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

Optimal. Leaf size=59 \[ -\frac{\log \left ((a+b x)^2+c\right )}{2 \left (a^2+c\right )}-\frac{a \tan ^{-1}\left (\frac{a+b x}{\sqrt{c}}\right )}{\sqrt{c} \left (a^2+c\right )}+\frac{\log (x)}{a^2+c} \]

[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))

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Rubi [A]  time = 0.0351043, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {371, 706, 31, 635, 203, 260} \[ -\frac{\log \left ((a+b x)^2+c\right )}{2 \left (a^2+c\right )}-\frac{a \tan ^{-1}\left (\frac{a+b x}{\sqrt{c}}\right )}{\sqrt{c} \left (a^2+c\right )}+\frac{\log (x)}{a^2+c} \]

Antiderivative was successfully verified.

[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 371

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 706

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]

Rule 31

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

Rule 635

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 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 260

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]

Rubi steps

\begin{align*} \int \frac{1}{x \left (c+(a+b x)^2\right )} \, dx &=\operatorname{Subst}\left (\int \frac{1}{(-a+x) \left (c+x^2\right )} \, dx,x,a+b x\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{-a+x} \, dx,x,a+b x\right )}{a^2+c}+\frac{\operatorname{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{\operatorname{Subst}\left (\int \frac{x}{c+x^2} \, dx,x,a+b x\right )}{a^2+c}-\frac{a \operatorname{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]  time = 0.0328941, size = 48, normalized size = 0.81 \[ -\frac{\log \left ((a+b x)^2+c\right )+\frac{2 a \tan ^{-1}\left (\frac{a+b x}{\sqrt{c}}\right )}{\sqrt{c}}-2 \log (b x)}{2 \left (a^2+c\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(c + (a + b*x)^2)),x]

[Out]

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

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Maple [A]  time = 0.006, size = 72, normalized size = 1.2 \begin{align*}{\frac{\ln \left ( x \right ) }{{a}^{2}+c}}-{\frac{\ln \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2}+c \right ) }{2\,{a}^{2}+2\,c}}-{\frac{a}{{a}^{2}+c}\arctan \left ({\frac{2\,{b}^{2}x+2\,ab}{2\,b}{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86333, size = 385, normalized size = 6.53 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [B]  time = 1.70819, size = 738, normalized size = 12.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.12384, size = 84, normalized size = 1.42 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)