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

Optimal. Leaf size=117 \[ \frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^2 \log \left (2-\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}-b c^2 \text {Li}_2\left (\frac {2}{\sqrt {x} c+1}-1\right )+b c^2 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {b c}{\sqrt {x}} \]

[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 = 0.36, 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 = {44, 1593, 5982, 5916, 325, 206, 5988, 5932, 2447} \[ -b c^2 \text {PolyLog}\left (2,\frac {2}{c \sqrt {x}+1}-1\right )+\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^2 \log \left (2-\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}+b c^2 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {b c}{\sqrt {x}} \]

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 44

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] && L
tQ[m + n + 2, 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 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 1593

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 2447

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 5916

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

Rule 5932

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 5982

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 5988

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 \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3-c^2 x^5} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {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 \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx,x,\sqrt {x}\right )+\left (2 c^2\right ) \operatorname {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) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )+\left (2 c^2\right ) \operatorname {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 ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-\left (2 b c^3\right ) \operatorname {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 = 0.34, size = 118, normalized size = 1.01 \[ 2 a c^2 \log \left (\sqrt {x}\right )-a c^2 \log \left (1-c^2 x\right )-\frac {a}{x}-b c^2 \left (-\tanh ^{-1}\left (c \sqrt {x}\right ) \left (-\frac {1-c^2 x}{c^2 x}+\tanh ^{-1}\left (c \sqrt {x}\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right )+\text {Li}_2\left (e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )+\frac {1}{c \sqrt {x}}\right ) \]

Warning: Unable to verify antiderivative.

[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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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \operatorname {artanh}\left (c \sqrt {x}\right ) + a}{c^{2} x^{3} - x^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \operatorname {artanh}\left (c \sqrt {x}\right ) + a}{{\left (c^{2} x - 1\right )} x^{2}}\,{d x} \]

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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maple [B]  time = 0.07, size = 315, normalized size = 2.69 \[ -\frac {a}{x}+2 c^{2} a \ln \left (c \sqrt {x}\right )-c^{2} a \ln \left (c \sqrt {x}-1\right )-c^{2} a \ln \left (1+c \sqrt {x}\right )-\frac {b \arctanh \left (c \sqrt {x}\right )}{x}+2 c^{2} b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}\right )-c^{2} b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )-c^{2} b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )-\frac {b c}{\sqrt {x}}-\frac {c^{2} b \ln \left (c \sqrt {x}-1\right )}{2}+\frac {c^{2} b \ln \left (1+c \sqrt {x}\right )}{2}-c^{2} b \dilog \left (c \sqrt {x}\right )-c^{2} b \dilog \left (1+c \sqrt {x}\right )-c^{2} b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )-\frac {c^{2} b \ln \left (c \sqrt {x}-1\right )^{2}}{4}+c^{2} b \dilog \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )+\frac {c^{2} b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{2}+\frac {c^{2} b \ln \left (1+c \sqrt {x}\right )^{2}}{4}-\frac {c^{2} b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {c^{2} b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{2} \]

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)

[Out]

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

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maxima [B]  time = 0.57, size = 248, normalized size = 2.12 \[ -{\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right )\right )} b c^{2} - {\left (\log \left (c \sqrt {x}\right ) \log \left (-c \sqrt {x} + 1\right ) + {\rm Li}_2\left (-c \sqrt {x} + 1\right )\right )} b c^{2} + {\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x}\right ) + {\rm Li}_2\left (c \sqrt {x} + 1\right )\right )} b c^{2} + \frac {1}{2} \, b c^{2} \log \left (c \sqrt {x} + 1\right ) - \frac {1}{2} \, b c^{2} \log \left (c \sqrt {x} - 1\right ) - {\left (c^{2} \log \left (c \sqrt {x} + 1\right ) + c^{2} \log \left (c \sqrt {x} - 1\right ) - c^{2} \log \relax (x) + \frac {1}{x}\right )} a - \frac {b c^{2} x \log \left (c \sqrt {x} + 1\right )^{2} - b c^{2} x \log \left (-c \sqrt {x} + 1\right )^{2} + 4 \, b c \sqrt {x} + 2 \, b \log \left (c \sqrt {x} + 1\right ) - 2 \, {\left (b c^{2} x \log \left (c \sqrt {x} + 1\right ) + b\right )} \log \left (-c \sqrt {x} + 1\right )}{4 \, x} \]

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

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

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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