3.64 \(\int \frac{(d+e x)^3 \sqrt{d^2-e^2 x^2}}{x^5} \, dx\)

Optimal. Leaf size=134 \[ -\frac{e^2 (13 d+8 e x) \sqrt{d^2-e^2 x^2}}{8 x^2}-\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+e^4 \left (-\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )\right )+\frac{13}{8} e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

-(e^2*(13*d + 8*e*x)*Sqrt[d^2 - e^2*x^2])/(8*x^2) - (d*(d^2 - e^2*x^2)^(3/2))/(4*x^4) - (e*(d^2 - e^2*x^2)^(3/
2))/x^3 - e^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] + (13*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/8

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Rubi [A]  time = 0.215096, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {1807, 811, 844, 217, 203, 266, 63, 208} \[ -\frac{e^2 (13 d+8 e x) \sqrt{d^2-e^2 x^2}}{8 x^2}-\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+e^4 \left (-\tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )\right )+\frac{13}{8} e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*Sqrt[d^2 - e^2*x^2])/x^5,x]

[Out]

-(e^2*(13*d + 8*e*x)*Sqrt[d^2 - e^2*x^2])/(8*x^2) - (d*(d^2 - e^2*x^2)^(3/2))/(4*x^4) - (e*(d^2 - e^2*x^2)^(3/
2))/x^3 - e^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] + (13*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/8

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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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)^3 \sqrt{d^2-e^2 x^2}}{x^5} \, dx &=-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{\int \frac{\sqrt{d^2-e^2 x^2} \left (-12 d^4 e-13 d^3 e^2 x-4 d^2 e^3 x^2\right )}{x^4} \, dx}{4 d^2}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}+\frac{\int \frac{\left (39 d^5 e^2+12 d^4 e^3 x\right ) \sqrt{d^2-e^2 x^2}}{x^3} \, dx}{12 d^4}\\ &=-\frac{e^2 (13 d+8 e x) \sqrt{d^2-e^2 x^2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac{\int \frac{78 d^7 e^4+48 d^6 e^5 x}{x \sqrt{d^2-e^2 x^2}} \, dx}{48 d^6}\\ &=-\frac{e^2 (13 d+8 e x) \sqrt{d^2-e^2 x^2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac{1}{8} \left (13 d e^4\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx-e^5 \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{e^2 (13 d+8 e x) \sqrt{d^2-e^2 x^2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac{1}{16} \left (13 d e^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )-e^5 \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )\\ &=-\frac{e^2 (13 d+8 e x) \sqrt{d^2-e^2 x^2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-e^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{1}{8} \left (13 d e^2\right ) \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 )\\ &=-\frac{e^2 (13 d+8 e x) \sqrt{d^2-e^2 x^2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-e^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{13}{8} e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\\ \end{align*}

Mathematica [C]  time = 0.261596, size = 196, normalized size = 1.46 \[ -\frac{e \sqrt{d^2-e^2 x^2} \left (2 e^3 x^3 \left (d^2-e^2 x^2\right ) \sqrt{1-\frac{e^2 x^2}{d^2}} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};1-\frac{e^2 x^2}{d^2}\right )+6 d^5 \sqrt{1-\frac{e^2 x^2}{d^2}}+9 d^4 e x \sqrt{1-\frac{e^2 x^2}{d^2}}+6 d^2 e^3 x^3 \sin ^{-1}\left (\frac{e x}{d}\right )-9 d^2 e^3 x^3 \tanh ^{-1}\left (\sqrt{1-\frac{e^2 x^2}{d^2}}\right )\right )}{6 d^3 x^3 \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*Sqrt[d^2 - e^2*x^2])/x^5,x]

[Out]

-(e*Sqrt[d^2 - e^2*x^2]*(6*d^5*Sqrt[1 - (e^2*x^2)/d^2] + 9*d^4*e*x*Sqrt[1 - (e^2*x^2)/d^2] + 6*d^2*e^3*x^3*Arc
Sin[(e*x)/d] - 9*d^2*e^3*x^3*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]] + 2*e^3*x^3*(d^2 - e^2*x^2)*Sqrt[1 - (e^2*x^2)/d
^2]*Hypergeometric2F1[3/2, 3, 5/2, 1 - (e^2*x^2)/d^2]))/(6*d^3*x^3*Sqrt[1 - (e^2*x^2)/d^2])

