\(\int \frac {1}{x^2 (a+b \sqrt {c+d x})} \, dx\) [116]

Optimal result

Integrand size = 19, antiderivative size = 130 \[ \int \frac {1}{x^2 \left (a+b \sqrt {c+d x}\right )} \, dx=-\frac {a-b \sqrt {c+d x}}{\left (a^2-b^2 c\right ) x}+\frac {b \left (a^2+b^2 c\right ) d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2-b^2 c\right )^2}+\frac {a b^2 d \log (x)}{\left (a^2-b^2 c\right )^2}-\frac {2 a b^2 d \log \left (a+b \sqrt {c+d x}\right )}{\left (a^2-b^2 c\right )^2} \] Output:

-(a-b*(d*x+c)^(1/2))/(-b^2*c+a^2)/x+b*(b^2*c+a^2)*d*arctanh((d*x+c)^(1/2)/ 
c^(1/2))/c^(1/2)/(-b^2*c+a^2)^2+a*b^2*d*ln(x)/(-b^2*c+a^2)^2-2*a*b^2*d*ln( 
a+b*(d*x+c)^(1/2))/(-b^2*c+a^2)^2
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^2 \left (a+b \sqrt {c+d x}\right )} \, dx=\frac {b \left (a^2+b^2 c\right ) d x \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\sqrt {c} \left (-\left (\left (a^2-b^2 c\right ) \left (a-b \sqrt {c+d x}\right )\right )+a b^2 d x \log (-d x)-2 a b^2 d x \log \left (a+b \sqrt {c+d x}\right )\right )}{\sqrt {c} \left (a^2-b^2 c\right )^2 x} \] Input:

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

Output:

(b*(a^2 + b^2*c)*d*x*ArcTanh[Sqrt[c + d*x]/Sqrt[c]] + Sqrt[c]*(-((a^2 - b^ 
2*c)*(a - b*Sqrt[c + d*x])) + a*b^2*d*x*Log[-(d*x)] - 2*a*b^2*d*x*Log[a + 
b*Sqrt[c + d*x]]))/(Sqrt[c]*(a^2 - b^2*c)^2*x)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {896, 1732, 593, 25, 657, 2009}

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.

\(\displaystyle \int \frac {1}{x^2 \left (a+b \sqrt {c+d x}\right )} \, dx\)

\(\Big \downarrow \) 896

\(\displaystyle d \int \frac {1}{d^2 x^2 \left (a+b \sqrt {c+d x}\right )}d(c+d x)\)

\(\Big \downarrow \) 1732

\(\displaystyle 2 d \int \frac {\sqrt {c+d x}}{d^2 x^2 \left (a+b \sqrt {c+d x}\right )}d\sqrt {c+d x}\)

\(\Big \downarrow \) 593

\(\displaystyle 2 d \left (-\frac {b \int \frac {a-b \sqrt {c+d x}}{d x \left (a+b \sqrt {c+d x}\right )}d\sqrt {c+d x}}{2 \left (a^2-b^2 c\right )}-\frac {a-b \sqrt {c+d x}}{2 d x \left (a^2-b^2 c\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 d \left (\frac {b \int -\frac {a-b \sqrt {c+d x}}{d x \left (a+b \sqrt {c+d x}\right )}d\sqrt {c+d x}}{2 \left (a^2-b^2 c\right )}-\frac {a-b \sqrt {c+d x}}{2 d x \left (a^2-b^2 c\right )}\right )\)

\(\Big \downarrow \) 657

\(\displaystyle 2 d \left (\frac {b \int \left (-\frac {2 a b^2}{\left (a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right )}-\frac {a^2-2 b \sqrt {c+d x} a+b^2 c}{\left (a^2-b^2 c\right ) d x}\right )d\sqrt {c+d x}}{2 \left (a^2-b^2 c\right )}-\frac {a-b \sqrt {c+d x}}{2 d x \left (a^2-b^2 c\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 d \left (\frac {b \left (\frac {\left (a^2+b^2 c\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2-b^2 c\right )}+\frac {a b \log (-d x)}{a^2-b^2 c}-\frac {2 a b \log \left (a+b \sqrt {c+d x}\right )}{a^2-b^2 c}\right )}{2 \left (a^2-b^2 c\right )}-\frac {a-b \sqrt {c+d x}}{2 d x \left (a^2-b^2 c\right )}\right )\)

Input:

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

Output:

2*d*(-1/2*(a - b*Sqrt[c + d*x])/((a^2 - b^2*c)*d*x) + (b*(((a^2 + b^2*c)*A 
rcTanh[Sqrt[c + d*x]/Sqrt[c]])/(Sqrt[c]*(a^2 - b^2*c)) + (a*b*Log[-(d*x)]) 
/(a^2 - b^2*c) - (2*a*b*Log[a + b*Sqrt[c + d*x]])/(a^2 - b^2*c)))/(2*(a^2 
- b^2*c)))
 

Defintions of rubi rules used

rule 25

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

rule 593

Int[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + 
a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a + b* 
x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] 
 && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 1732

