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

Optimal. Leaf size=117 \[ \frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{15 d+8 e x}{15 d^6 \sqrt{d^2-e^2 x^2}}+\frac{5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^6} \]

[Out]

(d + e*x)/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (5*d + 4*e*x)/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (15*d + 8*e*x)/(15*d^
6*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^6

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Rubi [A]  time = 0.103197, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {823, 12, 266, 63, 208} \[ \frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{15 d+8 e x}{15 d^6 \sqrt{d^2-e^2 x^2}}+\frac{5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x)/(5*d^2*(d^2 - e^2*x^2)^(5/2)) + (5*d + 4*e*x)/(15*d^4*(d^2 - e^2*x^2)^(3/2)) + (15*d + 8*e*x)/(15*d^
6*Sqrt[d^2 - e^2*x^2]) - ArcTanh[Sqrt[d^2 - e^2*x^2]/d]/d^6

Rule 823

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] + Di
st[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 12

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

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}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{\int \frac{5 d^3 e^2+4 d^2 e^3 x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^4 e^2}\\ &=\frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{15 d^5 e^4+8 d^4 e^5 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^8 e^4}\\ &=\frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+8 e x}{15 d^6 \sqrt{d^2-e^2 x^2}}+\frac{\int \frac{15 d^7 e^6}{x \sqrt{d^2-e^2 x^2}} \, dx}{15 d^{12} e^6}\\ &=\frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+8 e x}{15 d^6 \sqrt{d^2-e^2 x^2}}+\frac{\int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx}{d^5}\\ &=\frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+8 e x}{15 d^6 \sqrt{d^2-e^2 x^2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^5}\\ &=\frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+8 e x}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\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^2}\\ &=\frac{d+e x}{5 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{5 d+4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{15 d+8 e x}{15 d^6 \sqrt{d^2-e^2 x^2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{d^6}\\ \end{align*}

Mathematica [A]  time = 0.0679326, size = 131, normalized size = 1.12 \[ \frac{-27 d^2 e^2 x^2-15 (d-e x)^2 (d+e x) \sqrt{d^2-e^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )-8 d^3 e x+23 d^4+7 d e^3 x^3+8 e^4 x^4}{15 d^6 (d-e x)^2 (d+e x) \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23*d^4 - 8*d^3*e*x - 27*d^2*e^2*x^2 + 7*d*e^3*x^3 + 8*e^4*x^4 - 15*(d - e*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2]*
ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(15*d^6*(d - e*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2])

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Maple [A]  time = 0.055, size = 163, normalized size = 1.4 \begin{align*}{\frac{ex}{5\,{d}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{4\,ex}{15\,{d}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,ex}{15\,{d}^{6}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}+{\frac{1}{5\,d} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}}+{\frac{1}{3\,{d}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{1}{{d}^{5}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}}-{\frac{1}{{d}^{5}}\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}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*e*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/15*e/d^4*x/(-e^2*x^2+d^2)^(3/2)+8/15*e/d^6*x/(-e^2*x^2+d^2)^(1/2)+1/5/d/(-e
^2*x^2+d^2)^(5/2)+1/3/d^3/(-e^2*x^2+d^2)^(3/2)+1/d^5/(-e^2*x^2+d^2)^(1/2)-1/d^5/(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 [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)/x/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.97722, size = 493, normalized size = 4.21 \begin{align*} \frac{23 \, e^{5} x^{5} - 23 \, d e^{4} x^{4} - 46 \, d^{2} e^{3} x^{3} + 46 \, d^{3} e^{2} x^{2} + 23 \, d^{4} e x - 23 \, d^{5} + 15 \,{\left (e^{5} x^{5} - d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} + 2 \, d^{3} e^{2} x^{2} + d^{4} e x - d^{5}\right )} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (8 \, e^{4} x^{4} + 7 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} - 8 \, d^{3} e x + 23 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{6} e^{5} x^{5} - d^{7} e^{4} x^{4} - 2 \, d^{8} e^{3} x^{3} + 2 \, d^{9} e^{2} x^{2} + d^{10} e x - d^{11}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(23*e^5*x^5 - 23*d*e^4*x^4 - 46*d^2*e^3*x^3 + 46*d^3*e^2*x^2 + 23*d^4*e*x - 23*d^5 + 15*(e^5*x^5 - d*e^4*
x^4 - 2*d^2*e^3*x^3 + 2*d^3*e^2*x^2 + d^4*e*x - d^5)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (8*e^4*x^4 + 7*d*e^3
*x^3 - 27*d^2*e^2*x^2 - 8*d^3*e*x + 23*d^4)*sqrt(-e^2*x^2 + d^2))/(d^6*e^5*x^5 - d^7*e^4*x^4 - 2*d^8*e^3*x^3 +
2*d^9*e^2*x^2 + d^10*e*x - d^11)

