3.2.7 \(\int \frac {a+b \text {ArcTan}(c x^3)}{x^6} \, dx\) [107]

Optimal. Leaf size=115 \[ -\frac {3 b c}{10 x^2}-\frac {a+b \text {ArcTan}\left (c x^3\right )}{5 x^5}+\frac {1}{10} \sqrt {3} b c^{5/3} \text {ArcTan}\left (\frac {1-2 c^{2/3} x^2}{\sqrt {3}}\right )+\frac {1}{10} b c^{5/3} \log \left (1+c^{2/3} x^2\right )-\frac {1}{20} b c^{5/3} \log \left (1-c^{2/3} x^2+c^{4/3} x^4\right ) \]

[Out]

-3/10*b*c/x^2+1/5*(-a-b*arctan(c*x^3))/x^5+1/10*b*c^(5/3)*ln(1+c^(2/3)*x^2)-1/20*b*c^(5/3)*ln(1-c^(2/3)*x^2+c^
(4/3)*x^4)+1/10*b*c^(5/3)*arctan(1/3*(1-2*c^(2/3)*x^2)*3^(1/2))*3^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {4946, 281, 331, 298, 31, 648, 631, 210, 642} \begin {gather*} -\frac {a+b \text {ArcTan}\left (c x^3\right )}{5 x^5}+\frac {1}{10} \sqrt {3} b c^{5/3} \text {ArcTan}\left (\frac {1-2 c^{2/3} x^2}{\sqrt {3}}\right )+\frac {1}{10} b c^{5/3} \log \left (c^{2/3} x^2+1\right )-\frac {1}{20} b c^{5/3} \log \left (c^{4/3} x^4-c^{2/3} x^2+1\right )-\frac {3 b c}{10 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^3])/x^6,x]

[Out]

(-3*b*c)/(10*x^2) - (a + b*ArcTan[c*x^3])/(5*x^5) + (Sqrt[3]*b*c^(5/3)*ArcTan[(1 - 2*c^(2/3)*x^2)/Sqrt[3]])/10
+ (b*c^(5/3)*Log[1 + c^(2/3)*x^2])/10 - (b*c^(5/3)*Log[1 - c^(2/3)*x^2 + c^(4/3)*x^4])/20

Rule 31

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

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \tan ^{-1}\left (c x^3\right )}{x^6} \, dx &=-\frac {a+b \tan ^{-1}\left (c x^3\right )}{5 x^5}+\frac {1}{5} (3 b c) \int \frac {1}{x^3 \left (1+c^2 x^6\right )} \, dx\\ &=-\frac {a+b \tan ^{-1}\left (c x^3\right )}{5 x^5}+\frac {1}{10} (3 b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x^3\right )} \, dx,x,x^2\right )\\ &=-\frac {3 b c}{10 x^2}-\frac {a+b \tan ^{-1}\left (c x^3\right )}{5 x^5}-\frac {1}{10} \left (3 b c^3\right ) \text {Subst}\left (\int \frac {x}{1+c^2 x^3} \, dx,x,x^2\right )\\ &=-\frac {3 b c}{10 x^2}-\frac {a+b \tan ^{-1}\left (c x^3\right )}{5 x^5}+\frac {1}{10} \left (b c^{7/3}\right ) \text {Subst}\left (\int \frac {1}{1+c^{2/3} x} \, dx,x,x^2\right )-\frac {1}{10} \left (b c^{7/3}\right ) \text {Subst}\left (\int \frac {1+c^{2/3} x}{1-c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )\\ &=-\frac {3 b c}{10 x^2}-\frac {a+b \tan ^{-1}\left (c x^3\right )}{5 x^5}+\frac {1}{10} b c^{5/3} \log \left (1+c^{2/3} x^2\right )-\frac {1}{20} \left (b c^{5/3}\right ) \text {Subst}\left (\int \frac {-c^{2/3}+2 c^{4/3} x}{1-c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )-\frac {1}{20} \left (3 b c^{7/3}\right ) \text {Subst}\left (\int \frac {1}{1-c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )\\ &=-\frac {3 b c}{10 x^2}-\frac {a+b \tan ^{-1}\left (c x^3\right )}{5 x^5}+\frac {1}{10} b c^{5/3} \log \left (1+c^{2/3} x^2\right )-\frac {1}{20} b c^{5/3} \log \left (1-c^{2/3} x^2+c^{4/3} x^4\right )-\frac {1}{10} \left (3 b c^{5/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 c^{2/3} x^2\right )\\ &=-\frac {3 b c}{10 x^2}-\frac {a+b \tan ^{-1}\left (c x^3\right )}{5 x^5}+\frac {1}{10} \sqrt {3} b c^{5/3} \tan ^{-1}\left (\frac {1-2 c^{2/3} x^2}{\sqrt {3}}\right )+\frac {1}{10} b c^{5/3} \log \left (1+c^{2/3} x^2\right )-\frac {1}{20} b c^{5/3} \log \left (1-c^{2/3} x^2+c^{4/3} x^4\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 183, normalized size = 1.59 \begin {gather*} -\frac {a}{5 x^5}-\frac {3 b c}{10 x^2}-\frac {b \text {ArcTan}\left (c x^3\right )}{5 x^5}+\frac {1}{10} \sqrt {3} b c^{5/3} \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )+\frac {1}{10} \sqrt {3} b c^{5/3} \text {ArcTan}\left (\sqrt {3}+2 \sqrt [3]{c} x\right )+\frac {1}{10} b c^{5/3} \log \left (1+c^{2/3} x^2\right )-\frac {1}{20} b c^{5/3} \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )-\frac {1}{20} b c^{5/3} \log \left (1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^3])/x^6,x]

