\(\int \frac {1}{x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2}} \, dx\) [364]

Optimal result

Integrand size = 27, antiderivative size = 121 \[ \int \frac {1}{x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {d \sqrt {c^2-d^2 x^2}}{3 c^3 (c+d x)^2}-\frac {10 d \sqrt {c^2-d^2 x^2}}{3 c^4 (c+d x)}-\frac {\sqrt {c^2-d^2 x^2}}{c^3 x (c+d x)}+\frac {2 d \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )}{c^4} \] Output:

-1/3*d*(-d^2*x^2+c^2)^(1/2)/c^3/(d*x+c)^2-10/3*d*(-d^2*x^2+c^2)^(1/2)/c^4/ 
(d*x+c)-(-d^2*x^2+c^2)^(1/2)/c^3/x/(d*x+c)+2*d*arctanh((-d^2*x^2+c^2)^(1/2 
)/c)/c^4
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {\frac {c \sqrt {c^2-d^2 x^2} \left (3 c^2+14 c d x+10 d^2 x^2\right )}{x (c+d x)^2}-6 \sqrt {c^2} d \log (x)+6 \sqrt {c^2} d \log \left (\sqrt {c^2}-\sqrt {c^2-d^2 x^2}\right )}{3 c^5} \] Input:

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

Output:

-1/3*((c*Sqrt[c^2 - d^2*x^2]*(3*c^2 + 14*c*d*x + 10*d^2*x^2))/(x*(c + d*x) 
^2) - 6*Sqrt[c^2]*d*Log[x] + 6*Sqrt[c^2]*d*Log[Sqrt[c^2] - Sqrt[c^2 - d^2* 
x^2]])/c^5
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {570, 532, 25, 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.

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

\(\Big \downarrow \) 570

\(\displaystyle \int \frac {(c-d x)^2}{x^2 \left (c^2-d^2 x^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 532

\(\displaystyle -\frac {\int -\frac {3 c^2-6 d x c+4 d^2 x^2}{x^2 \left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c^2}-\frac {2 d (c-d x)}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {3 c^2-6 d x c+4 d^2 x^2}{x^2 \left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c^2}-\frac {2 d (c-d x)}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {-\frac {\int -\frac {3 c (c-2 d x)}{x^2 \sqrt {c^2-d^2 x^2}}dx}{c^2}-\frac {d (6 c-7 d x)}{c^2 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {2 d (c-d x)}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {c-2 d x}{x^2 \sqrt {c^2-d^2 x^2}}dx}{c}-\frac {d (6 c-7 d x)}{c^2 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {2 d (c-d x)}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {3 \left (-2 d \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx-\frac {\sqrt {c^2-d^2 x^2}}{c x}\right )}{c}-\frac {d (6 c-7 d x)}{c^2 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {2 d (c-d x)}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {3 \left (-d \int \frac {1}{x^2 \sqrt {c^2-d^2 x^2}}dx^2-\frac {\sqrt {c^2-d^2 x^2}}{c x}\right )}{c}-\frac {d (6 c-7 d x)}{c^2 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {2 d (c-d x)}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {3 \left (\frac {2 \int \frac {1}{\frac {c^2}{d^2}-\frac {x^4}{d^2}}d\sqrt {c^2-d^2 x^2}}{d}-\frac {\sqrt {c^2-d^2 x^2}}{c x}\right )}{c}-\frac {d (6 c-7 d x)}{c^2 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {2 d (c-d x)}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {3 \left (\frac {2 d \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )}{c}-\frac {\sqrt {c^2-d^2 x^2}}{c x}\right )}{c}-\frac {d (6 c-7 d x)}{c^2 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {2 d (c-d x)}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\)

Input:

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

Output:

(-2*d*(c - d*x))/(3*c^2*(c^2 - d^2*x^2)^(3/2)) + (-((d*(6*c - 7*d*x))/(c^2 
*Sqrt[c^2 - d^2*x^2])) + (3*(-(Sqrt[c^2 - d^2*x^2]/(c*x)) + (2*d*ArcTanh[S 
qrt[c^2 - d^2*x^2]/c])/c))/c)/(3*c^2)
 

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]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.27

method	result	size
risch	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}}{c^{4} x}+\frac {2 d \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{c^{3} \sqrt {c^{2}}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 d \,c^{3} \left (x +\frac {c}{d}\right )^{2}}-\frac {7 \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{4} \left (x +\frac {c}{d}\right )}\)	\(154\)
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}}{c^{4} x}+\frac {-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}}{c^{2}}+\frac {2 d \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{c^{3} \sqrt {c^{2}}}-\frac {2 \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{c^{4} \left (x +\frac {c}{d}\right )}\)	\(201\)

Input:

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

Output:

