2 x^2-\frac{90 e^3 x^3}{d}}{x^4 \sqrt{d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 d^6 x^3}+\frac{\int \frac{90 d^3 e+210 d^2 e^2 x+270 d e^3 x^2}{x^3 \sqrt{d^2-e^2 x^2}} \, dx}{45 d^8}\\ &=\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 d^6 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d^7 x^2}-\frac{\int \frac{-420 d^4 e^2-630 d^3 e^3 x}{x^2 \sqrt{d^2-e^2 x^2}} \, dx}{90 d^{10}}\\ &=\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 d^6 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d^7 x^2}-\frac{14 e^2 \sqrt{d^2-e^2 x^2}}{3 d^8 x}+\frac{\left (7 e^3\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^7}\\ &=\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 d^6 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d^7 x^2}-\frac{14 e^2 \sqrt{d^2-e^2 x^2}}{3 d^8 x}+\frac{\left (7 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^7}\\ &=\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 d^6 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d^7 x^2}-\frac{14 e^2 \sqrt{d^2-e^2 x^2}}{3 d^8 x}-\frac{(7 e) \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^7}\\ &=\frac{2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^3 (45 d+53 e x)}{15 d^8 \sqrt{d^2-e^2 x^2}}-\frac{\sqrt{d^2-e^2 x^2}}{3 d^6 x^3}-\frac{e \sqrt{d^2-e^2 x^2}}{d^7 x^2}-\frac{14 e^2 \sqrt{d^2-e^2 x^2}}{3 d^8 x}-\frac{7 e^3 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^8}\\ \end{align*}

Mathematica [C]  time = 0.0599333, size = 105, normalized size = 0.5 \[ \frac{6 d^5 e^3 x^3 \, _2F_1\left (-\frac{5}{2},2;-\frac{3}{2};1-\frac{e^2 x^2}{d^2}\right )-55 d^6 e^2 x^2+330 d^4 e^4 x^4-440 d^2 e^6 x^6-5 d^8+176 e^8 x^8}{15 d^8 x^3 \left (d^2-e^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(-5*d^8 - 55*d^6*e^2*x^2 + 330*d^4*e^4*x^4 - 440*d^2*e^6*x^6 + 176*e^8*x^8 + 6*d^5*e^3*x^3*Hypergeometric2F1[-
5/2, 2, -3/2, 1 - (e^2*x^2)/d^2])/(15*d^8*x^3*(d^2 - e^2*x^2)^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.069, size = 249, normalized size = 1.2 \begin{align*} -{\frac{1}{3\,{x}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-{\frac{11\,{e}^{2}}{3\,{d}^{2}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{22\,{e}^{4}x}{5\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{88\,{e}^{4}x}{15\,{d}^{6}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{176\,{e}^{4}x}{15\,{d}^{8}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{e}{d{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{7\,{e}^{3}}{5\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{7\,{e}^{3}}{3\,{d}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+7\,{\frac{{e}^{3}}{{d}^{7}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}-7\,{\frac{{e}^{3}}{{d}^{7}\sqrt{{d}^{2}}}\ln \left ({\frac{2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x)

[Out]

-1/3/x^3/(-e^2*x^2+d^2)^(5/2)-11/3/d^2*e^2/x/(-e^2*x^2+d^2)^(5/2)+22/5/d^4*e^4*x/(-e^2*x^2+d^2)^(5/2)+88/15/d^
6*e^4*x/(-e^2*x^2+d^2)^(3/2)+176/15/d^8*e^4*x/(-e^2*x^2+d^2)^(1/2)-1/d*e/x^2/(-e^2*x^2+d^2)^(5/2)+7/5/d^3*e^3/
(-e^2*x^2+d^2)^(5/2)+7/3/d^5*e^3/(-e^2*x^2+d^2)^(3/2)+7/d^7*e^3/(-e^2*x^2+d^2)^(1/2)-7/d^7*e^3/(d^2)^(1/2)*ln(
(2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

________________________________________________________________________________________

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^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.27404, size = 471, normalized size = 2.25 \begin{align*} \frac{116 \, e^{7} x^{7} - 232 \, d e^{6} x^{6} + 232 \, d^{3} e^{4} x^{4} - 116 \, d^{4} e^{3} x^{3} + 105 \,{\left (e^{7} x^{7} - 2 \, d e^{6} x^{6} + 2 \, d^{3} e^{4} x^{4} - d^{4} e^{3} x^{3}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (176 \, e^{6} x^{6} - 247 \, d e^{5} x^{5} - 122 \, d^{2} e^{4} x^{4} + 246 \, d^{3} e^{3} x^{3} - 40 \, d^{4} e^{2} x^{2} - 5 \, d^{5} e x - 5 \, d^{6}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{8} e^{4} x^{7} - 2 \, d^{9} e^{3} x^{6} + 2 \, d^{11} e x^{4} - d^{12} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(116*e^7*x^7 - 232*d*e^6*x^6 + 232*d^3*e^4*x^4 - 116*d^4*e^3*x^3 + 105*(e^7*x^7 - 2*d*e^6*x^6 + 2*d^3*e^4
*x^4 - d^4*e^3*x^3)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (176*e^6*x^6 - 247*d*e^5*x^5 - 122*d^2*e^4*x^4 + 246*
d^3*e^3*x^3 - 40*d^4*e^2*x^2 - 5*d^5*e*x - 5*d^6)*sqrt(-e^2*x^2 + d^2))/(d^8*e^4*x^7 - 2*d^9*e^3*x^6 + 2*d^11*
e*x^4 - d^12*x^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{2}}{x^{4} \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**4/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**2/(x**4*(-(-d + e*x)*(d + e*x))**(7/2)), x)

________________________________________________________________________________________

Giac [A]  time = 1.17848, size = 439, normalized size = 2.1 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left ({\left (2 \, x{\left (\frac{53 \, x e^{8}}{d^{8}} + \frac{45 \, e^{7}}{d^{7}}\right )} - \frac{235 \, e^{6}}{d^{6}}\right )} x - \frac{200 \, e^{5}}{d^{5}}\right )} x + \frac{135 \, e^{4}}{d^{4}}\right )} x + \frac{116 \, e^{3}}{d^{3}}\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} + \frac{x^{3}{\left (\frac{6 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{6}}{x} + \frac{57 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{4}}{x^{2}} + e^{8}\right )} e}{24 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{8}} - \frac{7 \, e^{3} \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^{8}} - \frac{{\left (\frac{57 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{16} e^{16}}{x} + \frac{6 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{16} e^{14}}{x^{2}} + \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{16} e^{12}}{x^{3}}\right )} e^{\left (-15\right )}}{24 \, d^{24}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*((((2*x*(53*x*e^8/d^8 + 45*e^7/d^7) - 235*e^6/d^6)*x - 200*e^5/d^5)*x + 135*e^4/d^4
)*x + 116*e^3/d^3)/(x^2*e^2 - d^2)^3 + 1/24*x^3*(6*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^6/x + 57*(d*e + sqrt(-x^2*
e^2 + d^2)*e)^2*e^4/x^2 + e^8)*e/((d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^8) - 7*e^3*log(1/2*abs(-2*d*e - 2*sqrt(-x
^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^8 - 1/24*(57*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^16*e^16/x + 6*(d*e + sqrt(-x^2
*e^2 + d^2)*e)^2*d^16*e^14/x^2 + (d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^16*e^12/x^3)*e^(-15)/d^24