3.1.51 \(\int \frac {\text {ArcTan}(a+b x)}{x^4} \, dx\) [51]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5153, 378, 724, 815, 649, 209, 266} \begin {gather*} \frac {a \left (3-a^2\right ) b^3 \text {ArcTan}(a+b x)}{3 \left (a^2+1\right )^3}-\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 {2 a b^2}{3 \left (a^2+1\right )^2 x}-\frac {b}{6 \left (a^2+1\right ) x^2}-\frac {\text {ArcTan}(a+b x)}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/x^4,x]

[Out]

-1/6*b/((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) - Arc
Tan[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 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 724

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 815

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 5153

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^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 \text {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 \text {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 \text {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 \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.10, size = 128, normalized size = 0.99 \begin {gather*} \frac {-2 \left (1+a^2\right )^3 \text {ArcTan}(a+b x)+2 \left (-1+3 a^2\right ) b^3 x^3 \log (x)+i (i+a)^3 b^3 x^3 \log (i-a-b x)-(-i+a) b x \left ((i+a) \left (1+a^2-4 a b x\right )+i (-i+a)^2 b^2 x^2 \log (i+a+b x)\right )}{6 \left (1+a^2\right )^3 x^3} \end {gather*}

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)

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Maple [A]
time = 0.12, size = 115, normalized size = 0.89

