3.52 \(\int \frac{(d+e x)^2}{x^3 (d^2-e^2 x^2)^{7/2}} \, dx\)

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

[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)

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Rubi [A]  time = 0.358016, antiderivative size = 182, normalized size of antiderivative = 1., 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 = {1805, 1807, 807, 266, 63, 208} \[ \frac{2 e^2 (10 d+11 e x)}{5 d^7 \sqrt{d^2-e^2 x^2}}+\frac{e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{2 e \sqrt{d^2-e^2 x^2}}{d^7 x}-\frac{\sqrt{d^2-e^2 x^2}}{2 d^6 x^2}-\frac{9 e^2 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{2 d^7} \]

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 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^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 ) \operatorname{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 \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 )}{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*}

Mathematica [C]  time = 0.0612163, size = 117, normalized size = 0.64 \[ \frac{e \left (d^5 e x \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};1-\frac{e^2 x^2}{d^2}\right )+d^5 e x \, _2F_1\left (-\frac{5}{2},2;-\frac{3}{2};1-\frac{e^2 x^2}{d^2}\right )+60 d^4 e^2 x^2-80 d^2 e^4 x^4-10 d^6+32 e^6 x^6\right )}{5 d^7 x \left (d^2-e^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(-10*d^6 + 60*d^4*e^2*x^2 - 80*d^2*e^4*x^4 + 32*e^6*x^6 + d^5*e*x*Hypergeometric2F1[-5/2, 1, -3/2, 1 - (e^2
*x^2)/d^2] + d^5*e*x*Hypergeometric2F1[-5/2, 2, -3/2, 1 - (e^2*x^2)/d^2]))/(5*d^7*x*(d^2 - e^2*x^2)^(5/2))

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Maple [A]  time = 0.069, size = 224, normalized size = 1.2 \begin{align*}{\frac{9\,{e}^{2}}{10\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{3\,{e}^{2}}{2\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{9\,{e}^{2}}{2\,{d}^{6}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{9\,{e}^{2}}{2\,{d}^{6}}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{2}}}}}-{\frac{1}{2\,{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}-2\,{\frac{e}{dx \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}}+{\frac{12\,{e}^{3}x}{5\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{16\,{e}^{3}x}{5\,{d}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{32\,{e}^{3}x}{5\,{d}^{7}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \end{align*}

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)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0328, size = 440, normalized size = 2.42 \begin{align*} \frac{54 \, e^{6} x^{6} - 108 \, d e^{5} x^{5} + 108 \, d^{3} e^{3} x^{3} - 54 \, d^{4} e^{2} x^{2} + 45 \,{\left (e^{6} x^{6} - 2 \, d e^{5} x^{5} + 2 \, d^{3} e^{3} x^{3} - d^{4} e^{2} x^{2}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (64 \, e^{5} x^{5} - 83 \, d e^{4} x^{4} - 58 \, d^{2} e^{3} x^{3} + 94 \, d^{3} e^{2} x^{2} - 10 \, d^{4} e x - 5 \, d^{5}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{10 \,{\left (d^{7} e^{4} x^{6} - 2 \, d^{8} e^{3} x^{5} + 2 \, d^{10} e x^{3} - d^{11} x^{2}\right )}} \end{align*}

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

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

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 [A]  time = 1.16959, size = 351, normalized size = 1.93 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left (2 \,{\left (x{\left (\frac{11 \, x e^{7}}{d^{7}} + \frac{10 \, e^{6}}{d^{6}}\right )} - \frac{25 \, e^{5}}{d^{5}}\right )} x - \frac{45 \, e^{4}}{d^{4}}\right )} x + \frac{30 \, e^{3}}{d^{3}}\right )} x + \frac{27 \, e^{2}}{d^{2}}\right )}}{5 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac{9 \, e^{2} \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 )}{2 \, d^{7}} + \frac{x^{2}{\left (\frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{4}}{x} + e^{6}\right )}}{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{7}} - \frac{{\left (\frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{7} e^{8}}{x} + \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{7} e^{6}}{x^{2}}\right )} e^{\left (-8\right )}}{8 \, d^{14}} \end{align*}

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]

-1/5*sqrt(-x^2*e^2 + d^2)*(((2*(x*(11*x*e^7/d^7 + 10*e^6/d^6) - 25*e^5/d^5)*x - 45*e^4/d^4)*x + 30*e^3/d^3)*x
+ 27*e^2/d^2)/(x^2*e^2 - d^2)^3 - 9/2*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^7 +
1/8*x^2*(8*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^4/x + e^6)/((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^7) - 1/8*(8*(d*e +
sqrt(-x^2*e^2 + d^2)*e)*d^7*e^8/x + (d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^7*e^6/x^2)*e^(-8)/d^14