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*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

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Rubi [A]  time = 0.267174, antiderivative size = 174, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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 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 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])

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 = 0.0516986, 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]

-a/(4*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

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Maple [A]  time = 0.013, size = 172, normalized size = 1. \begin{align*} -{\frac{a}{4\,{x}^{4}}}-{\frac{b{\it Artanh} \left ( c{x}^{3} \right ) }{4\,{x}^{4}}}-{\frac{3\,bc}{4\,x}}-{\frac{bc}{8}\ln \left ( x-\sqrt [3]{{c}^{-1}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}+{\frac{bc}{16}\ln \left ({x}^{2}+\sqrt [3]{{c}^{-1}}x+ \left ({c}^{-1} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}-{\frac{bc\sqrt{3}}{8}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{x}{\sqrt [3]{{c}^{-1}}}}+1 \right ) } \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}+{\frac{bc}{8}\ln \left ( x+\sqrt [3]{{c}^{-1}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}-{\frac{bc}{16}\ln \left ({x}^{2}-\sqrt [3]{{c}^{-1}}x+ \left ({c}^{-1} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}}-{\frac{bc\sqrt{3}}{8}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{x}{\sqrt [3]{{c}^{-1}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{c}^{-1}}}}} \end{align*}

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))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.67314, size = 547, normalized size = 3.14 \begin{align*} -\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}} \end{align*}

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: KeyError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: KeyError

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Giac [A]  time = 2.08691, size = 261, normalized size = 1.5 \begin{align*} -\frac{1}{16} \, b c^{3}{\left (\frac{2 \, \sqrt{3}{\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 )}{c^{2}} + \frac{2 \, \sqrt{3}{\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 )}{c^{2}} - \frac{{\left | c \right |}^{\frac{1}{3}} \log \left (x^{2} + \frac{x}{{\left | c \right |}^{\frac{1}{3}}} + \frac{1}{{\left | c \right |}^{\frac{2}{3}}}\right )}{c^{2}} + \frac{{\left | c \right |}^{\frac{1}{3}} \log \left (x^{2} - \frac{x}{{\left | c \right |}^{\frac{1}{3}}} + \frac{1}{{\left | c \right |}^{\frac{2}{3}}}\right )}{c^{2}} - \frac{2 \,{\left | c \right |}^{\frac{1}{3}} \log \left ({\left | x + \frac{1}{{\left | c \right |}^{\frac{1}{3}}} \right |}\right )}{c^{2}} + \frac{2 \,{\left | c \right |}^{\frac{1}{3}} \log \left ({\left | x - \frac{1}{{\left | c \right |}^{\frac{1}{3}}} \right |}\right )}{c^{2}}\right )} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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