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

Optimal. Leaf size=41 \[ \frac {\log \left ((a+b x)^2+c\right )}{2 b^2}-\frac {a \tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{b^2 \sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {371, 635, 203, 260} \[ \frac {\log \left ((a+b x)^2+c\right )}{2 b^2}-\frac {a \tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{b^2 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[x/(c + (a + b*x)^2),x]

[Out]

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

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]

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 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)]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 38, normalized size = 0.93 \[ \frac {\log \left ((a+b x)^2+c\right )-\frac {2 a \tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{\sqrt {c}}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(c + (a + b*x)^2),x]

[Out]

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

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fricas [A]  time = 0.82, size = 136, normalized size = 3.32 \[ \left [-\frac {a \sqrt {-c} \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 2 \, {\left (b x + a\right )} \sqrt {-c} - c}{b^{2} x^{2} + 2 \, a b x + a^{2} + c}\right ) - c \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{2 \, b^{2} c}, -\frac {2 \, a \sqrt {c} \arctan \left (\frac {b x + a}{\sqrt {c}}\right ) - c \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{2 \, b^{2} c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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))/(b^2*c), -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))/(b^2*c)]

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giac [A]  time = 0.36, size = 43, normalized size = 1.05 \[ -\frac {a \arctan \left (\frac {b x + a}{\sqrt {c}}\right )}{b^{2} \sqrt {c}} + \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 54, normalized size = 1.32 \[ -\frac {a \arctan \left (\frac {2 b^{2} x +2 b a}{2 b \sqrt {c}}\right )}{b^{2} \sqrt {c}}+\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+c \right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 50, normalized size = 1.22 \[ -\frac {a \arctan \left (\frac {b^{2} x + a b}{b \sqrt {c}}\right )}{b^{2} \sqrt {c}} + \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.08, size = 46, normalized size = 1.12 \[ \frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+c\right )}{2\,b^2}-\frac {a\,\mathrm {atan}\left (\frac {a}{\sqrt {c}}+\frac {b\,x}{\sqrt {c}}\right )}{b^2\,\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(c + (a + b*x)^2),x)

[Out]

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

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sympy [B]  time = 0.25, size = 124, normalized size = 3.02 \[ \left (- \frac {a \sqrt {- c}}{2 b^{2} c} + \frac {1}{2 b^{2}}\right ) \log {\left (x + \frac {a^{2} - 2 b^{2} c \left (- \frac {a \sqrt {- c}}{2 b^{2} c} + \frac {1}{2 b^{2}}\right ) + c}{a b} \right )} + \left (\frac {a \sqrt {- c}}{2 b^{2} c} + \frac {1}{2 b^{2}}\right ) \log {\left (x + \frac {a^{2} - 2 b^{2} c \left (\frac {a \sqrt {- c}}{2 b^{2} c} + \frac {1}{2 b^{2}}\right ) + c}{a b} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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