3.1.41 \(\int \frac {a+b \tanh ^{-1}(c \sqrt {x})}{x^2 (1-c^2 x)} \, dx\) [41]

Optimal. Leaf size=117 \[ -\frac {b c}{\sqrt {x}}+b c^2 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}+\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (2-\frac {2}{1+c \sqrt {x}}\right )-b c^2 \text {PolyLog}\left (2,-1+\frac {2}{1+c \sqrt {x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {46, 1607, 6129, 6037, 331, 212, 6135, 6079, 2497} \begin {gather*} \frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^2 \log \left (2-\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}-b c^2 \text {Li}_2\left (\frac {2}{\sqrt {x} c+1}-1\right )+b c^2 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {b c}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/(x^2*(1 - c^2*x)),x]

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

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

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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 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 6129

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

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x^2 \left (1-c^2 x\right )} \, dx &=2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3-c^2 x^5} \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx,x,\sqrt {x}\right )+\left (2 c^2\right ) \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )\\ &=-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}+\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+(b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )+\left (2 c^2\right ) \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx,x,\sqrt {x}\right )\\ &=-\frac {b c}{\sqrt {x}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}+\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (2-\frac {2}{1+c \sqrt {x}}\right )+\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-\left (2 b c^3\right ) \text {Subst}\left (\int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {b c}{\sqrt {x}}+b c^2 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}+\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (2-\frac {2}{1+c \sqrt {x}}\right )-b c^2 \text {Li}_2\left (-1+\frac {2}{1+c \sqrt {x}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 118, normalized size = 1.01 \begin {gather*} -\frac {a}{x}+2 a c^2 \log \left (\sqrt {x}\right )-a c^2 \log \left (1-c^2 x\right )-b c^2 \left (\frac {1}{c \sqrt {x}}-\tanh ^{-1}\left (c \sqrt {x}\right ) \left (-\frac {1-c^2 x}{c^2 x}+\tanh ^{-1}\left (c \sqrt {x}\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right )+\text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/(x^2*(1 - c^2*x)),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(277\) vs. \(2(105)=210\).
time = 0.26, size = 278, normalized size = 2.38

method	result	size
derivativedivides	\(-2 c^{2} \left (\frac {a}{2 c^{2} x}-a \ln \left (c \sqrt {x}\right )+\frac {a \ln \left (1+c \sqrt {x}\right )}{2}+\frac {a \ln \left (c \sqrt {x}-1\right )}{2}+\frac {b \arctanh \left (c \sqrt {x}\right )}{2 c^{2} x}-b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}\right )+\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {b}{2 c \sqrt {x}}-\frac {b \ln \left (1+c \sqrt {x}\right )}{4}+\frac {b \ln \left (c \sqrt {x}-1\right )}{4}+\frac {b \dilog \left (1+c \sqrt {x}\right )}{2}+\frac {b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b \dilog \left (c \sqrt {x}\right )}{2}+\frac {b \ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {b \dilog \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}-\frac {b \ln \left (1+c \sqrt {x}\right )^{2}}{8}-\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{4}\right )\)	\(278\)
default	\(-2 c^{2} \left (\frac {a}{2 c^{2} x}-a \ln \left (c \sqrt {x}\right )+\frac {a \ln \left (1+c \sqrt {x}\right )}{2}+\frac {a \ln \left (c \sqrt {x}-1\right )}{2}+\frac {b \arctanh \left (c \sqrt {x}\right )}{2 c^{2} x}-b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}\right )+\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {b}{2 c \sqrt {x}}-\frac {b \ln \left (1+c \sqrt {x}\right )}{4}+\frac {b \ln \left (c \sqrt {x}-1\right )}{4}+\frac {b \dilog \left (1+c \sqrt {x}\right )}{2}+\frac {b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b \dilog \left (c \sqrt {x}\right )}{2}+\frac {b \ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {b \dilog \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}-\frac {b \ln \left (1+c \sqrt {x}\right )^{2}}{8}-\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{4}\right )\)	\(278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))/x^2/(-c^2*x+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (102) = 204\).
time = 0.41, size = 248, normalized size = 2.12 \begin {gather*} -{\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right )\right )} b c^{2} - {\left (\log \left (c \sqrt {x}\right ) \log \left (-c \sqrt {x} + 1\right ) + {\rm Li}_2\left (-c \sqrt {x} + 1\right )\right )} b c^{2} + {\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x}\right ) + {\rm Li}_2\left (c \sqrt {x} + 1\right )\right )} b c^{2} + \frac {1}{2} \, b c^{2} \log \left (c \sqrt {x} + 1\right ) - \frac {1}{2} \, b c^{2} \log \left (c \sqrt {x} - 1\right ) - {\left (c^{2} \log \left (c \sqrt {x} + 1\right ) + c^{2} \log \left (c \sqrt {x} - 1\right ) - c^{2} \log \left (x\right ) + \frac {1}{x}\right )} a - \frac {b c^{2} x \log \left (c \sqrt {x} + 1\right )^{2} - b c^{2} x \log \left (-c \sqrt {x} + 1\right )^{2} + 4 \, b c \sqrt {x} + 2 \, b \log \left (c \sqrt {x} + 1\right ) - 2 \, {\left (b c^{2} x \log \left (c \sqrt {x} + 1\right ) + b\right )} \log \left (-c \sqrt {x} + 1\right )}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^2/(-c^2*x+1),x, algorithm="maxima")

[Out]

-(log(c*sqrt(x) + 1)*log(-1/2*c*sqrt(x) + 1/2) + dilog(1/2*c*sqrt(x) + 1/2))*b*c^2 - (log(c*sqrt(x))*log(-c*sq
rt(x) + 1) + dilog(-c*sqrt(x) + 1))*b*c^2 + (log(c*sqrt(x) + 1)*log(-c*sqrt(x)) + dilog(c*sqrt(x) + 1))*b*c^2
+ 1/2*b*c^2*log(c*sqrt(x) + 1) - 1/2*b*c^2*log(c*sqrt(x) - 1) - (c^2*log(c*sqrt(x) + 1) + c^2*log(c*sqrt(x) -
1) - c^2*log(x) + 1/x)*a - 1/4*(b*c^2*x*log(c*sqrt(x) + 1)^2 - b*c^2*x*log(-c*sqrt(x) + 1)^2 + 4*b*c*sqrt(x) +
2*b*log(c*sqrt(x) + 1) - 2*(b*c^2*x*log(c*sqrt(x) + 1) + b)*log(-c*sqrt(x) + 1))/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^(1/2)))/x^2/(-c^2*x+1),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(1/2)))/x**2/(-c**2*x+1),x)

[Out]

-Integral(a/(c**2*x**3 - x**2), x) - Integral(b*atanh(c*sqrt(x))/(c**2*x**3 - x**2), 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^(1/2)))/x^2/(-c^2*x+1),x, algorithm="giac")

[Out]

integrate(-(b*arctanh(c*sqrt(x)) + a)/((c^2*x - 1)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a + b*atanh(c*x^(1/2)))/(x^2*(c^2*x - 1)),x)

[Out]

-int((a + b*atanh(c*x^(1/2)))/(x^2*(c^2*x - 1)), x)

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