[Out]

-1/5*a/x^5 - (3*b*c)/(10*x^2) - (b*ArcTan[c*x^3])/(5*x^5) + (Sqrt[3]*b*c^(5/3)*ArcTan[Sqrt[3] - 2*c^(1/3)*x])/
10 + (Sqrt[3]*b*c^(5/3)*ArcTan[Sqrt[3] + 2*c^(1/3)*x])/10 + (b*c^(5/3)*Log[1 + c^(2/3)*x^2])/10 - (b*c^(5/3)*L
og[1 - Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/20 - (b*c^(5/3)*Log[1 + Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/20

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 105, normalized size = 0.91

method	result	size
default	\(-\frac {a}{5 x^{5}}-\frac {b \arctan \left (c \,x^{3}\right )}{5 x^{5}}-\frac {3 b c}{10 x^{2}}+\frac {b c \ln \left (x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}-\frac {b c \ln \left (x^{4}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}} x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}-\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{2}}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{10 \left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}\)	\(105\)
risch	\(\frac {i b \ln \left (i c \,x^{3}+1\right )}{10 x^{5}}-\frac {i b \ln \left (-i c \,x^{3}+1\right )}{10 x^{5}}-\frac {3 b c}{10 x^{2}}-\frac {b c \ln \left (x +\left (\frac {i}{c}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {i}{c}\right )^{\frac {2}{3}}}+\frac {b c \ln \left (x^{2}-\left (\frac {i}{c}\right )^{\frac {1}{3}} x +\left (\frac {i}{c}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {i}{c}\right )^{\frac {2}{3}}}-\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {i}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{10 \left (\frac {i}{c}\right )^{\frac {2}{3}}}-\frac {a}{5 x^{5}}-\frac {b c \ln \left (x -\left (\frac {i}{c}\right )^{\frac {1}{3}}\right )}{10 \left (\frac {i}{c}\right )^{\frac {2}{3}}}+\frac {b c \ln \left (x^{2}+\left (\frac {i}{c}\right )^{\frac {1}{3}} x +\left (\frac {i}{c}\right )^{\frac {2}{3}}\right )}{20 \left (\frac {i}{c}\right )^{\frac {2}{3}}}+\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {i}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{10 \left (\frac {i}{c}\right )^{\frac {2}{3}}}\)	\(236\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^3))/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/5*a/x^5-1/5*b/x^5*arctan(c*x^3)-3/10*b*c/x^2+1/10*b*c/(1/c^2)^(1/3)*ln(x^2+(1/c^2)^(1/3))-1/20*b*c/(1/c^2)^
(1/3)*ln(x^4-(1/c^2)^(1/3)*x^2+(1/c^2)^(2/3))-1/10*b*c*3^(1/2)/(1/c^2)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c^2)^(1/
3)*x^2-1))

