3.110 \(\int \frac{a+b \tanh ^{-1}(c x^3)}{x^6} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0945154, antiderivative size = 115, normalized size of antiderivative = 1., 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 = {6097, 275, 325, 292, 31, 634, 617, 204, 628} \[ -\frac{a+b \tanh ^{-1}\left (c x^3\right )}{5 x^5}-\frac{1}{10} b c^{5/3} \log \left (1-c^{2/3} x^2\right )+\frac{1}{20} b c^{5/3} \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )-\frac{1}{10} \sqrt{3} b c^{5/3} \tan ^{-1}\left (\frac{2 c^{2/3} x^2+1}{\sqrt{3}}\right )-\frac{3 b c}{10 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^3])/x^6,x]

[Out]

(-3*b*c)/(10*x^2) - (Sqrt[3]*b*c^(5/3)*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/10 - (a + b*ArcTanh[c*x^3])/(5*x^5
) - (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 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 275

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 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 292

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 31

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

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 617

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0496929, size = 196, normalized size = 1.7 \[ -\frac{a}{5 x^5}+\frac{1}{20} b c^{5/3} \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )+\frac{1}{20} b c^{5/3} \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )-\frac{1}{10} b c^{5/3} \log \left (1-\sqrt [3]{c} x\right )-\frac{1}{10} b c^{5/3} \log \left (\sqrt [3]{c} x+1\right )-\frac{1}{10} \sqrt{3} b c^{5/3} \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x-1}{\sqrt{3}}\right )+\frac{1}{10} \sqrt{3} b c^{5/3} \tan ^{-1}\left (\frac{2 \sqrt [3]{c} x+1}{\sqrt{3}}\right )-\frac{3 b c}{10 x^2}-\frac{b \tanh ^{-1}\left (c x^3\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 172, normalized size = 1.5 \begin{align*} -{\frac{a}{5\,{x}^{5}}}-{\frac{b{\it Artanh} \left ( c{x}^{3} \right ) }{5\,{x}^{5}}}-{\frac{3\,bc}{10\,{x}^{2}}}-{\frac{bc}{10}\ln \left ( x-\sqrt [3]{{c}^{-1}} \right ) \left ({c}^{-1} \right ) ^{-{\frac{2}{3}}}}+{\frac{bc}{20}\ln \left ({x}^{2}+\sqrt [3]{{c}^{-1}}x+ \left ({c}^{-1} \right ) ^{{\frac{2}{3}}} \right ) \left ({c}^{-1} \right ) ^{-{\frac{2}{3}}}}+{\frac{bc\sqrt{3}}{10}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{x}{\sqrt [3]{{c}^{-1}}}}+1 \right ) } \right ) \left ({c}^{-1} \right ) ^{-{\frac{2}{3}}}}-{\frac{bc}{10}\ln \left ( x+\sqrt [3]{{c}^{-1}} \right ) \left ({c}^{-1} \right ) ^{-{\frac{2}{3}}}}+{\frac{bc}{20}\ln \left ({x}^{2}-\sqrt [3]{{c}^{-1}}x+ \left ({c}^{-1} \right ) ^{{\frac{2}{3}}} \right ) \left ({c}^{-1} \right ) ^{-{\frac{2}{3}}}}-{\frac{bc\sqrt{3}}{10}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{x}{\sqrt [3]{{c}^{-1}}}}-1 \right ) } \right ) \left ({c}^{-1} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^3))/x^6,x)

[Out]

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

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Maxima [A]  time = 1.44875, size = 144, normalized size = 1.25 \begin{align*} -\frac{1}{20} \,{\left ({\left (2 \, \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{3}} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{3}}{\left (2 \, x^{2} + \frac{1}{c^{2}}^{\frac{1}{3}}\right )}\right ) -{\left (c^{2}\right )}^{\frac{1}{3}} \log \left (x^{4} + \frac{1}{c^{2}}^{\frac{1}{3}} x^{2} + \frac{1}{c^{2}}^{\frac{2}{3}}\right ) + 2 \,{\left (c^{2}\right )}^{\frac{1}{3}} \log \left (x^{2} - \frac{1}{c^{2}}^{\frac{1}{3}}\right ) + \frac{6}{x^{2}}\right )} c + \frac{4 \, \operatorname{artanh}\left (c x^{3}\right )}{x^{5}}\right )} b - \frac{a}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76675, size = 367, normalized size = 3.19 \begin{align*} -\frac{2 \, \sqrt{3} \left (-c^{2}\right )^{\frac{1}{3}} b 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 ) + \left (-c^{2}\right )^{\frac{1}{3}} b 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 \, \left (-c^{2}\right )^{\frac{1}{3}} b c x^{5} \log \left (c^{2} x^{2} - \left (-c^{2}\right )^{\frac{2}{3}}\right ) + 6 \, b c x^{3} + 2 \, b \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right ) + 4 \, a}{20 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.21764, size = 169, normalized size = 1.47 \begin{align*} -\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 ({\left | x^{2} - \frac{1}{{\left | c \right |}^{\frac{2}{3}}} \right |}\right )}{{\left | c \right |}^{\frac{4}{3}}}\right )} - \frac{b \log \left (-\frac{c x^{3} + 1}{c x^{3} - 1}\right )}{10 \, x^{5}} - \frac{3 \, b c x^{3} + 2 \, a}{10 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(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(abs(x^2 - 1/abs(c)^(2/3)))/abs(c)^(4/3)) - 1/10*b*
log(-(c*x^3 + 1)/(c*x^3 - 1))/x^5 - 1/10*(3*b*c*x^3 + 2*a)/x^5