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Sympy [C]  time = 22.3219, size = 2382, normalized size = 20.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*Piecewise((-46*I*d**6*sqrt(-1 + e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**
7*e**6*x**6) - 15*d**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*
x**6) + 30*d**6*log(e*x/d)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 30*I*d**
6*asin(d/(e*x))/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 70*I*d**4*e**2*x**2
*sqrt(-1 + e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 45*d**4*
e**2*x**2*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 90*d*
*4*e**2*x**2*log(e*x/d)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 90*I*d**4*e
**2*x**2*asin(d/(e*x))/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 30*I*d**2*e*
*4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 4
5*d**2*e**4*x**4*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6)
+ 90*d**2*e**4*x**4*log(e*x/d)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 90*I
*d**2*e**4*x**4*asin(d/(e*x))/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 15*e*
*6*x**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 30*e**6
*x**6*log(e*x/d)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 30*I*e**6*x**6*asi
n(d/(e*x))/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6), Abs(e**2*x**2)/Abs(d**2)
> 1), (-46*d**6*sqrt(1 - e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x*
*6) - 15*d**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 3
0*d**6*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x*
*6) - 15*I*pi*d**6/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 70*d**4*e**2*x**
2*sqrt(1 - e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 45*d**4*
e**2*x**2*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 90*d*
*4*e**2*x**2*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e
**6*x**6) + 45*I*pi*d**4*e**2*x**2/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) -
30*d**2*e**4*x**4*sqrt(1 - e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*
x**6) - 45*d**2*e**4*x**4*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e*
*6*x**6) + 90*d**2*e**4*x**4*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*
x**4 + 30*d**7*e**6*x**6) - 45*I*pi*d**2*e**4*x**4/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d*
*7*e**6*x**6) + 15*e**6*x**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7
*e**6*x**6) - 30*e**6*x**6*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x*
*4 + 30*d**7*e**6*x**6) + 15*I*pi*e**6*x**6/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6
*x**6), True)) + e*Piecewise((-15*I*d**4*x/(15*d**11*sqrt(-1 + e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(-1 + e
**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)) + 20*I*d**2*e**2*x**3/(15*d**11*sqrt(-1 + e**2*x
**2/d**2) - 30*d**9*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)) - 8*I*e
**4*x**5/(15*d**11*sqrt(-1 + e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(-1 + e**2*x**2/d**2) + 15*d**7*e**4*x**4
*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (15*d**4*x/(15*d**11*sqrt(1 - e**2*x**2/d**2) - 30
*d**9*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(1 - e**2*x**2/d**2)) - 20*d**2*e**2*x**3/(15
*d**11*sqrt(1 - e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 15*d**7*e**4*x**4*sqrt(1 - e**2
*x**2/d**2)) + 8*e**4*x**5/(15*d**11*sqrt(1 - e**2*x**2/d**2) - 30*d**9*e**2*x**2*sqrt(1 - e**2*x**2/d**2) + 1
5*d**7*e**4*x**4*sqrt(1 - e**2*x**2/d**2)), True))

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Giac [A]  time = 1.1866, size = 165, normalized size = 1.41 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left ({\left (x{\left (\frac{8 \, x e^{5}}{d^{6}} + \frac{15 \, e^{4}}{d^{5}}\right )} - \frac{20 \, e^{3}}{d^{4}}\right )} x - \frac{35 \, e^{2}}{d^{3}}\right )} x + \frac{15 \, e}{d^{2}}\right )} x + \frac{23}{d}\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*((((x*(8*x*e^5/d^6 + 15*e^4/d^5) - 20*e^3/d^4)*x - 35*e^2/d^3)*x + 15*e/d^2)*x + 23
/d)/(x^2*e^2 - d^2)^3 - log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^6