\(\int \frac {1}{x (c+(a+b x)^2)} \, dx\) [82]

Optimal result

Integrand size = 15, antiderivative size = 59 \[ \int \frac {1}{x \left (c+(a+b x)^2\right )} \, dx=-\frac {a \arctan \left (\frac {a+b x}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2+c\right )}+\frac {\log (x)}{a^2+c}-\frac {\log \left (c+(a+b x)^2\right )}{2 \left (a^2+c\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {378, 720, 31, 649, 209, 266} \[ \int \frac {1}{x \left (c+(a+b x)^2\right )} \, dx=-\frac {a \arctan \left (\frac {a+b x}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2+c\right )}-\frac {\log \left ((a+b x)^2+c\right )}{2 \left (a^2+c\right )}+\frac {\log (x)}{a^2+c} \]

[In]

Int[1/(x*(c + (a + b*x)^2)),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 378

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(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 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 720

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{(-a+x) \left (c+x^2\right )} \, dx,x,a+b x\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{-a+x} \, dx,x,a+b x\right )}{a^2+c}+\frac {\text {Subst}\left (\int \frac {-a-x}{c+x^2} \, dx,x,a+b x\right )}{a^2+c} \\ & = \frac {\log (x)}{a^2+c}-\frac {\text {Subst}\left (\int \frac {x}{c+x^2} \, dx,x,a+b x\right )}{a^2+c}-\frac {a \text {Subst}\left (\int \frac {1}{c+x^2} \, dx,x,a+b x\right )}{a^2+c} \\ & = -\frac {a \tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2+c\right )}+\frac {\log (x)}{a^2+c}-\frac {\log \left (c+(a+b x)^2\right )}{2 \left (a^2+c\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x \left (c+(a+b x)^2\right )} \, dx=-\frac {\frac {2 a \arctan \left (\frac {a+b x}{\sqrt {c}}\right )}{\sqrt {c}}-2 \log (b x)+\log \left (c+(a+b x)^2\right )}{2 \left (a^2+c\right )} \]

[In]

Integrate[1/(x*(c + (a + b*x)^2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {\ln \left (x \right )}{a^{2}+c}-\frac {b \left (\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+c \right )}{2 b}+\frac {a \arctan \left (\frac {2 b^{2} x +2 a b}{2 \sqrt {c}\, b}\right )}{\sqrt {c}\, b}\right )}{a^{2}+c}\)	\(74\)
risch	\(\frac {\ln \left (x \right )}{a^{2}+c}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (c \,a^{2}+c^{2}\right ) \textit {\_Z}^{2}+2 c \textit {\_Z} \right )}{\sum }\textit {\_R} \ln \left (\left (\left (-a^{2} b +3 b c \right ) \textit {\_R} +3 b \right ) x +\left (-a^{3}-a c \right ) \textit {\_R} +2 a \right )\right )}{2}\)	\(75\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.61 \[ \int \frac {1}{x \left (c+(a+b x)^2\right )} \, dx=\left [-\frac {a \sqrt {-c} \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 2 \, {\left (b x + a\right )} \sqrt {-c} - c}{b^{2} x^{2} + 2 \, a b x + a^{2} + c}\right ) + c \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right ) - 2 \, c \log \left (x\right )}{2 \, {\left (a^{2} c + c^{2}\right )}}, -\frac {2 \, a \sqrt {c} \arctan \left (\frac {b x + a}{\sqrt {c}}\right ) + c \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right ) - 2 \, c \log \left (x\right )}{2 \, {\left (a^{2} c + c^{2}\right )}}\right ] \]

[In]

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

[Out]

[-1/2*(a*sqrt(-c)*log((b^2*x^2 + 2*a*b*x + a^2 + 2*(b*x + a)*sqrt(-c) - c)/(b^2*x^2 + 2*a*b*x + a^2 + c)) + c*
log(b^2*x^2 + 2*a*b*x + a^2 + c) - 2*c*log(x))/(a^2*c + c^2), -1/2*(2*a*sqrt(c)*arctan((b*x + a)/sqrt(c)) + c*
log(b^2*x^2 + 2*a*b*x + a^2 + c) - 2*c*log(x))/(a^2*c + c^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 738 vs. \(2 (49) = 98\).