________________________________________________________________________________________

Maxima [A]
time = 0.47, size = 102, normalized size = 0.89 \begin {gather*} -\frac {1}{20} \, {\left ({\left (2 \, \sqrt {3} c^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {4}{3}} x^{2} - c^{\frac {2}{3}}\right )}}{3 \, c^{\frac {2}{3}}}\right ) + c^{\frac {2}{3}} \log \left (c^{\frac {4}{3}} x^{4} - c^{\frac {2}{3}} x^{2} + 1\right ) - 2 \, c^{\frac {2}{3}} \log \left (\frac {c^{\frac {2}{3}} x^{2} + 1}{c^{\frac {2}{3}}}\right ) + \frac {6}{x^{2}}\right )} c + \frac {4 \, \arctan \left (c x^{3}\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^6,x, algorithm="maxima")

[Out]

-1/20*((2*sqrt(3)*c^(2/3)*arctan(1/3*sqrt(3)*(2*c^(4/3)*x^2 - c^(2/3))/c^(2/3)) + c^(2/3)*log(c^(4/3)*x^4 - c^
(2/3)*x^2 + 1) - 2*c^(2/3)*log((c^(2/3)*x^2 + 1)/c^(2/3)) + 6/x^2)*c + 4*arctan(c*x^3)/x^5)*b - 1/5*a/x^5

________________________________________________________________________________________

Fricas [A]
time = 1.16, size = 121, normalized size = 1.05 \begin {gather*} -\frac {2 \, \sqrt {3} b {\left (c^{2}\right )}^{\frac {1}{3}} c x^{5} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (c^{2}\right )}^{\frac {1}{3}} x^{2} - \frac {1}{3} \, \sqrt {3}\right ) + b {\left (c^{2}\right )}^{\frac {1}{3}} c x^{5} \log \left (c^{2} x^{4} - {\left (c^{2}\right )}^{\frac {2}{3}} x^{2} + {\left (c^{2}\right )}^{\frac {1}{3}}\right ) - 2 \, b {\left (c^{2}\right )}^{\frac {1}{3}} c x^{5} \log \left (c^{2} x^{2} + {\left (c^{2}\right )}^{\frac {2}{3}}\right ) + 6 \, b c x^{3} + 4 \, b \arctan \left (c x^{3}\right ) + 4 \, a}{20 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^6,x, algorithm="fricas")

[Out]

-1/20*(2*sqrt(3)*b*(c^2)^(1/3)*c*x^5*arctan(2/3*sqrt(3)*(c^2)^(1/3)*x^2 - 1/3*sqrt(3)) + b*(c^2)^(1/3)*c*x^5*l
og(c^2*x^4 - (c^2)^(2/3)*x^2 + (c^2)^(1/3)) - 2*b*(c^2)^(1/3)*c*x^5*log(c^2*x^2 + (c^2)^(2/3)) + 6*b*c*x^3 + 4
*b*arctan(c*x^3) + 4*a)/x^5

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (112) = 224\).
time = 58.14, size = 286, normalized size = 2.49 \begin {gather*} \begin {cases} - \frac {a}{5 x^{5}} + \frac {b c^{2} \sqrt [6]{- \frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{3} \right )}}{5} - \frac {b c \log {\left (x - \sqrt [6]{- \frac {1}{c^{2}}} \right )}}{5 \sqrt [3]{- \frac {1}{c^{2}}}} + \frac {3 b c \log {\left (4 x^{2} - 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{20 \sqrt [3]{- \frac {1}{c^{2}}}} - \frac {b c \log {\left (4 x^{2} + 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{20 \sqrt [3]{- \frac {1}{c^{2}}}} - \frac {\sqrt {3} b c \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [6]{- \frac {1}{c^{2}}}} - \frac {\sqrt {3}}{3} \right )}}{10 \sqrt [3]{- \frac {1}{c^{2}}}} + \frac {\sqrt {3} b c \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [6]{- \frac {1}{c^{2}}}} + \frac {\sqrt {3}}{3} \right )}}{10 \sqrt [3]{- \frac {1}{c^{2}}}} - \frac {3 b c}{10 x^{2}} - \frac {b \operatorname {atan}{\left (c x^{3} \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a}{5 x^{5}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**3))/x**6,x)

