3.2.86 \(\int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [186]

3.2.86.1 Optimal result

Integrand size = 27, antiderivative size = 146 \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {e (5 d-7 e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {e (15 d-19 e x)}{5 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{d^5 x}+\frac {3 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \]

output

-4/5*e*(-e*x+d)/d/(-e^2*x^2+d^2)^(5/2)-1/5*e*(-7*e*x+5*d)/d^3/(-e^2*x^2+d^ 
2)^(3/2)+3*e*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^5-1/5*e*(-19*e*x+15*d)/d^5/ 
(-e^2*x^2+d^2)^(1/2)-(-e^2*x^2+d^2)^(1/2)/d^5/x
 

3.2.86.2 Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\frac {d \sqrt {d^2-e^2 x^2} \left (5 d^3+39 d^2 e x+57 d e^2 x^2+24 e^3 x^3\right )}{x (d+e x)^3}-15 \sqrt {d^2} e \log (x)+15 \sqrt {d^2} e \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{5 d^6} \]

input

Integrate[1/(x^2*(d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]
 

output

-1/5*((d*Sqrt[d^2 - e^2*x^2]*(5*d^3 + 39*d^2*e*x + 57*d*e^2*x^2 + 24*e^3*x 
^3))/(x*(d + e*x)^3) - 15*Sqrt[d^2]*e*Log[x] + 15*Sqrt[d^2]*e*Log[Sqrt[d^2 
] - Sqrt[d^2 - e^2*x^2]])/d^6
 

3.2.86.3 Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {570, 532, 25, 2336, 27, 2336, 27, 534, 243, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\)

\(\Big \downarrow \) 570

\(\displaystyle \int \frac {(d-e x)^3}{x^2 \left (d^2-e^2 x^2\right )^{7/2}}dx\)

\(\Big \downarrow \) 532

\(\displaystyle -\frac {\int -\frac {5 d^3-15 e x d^2+16 e^2 x^2 d}{x^2 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {5 d^3-15 e x d^2+16 e^2 x^2 d}{x^2 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {-\frac {\int -\frac {3 \left (5 d^3-15 e x d^2+14 e^2 x^2 d\right )}{x^2 \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}-\frac {e (5 d-7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {5 d^3-15 e x d^2+14 e^2 x^2 d}{x^2 \left (d^2-e^2 x^2\right )^{3/2}}dx}{d^2}-\frac {e (5 d-7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {\frac {-\frac {\int -\frac {5 d^2 (d-3 e x)}{x^2 \sqrt {d^2-e^2 x^2}}dx}{d^2}-\frac {e (15 d-19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}-\frac {e (5 d-7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 \int \frac {d-3 e x}{x^2 \sqrt {d^2-e^2 x^2}}dx-\frac {e (15 d-19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}-\frac {e (5 d-7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {5 \left (-3 e \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )-\frac {e (15 d-19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}-\frac {e (5 d-7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {5 \left (-\frac {3}{2} e \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )-\frac {e (15 d-19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}-\frac {e (5 d-7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {5 \left (\frac {3 \int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{e}-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )-\frac {e (15 d-19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}-\frac {e (5 d-7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {5 \left (\frac {3 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )-\frac {e (15 d-19 e x)}{d \sqrt {d^2-e^2 x^2}}}{d^2}-\frac {e (5 d-7 e x)}{d \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}-\frac {4 e (d-e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}\)

input

Int[1/(x^2*(d + e*x)^3*Sqrt[d^2 - e^2*x^2]),x]
 

output

(-4*e*(d - e*x))/(5*d*(d^2 - e^2*x^2)^(5/2)) + (-((e*(5*d - 7*e*x))/(d*(d^ 
2 - e^2*x^2)^(3/2))) + (-((e*(15*d - 19*e*x))/(d*Sqrt[d^2 - e^2*x^2])) + 5 
*(-(Sqrt[d^2 - e^2*x^2]/(d*x)) + (3*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d))/ 
d^2)/(5*d^2)
 

3.2.86.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 570

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, -1] &&  !(IGtQ[m, 0] && ILtQ[m + n, 0] &&  !GtQ[p, 1])
 

rule 2336

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[PolynomialRema 
inder[(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] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

