3.49 \(\int \frac {\tan ^{-1}(a+b x)}{x^2} \, dx\)

Optimal. Leaf size=62 \[ \frac {b \log (x)}{a^2+1}-\frac {b \log \left ((a+b x)^2+1\right )}{2 \left (a^2+1\right )}-\frac {a b \tan ^{-1}(a+b x)}{a^2+1}-\frac {\tan ^{-1}(a+b x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5045, 371, 706, 31, 635, 203, 260} \[ \frac {b \log (x)}{a^2+1}-\frac {b \log \left ((a+b x)^2+1\right )}{2 \left (a^2+1\right )}-\frac {a b \tan ^{-1}(a+b x)}{a^2+1}-\frac {\tan ^{-1}(a+b x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/x^2,x]

[Out]

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

Rule 31

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

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)]

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 5045

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*(a + b*ArcTan[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*Arc
Tan[c + d*x])^(p - 1))/(1 + (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -1]

Rubi steps

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

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Mathematica [C]  time = 0.06, size = 67, normalized size = 1.08 \[ -\frac {\tan ^{-1}(a+b x)}{x}+\frac {b (i (a+i) \log (-a-b x+i)+(-1-i a) \log (a+b x+i)+2 \log (x))}{2 \left (a^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[a + b*x]/x^2,x]

[Out]

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

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fricas [A]  time = 0.41, size = 57, normalized size = 0.92 \[ -\frac {b x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, b x \log \relax (x) + 2 \, {\left (a b x + a^{2} + 1\right )} \arctan \left (b x + a\right )}{2 \, {\left (a^{2} + 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 63, normalized size = 1.02 \[ -\frac {\arctan \left (b x +a \right )}{x}+\frac {b \ln \left (b x \right )}{a^{2}+1}-\frac {b \ln \left (1+\left (b x +a \right )^{2}\right )}{2 \left (a^{2}+1\right )}-\frac {a b \arctan \left (b x +a \right )}{a^{2}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 77, normalized size = 1.24 \[ -\frac {1}{2} \, b {\left (\frac {2 \, a \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{2} + 1} + \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{2} + 1} - \frac {2 \, \log \relax (x)}{a^{2} + 1}\right )} - \frac {\arctan \left (b x + a\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.04, size = 63, normalized size = 1.02 \[ -\frac {\mathrm {atan}\left (a+b\,x\right )}{x}-\frac {\frac {b\,x\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2}-b\,x\,\ln \relax (x)+a\,b\,x\,\mathrm {atan}\left (a+b\,x\right )}{x\,\left (a^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a + b*x)/x^2,x)

[Out]

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

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sympy [B]  time = 1.89, size = 168, normalized size = 2.71 \[ \begin {cases} - \frac {i b \operatorname {atan}{\left (b x - i \right )}}{2} - \frac {\operatorname {atan}{\left (b x - i \right )}}{x} - \frac {i}{2 x} & \text {for}\: a = - i \\\frac {i b \operatorname {atan}{\left (b x + i \right )}}{2} - \frac {\operatorname {atan}{\left (b x + i \right )}}{x} + \frac {i}{2 x} & \text {for}\: a = i \\- \frac {2 a^{2} \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} - \frac {2 a b x \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} + \frac {2 b x \log {\relax (x )}}{2 a^{2} x + 2 x} - \frac {b x \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 a^{2} x + 2 x} - \frac {2 \operatorname {atan}{\left (a + b x \right )}}{2 a^{2} x + 2 x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*b*atan(b*x - I)/2 - atan(b*x - I)/x - I/(2*x), Eq(a, -I)), (I*b*atan(b*x + I)/2 - atan(b*x + I)/
x + I/(2*x), Eq(a, I)), (-2*a**2*atan(a + b*x)/(2*a**2*x + 2*x) - 2*a*b*x*atan(a + b*x)/(2*a**2*x + 2*x) + 2*b
*x*log(x)/(2*a**2*x + 2*x) - b*x*log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*a**2*x + 2*x) - 2*atan(a + b*x)/(2*a**
2*x + 2*x), True))

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