[Out]

Piecewise((-a/(5*x**5) + b*c**2*(-1/c**2)**(1/6)*atan(c*x**3)/5 - b*c*log(x - (-1/c**2)**(1/6))/(5*(-1/c**2)**
(1/3)) + 3*b*c*log(4*x**2 - 4*x*(-1/c**2)**(1/6) + 4*(-1/c**2)**(1/3))/(20*(-1/c**2)**(1/3)) - b*c*log(4*x**2
+ 4*x*(-1/c**2)**(1/6) + 4*(-1/c**2)**(1/3))/(20*(-1/c**2)**(1/3)) - sqrt(3)*b*c*atan(2*sqrt(3)*x/(3*(-1/c**2)
**(1/6)) - sqrt(3)/3)/(10*(-1/c**2)**(1/3)) + sqrt(3)*b*c*atan(2*sqrt(3)*x/(3*(-1/c**2)**(1/6)) + sqrt(3)/3)/(
10*(-1/c**2)**(1/3)) - 3*b*c/(10*x**2) - b*atan(c*x**3)/(5*x**5), Ne(c, 0)), (-a/(5*x**5), True))

________________________________________________________________________________________

Giac [A]
time = 0.49, size = 108, normalized size = 0.94 \begin {gather*} -\frac {1}{20} \, b c^{3} {\left (\frac {2 \, \sqrt {3} {\left | c \right |}^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )} {\left | c \right |}^{\frac {2}{3}}\right )}{c^{2}} + \frac {{\left | c \right |}^{\frac {2}{3}} \log \left (x^{4} - \frac {x^{2}}{{\left | c \right |}^{\frac {2}{3}}} + \frac {1}{{\left | c \right |}^{\frac {4}{3}}}\right )}{c^{2}} - \frac {2 \, \log \left (x^{2} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{{\left | c \right |}^{\frac {4}{3}}}\right )} - \frac {3 \, b c x^{3} + 2 \, b \arctan \left (c x^{3}\right ) + 2 \, a}{10 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x^6,x, algorithm="giac")

[Out]

-1/20*b*c^3*(2*sqrt(3)*abs(c)^(2/3)*arctan(1/3*sqrt(3)*(2*x^2 - 1/abs(c)^(2/3))*abs(c)^(2/3))/c^2 + abs(c)^(2/
3)*log(x^4 - x^2/abs(c)^(2/3) + 1/abs(c)^(4/3))/c^2 - 2*log(x^2 + 1/abs(c)^(2/3))/abs(c)^(4/3)) - 1/10*(3*b*c*
x^3 + 2*b*arctan(c*x^3) + 2*a)/x^5

________________________________________________________________________________________

Mupad [B]
time = 2.57, size = 118, normalized size = 1.03 \begin {gather*} \frac {b\,c^{5/3}\,\ln \left (c^{2/3}\,x^2+1\right )}{10}-\frac {\frac {3\,b\,c\,x^3}{2}+a}{5\,x^5}-\frac {b\,\mathrm {atan}\left (c\,x^3\right )}{5\,x^5}-\frac {b\,c^{5/3}\,\ln \left (\sqrt {3}\,c^{2/3}\,x^2-c^{2/3}\,x^2\,1{}\mathrm {i}+2{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{20}+\frac {b\,c^{5/3}\,\ln \left (-c^{2/3}\,x^2\,1{}\mathrm {i}-\sqrt {3}\,c^{2/3}\,x^2+2{}\mathrm {i}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))/x^6,x)

[Out]

(b*c^(5/3)*log(c^(2/3)*x^2 + 1))/10 - (a + (3*b*c*x^3)/2)/(5*x^5) - (b*atan(c*x^3))/(5*x^5) - (b*c^(5/3)*log(3
^(1/2)*c^(2/3)*x^2 - c^(2/3)*x^2*1i + 2i)*(3^(1/2)*1i + 1))/20 + (b*c^(5/3)*log(2i - 3^(1/2)*c^(2/3)*x^2 - c^(
2/3)*x^2*1i)*(3^(1/2)*1i - 1))/20

________________________________________________________________________________________