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

Optimal. Leaf size=143 \[ \frac{e^4 \sqrt{d^2-e^2 x^2}}{16 d x^2}-\frac{e^2 \left (d^2-e^2 x^2\right )^{3/2}}{24 d x^4}-\frac{e \left (d^2-e^2 x^2\right )^{5/2}}{5 d^2 x^5}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{e^6 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{16 d^2} \]

[Out]

(e^4*Sqrt[d^2 - e^2*x^2])/(16*d*x^2) - (e^2*(d^2 - e^2*x^2)^(3/2))/(24*d*x^4) - (d^2 - e^2*x^2)^(5/2)/(6*d*x^6
) - (e*(d^2 - e^2*x^2)^(5/2))/(5*d^2*x^5) - (e^6*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(16*d^2)

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Rubi [A]  time = 0.0948237, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {835, 807, 266, 47, 63, 208} \[ \frac{e^4 \sqrt{d^2-e^2 x^2}}{16 d x^2}-\frac{e^2 \left (d^2-e^2 x^2\right )^{3/2}}{24 d x^4}-\frac{e \left (d^2-e^2 x^2\right )^{5/2}}{5 d^2 x^5}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{e^6 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{16 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*Sqrt[d^2 - e^2*x^2])/(16*d*x^2) - (e^2*(d^2 - e^2*x^2)^(3/2))/(24*d*x^4) - (d^2 - e^2*x^2)^(5/2)/(6*d*x^6
) - (e*(d^2 - e^2*x^2)^(5/2))/(5*d^2*x^5) - (e^6*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(16*d^2)

Rule 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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) \left (d^2-e^2 x^2\right )^{3/2}}{x^7} \, dx &=-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{\int \frac{\left (-6 d^2 e-d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^6} \, dx}{6 d^2}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{e \left (d^2-e^2 x^2\right )^{5/2}}{5 d^2 x^5}+\frac{e^2 \int \frac{\left (d^2-e^2 x^2\right )^{3/2}}{x^5} \, dx}{6 d}\\ &=-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{e \left (d^2-e^2 x^2\right )^{5/2}}{5 d^2 x^5}+\frac{e^2 \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )}{12 d}\\ &=-\frac{e^2 \left (d^2-e^2 x^2\right )^{3/2}}{24 d x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{e \left (d^2-e^2 x^2\right )^{5/2}}{5 d^2 x^5}-\frac{e^4 \operatorname{Subst}\left (\int \frac{\sqrt{d^2-e^2 x}}{x^2} \, dx,x,x^2\right )}{16 d}\\ &=\frac{e^4 \sqrt{d^2-e^2 x^2}}{16 d x^2}-\frac{e^2 \left (d^2-e^2 x^2\right )^{3/2}}{24 d x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{e \left (d^2-e^2 x^2\right )^{5/2}}{5 d^2 x^5}+\frac{e^6 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )}{32 d}\\ &=\frac{e^4 \sqrt{d^2-e^2 x^2}}{16 d x^2}-\frac{e^2 \left (d^2-e^2 x^2\right )^{3/2}}{24 d x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{e \left (d^2-e^2 x^2\right )^{5/2}}{5 d^2 x^5}-\frac{e^4 \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 )}{16 d}\\ &=\frac{e^4 \sqrt{d^2-e^2 x^2}}{16 d x^2}-\frac{e^2 \left (d^2-e^2 x^2\right )^{3/2}}{24 d x^4}-\frac{\left (d^2-e^2 x^2\right )^{5/2}}{6 d x^6}-\frac{e \left (d^2-e^2 x^2\right )^{5/2}}{5 d^2 x^5}-\frac{e^6 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )}{16 d^2}\\ \end{align*}

Mathematica [C]  time = 0.024719, size = 59, normalized size = 0.41 \[ -\frac{e \left (d^2-e^2 x^2\right )^{5/2} \left (e^5 x^5 \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};1-\frac{e^2 x^2}{d^2}\right )+d^5\right )}{5 d^7 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*(d^2 - e^2*x^2)^(5/2)*(d^5 + e^5*x^5*Hypergeometric2F1[5/2, 4, 7/2, 1 - (e^2*x^2)/d^2]))/(5*d^7*x^5)

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Maple [A]  time = 0.079, size = 186, normalized size = 1.3 \begin{align*} -{\frac{e}{5\,{d}^{2}{x}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{1}{6\,d{x}^{6}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{2}}{24\,{d}^{3}{x}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{{e}^{4}}{48\,{d}^{5}{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{{e}^{6}}{48\,{d}^{5}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{{e}^{6}}{16\,{d}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{{e}^{6}}{16\,d}\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)*(-e^2*x^2+d^2)^(3/2)/x^7,x)

[Out]

