3.115 \(\int \frac {a+b \tanh ^{-1}(c x^3)}{x^5} \, dx\)

Optimal. Leaf size=174 \[ -\frac {a+b \tanh ^{-1}\left (c x^3\right )}{4 x^4}-\frac {1}{16} b c^{4/3} \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )+\frac {1}{16} b c^{4/3} \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )+\frac {1}{8} \sqrt {3} b c^{4/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )-\frac {1}{8} \sqrt {3} b c^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )+\frac {1}{4} b c^{4/3} \tanh ^{-1}\left (\sqrt [3]{c} x\right )-\frac {3 b c}{4 x} \]

[Out]

-3/4*b*c/x+1/4*b*c^(4/3)*arctanh(c^(1/3)*x)+1/4*(-a-b*arctanh(c*x^3))/x^4-1/16*b*c^(4/3)*ln(1-c^(1/3)*x+c^(2/3
)*x^2)+1/16*b*c^(4/3)*ln(1+c^(1/3)*x+c^(2/3)*x^2)-1/8*b*c^(4/3)*arctan(-1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(
1/2)-1/8*b*c^(4/3)*arctan(1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6097, 325, 296, 634, 618, 204, 628, 206} \[ -\frac {a+b \tanh ^{-1}\left (c x^3\right )}{4 x^4}-\frac {1}{16} b c^{4/3} \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )+\frac {1}{16} b c^{4/3} \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )+\frac {1}{8} \sqrt {3} b c^{4/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )-\frac {1}{8} \sqrt {3} b c^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )+\frac {1}{4} b c^{4/3} \tanh ^{-1}\left (\sqrt [3]{c} x\right )-\frac {3 b c}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^3])/x^5,x]

[Out]

(-3*b*c)/(4*x) + (Sqrt[3]*b*c^(4/3)*ArcTan[1/Sqrt[3] - (2*c^(1/3)*x)/Sqrt[3]])/8 - (Sqrt[3]*b*c^(4/3)*ArcTan[1
/Sqrt[3] + (2*c^(1/3)*x)/Sqrt[3]])/8 + (b*c^(4/3)*ArcTanh[c^(1/3)*x])/4 - (a + b*ArcTanh[c*x^3])/(4*x^4) - (b*
c^(4/3)*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/16 + (b*c^(4/3)*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/16

Rule 204

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

Rule 206

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

Rule 296

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[(2*k*m*Pi)/n] - s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi
)/n]*x + s^2*x^2), x] + Int[(r*Cos[(2*k*m*Pi)/n] + s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x
+ s^2*x^2), x]; (2*r^(m + 2)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

Rule 325

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 618

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

Rule 628

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 634

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 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}\left (c x^3\right )}{x^5} \, dx &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{4 x^4}+\frac {1}{4} (3 b c) \int \frac {1}{x^2 \left (1-c^2 x^6\right )} \, dx\\ &=-\frac {3 b c}{4 x}-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{4 x^4}+\frac {1}{4} \left (3 b c^3\right ) \int \frac {x^4}{1-c^2 x^6} \, dx\\ &=-\frac {3 b c}{4 x}-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{4 x^4}+\frac {1}{4} \left (b c^{5/3}\right ) \int \frac {1}{1-c^{2/3} x^2} \, dx+\frac {1}{4} \left (b c^{5/3}\right ) \int \frac {-\frac {1}{2}-\frac {\sqrt [3]{c} x}{2}}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx+\frac {1}{4} \left (b c^{5/3}\right ) \int \frac {-\frac {1}{2}+\frac {\sqrt [3]{c} x}{2}}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx\\ &=-\frac {3 b c}{4 x}+\frac {1}{4} b c^{4/3} \tanh ^{-1}\left (\sqrt [3]{c} x\right )-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{4 x^4}-\frac {1}{16} \left (b c^{4/3}\right ) \int \frac {-\sqrt [3]{c}+2 c^{2/3} x}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx+\frac {1}{16} \left (b c^{4/3}\right ) \int \frac {\sqrt [3]{c}+2 c^{2/3} x}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx-\frac {1}{16} \left (3 b c^{5/3}\right ) \int \frac {1}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx-\frac {1}{16} \left (3 b c^{5/3}\right ) \int \frac {1}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx\\ &=-\frac {3 b c}{4 x}+\frac {1}{4} b c^{4/3} \tanh ^{-1}\left (\sqrt [3]{c} x\right )-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{4 x^4}-\frac {1}{16} b c^{4/3} \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )+\frac {1}{16} b c^{4/3} \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )-\frac {1}{8} \left (3 b c^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{c} x\right )+\frac {1}{8} \left (3 b c^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{c} x\right )\\ &=-\frac {3 b c}{4 x}+\frac {1}{8} \sqrt {3} b c^{4/3} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )-\frac {1}{8} \sqrt {3} b c^{4/3} \tan ^{-1}\left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )+\frac {1}{4} b c^{4/3} \tanh ^{-1}\left (\sqrt [3]{c} x\right )-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{4 x^4}-\frac {1}{16} b c^{4/3} \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )+\frac {1}{16} b c^{4/3} \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 196, normalized size = 1.13 \[ -\frac {a}{4 x^4}-\frac {1}{16} b c^{4/3} \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )+\frac {1}{16} b c^{4/3} \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )-\frac {1}{8} b c^{4/3} \log \left (1-\sqrt [3]{c} x\right )+\frac {1}{8} b c^{4/3} \log \left (\sqrt [3]{c} x+1\right )-\frac {1}{8} \sqrt {3} b c^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x-1}{\sqrt {3}}\right )-\frac {1}{8} \sqrt {3} b c^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x+1}{\sqrt {3}}\right )-\frac {b \tanh ^{-1}\left (c x^3\right )}{4 x^4}-\frac {3 b c}{4 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^3])/x^5,x]