-(-d^2*x^2+c^2)^(1/2)/c^4/x+2/c^3*d/(c^2)^(1/2)*ln((2*c^2+2*(c^2)^(1/2)*(- 
d^2*x^2+c^2)^(1/2))/x)-1/3/d/c^3/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^ 
(1/2)-7/3/c^4/(x+c/d)*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {8 \, d^{3} x^{3} + 16 \, c d^{2} x^{2} + 8 \, c^{2} d x + 6 \, {\left (d^{3} x^{3} + 2 \, c d^{2} x^{2} + c^{2} d x\right )} \log \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{x}\right ) + {\left (10 \, d^{2} x^{2} + 14 \, c d x + 3 \, c^{2}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{3 \, {\left (c^{4} d^{2} x^{3} + 2 \, c^{5} d x^{2} + c^{6} x\right )}} \] Input:

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

Output:

-1/3*(8*d^3*x^3 + 16*c*d^2*x^2 + 8*c^2*d*x + 6*(d^3*x^3 + 2*c*d^2*x^2 + c^ 
2*d*x)*log(-(c - sqrt(-d^2*x^2 + c^2))/x) + (10*d^2*x^2 + 14*c*d*x + 3*c^2 
)*sqrt(-d^2*x^2 + c^2))/(c^4*d^2*x^3 + 2*c^5*d*x^2 + c^6*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.22 \[ \int \frac {1}{x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2}} \, dx=\frac {d^{5} {\left (\frac {12 \, \log \left (\sqrt {\frac {2 \, c}{d x + c} - 1} + 1\right )}{c^{4} d^{3} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )} - \frac {12 \, \log \left ({\left | \sqrt {\frac {2 \, c}{d x + c} - 1} - 1 \right |}\right )}{c^{4} d^{3} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )} + \frac {6 \, \sqrt {\frac {2 \, c}{d x + c} - 1}}{c^{4} d^{3} {\left (\frac {c}{d x + c} - 1\right )} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )} - \frac {c^{8} d^{6} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{2} \mathrm {sgn}\left (d\right )^{2} + 15 \, c^{8} d^{6} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{2} \mathrm {sgn}\left (d\right )^{2}}{c^{12} d^{9} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{3} \mathrm {sgn}\left (d\right )^{3}}\right )} + \frac {2 \, {\left (3 \, d^{2} \log \left (2\right ) - 6 \, d^{2} \log \left (i + 1\right ) + 10 i \, d^{2}\right )} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{4}}}{6 \, {\left | d \right |}} \] Input:

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

Output:

1/6*(d^5*(12*log(sqrt(2*c/(d*x + c) - 1) + 1)/(c^4*d^3*sgn(1/(d*x + c))*sg 
n(d)) - 12*log(abs(sqrt(2*c/(d*x + c) - 1) - 1))/(c^4*d^3*sgn(1/(d*x + c)) 
*sgn(d)) + 6*sqrt(2*c/(d*x + c) - 1)/(c^4*d^3*(c/(d*x + c) - 1)*sgn(1/(d*x 
 + c))*sgn(d)) - (c^8*d^6*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))^2*sgn 
(d)^2 + 15*c^8*d^6*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))^2*sgn(d)^2)/(c 
^12*d^9*sgn(1/(d*x + c))^3*sgn(d)^3)) + 2*(3*d^2*log(2) - 6*d^2*log(I + 1) 
 + 10*I*d^2)*sgn(1/(d*x + c))*sgn(d)/c^4)/abs(d)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.93 \[ \int \frac {1}{x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2}} \, dx=\frac {-3 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}-14 \sqrt {-d^{2} x^{2}+c^{2}}\, c d x -10 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}-3 \,\mathrm {log}\left (\sqrt {-d^{2} x^{2}+c^{2}}-c \right ) c^{2} d x -6 \,\mathrm {log}\left (\sqrt {-d^{2} x^{2}+c^{2}}-c \right ) c \,d^{2} x^{2}-3 \,\mathrm {log}\left (\sqrt {-d^{2} x^{2}+c^{2}}-c \right ) d^{3} x^{3}+3 \,\mathrm {log}\left (\sqrt {-d^{2} x^{2}+c^{2}}+c \right ) c^{2} d x +6 \,\mathrm {log}\left (\sqrt {-d^{2} x^{2}+c^{2}}+c \right ) c \,d^{2} x^{2}+3 \,\mathrm {log}\left (\sqrt {-d^{2} x^{2}+c^{2}}+c \right ) d^{3} x^{3}}{3 c^{4} x \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

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

Output:

( - 3*sqrt(c**2 - d**2*x**2)*c**2 - 14*sqrt(c**2 - d**2*x**2)*c*d*x - 10*s 
qrt(c**2 - d**2*x**2)*d**2*x**2 - 3*log(sqrt(c**2 - d**2*x**2) - c)*c**2*d 
*x - 6*log(sqrt(c**2 - d**2*x**2) - c)*c*d**2*x**2 - 3*log(sqrt(c**2 - d** 
2*x**2) - c)*d**3*x**3 + 3*log(sqrt(c**2 - d**2*x**2) + c)*c**2*d*x + 6*lo 
g(sqrt(c**2 - d**2*x**2) + c)*c*d**2*x**2 + 3*log(sqrt(c**2 - d**2*x**2) + 
 c)*d**3*x**3)/(3*c**4*x*(c**2 + 2*c*d*x + d**2*x**2))