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

Optimal. Leaf size=120 \[ -\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b c^4}+\frac {2 \log \left (\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^4}-\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^2}+\frac {b \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )}{c^4}+\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^4}-\frac {b \sqrt {x}}{c^3} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {43, 5980, 5916, 321, 206, 5984, 5918, 2402, 2315} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )}{c^4}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b c^4}-\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^2}+\frac {2 \log \left (\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^4}-\frac {b \sqrt {x}}{c^3}+\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[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 5980

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

Rule 5984

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

Rubi steps

\begin {align*} \int \frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{1-c^2 x} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {2 \operatorname {Subst}\left (\int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx,x,\sqrt {x}\right )}{c^2}+\frac {2 \operatorname {Subst}\left (\int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{c^2}\\ &=-\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^2}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b c^4}+\frac {2 \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx,x,\sqrt {x}\right )}{c^3}+\frac {b \operatorname {Subst}\left (\int \frac {x^2}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{c}\\ &=-\frac {b \sqrt {x}}{c^3}-\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^2}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b c^4}+\frac {2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1-c \sqrt {x}}\right )}{c^4}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{c^3}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{c^3}\\ &=-\frac {b \sqrt {x}}{c^3}+\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^4}-\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^2}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b c^4}+\frac {2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1-c \sqrt {x}}\right )}{c^4}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c \sqrt {x}}\right )}{c^4}\\ &=-\frac {b \sqrt {x}}{c^3}+\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^4}-\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^2}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b c^4}+\frac {2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1-c \sqrt {x}}\right )}{c^4}+\frac {b \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )}{c^4}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 96, normalized size = 0.80 \[ -\frac {a c^2 x+a \log \left (1-c^2 x\right )+b \tanh ^{-1}\left (c \sqrt {x}\right ) \left (c^2 x-2 \log \left (e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}+1\right )-1\right )+b \text {Li}_2\left (-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )+b c \sqrt {x}-b \tanh ^{-1}\left (c \sqrt {x}\right )^2}{c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x \operatorname {artanh}\left (c \sqrt {x}\right ) + a x}{c^{2} x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 243, normalized size = 2.02 \[ -\frac {x a}{c^{2}}-\frac {a \ln \left (c \sqrt {x}-1\right )}{c^{4}}-\frac {a \ln \left (1+c \sqrt {x}\right )}{c^{4}}-\frac {b \arctanh \left (c \sqrt {x}\right ) x}{c^{2}}-\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{c^{4}}-\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{c^{4}}-\frac {b \sqrt {x}}{c^{3}}-\frac {b \ln \left (c \sqrt {x}-1\right )}{2 c^{4}}+\frac {b \ln \left (1+c \sqrt {x}\right )}{2 c^{4}}-\frac {b \ln \left (c \sqrt {x}-1\right )^{2}}{4 c^{4}}+\frac {b \dilog \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{c^{4}}+\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{2 c^{4}}+\frac {b \ln \left (1+c \sqrt {x}\right )^{2}}{4 c^{4}}-\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{2 c^{4}}+\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{2 c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 166, normalized size = 1.38 \[ -a {\left (\frac {x}{c^{2}} + \frac {\log \left (c^{2} x - 1\right )}{c^{4}}\right )} - \frac {{\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right )\right )} b}{c^{4}} + \frac {b \log \left (c \sqrt {x} + 1\right )}{2 \, c^{4}} - \frac {b \log \left (c \sqrt {x} - 1\right )}{2 \, c^{4}} - \frac {2 \, b c^{2} x \log \left (c \sqrt {x} + 1\right ) + b \log \left (c \sqrt {x} + 1\right )^{2} - b \log \left (-c \sqrt {x} + 1\right )^{2} + 4 \, b c \sqrt {x} - 2 \, {\left (b c^{2} x + b \log \left (c \sqrt {x} + 1\right )\right )} \log \left (-c \sqrt {x} + 1\right )}{4 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(x/c^2 + log(c^2*x - 1)/c^4) - (log(c*sqrt(x) + 1)*log(-1/2*c*sqrt(x) + 1/2) + dilog(1/2*c*sqrt(x) + 1/2))*
b/c^4 + 1/2*b*log(c*sqrt(x) + 1)/c^4 - 1/2*b*log(c*sqrt(x) - 1)/c^4 - 1/4*(2*b*c^2*x*log(c*sqrt(x) + 1) + b*lo
g(c*sqrt(x) + 1)^2 - b*log(-c*sqrt(x) + 1)^2 + 4*b*c*sqrt(x) - 2*(b*c^2*x + b*log(c*sqrt(x) + 1))*log(-c*sqrt(
x) + 1))/c^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a*x/(c**2*x - 1), x) - Integral(b*x*atanh(c*sqrt(x))/(c**2*x - 1), x)

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