[Out]

-1/4*a/x^4 - (3*b*c)/(4*x) - (Sqrt[3]*b*c^(4/3)*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]])/8 - (Sqrt[3]*b*c^(4/3)*Arc
Tan[(1 + 2*c^(1/3)*x)/Sqrt[3]])/8 - (b*ArcTanh[c*x^3])/(4*x^4) - (b*c^(4/3)*Log[1 - c^(1/3)*x])/8 + (b*c^(4/3)
*Log[1 + c^(1/3)*x])/8 - (b*c^(4/3)*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/16 + (b*c^(4/3)*Log[1 + c^(1/3)*x + c^(2
/3)*x^2])/16

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 196, normalized size = 1.13 \[ -\frac {2 \, \sqrt {3} b \left (-c\right )^{\frac {1}{3}} c x^{4} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (-c\right )^{\frac {1}{3}} x - \frac {1}{3} \, \sqrt {3}\right ) + 2 \, \sqrt {3} b c^{\frac {4}{3}} x^{4} \arctan \left (\frac {2}{3} \, \sqrt {3} c^{\frac {1}{3}} x - \frac {1}{3} \, \sqrt {3}\right ) + b \left (-c\right )^{\frac {1}{3}} c x^{4} \log \left (c x^{2} + \left (-c\right )^{\frac {2}{3}} x - \left (-c\right )^{\frac {1}{3}}\right ) + b c^{\frac {4}{3}} x^{4} \log \left (c x^{2} - c^{\frac {2}{3}} x + c^{\frac {1}{3}}\right ) - 2 \, b \left (-c\right )^{\frac {1}{3}} c x^{4} \log \left (c x - \left (-c\right )^{\frac {2}{3}}\right ) - 2 \, b c^{\frac {4}{3}} x^{4} \log \left (c x + c^{\frac {2}{3}}\right ) + 12 \, b c x^{3} + 2 \, b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 4 \, a}{16 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^5,x, algorithm="fricas")

[Out]

-1/16*(2*sqrt(3)*b*(-c)^(1/3)*c*x^4*arctan(2/3*sqrt(3)*(-c)^(1/3)*x - 1/3*sqrt(3)) + 2*sqrt(3)*b*c^(4/3)*x^4*a
rctan(2/3*sqrt(3)*c^(1/3)*x - 1/3*sqrt(3)) + b*(-c)^(1/3)*c*x^4*log(c*x^2 + (-c)^(2/3)*x - (-c)^(1/3)) + b*c^(
4/3)*x^4*log(c*x^2 - c^(2/3)*x + c^(1/3)) - 2*b*(-c)^(1/3)*c*x^4*log(c*x - (-c)^(2/3)) - 2*b*c^(4/3)*x^4*log(c
*x + c^(2/3)) + 12*b*c*x^3 + 2*b*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 4*a)/x^4

________________________________________________________________________________________

giac [A]  time = 0.44, size = 187, normalized size = 1.07 \[ -\frac {1}{8} \, \sqrt {3} b c {\left | c \right |}^{\frac {1}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right ) - \frac {1}{8} \, \sqrt {3} b c {\left | c \right |}^{\frac {1}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right ) + \frac {b c^{3} \log \left (x^{2} + \frac {x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{16 \, {\left | c \right |}^{\frac {5}{3}}} - \frac {b c^{3} \log \left (x^{2} - \frac {x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{16 \, {\left | c \right |}^{\frac {5}{3}}} + \frac {1}{8} \, b c {\left | c \right |}^{\frac {1}{3}} \log \left ({\left | x + \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right ) - \frac {b c^{3} \log \left ({\left | x - \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right )}{8 \, {\left | c \right |}^{\frac {5}{3}}} - \frac {b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )}{8 \, x^{4}} - \frac {3 \, b c x^{3} + a}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^5,x, algorithm="giac")

[Out]

-1/8*sqrt(3)*b*c*abs(c)^(1/3)*arctan(1/3*sqrt(3)*(2*x + 1/abs(c)^(1/3))*abs(c)^(1/3)) - 1/8*sqrt(3)*b*c*abs(c)
^(1/3)*arctan(1/3*sqrt(3)*(2*x - 1/abs(c)^(1/3))*abs(c)^(1/3)) + 1/16*b*c^3*log(x^2 + x/abs(c)^(1/3) + 1/abs(c
)^(2/3))/abs(c)^(5/3) - 1/16*b*c^3*log(x^2 - x/abs(c)^(1/3) + 1/abs(c)^(2/3))/abs(c)^(5/3) + 1/8*b*c*abs(c)^(1
/3)*log(abs(x + 1/abs(c)^(1/3))) - 1/8*b*c^3*log(abs(x - 1/abs(c)^(1/3)))/abs(c)^(5/3) - 1/8*b*log(-(c*x^3 + 1
)/(c*x^3 - 1))/x^4 - 1/4*(3*b*c*x^3 + a)/x^4

