3.51 \(\int \frac{\tan ^{-1}(a+b x)}{x^4} \, dx\)

Optimal. Leaf size=129 \[ \frac{2 a b^2}{3 \left (a^2+1\right )^2 x}-\frac{\left (1-3 a^2\right ) b^3 \log (x)}{3 \left (a^2+1\right )^3}+\frac{\left (1-3 a^2\right ) b^3 \log \left ((a+b x)^2+1\right )}{6 \left (a^2+1\right )^3}+\frac{a \left (3-a^2\right ) b^3 \tan ^{-1}(a+b x)}{3 \left (a^2+1\right )^3}-\frac{b}{6 \left (a^2+1\right ) x^2}-\frac{\tan ^{-1}(a+b x)}{3 x^3} \]

[Out]

-b/(6*(1 + a^2)*x^2) + (2*a*b^2)/(3*(1 + a^2)^2*x) + (a*(3 - a^2)*b^3*ArcTan[a + b*x])/(3*(1 + a^2)^3) - ArcTa
n[a + b*x]/(3*x^3) - ((1 - 3*a^2)*b^3*Log[x])/(3*(1 + a^2)^3) + ((1 - 3*a^2)*b^3*Log[1 + (a + b*x)^2])/(6*(1 +
a^2)^3)

________________________________________________________________________________________

Rubi [A]  time = 0.114371, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5045, 371, 710, 801, 635, 203, 260} \[ \frac{2 a b^2}{3 \left (a^2+1\right )^2 x}-\frac{\left (1-3 a^2\right ) b^3 \log (x)}{3 \left (a^2+1\right )^3}+\frac{\left (1-3 a^2\right ) b^3 \log \left ((a+b x)^2+1\right )}{6 \left (a^2+1\right )^3}+\frac{a \left (3-a^2\right ) b^3 \tan ^{-1}(a+b x)}{3 \left (a^2+1\right )^3}-\frac{b}{6 \left (a^2+1\right ) x^2}-\frac{\tan ^{-1}(a+b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/x^4,x]

[Out]

-b/(6*(1 + a^2)*x^2) + (2*a*b^2)/(3*(1 + a^2)^2*x) + (a*(3 - a^2)*b^3*ArcTan[a + b*x])/(3*(1 + a^2)^3) - ArcTa
n[a + b*x]/(3*x^3) - ((1 - 3*a^2)*b^3*Log[x])/(3*(1 + a^2)^3) + ((1 - 3*a^2)*b^3*Log[1 + (a + b*x)^2])/(6*(1 +
a^2)^3)

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]

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 710

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[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 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{\tan ^{-1}(a+b x)}{x^4} \, dx &=-\frac{\tan ^{-1}(a+b x)}{3 x^3}+\frac{1}{3} b \int \frac{1}{x^3 \left (1+(a+b x)^2\right )} \, dx\\ &=-\frac{\tan ^{-1}(a+b x)}{3 x^3}+\frac{1}{3} b^3 \operatorname{Subst}\left (\int \frac{1}{(-a+x)^3 \left (1+x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac{b}{6 \left (1+a^2\right ) x^2}-\frac{\tan ^{-1}(a+b x)}{3 x^3}+\frac{b^3 \operatorname{Subst}\left (\int \frac{-a-x}{(-a+x)^2 \left (1+x^2\right )} \, dx,x,a+b x\right )}{3 \left (1+a^2\right )}\\ &=-\frac{b}{6 \left (1+a^2\right ) x^2}-\frac{\tan ^{-1}(a+b x)}{3 x^3}+\frac{b^3 \operatorname{Subst}\left (\int \left (-\frac{2 a}{\left (1+a^2\right ) (a-x)^2}+\frac{1-3 a^2}{\left (1+a^2\right )^2 (a-x)}+\frac{a \left (3-a^2\right )+\left (1-3 a^2\right ) x}{\left (1+a^2\right )^2 \left (1+x^2\right )}\right ) \, dx,x,a+b x\right )}{3 \left (1+a^2\right )}\\ &=-\frac{b}{6 \left (1+a^2\right ) x^2}+\frac{2 a b^2}{3 \left (1+a^2\right )^2 x}-\frac{\tan ^{-1}(a+b x)}{3 x^3}-\frac{\left (1-3 a^2\right ) b^3 \log (x)}{3 \left (1+a^2\right )^3}+\frac{b^3 \operatorname{Subst}\left (\int \frac{a \left (3-a^2\right )+\left (1-3 a^2\right ) x}{1+x^2} \, dx,x,a+b x\right )}{3 \left (1+a^2\right )^3}\\ &=-\frac{b}{6 \left (1+a^2\right ) x^2}+\frac{2 a b^2}{3 \left (1+a^2\right )^2 x}-\frac{\tan ^{-1}(a+b x)}{3 x^3}-\frac{\left (1-3 a^2\right ) b^3 \log (x)}{3 \left (1+a^2\right )^3}+\frac{\left (\left (1-3 a^2\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,a+b x\right )}{3 \left (1+a^2\right )^3}+\frac{\left (a \left (3-a^2\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,a+b x\right )}{3 \left (1+a^2\right )^3}\\ &=-\frac{b}{6 \left (1+a^2\right ) x^2}+\frac{2 a b^2}{3 \left (1+a^2\right )^2 x}+\frac{a \left (3-a^2\right ) b^3 \tan ^{-1}(a+b x)}{3 \left (1+a^2\right )^3}-\frac{\tan ^{-1}(a+b x)}{3 x^3}-\frac{\left (1-3 a^2\right ) b^3 \log (x)}{3 \left (1+a^2\right )^3}+\frac{\left (1-3 a^2\right ) b^3 \log \left (1+(a+b x)^2\right )}{6 \left (1+a^2\right )^3}\\ \end{align*}