Time = 1.85 (sec) , antiderivative size = 738, normalized size of antiderivative = 12.51 \[ \int \frac {1}{x \left (c+(a+b x)^2\right )} \, dx=\left (- \frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right ) \log {\left (x + \frac {- 4 a^{6} c \left (- \frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right )^{2} + 4 a^{4} c^{2} \left (- \frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right )^{2} - 6 a^{4} c \left (- \frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right ) + 20 a^{2} c^{3} \left (- \frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right )^{2} - 12 a^{2} c^{2} \left (- \frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right ) + 10 a^{2} c + 12 c^{4} \left (- \frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right )^{2} - 6 c^{3} \left (- \frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right ) - 6 c^{2}}{a^{3} b + 9 a b c} \right )} + \left (\frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right ) \log {\left (x + \frac {- 4 a^{6} c \left (\frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right )^{2} + 4 a^{4} c^{2} \left (\frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right )^{2} - 6 a^{4} c \left (\frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right ) + 20 a^{2} c^{3} \left (\frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right )^{2} - 12 a^{2} c^{2} \left (\frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right ) + 10 a^{2} c + 12 c^{4} \left (\frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right )^{2} - 6 c^{3} \left (\frac {a \sqrt {- c}}{2 c \left (a^{2} + c\right )} - \frac {1}{2 \left (a^{2} + c\right )}\right ) - 6 c^{2}}{a^{3} b + 9 a b c} \right )} + \frac {\log {\left (x + \frac {- \frac {4 a^{6} c}{\left (a^{2} + c\right )^{2}} + \frac {4 a^{4} c^{2}}{\left (a^{2} + c\right )^{2}} - \frac {6 a^{4} c}{a^{2} + c} + \frac {20 a^{2} c^{3}}{\left (a^{2} + c\right )^{2}} - \frac {12 a^{2} c^{2}}{a^{2} + c} + 10 a^{2} c + \frac {12 c^{4}}{\left (a^{2} + c\right )^{2}} - \frac {6 c^{3}}{a^{2} + c} - 6 c^{2}}{a^{3} b + 9 a b c} \right )}}{a^{2} + c} \]

[In]

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

[Out]

(-a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c)))*log(x + (-4*a**6*c*(-a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**
2 + c)))**2 + 4*a**4*c**2*(-a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c)))**2 - 6*a**4*c*(-a*sqrt(-c)/(2*c*(a
**2 + c)) - 1/(2*(a**2 + c))) + 20*a**2*c**3*(-a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c)))**2 - 12*a**2*c*
*2*(-a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c))) + 10*a**2*c + 12*c**4*(-a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(
2*(a**2 + c)))**2 - 6*c**3*(-a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c))) - 6*c**2)/(a**3*b + 9*a*b*c)) + (
a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c)))*log(x + (-4*a**6*c*(a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 +
c)))**2 + 4*a**4*c**2*(a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c)))**2 - 6*a**4*c*(a*sqrt(-c)/(2*c*(a**2 +
c)) - 1/(2*(a**2 + c))) + 20*a**2*c**3*(a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c)))**2 - 12*a**2*c**2*(a*
sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c))) + 10*a**2*c + 12*c**4*(a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2
+ c)))**2 - 6*c**3*(a*sqrt(-c)/(2*c*(a**2 + c)) - 1/(2*(a**2 + c))) - 6*c**2)/(a**3*b + 9*a*b*c)) + log(x + (-
4*a**6*c/(a**2 + c)**2 + 4*a**4*c**2/(a**2 + c)**2 - 6*a**4*c/(a**2 + c) + 20*a**2*c**3/(a**2 + c)**2 - 12*a**
2*c**2/(a**2 + c) + 10*a**2*c + 12*c**4/(a**2 + c)**2 - 6*c**3/(a**2 + c) - 6*c**2)/(a**3*b + 9*a*b*c))/(a**2
+ c)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x \left (c+(a+b x)^2\right )} \, dx=-\frac {a \arctan \left (\frac {b^{2} x + a b}{b \sqrt {c}}\right )}{{\left (a^{2} + c\right )} \sqrt {c}} - \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{2 \, {\left (a^{2} + c\right )}} + \frac {\log \left (x\right )}{a^{2} + c} \]

[In]

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

[Out]

-a*arctan((b^2*x + a*b)/(b*sqrt(c)))/((a^2 + c)*sqrt(c)) - 1/2*log(b^2*x^2 + 2*a*b*x + a^2 + c)/(a^2 + c) + lo
g(x)/(a^2 + c)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x \left (c+(a+b x)^2\right )} \, dx=-\frac {a \arctan \left (\frac {b x + a}{\sqrt {c}}\right )}{{\left (a^{2} + c\right )} \sqrt {c}} - \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{2 \, {\left (a^{2} + c\right )}} + \frac {\log \left ({\left | x \right |}\right )}{a^{2} + c} \]

[In]

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

[Out]

-a*arctan((b*x + a)/sqrt(c))/((a^2 + c)*sqrt(c)) - 1/2*log(b^2*x^2 + 2*a*b*x + a^2 + c)/(a^2 + c) + log(abs(x)
)/(a^2 + c)

Mupad [B] (verification not implemented)

Time = 10.58 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.93 \[ \int \frac {1}{x \left (c+(a+b x)^2\right )} \, dx=\frac {\ln \left (x\right )}{a^2+c}-\frac {\ln \left (2\,a\,b^3+3\,b^4\,x+\frac {b^3\,\left (c+a\,\sqrt {-c}\right )\,\left (a^3+b\,x\,a^2+c\,a-3\,b\,c\,x\right )}{c\,\left (a^2+c\right )}\right )\,\left (c+a\,\sqrt {-c}\right )}{2\,\left (a^2\,c+c^2\right )}-\frac {\ln \left (2\,a\,b^3+3\,b^4\,x+\frac {b^3\,\left (c-a\,\sqrt {-c}\right )\,\left (a^3+b\,x\,a^2+c\,a-3\,b\,c\,x\right )}{c\,\left (a^2+c\right )}\right )\,\left (c-a\,\sqrt {-c}\right )}{2\,\left (a^2\,c+c^2\right )} \]

[In]

int(1/(x*(c + (a + b*x)^2)),x)

[Out]

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