-1/5*e*(-e^2*x^2+d^2)^(5/2)/d^2/x^5-1/6*(-e^2*x^2+d^2)^(5/2)/d/x^6-1/24*e^2/d^3/x^4*(-e^2*x^2+d^2)^(5/2)+1/48*
e^4/d^5/x^2*(-e^2*x^2+d^2)^(5/2)+1/48*e^6/d^5*(-e^2*x^2+d^2)^(3/2)+1/16*e^6/d^3*(-e^2*x^2+d^2)^(1/2)-1/16*e^6/
d/(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)*(-e^2*x^2+d^2)^(3/2)/x^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7701, size = 232, normalized size = 1.62 \begin{align*} \frac{15 \, e^{6} x^{6} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (48 \, e^{5} x^{5} + 15 \, d e^{4} x^{4} - 96 \, d^{2} e^{3} x^{3} - 70 \, d^{3} e^{2} x^{2} + 48 \, d^{4} e x + 40 \, d^{5}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{240 \, d^{2} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*e^6*x^6*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (48*e^5*x^5 + 15*d*e^4*x^4 - 96*d^2*e^3*x^3 - 70*d^3*e^
2*x^2 + 48*d^4*e*x + 40*d^5)*sqrt(-e^2*x^2 + d^2))/(d^2*x^6)

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Sympy [C]  time = 13.967, size = 930, normalized size = 6.5 \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)*(-e**2*x**2+d**2)**(3/2)/x**7,x)

[Out]

d**3*Piecewise((-d**2/(6*e*x**7*sqrt(d**2/(e**2*x**2) - 1)) + 5*e/(24*x**5*sqrt(d**2/(e**2*x**2) - 1)) + e**3/
(48*d**2*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**5/(16*d**4*x*sqrt(d**2/(e**2*x**2) - 1)) + e**6*acosh(d/(e*x))/
(16*d**5), Abs(d**2)/(Abs(e**2)*Abs(x**2)) > 1), (I*d**2/(6*e*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e/(24*x*
*5*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**3/(48*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**5/(16*d**4*x*sqrt(-
d**2/(e**2*x**2) + 1)) - I*e**6*asin(d/(e*x))/(16*d**5), True)) + d**2*e*Piecewise((3*I*d**3*sqrt(-1 + e**2*x*
*2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*I*d*e**2*x**2*sqrt(-1 + e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x
**7) + 2*I*e**6*x**6*sqrt(-1 + e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d**3*e**2*x**7) - I*e**4*x**4*sqrt(-1 + e**
2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), Abs(e**2*x**2)/Abs(d**2) > 1), (3*d**3*sqrt(1 - e**2*x**2/d**2)
/(-15*d**2*x**5 + 15*e**2*x**7) - 4*d*e**2*x**2*sqrt(1 - e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) + 2*e*
*6*x**6*sqrt(1 - e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d**3*e**2*x**7) - e**4*x**4*sqrt(1 - e**2*x**2/d**2)/(-15
*d**3*x**5 + 15*d*e**2*x**7), True)) - d*e**2*Piecewise((-d**2/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*
x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**3/(8*d**2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**4*acosh(d/(e*x))/(8*d**3),
Abs(d**2)/(Abs(e**2)*Abs(x**2)) > 1), (I*d**2/(4*e*x**5*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e/(8*x**3*sqrt(-d**
2/(e**2*x**2) + 1)) + I*e**3/(8*d**2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True)) -
e**3*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(3*x**2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**2), Abs(d**2)/(
Abs(e**2)*Abs(x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*x**2) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*
d**2), True))

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Giac [B]  time = 1.27681, size = 582, normalized size = 4.07 \begin{align*} \frac{x^{6}{\left (\frac{12 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{12}}{x} - \frac{15 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{10}}{x^{2}} - \frac{60 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{8}}{x^{3}} - \frac{15 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{6}}{x^{4}} + \frac{120 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{5} e^{4}}{x^{5}} + 5 \, e^{14}\right )} e^{4}}{1920 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{6} d^{2}} - \frac{e^{6} \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 )}{16 \, d^{2}} - \frac{{\left (\frac{120 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{10} e^{52}}{x} - \frac{15 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{10} e^{50}}{x^{2}} - \frac{60 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{10} e^{48}}{x^{3}} - \frac{15 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{10} e^{46}}{x^{4}} + \frac{12 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{10} e^{44}}{x^{5}} + \frac{5 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{6} d^{10} e^{42}}{x^{6}}\right )} e^{\left (-48\right )}}{1920 \, d^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*x^6*(12*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^12/x - 15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^10/x^2 - 60*(d*e
+ sqrt(-x^2*e^2 + d^2)*e)^3*e^8/x^3 - 15*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^6/x^4 + 120*(d*e + sqrt(-x^2*e^2 +
d^2)*e)^5*e^4/x^5 + 5*e^14)*e^4/((d*e + sqrt(-x^2*e^2 + d^2)*e)^6*d^2) - 1/16*e^6*log(1/2*abs(-2*d*e - 2*sqrt
(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^2 - 1/1920*(120*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^10*e^52/x - 15*(d*e + sq
rt(-x^2*e^2 + d^2)*e)^2*d^10*e^50/x^2 - 60*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^10*e^48/x^3 - 15*(d*e + sqrt(-x^
2*e^2 + d^2)*e)^4*d^10*e^46/x^4 + 12*(d*e + sqrt(-x^2*e^2 + d^2)*e)^5*d^10*e^44/x^5 + 5*(d*e + sqrt(-x^2*e^2 +
d^2)*e)^6*d^10*e^42/x^6)*e^(-48)/d^12