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb 
ol] :> With[{g = Denominator[n]}, Simp[g   Subst[Int[x^(g - 1)*(d + e*x^(g* 
n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} 
, x] && EqQ[n2, 2*n] && FractionQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(2 d \left (-\frac {a \,b^{2} \ln \left (a +b \sqrt {d x +c}\right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {-\frac {\left (\frac {1}{2} b^{3} c -\frac {1}{2} a^{2} b \right ) \sqrt {d x +c}-\frac {c a \,b^{2}}{2}+\frac {a^{3}}{2}}{d x}+\frac {b \left (a b \ln \left (-d x \right )+\frac {\left (b^{2} c +a^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{2}}{\left (-b^{2} c +a^{2}\right )^{2}}\right )\)	\(128\)
default	\(2 d \left (-\frac {a \,b^{2} \ln \left (a +b \sqrt {d x +c}\right )}{\left (-b^{2} c +a^{2}\right )^{2}}+\frac {-\frac {\left (\frac {1}{2} b^{3} c -\frac {1}{2} a^{2} b \right ) \sqrt {d x +c}-\frac {c a \,b^{2}}{2}+\frac {a^{3}}{2}}{d x}+\frac {b \left (a b \ln \left (-d x \right )+\frac {\left (b^{2} c +a^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{2}}{\left (-b^{2} c +a^{2}\right )^{2}}\right )\)	\(128\)

Input:

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

Output:

2*d*(-a*b^2/(-b^2*c+a^2)^2*ln(a+b*(d*x+c)^(1/2))+1/(-b^2*c+a^2)^2*(-((1/2* 
b^3*c-1/2*a^2*b)*(d*x+c)^(1/2)-1/2*c*a*b^2+1/2*a^3)/d/x+1/2*b*(a*b*ln(-d*x 
)+(b^2*c+a^2)/c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1/2)))))
 

Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.15 \[ \int \frac {1}{x^2 \left (a+b \sqrt {c+d x}\right )} \, dx=\left [-\frac {4 \, a b^{2} c d x \log \left (\sqrt {d x + c} b + a\right ) - 2 \, a b^{2} c d x \log \left (x\right ) - 2 \, a b^{2} c^{2} - {\left (b^{3} c + a^{2} b\right )} \sqrt {c} d x \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, a^{3} c + 2 \, {\left (b^{3} c^{2} - a^{2} b c\right )} \sqrt {d x + c}}{2 \, {\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )} x}, -\frac {2 \, a b^{2} c d x \log \left (\sqrt {d x + c} b + a\right ) - a b^{2} c d x \log \left (x\right ) - a b^{2} c^{2} + {\left (b^{3} c + a^{2} b\right )} \sqrt {-c} d x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + a^{3} c + {\left (b^{3} c^{2} - a^{2} b c\right )} \sqrt {d x + c}}{{\left (b^{4} c^{3} - 2 \, a^{2} b^{2} c^{2} + a^{4} c\right )} x}\right ] \] Input:

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

Output:

[-1/2*(4*a*b^2*c*d*x*log(sqrt(d*x + c)*b + a) - 2*a*b^2*c*d*x*log(x) - 2*a 
*b^2*c^2 - (b^3*c + a^2*b)*sqrt(c)*d*x*log((d*x + 2*sqrt(d*x + c)*sqrt(c) 
+ 2*c)/x) + 2*a^3*c + 2*(b^3*c^2 - a^2*b*c)*sqrt(d*x + c))/((b^4*c^3 - 2*a 
^2*b^2*c^2 + a^4*c)*x), -(2*a*b^2*c*d*x*log(sqrt(d*x + c)*b + a) - a*b^2*c 
*d*x*log(x) - a*b^2*c^2 + (b^3*c + a^2*b)*sqrt(-c)*d*x*arctan(sqrt(-c)/sqr 
t(d*x + c)) + a^3*c + (b^3*c^2 - a^2*b*c)*sqrt(d*x + c))/((b^4*c^3 - 2*a^2 
*b^2*c^2 + a^4*c)*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.47 \[ \int \frac {1}{x^2 \left (a+b \sqrt {c+d x}\right )} \, dx=\frac {1}{2} \, {\left (\frac {2 \, a b^{2} \log \left (d x\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {4 \, a b^{2} \log \left (\sqrt {d x + c} b + a\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {{\left (b^{3} c + a^{2} b\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{{\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} \sqrt {c}} + \frac {2 \, {\left (\sqrt {d x + c} b - a\right )}}{b^{2} c^{2} - a^{2} c - {\left (b^{2} c - a^{2}\right )} {\left (d x + c\right )}}\right )} d \] Input:

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

Output:

1/2*(2*a*b^2*log(d*x)/(b^4*c^2 - 2*a^2*b^2*c + a^4) - 4*a*b^2*log(sqrt(d*x 
 + c)*b + a)/(b^4*c^2 - 2*a^2*b^2*c + a^4) - (b^3*c + a^2*b)*log((sqrt(d*x 
 + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/((b^4*c^2 - 2*a^2*b^2*c + a^4) 
*sqrt(c)) + 2*(sqrt(d*x + c)*b - a)/(b^2*c^2 - a^2*c - (b^2*c - a^2)*(d*x 
+ c)))*d
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.47 \[ \int \frac {1}{x^2 \left (a+b \sqrt {c+d x}\right )} \, dx=-\frac {2 \, a b^{3} d \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{5} c^{2} - 2 \, a^{2} b^{3} c + a^{4} b} + \frac {a b^{2} d \log \left (-d x\right )}{b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}} - \frac {{\left (b^{3} c d + a^{2} b d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{{\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} \sqrt {-c}} + \frac {a b^{2} c d - a^{3} d - {\left (b^{3} c d - a^{2} b d\right )} \sqrt {d x + c}}{{\left (b^{2} c - a^{2}\right )}^{2} d x} \] Input:

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

Output:

-2*a*b^3*d*log(abs(sqrt(d*x + c)*b + a))/(b^5*c^2 - 2*a^2*b^3*c + a^4*b) + 
 a*b^2*d*log(-d*x)/(b^4*c^2 - 2*a^2*b^2*c + a^4) - (b^3*c*d + a^2*b*d)*arc 
tan(sqrt(d*x + c)/sqrt(-c))/((b^4*c^2 - 2*a^2*b^2*c + a^4)*sqrt(-c)) + (a* 
b^2*c*d - a^3*d - (b^3*c*d - a^2*b*d)*sqrt(d*x + c))/((b^2*c - a^2)^2*d*x)
 

Mupad [B] (verification not implemented)

Time = 9.70 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.69 \[ \int \frac {1}{x^2 \left (a+b \sqrt {c+d x}\right )} \, dx=\frac {\ln \left (\sqrt {c+d\,x}-\sqrt {c}\right )\,\left (4\,a\,b^2\,c\,d-b\,\sqrt {c}\,d\,\left (2\,a^2+2\,c\,b^2\right )\right )}{4\,a^4\,c-8\,a^2\,b^2\,c^2+4\,b^4\,c^3}+\frac {\ln \left (\sqrt {c+d\,x}+\sqrt {c}\right )\,\left (4\,a\,b^2\,c\,d+b\,\sqrt {c}\,d\,\left (2\,a^2+2\,c\,b^2\right )\right )}{4\,a^4\,c-8\,a^2\,b^2\,c^2+4\,b^4\,c^3}+\frac {\frac {a\,d}{b^2\,c-a^2}-\frac {b\,d\,\sqrt {c+d\,x}}{b^2\,c-a^2}}{d\,x}-\frac {2\,a\,b^2\,d\,\ln \left (a+b\,\sqrt {c+d\,x}\right )}{{\left (b^2\,c-a^2\right )}^2} \] Input:

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

Output:

(log((c + d*x)^(1/2) - c^(1/2))*(4*a*b^2*c*d - b*c^(1/2)*d*(2*b^2*c + 2*a^ 
2)))/(4*a^4*c + 4*b^4*c^3 - 8*a^2*b^2*c^2) + (log((c + d*x)^(1/2) + c^(1/2 
))*(4*a*b^2*c*d + b*c^(1/2)*d*(2*b^2*c + 2*a^2)))/(4*a^4*c + 4*b^4*c^3 - 8 
*a^2*b^2*c^2) + ((a*d)/(b^2*c - a^2) - (b*d*(c + d*x)^(1/2))/(b^2*c - a^2) 
)/(d*x) - (2*a*b^2*d*log(a + b*(c + d*x)^(1/2)))/(b^2*c - a^2)^2
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.78 \[ \int \frac {1}{x^2 \left (a+b \sqrt {c+d x}\right )} \, dx=\frac {2 \sqrt {d x +c}\, a^{2} b c -2 \sqrt {d x +c}\, b^{3} c^{2}-\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} b d x -\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b^{3} c d x +\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} b d x +\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b^{3} c d x +2 \,\mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,b^{2} c d x +2 \,\mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,b^{2} c d x -4 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a \,b^{2} c d x -2 a^{3} c -2 a^{3} d x +2 a \,b^{2} c^{2}+2 a \,b^{2} c d x}{2 c x \left (b^{4} c^{2}-2 a^{2} b^{2} c +a^{4}\right )} \] Input:

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

Output:

(2*sqrt(c + d*x)*a**2*b*c - 2*sqrt(c + d*x)*b**3*c**2 - sqrt(c)*log(sqrt(c 
 + d*x) - sqrt(c))*a**2*b*d*x - sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*b**3* 
c*d*x + sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**2*b*d*x + sqrt(c)*log(sqrt 
(c + d*x) + sqrt(c))*b**3*c*d*x + 2*log(sqrt(c + d*x) - sqrt(c))*a*b**2*c* 
d*x + 2*log(sqrt(c + d*x) + sqrt(c))*a*b**2*c*d*x - 4*log(sqrt(c + d*x)*b 
+ a)*a*b**2*c*d*x - 2*a**3*c - 2*a**3*d*x + 2*a*b**2*c**2 + 2*a*b**2*c*d*x 
)/(2*c*x*(a**4 - 2*a**2*b**2*c + b**4*c**2))