Mathematica [C]  time = 0.13592, size = 128, normalized size = 0.99 \[ \frac{2 \left (3 a^2-1\right ) b^3 x^3 \log (x)-(a-i) b x \left ((a+i) \left (a^2-4 a b x+1\right )+i (a-i)^2 b^2 x^2 \log (a+b x+i)\right )-2 \left (a^2+1\right )^3 \tan ^{-1}(a+b x)+i (a+i)^3 b^3 x^3 \log (-a-b x+i)}{6 \left (a^2+1\right )^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[a + b*x]/x^4,x]

[Out]

(-2*(1 + a^2)^3*ArcTan[a + b*x] + 2*(-1 + 3*a^2)*b^3*x^3*Log[x] + I*(I + a)^3*b^3*x^3*Log[I - a - b*x] - (-I +
a)*b*x*((I + a)*(1 + a^2 - 4*a*b*x) + I*(-I + a)^2*b^2*x^2*Log[I + a + b*x]))/(6*(1 + a^2)^3*x^3)

________________________________________________________________________________________

Maple [A]  time = 0.043, size = 162, normalized size = 1.3 \begin{align*} -{\frac{\arctan \left ( bx+a \right ) }{3\,{x}^{3}}}-{\frac{{b}^{3}\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ){a}^{2}}{2\, \left ({a}^{2}+1 \right ) ^{3}}}+{\frac{{b}^{3}\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{6\, \left ({a}^{2}+1 \right ) ^{3}}}-{\frac{{b}^{3}\arctan \left ( bx+a \right ){a}^{3}}{3\, \left ({a}^{2}+1 \right ) ^{3}}}+{\frac{{b}^{3}\arctan \left ( bx+a \right ) a}{ \left ({a}^{2}+1 \right ) ^{3}}}-{\frac{b}{ \left ( 6\,{a}^{2}+6 \right ){x}^{2}}}+{\frac{{b}^{3}\ln \left ( bx \right ){a}^{2}}{ \left ({a}^{2}+1 \right ) ^{3}}}-{\frac{{b}^{3}\ln \left ( bx \right ) }{3\, \left ({a}^{2}+1 \right ) ^{3}}}+{\frac{2\,a{b}^{2}}{3\, \left ({a}^{2}+1 \right ) ^{2}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(b*x+a)/x^4,x)

[Out]

-1/3*arctan(b*x+a)/x^3-1/2*b^3/(a^2+1)^3*ln(1+(b*x+a)^2)*a^2+1/6*b^3/(a^2+1)^3*ln(1+(b*x+a)^2)-1/3*b^3/(a^2+1)
^3*arctan(b*x+a)*a^3+b^3/(a^2+1)^3*arctan(b*x+a)*a-1/6*b/(a^2+1)/x^2+b^3/(a^2+1)^3*ln(b*x)*a^2-1/3*b^3/(a^2+1)
^3*ln(b*x)+2/3*a*b^2/(a^2+1)^2/x

________________________________________________________________________________________

Maxima [A]  time = 1.54328, size = 223, normalized size = 1.73 \begin{align*} -\frac{1}{6} \,{\left (\frac{2 \,{\left (a^{3} - 3 \, a\right )} b^{2} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac{{\left (3 \, a^{2} - 1\right )} b^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} - \frac{2 \,{\left (3 \, a^{2} - 1\right )} b^{2} \log \left (x\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} - \frac{4 \, a b x - a^{2} - 1}{{\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}}\right )} b - \frac{\arctan \left (b x + a\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(b*x+a)/x^4,x, algorithm="maxima")

[Out]

-1/6*(2*(a^3 - 3*a)*b^2*arctan((b^2*x + a*b)/b)/(a^6 + 3*a^4 + 3*a^2 + 1) + (3*a^2 - 1)*b^2*log(b^2*x^2 + 2*a*
b*x + a^2 + 1)/(a^6 + 3*a^4 + 3*a^2 + 1) - 2*(3*a^2 - 1)*b^2*log(x)/(a^6 + 3*a^4 + 3*a^2 + 1) - (4*a*b*x - a^2
- 1)/((a^4 + 2*a^2 + 1)*x^2))*b - 1/3*arctan(b*x + a)/x^3