3.2.86.4 Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.36

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{5} x}+\frac {3 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{4} \sqrt {d^{2}}}-\frac {4 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d^{4} e \left (x +\frac {d}{e}\right )^{2}}-\frac {19 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d^{5} \left (x +\frac {d}{e}\right )}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d^{3} e^{2} \left (x +\frac {d}{e}\right )^{3}}\)	\(199\)
default	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{5} x}+\frac {3 e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{4} \sqrt {d^{2}}}+\frac {-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}}{e \,d^{2}}+\frac {-\frac {2 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {2 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}}{d^{3}}-\frac {3 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{d^{5} \left (x +\frac {d}{e}\right )}\)	\(349\)

input

int(1/x^2/(e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)
 

output

-(-e^2*x^2+d^2)^(1/2)/d^5/x+3/d^4*e/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(- 
e^2*x^2+d^2)^(1/2))/x)-4/5/d^4/e/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^ 
(1/2)-19/5/d^5/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)-1/5/d^3/e^2/(x 
+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)
 

3.2.86.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {24 \, e^{4} x^{4} + 72 \, d e^{3} x^{3} + 72 \, d^{2} e^{2} x^{2} + 24 \, d^{3} e x + 15 \, {\left (e^{4} x^{4} + 3 \, d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + d^{3} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (24 \, e^{3} x^{3} + 57 \, d e^{2} x^{2} + 39 \, d^{2} e x + 5 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d^{5} e^{3} x^{4} + 3 \, d^{6} e^{2} x^{3} + 3 \, d^{7} e x^{2} + d^{8} x\right )}} \]

input

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

output

-1/5*(24*e^4*x^4 + 72*d*e^3*x^3 + 72*d^2*e^2*x^2 + 24*d^3*e*x + 15*(e^4*x^ 
4 + 3*d*e^3*x^3 + 3*d^2*e^2*x^2 + d^3*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2)) 
/x) + (24*e^3*x^3 + 57*d*e^2*x^2 + 39*d^2*e*x + 5*d^3)*sqrt(-e^2*x^2 + d^2 
))/(d^5*e^3*x^4 + 3*d^6*e^2*x^3 + 3*d^7*e*x^2 + d^8*x)
 

3.2.86.6 Sympy [F]

\[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {1}{x^{2} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{3}}\, dx \]

input

integrate(1/x**2/(e*x+d)**3/(-e**2*x**2+d**2)**(1/2),x)
 

output

Integral(1/(x**2*sqrt(-(-d + e*x)*(d + e*x))*(d + e*x)**3), x)
 

3.2.86.7 Maxima [F]

\[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int { \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} {\left (e x + d\right )}^{3} x^{2}} \,d x } \]

input

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

output

integrate(1/(sqrt(-e^2*x^2 + d^2)*(e*x + d)^3*x^2), x)
 

3.2.86.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (131) = 262\).

Time = 0.31 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.08 \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {3 \, e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{d^{5} {\left | e \right |}} - \frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{2 \, d^{5} x {\left | e \right |}} + \frac {{\left (5 \, e^{2} + \frac {121 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{x} + \frac {410 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{2} x^{2}} + \frac {610 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{4} x^{3}} + \frac {425 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{6} x^{4}} + \frac {125 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5}}{e^{8} x^{5}}\right )} e^{2} x}{10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{5} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

input

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

output

3*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d 
^5*abs(e)) - 1/2*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(d^5*x*abs(e)) + 1/10 
*(5*e^2 + 121*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/x + 410*(d*e + sqrt(-e^2 
*x^2 + d^2)*abs(e))^2/(e^2*x^2) + 610*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^ 
3/(e^4*x^3) + 425*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/(e^6*x^4) + 125*(d 
*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5/(e^8*x^5))*e^2*x/((d*e + sqrt(-e^2*x^2 
 + d^2)*abs(e))*d^5*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*ab 
s(e))
 

3.2.86.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 (d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\int \frac {1}{x^2\,\sqrt {d^2-e^2\,x^2}\,{\left (d+e\,x\right )}^3} \,d x \]

input

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

output

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