3.1.80 \(\int \frac {(a+b \tanh ^{-1}(c x^2))^3}{x^3} \, dx\) [80]

Optimal. Leaf size=125 \[ \frac {1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{2 x^2}+\frac {3}{2} b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \log \left (2-\frac {2}{1+c x^2}\right )-\frac {3}{2} b^2 c \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+c x^2}\right )-\frac {3}{4} b^3 c \text {PolyLog}\left (3,-1+\frac {2}{1+c x^2}\right ) \]

[Out]

1/2*c*(a+b*arctanh(c*x^2))^3-1/2*(a+b*arctanh(c*x^2))^3/x^2+3/2*b*c*(a+b*arctanh(c*x^2))^2*ln(2-2/(c*x^2+1))-3
/2*b^2*c*(a+b*arctanh(c*x^2))*polylog(2,-1+2/(c*x^2+1))-3/4*b^3*c*polylog(3,-1+2/(c*x^2+1))

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Rubi [A]
time = 0.23, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6039, 6037, 6135, 6079, 6095, 6203, 6745} \begin {gather*} -\frac {3}{2} b^2 c \text {Li}_2\left (\frac {2}{c x^2+1}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{2 x^2}+\frac {3}{2} b c \log \left (2-\frac {2}{c x^2+1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {3}{4} b^3 c \text {Li}_3\left (\frac {2}{c x^2+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])^3/x^3,x]

[Out]

(c*(a + b*ArcTanh[c*x^2])^3)/2 - (a + b*ArcTanh[c*x^2])^3/(2*x^2) + (3*b*c*(a + b*ArcTanh[c*x^2])^2*Log[2 - 2/
(1 + c*x^2)])/2 - (3*b^2*c*(a + b*ArcTanh[c*x^2])*PolyLog[2, -1 + 2/(1 + c*x^2)])/2 - (3*b^3*c*PolyLog[3, -1 +
2/(1 + c*x^2)])/4

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[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
]

Rule 6039

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[S
implify[(m + 1)/n]]

Rule 6079

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTanh[c*x
])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/
d))]/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 6095

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

Rule 6135

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]

Rule 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x] - Dist[b*(p/2), Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 - u]/
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2
/(1 + c*x))^2, 0]

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{x^3} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{8 x^3}+\frac {3 b \left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{8 x^3}-\frac {3 b^2 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{8 x^3}+\frac {b^3 \log ^3\left (1+c x^2\right )}{8 x^3}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{x^3} \, dx+\frac {1}{8} (3 b) \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{x^3} \, dx-\frac {1}{8} \left (3 b^2\right ) \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{x^3} \, dx+\frac {1}{8} b^3 \int \frac {\log ^3\left (1+c x^2\right )}{x^3} \, dx\\ &=\frac {1}{16} \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^3}{x^2} \, dx,x,x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{16} b^3 \text {Subst}\left (\int \frac {\log ^3(1+c x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{16} (3 b c) \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x} \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(1+c x)}{x} \, dx,x,x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (c x) (2 a-b \log (1-c x))}{1-c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^3 c^2\right ) \text {Subst}\left (\int \frac {\log (-c x) \log (1+c x)}{1+c x} \, dx,x,x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2 c\right ) \text {Subst}\left (\int \frac {(2 a-b \log (x)) \log \left (c \left (\frac {1}{c}-\frac {x}{c}\right )\right )}{x} \, dx,x,1-c x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log (x) \log \left (-c \left (-\frac {1}{c}+\frac {x}{c}\right )\right )}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}-\frac {3}{8} b^2 c \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (1-c x^2\right )+\frac {3}{8} b^3 c \log \left (1+c x^2\right ) \text {Li}_2\left (1+c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}-\frac {3}{8} b^2 c \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (1-c x^2\right )+\frac {3}{8} b^3 c \log \left (1+c x^2\right ) \text {Li}_2\left (1+c x^2\right )-\frac {3}{8} b^3 c \text {Li}_3\left (1-c x^2\right )-\frac {3}{8} b^3 c \text {Li}_3\left (1+c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.29, size = 222, normalized size = 1.78 \begin {gather*} \frac {1}{4} \left (-\frac {2 a^3}{x^2}-\frac {6 a^2 b \tanh ^{-1}\left (c x^2\right )}{x^2}+12 a^2 b c \log (x)-3 a^2 b c \log \left (1-c^2 x^4\right )+6 a b^2 c \left (\tanh ^{-1}\left (c x^2\right ) \left (\left (1-\frac {1}{c x^2}\right ) \tanh ^{-1}\left (c x^2\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )-\text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )+2 b^3 c \left (\frac {i \pi ^3}{8}-\tanh ^{-1}\left (c x^2\right )^3-\frac {\tanh ^{-1}\left (c x^2\right )^3}{c x^2}+3 \tanh ^{-1}\left (c x^2\right )^2 \log \left (1-e^{2 \tanh ^{-1}\left (c x^2\right )}\right )+3 \tanh ^{-1}\left (c x^2\right ) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}\left (c x^2\right )}\right )-\frac {3}{2} \text {PolyLog}\left (3,e^{2 \tanh ^{-1}\left (c x^2\right )}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])^3/x^3,x]