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Maple [A]  time = 0.064, size = 212, normalized size = 1.6 \begin{align*} -{\frac{e}{{x}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{d}{4\,{x}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{13\,{e}^{2}}{8\,d{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{13\,{e}^{4}}{8\,d}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{13\,d{e}^{4}}{8}\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{{e}^{3}}{{d}^{2}x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{5}x}{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{{e}^{5}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(-e^2*x^2+d^2)^(1/2)/x^5,x)

[Out]

-e*(-e^2*x^2+d^2)^(3/2)/x^3-1/4*d*(-e^2*x^2+d^2)^(3/2)/x^4-13/8/d*e^2/x^2*(-e^2*x^2+d^2)^(3/2)-13/8*e^4/d*(-e^
2*x^2+d^2)^(1/2)+13/8*d*e^4/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-e^3/d^2/x*(-e^2*x^2+d
^2)^(3/2)-e^5/d^2*x*(-e^2*x^2+d^2)^(1/2)-e^5/(e^2)^(1/2)*arctan((e^2)^(1/2)*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)^3*(-e^2*x^2+d^2)^(1/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76287, size = 230, normalized size = 1.72 \begin{align*} \frac{16 \, e^{4} x^{4} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) - 13 \, e^{4} x^{4} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) -{\left (11 \, d e^{2} x^{2} + 8 \, d^{2} e x + 2 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{8 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(-e^2*x^2+d^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

1/8*(16*e^4*x^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 13*e^4*x^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (1
1*d*e^2*x^2 + 8*d^2*e*x + 2*d^3)*sqrt(-e^2*x^2 + d^2))/x^4

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Sympy [C]  time = 9.60595, size = 556, normalized size = 4.15 \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)**3*(-e**2*x**2+d**2)**(1/2)/x**5,x)

[Out]

d**3*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*sq
rt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True)) + 3*d**2*e*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)) + 3*d*e**2*Piecewise((-
d**2/(2*e*x**3*sqrt(d**2/(e**2*x**2) - 1)) + e/(2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**2*acosh(d/(e*x))/(2*d), A
bs(d**2)/(Abs(e**2)*Abs(x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(2*x) - I*e**2*asin(d/(e*x))/(2*d), Tru
e)) + e**3*Piecewise((I*d/(x*sqrt(-1 + e**2*x**2/d**2)) + I*e*acosh(e*x/d) - I*e**2*x/(d*sqrt(-1 + e**2*x**2/d
**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (-d/(x*sqrt(1 - e**2*x**2/d**2)) - e*asin(e*x/d) + e**2*x/(d*sqrt(1 - e*
*2*x**2/d**2)), True))

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Giac [B]  time = 1.17322, size = 398, normalized size = 2.97 \begin{align*} -\arcsin \left (\frac{x e}{d}\right ) e^{4} \mathrm{sgn}\left (d\right ) + \frac{x^{4}{\left (\frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{8}}{x} + \frac{24 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{6}}{x^{2}} + \frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{4}}{x^{3}} + e^{10}\right )} e^{2}}{64 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{4}} - \frac{1}{64} \,{\left (\frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} e^{26}}{x} + \frac{24 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{24}}{x^{2}} + \frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{22}}{x^{3}} + \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{20}}{x^{4}}\right )} e^{\left (-24\right )} + \frac{13}{8} \, e^{4} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(-e^2*x^2+d^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^4*sgn(d) + 1/64*x^4*(8*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^8/x + 24*(d*e + sqrt(-x^2*e^2 + d^2)*
e)^2*e^6/x^2 + 8*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^4/x^3 + e^10)*e^2/(d*e + sqrt(-x^2*e^2 + d^2)*e)^4 - 1/64*
(8*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^26/x + 24*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^24/x^2 + 8*(d*e + sqrt(-x^2*e
^2 + d^2)*e)^3*e^22/x^3 + (d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^20/x^4)*e^(-24) + 13/8*e^4*log(1/2*abs(-2*d*e - 2
*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))