________________________________________________________________________________________

Fricas [A]  time = 1.65264, size = 321, normalized size = 2.49 \begin{align*} -\frac{{\left (3 \, a^{2} - 1\right )} b^{3} x^{3} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \,{\left (3 \, a^{2} - 1\right )} b^{3} x^{3} \log \left (x\right ) - 4 \,{\left (a^{3} + a\right )} b^{2} x^{2} +{\left (a^{4} + 2 \, a^{2} + 1\right )} b x + 2 \,{\left ({\left (a^{3} - 3 \, a\right )} b^{3} x^{3} + a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} \arctan \left (b x + a\right )}{6 \,{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(b*x+a)/x^4,x, algorithm="fricas")

[Out]

-1/6*((3*a^2 - 1)*b^3*x^3*log(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*(3*a^2 - 1)*b^3*x^3*log(x) - 4*(a^3 + a)*b^2*x^
2 + (a^4 + 2*a^2 + 1)*b*x + 2*((a^3 - 3*a)*b^3*x^3 + a^6 + 3*a^4 + 3*a^2 + 1)*arctan(b*x + a))/((a^6 + 3*a^4 +
3*a^2 + 1)*x^3)

________________________________________________________________________________________

Sympy [B]  time = 26.9557, size = 1127, normalized size = 8.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(b*x+a)/x**4,x)

[Out]

Piecewise((3*I*b**4*x**4*atan(b*x - I)/(72*b*x**4 - 144*I*x**3) + 6*b**3*x**3*atan(b*x - I)/(72*b*x**4 - 144*I
*x**3) + 3*I*b**3*x**3/(72*b*x**4 - 144*I*x**3) + 3*b**2*x**2/(72*b*x**4 - 144*I*x**3) - 24*b*x*atan(b*x - I)/
(72*b*x**4 - 144*I*x**3) + 2*I*b*x/(72*b*x**4 - 144*I*x**3) + 48*I*atan(b*x - I)/(72*b*x**4 - 144*I*x**3) - 8/
(72*b*x**4 - 144*I*x**3), Eq(a, -I)), (-3*I*b**4*x**4*atan(b*x + I)/(72*b*x**4 + 144*I*x**3) + 6*b**3*x**3*ata
n(b*x + I)/(72*b*x**4 + 144*I*x**3) - 3*I*b**3*x**3/(72*b*x**4 + 144*I*x**3) + 3*b**2*x**2/(72*b*x**4 + 144*I*
x**3) - 24*b*x*atan(b*x + I)/(72*b*x**4 + 144*I*x**3) - 2*I*b*x/(72*b*x**4 + 144*I*x**3) - 48*I*atan(b*x + I)/
(72*b*x**4 + 144*I*x**3) - 8/(72*b*x**4 + 144*I*x**3), Eq(a, I)), (-2*a**6*atan(a + b*x)/(6*a**6*x**3 + 18*a**
4*x**3 + 18*a**2*x**3 + 6*x**3) - a**4*b*x/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 6*a**4*atan(
a + b*x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*a**3*b**3*x**3*atan(a + b*x)/(6*a**6*x**3 +
18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) + 4*a**3*b**2*x**2/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3)
+ 6*a**2*b**3*x**3*log(x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 3*a**2*b**3*x**3*log(a**2 + 2
*a*b*x + b**2*x**2 + 1)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*a**2*b**3*x**3/(6*a**6*x**3 +
18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*a**2*b*x/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 6*a
**2*atan(a + b*x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) + 6*a*b**3*x**3*atan(a + b*x)/(6*a**6*x
**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) + 4*a*b**2*x**2/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**
3) - 2*b**3*x**3*log(x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) + b**3*x**3*log(a**2 + 2*a*b*x +
b**2*x**2 + 1)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*b**3*x**3/(6*a**6*x**3 + 18*a**4*x**3
+ 18*a**2*x**3 + 6*x**3) - b*x/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*atan(a + b*x)/(6*a**6*
x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3), True))

________________________________________________________________________________________

Giac [A]  time = 1.1008, size = 239, normalized size = 1.85 \begin{align*} -\frac{1}{6} \, b{\left (\frac{{\left (3 \, a^{2} b^{2} - b^{2}\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} - \frac{2 \,{\left (3 \, a^{2} b^{2} - b^{2}\right )} \log \left ({\left | x \right |}\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac{2 \,{\left (a^{3} b^{3} - 3 \, a b^{3}\right )} \arctan \left (b x + a\right )}{{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} b} + \frac{a^{4} + 2 \, a^{2} - 4 \,{\left (a^{3} b + a b\right )} x + 1}{{\left (a^{2} + 1\right )}^{3} x^{2}}\right )} - \frac{\arctan \left (b x + a\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(b*x+a)/x^4,x, algorithm="giac")

[Out]

-1/6*b*((3*a^2*b^2 - b^2)*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^6 + 3*a^4 + 3*a^2 + 1) - 2*(3*a^2*b^2 - b^2)*log
(abs(x))/(a^6 + 3*a^4 + 3*a^2 + 1) + 2*(a^3*b^3 - 3*a*b^3)*arctan(b*x + a)/((a^6 + 3*a^4 + 3*a^2 + 1)*b) + (a^
4 + 2*a^2 - 4*(a^3*b + a*b)*x + 1)/((a^2 + 1)^3*x^2)) - 1/3*arctan(b*x + a)/x^3