[Out]

((-2*a^3)/x^2 - (6*a^2*b*ArcTanh[c*x^2])/x^2 + 12*a^2*b*c*Log[x] - 3*a^2*b*c*Log[1 - c^2*x^4] + 6*a*b^2*c*(Arc
Tanh[c*x^2]*((1 - 1/(c*x^2))*ArcTanh[c*x^2] + 2*Log[1 - E^(-2*ArcTanh[c*x^2])]) - PolyLog[2, E^(-2*ArcTanh[c*x
^2])]) + 2*b^3*c*((I/8)*Pi^3 - ArcTanh[c*x^2]^3 - ArcTanh[c*x^2]^3/(c*x^2) + 3*ArcTanh[c*x^2]^2*Log[1 - E^(2*A
rcTanh[c*x^2])] + 3*ArcTanh[c*x^2]*PolyLog[2, E^(2*ArcTanh[c*x^2])] - (3*PolyLog[3, E^(2*ArcTanh[c*x^2])])/2))
/4

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctanh \left (c \,x^{2}\right )\right )^{3}}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))^3/x^3,x)

[Out]

int((a+b*arctanh(c*x^2))^3/x^3,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))^3/x^3,x, algorithm="maxima")

[Out]

-3/4*(c*(log(c^2*x^4 - 1) - log(x^4)) + 2*arctanh(c*x^2)/x^2)*a^2*b - 1/2*a^3/x^2 - 1/16*((b^3*c*x^2 - b^3)*lo
g(-c*x^2 + 1)^3 + 3*(2*a*b^2 + (b^3*c*x^2 + b^3)*log(c*x^2 + 1))*log(-c*x^2 + 1)^2)/x^2 - integrate(-1/8*((b^3
*c*x^2 - b^3)*log(c*x^2 + 1)^3 + 6*(a*b^2*c*x^2 - a*b^2)*log(c*x^2 + 1)^2 + 3*(4*a*b^2*c*x^2 - (b^3*c*x^2 - b^
3)*log(c*x^2 + 1)^2 + 2*(b^3*c^2*x^4 + 2*a*b^2 - (2*a*b^2*c - b^3*c)*x^2)*log(c*x^2 + 1))*log(-c*x^2 + 1))/(c*
x^5 - x^3), x)

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Fricas [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((a+b*arctanh(c*x^2))^3/x^3,x, algorithm="fricas")

[Out]

integral((b^3*arctanh(c*x^2)^3 + 3*a*b^2*arctanh(c*x^2)^2 + 3*a^2*b*arctanh(c*x^2) + a^3)/x^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**2))**3/x**3,x)

[Out]

Integral((a + b*atanh(c*x**2))**3/x**3, x)

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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((a+b*arctanh(c*x^2))^3/x^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^2) + a)^3/x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))^3/x^3,x)

[Out]

int((a + b*atanh(c*x^2))^3/x^3, x)

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