method	result	size
derivativedivides	\(b^{3} \left (-\frac {\arctan \left (b x +a \right )}{3 b^{3} x^{3}}-\frac {\frac {\left (3 a^{2}-1\right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{2}+\left (a^{3}-3 a \right ) \arctan \left (b x +a \right )}{3 \left (a^{2}+1\right )^{3}}-\frac {\left (-3 a^{2}+1\right ) \ln \left (-b x \right )}{3 \left (a^{2}+1\right )^{3}}-\frac {1}{6 \left (a^{2}+1\right ) b^{2} x^{2}}+\frac {2 a}{3 \left (a^{2}+1\right )^{2} b x}\right )\)	\(115\)
default	\(b^{3} \left (-\frac {\arctan \left (b x +a \right )}{3 b^{3} x^{3}}-\frac {\frac {\left (3 a^{2}-1\right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{2}+\left (a^{3}-3 a \right ) \arctan \left (b x +a \right )}{3 \left (a^{2}+1\right )^{3}}-\frac {\left (-3 a^{2}+1\right ) \ln \left (-b x \right )}{3 \left (a^{2}+1\right )^{3}}-\frac {1}{6 \left (a^{2}+1\right ) b^{2} x^{2}}+\frac {2 a}{3 \left (a^{2}+1\right )^{2} b x}\right )\)	\(115\)
risch	\(\frac {i \ln \left (1+i \left (b x +a \right )\right )}{6 x^{3}}-\frac {i \left (a^{6} \ln \left (1-i \left (b x +a \right )\right )+3 a^{4} \ln \left (1-i \left (b x +a \right )\right )+3 a^{2} \ln \left (1-i \left (b x +a \right )\right )+\ln \left (1-i \left (b x +a \right )\right )-\ln \left (\left (a^{10} b +5 i a^{9} b +24 a^{8} b -56 i a^{7} b -110 a^{6} b +138 i a^{5} b +96 a^{4} b -48 i a^{3} b -27 a^{2} b +9 i a b \right ) x +9 a +234 a^{5}-80 i a^{8}-75 a^{3}+36 i a^{2}+a^{11}+29 a^{9}-166 a^{7}+4 i a^{10}-144 i a^{4}+248 i a^{6}\right ) a^{3} b^{3} x^{3}+\ln \left (\left (a^{10} b -5 i a^{9} b +24 a^{8} b +56 i a^{7} b -110 a^{6} b -138 i a^{5} b +96 a^{4} b +48 i a^{3} b -27 a^{2} b -9 i a b \right ) x +9 a +234 a^{5}+80 i a^{8}-75 a^{3}-36 i a^{2}+a^{11}+29 a^{9}-166 a^{7}-4 i a^{10}+144 i a^{4}-248 i a^{6}\right ) a^{3} b^{3} x^{3}+3 \ln \left (\left (a^{10} b +5 i a^{9} b +24 a^{8} b -56 i a^{7} b -110 a^{6} b +138 i a^{5} b +96 a^{4} b -48 i a^{3} b -27 a^{2} b +9 i a b \right ) x +9 a +234 a^{5}-80 i a^{8}-75 a^{3}+36 i a^{2}+a^{11}+29 a^{9}-166 a^{7}+4 i a^{10}-144 i a^{4}+248 i a^{6}\right ) a \,b^{3} x^{3}+4 i a \,b^{2} x^{2}-3 \ln \left (\left (a^{10} b -5 i a^{9} b +24 a^{8} b +56 i a^{7} b -110 a^{6} b -138 i a^{5} b +96 a^{4} b +48 i a^{3} b -27 a^{2} b -9 i a b \right ) x +9 a +234 a^{5}+80 i a^{8}-75 a^{3}-36 i a^{2}+a^{11}+29 a^{9}-166 a^{7}-4 i a^{10}+144 i a^{4}-248 i a^{6}\right ) a \,b^{3} x^{3}-2 i x \,a^{2} b +6 i \ln \left (\left (a^{6} b +75 a^{4} b -45 a^{2} b +9 b \right ) x \right ) a^{2} b^{3} x^{3}-3 i \ln \left (\left (a^{10} b -5 i a^{9} b +24 a^{8} b +56 i a^{7} b -110 a^{6} b -138 i a^{5} b +96 a^{4} b +48 i a^{3} b -27 a^{2} b -9 i a b \right ) x +9 a +234 a^{5}+80 i a^{8}-75 a^{3}-36 i a^{2}+a^{11}+29 a^{9}-166 a^{7}-4 i a^{10}+144 i a^{4}-248 i a^{6}\right ) a^{2} b^{3} x^{3}-3 i \ln \left (\left (a^{10} b +5 i a^{9} b +24 a^{8} b -56 i a^{7} b -110 a^{6} b +138 i a^{5} b +96 a^{4} b -48 i a^{3} b -27 a^{2} b +9 i a b \right ) x +9 a +234 a^{5}-80 i a^{8}-75 a^{3}+36 i a^{2}+a^{11}+29 a^{9}-166 a^{7}+4 i a^{10}-144 i a^{4}+248 i a^{6}\right ) a^{2} b^{3} x^{3}-i a^{4} b x +4 i a^{3} b^{2} x^{2}-2 i \ln \left (\left (a^{6} b +75 a^{4} b -45 a^{2} b +9 b \right ) x \right ) b^{3} x^{3}-i b x +i \ln \left (\left (a^{10} b +5 i a^{9} b +24 a^{8} b -56 i a^{7} b -110 a^{6} b +138 i a^{5} b +96 a^{4} b -48 i a^{3} b -27 a^{2} b +9 i a b \right ) x +9 a +234 a^{5}-80 i a^{8}-75 a^{3}+36 i a^{2}+a^{11}+29 a^{9}-166 a^{7}+4 i a^{10}-144 i a^{4}+248 i a^{6}\right ) b^{3} x^{3}+i \ln \left (\left (a^{10} b -5 i a^{9} b +24 a^{8} b +56 i a^{7} b -110 a^{6} b -138 i a^{5} b +96 a^{4} b +48 i a^{3} b -27 a^{2} b -9 i a b \right ) x +9 a +234 a^{5}+80 i a^{8}-75 a^{3}-36 i a^{2}+a^{11}+29 a^{9}-166 a^{7}-4 i a^{10}+144 i a^{4}-248 i a^{6}\right ) b^{3} x^{3}\right )}{6 x^{3} \left (a^{3}-3 i a^{2}-3 a +i\right ) \left (a^{3}+3 i a^{2}-3 a -i\right )}\)	\(1295\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 165, normalized size = 1.28 \begin {gather*} -\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 {gather*}

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

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Fricas [A]
time = 2.22, size = 135, normalized size = 1.05 \begin {gather*} -\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 {gather*}

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)