________________________________________________________________________________________

maple [A]  time = 0.04, size = 172, normalized size = 0.99 \[ -\frac {a}{4 x^{4}}-\frac {b \arctanh \left (c \,x^{3}\right )}{4 x^{4}}-\frac {3 b c}{4 x}-\frac {b c \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{8 \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b c \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{16 \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{8 \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b c \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{8 \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b c \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{16 \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b c \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{8 \left (\frac {1}{c}\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^3))/x^5,x)

[Out]

-1/4*a/x^4-1/4*b/x^4*arctanh(c*x^3)-3/4*b*c/x-1/8*b*c/(1/c)^(1/3)*ln(x-(1/c)^(1/3))+1/16*b*c/(1/c)^(1/3)*ln(x^
2+(1/c)^(1/3)*x+(1/c)^(2/3))-1/8*b*c*3^(1/2)/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))+1/8*b*c/(1/c)
^(1/3)*ln(x+(1/c)^(1/3))-1/16*b*c/(1/c)^(1/3)*ln(x^2-(1/c)^(1/3)*x+(1/c)^(2/3))-1/8*b*c*3^(1/2)/(1/c)^(1/3)*ar
ctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x-1))

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 160, normalized size = 0.92 \[ -\frac {1}{16} \, {\left ({\left (2 \, \sqrt {3} c^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x + c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right ) + 2 \, \sqrt {3} c^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x - c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right ) - c^{\frac {1}{3}} \log \left (c^{\frac {2}{3}} x^{2} + c^{\frac {1}{3}} x + 1\right ) + c^{\frac {1}{3}} \log \left (c^{\frac {2}{3}} x^{2} - c^{\frac {1}{3}} x + 1\right ) - 2 \, c^{\frac {1}{3}} \log \left (\frac {c^{\frac {1}{3}} x + 1}{c^{\frac {1}{3}}}\right ) + 2 \, c^{\frac {1}{3}} \log \left (\frac {c^{\frac {1}{3}} x - 1}{c^{\frac {1}{3}}}\right ) + \frac {12}{x}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x^{3}\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^5,x, algorithm="maxima")

[Out]

-1/16*((2*sqrt(3)*c^(1/3)*arctan(1/3*sqrt(3)*(2*c^(2/3)*x + c^(1/3))/c^(1/3)) + 2*sqrt(3)*c^(1/3)*arctan(1/3*s
qrt(3)*(2*c^(2/3)*x - c^(1/3))/c^(1/3)) - c^(1/3)*log(c^(2/3)*x^2 + c^(1/3)*x + 1) + c^(1/3)*log(c^(2/3)*x^2 -
c^(1/3)*x + 1) - 2*c^(1/3)*log((c^(1/3)*x + 1)/c^(1/3)) + 2*c^(1/3)*log((c^(1/3)*x - 1)/c^(1/3)) + 12/x)*c +
4*arctanh(c*x^3)/x^4)*b - 1/4*a/x^4

________________________________________________________________________________________

mupad [B]  time = 1.28, size = 125, normalized size = 0.72 \[ \frac {b\,\ln \left (1-c\,x^3\right )}{8\,x^4}-\frac {b\,c^{4/3}\,\left (-\frac {\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}-\mathrm {i}\right )}{2}\right )}{2}+\frac {\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}+1{}\mathrm {i}\right )}{2}\right )}{2}+\mathrm {atan}\left (c^{1/3}\,x\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{4}-\frac {3\,b\,c}{4\,x}-\frac {b\,\ln \left (c\,x^3+1\right )}{8\,x^4}-\frac {a}{4\,x^4}-\frac {\sqrt {3}\,b\,c^{4/3}\,\left (\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}-\mathrm {i}\right )}{2}\right )+\mathrm {atan}\left (\frac {c^{1/3}\,x\,\left (\sqrt {3}+1{}\mathrm {i}\right )}{2}\right )\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^3))/x^5,x)

[Out]

(b*log(1 - c*x^3))/(8*x^4) - (b*c^(4/3)*(atan((c^(1/3)*x*(3^(1/2) + 1i))/2)/2 - atan((c^(1/3)*x*(3^(1/2) - 1i)
)/2)/2 + atan(c^(1/3)*x*1i))*1i)/4 - (3*b*c)/(4*x) - (b*log(c*x^3 + 1))/(8*x^4) - a/(4*x^4) - (3^(1/2)*b*c^(4/
3)*(atan((c^(1/3)*x*(3^(1/2) - 1i))/2) + atan((c^(1/3)*x*(3^(1/2) + 1i))/2)))/8

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**3))/x**5,x)

[Out]

Timed out

________________________________________________________________________________________