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Sympy [C] Result contains complex when optimal does not.
time = 1.63, size = 760, normalized size = 5.89 \begin {gather*} \begin {cases} \frac {i b^{3} \operatorname {atan}{\left (b x - i \right )}}{24} + \frac {i b^{2}}{24 x} - \frac {b}{24 x^{2}} - \frac {\operatorname {atan}{\left (b x - i \right )}}{3 x^{3}} - \frac {i}{18 x^{3}} & \text {for}\: a = - i \\- \frac {i b^{3} \operatorname {atan}{\left (b x + i \right )}}{24} - \frac {i b^{2}}{24 x} - \frac {b}{24 x^{2}} - \frac {\operatorname {atan}{\left (b x + i \right )}}{3 x^{3}} + \frac {i}{18 x^{3}} & \text {for}\: a = i \\- \frac {2 a^{6} \operatorname {atan}{\left (a + b x \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {a^{4} b x}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {6 a^{4} \operatorname {atan}{\left (a + b x \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {2 a^{3} b^{3} x^{3} \operatorname {atan}{\left (a + b x \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} + \frac {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}} + \frac {6 a^{2} b^{3} x^{3} \log {\left (x \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {3 a^{2} b^{3} x^{3} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {2 a^{2} b x}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {6 a^{2} \operatorname {atan}{\left (a + b x \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} + \frac {6 a b^{3} x^{3} \operatorname {atan}{\left (a + b x \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} + \frac {4 a b^{2} x^{2}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {2 b^{3} x^{3} \log {\left (x \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} + \frac {b^{3} x^{3} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {b x}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} - \frac {2 \operatorname {atan}{\left (a + b x \right )}}{6 a^{6} x^{3} + 18 a^{4} x^{3} + 18 a^{2} x^{3} + 6 x^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*b**3*atan(b*x - I)/24 + I*b**2/(24*x) - b/(24*x**2) - atan(b*x - I)/(3*x**3) - I/(18*x**3), Eq(a,
-I)), (-I*b**3*atan(b*x + I)/24 - I*b**2/(24*x) - b/(24*x**2) - atan(b*x + I)/(3*x**3) + I/(18*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*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) - b*x/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3) - 2*ata
n(a + b*x)/(6*a**6*x**3 + 18*a**4*x**3 + 18*a**2*x**3 + 6*x**3), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [B]
time = 1.05, size = 288, normalized size = 2.23 \begin {gather*} -\frac {\frac {b\,x}{6}+\mathrm {atan}\left (a+b\,x\right )\,\left (\frac {a^2}{3}+\frac {1}{3}\right )+\frac {b^2\,x^2\,\mathrm {atan}\left (a+b\,x\right )}{3}+\frac {x^3\,\left (b^3-7\,a^2\,b^3\right )}{6\,\left (a^4+2\,a^2+1\right )}-\frac {a\,b^2\,x^2}{3\,\left (a^2+1\right )}-\frac {2\,a\,b^4\,x^4}{3\,{\left (a^2+1\right )}^2}+\frac {2\,a\,b\,x\,\mathrm {atan}\left (a+b\,x\right )}{3}}{a^2\,x^3+2\,a\,b\,x^4+b^2\,x^5+x^3}-\frac {\ln \left (x\right )\,\left (\frac {b^3}{3}-a^2\,b^3\right )}{a^6+3\,a^4+3\,a^2+1}-\frac {b^3\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )\,\left (3\,a^2-1\right )}{6\,\left (a^6+3\,a^4+3\,a^2+1\right )}-\frac {a\,\mathrm {atan}\left (\frac {2\,x\,b^2+2\,a\,b}{2\,\sqrt {b^2\,\left (a^2+1\right )-a^2\,b^2}}\right )\,\left (a^2-3\right )\,{\left (b^2\right )}^{3/2}}{3\,\left (a^6+3\,a^4+3\,a^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a + b*x)/x^4,x)

[Out]

- ((b*x)/6 + atan(a + b*x)*(a^2/3 + 1/3) + (b^2*x^2*atan(a + b*x))/3 + (x^3*(b^3 - 7*a^2*b^3))/(6*(2*a^2 + a^4
+ 1)) - (a*b^2*x^2)/(3*(a^2 + 1)) - (2*a*b^4*x^4)/(3*(a^2 + 1)^2) + (2*a*b*x*atan(a + b*x))/3)/(x^3 + a^2*x^3
+ b^2*x^5 + 2*a*b*x^4) - (log(x)*(b^3/3 - a^2*b^3))/(3*a^2 + 3*a^4 + a^6 + 1) - (b^3*log(a^2 + b^2*x^2 + 2*a*
b*x + 1)*(3*a^2 - 1))/(6*(3*a^2 + 3*a^4 + a^6 + 1)) - (a*atan((2*a*b + 2*b^2*x)/(2*(b^2*(a^2 + 1) - a^2*b^2)^(
1/2)))*(a^2 - 3)*(b^2)^(3/2))/(3*(3*a^2 + 3*a^4 + a^6 + 1))

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