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

Optimal. Leaf size=84 \[ \frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c d}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5918, 5948, 6056, 6610} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c d}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 c d}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5948

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 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[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 6610

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

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Mathematica [A]  time = 0.21, size = 102, normalized size = 1.21 \[ \frac {2 a^2 \log (c x+1)+2 b \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right ) \left (a+b \tanh ^{-1}(c x)\right )-4 a b \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+b^2 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )-2 b^2 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )}{2 c d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b \operatorname {artanh}\left (c x\right ) + a^{2}}{c d x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.40, size = 822, normalized size = 9.79 \[ \frac {a^{2} \ln \left (c x +1\right )}{c d}+\frac {b^{2} \arctanh \left (c x \right )^{2} \ln \left (c x +1\right )}{c d}-\frac {2 b^{2} \arctanh \left (c x \right )^{2} \ln \left (\frac {c x +1}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d}+\frac {2 b^{2} \arctanh \left (c x \right )^{3}}{3 c d}-\frac {i b^{2} \pi \mathrm {csgn}\left (\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )^{2} \mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{c^{2} x^{2}-1}\right ) \arctanh \left (c x \right )^{2}}{2 c d}-\frac {i b^{2} \pi \mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{\left (c^{2} x^{2}-1\right ) \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}\right )^{3} \arctanh \left (c x \right )^{2}}{2 c d}+\frac {i b^{2} \pi \,\mathrm {csgn}\left (\frac {i}{1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{c^{2} x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{\left (c^{2} x^{2}-1\right ) \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}\right ) \arctanh \left (c x \right )^{2}}{2 c d}+\frac {i b^{2} \pi \,\mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{c^{2} x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{\left (c^{2} x^{2}-1\right ) \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}\right )^{2} \arctanh \left (c x \right )^{2}}{2 c d}-\frac {i b^{2} \pi \mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{c^{2} x^{2}-1}\right )^{3} \arctanh \left (c x \right )^{2}}{2 c d}-\frac {i b^{2} \pi \,\mathrm {csgn}\left (\frac {i}{1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{\left (c^{2} x^{2}-1\right ) \left (1+\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}\right )^{2} \arctanh \left (c x \right )^{2}}{2 c d}-\frac {i b^{2} \pi \,\mathrm {csgn}\left (\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (c x +1\right )^{2}}{c^{2} x^{2}-1}\right )^{2} \arctanh \left (c x \right )^{2}}{c d}-\frac {b^{2} \arctanh \left (c x \right )^{2} \ln \relax (2)}{c d}-\frac {b^{2} \arctanh \left (c x \right ) \polylog \left (2, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}{c d}+\frac {b^{2} \polylog \left (3, -\frac {\left (c x +1\right )^{2}}{-c^{2} x^{2}+1}\right )}{2 c d}+\frac {2 a b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{c d}-\frac {a b \ln \left (c x +1\right )^{2}}{2 c d}+\frac {a b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{c d}-\frac {a b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{c d}-\frac {a b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*a^2/d*ln(c*x+1)+1/c*b^2/d*arctanh(c*x)^2*ln(c*x+1)-2/c*b^2/d*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))
+2/3/c*b^2/d*arctanh(c*x)^3-1/2*I/c*b^2/d*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1)
)*arctanh(c*x)^2+1/2*I/c*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^
2*x^2+1)))^2*arctanh(c*x)^2-1/2*I/c*b^2/d*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(
1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2-1/2*I/c*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2-1/
2*I/c*b^2/d*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*arctanh(c*x)^2+1/2*I/c*b^2/d*Pi*csgn
(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x
^2+1)))*arctanh(c*x)^2-I/c*b^2/d*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*arctanh
(c*x)^2-1/c*b^2/d*arctanh(c*x)^2*ln(2)-1/c*b^2/d*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+1/2/c*b^2/d*p
olylog(3,-(c*x+1)^2/(-c^2*x^2+1))+2/c*a*b/d*arctanh(c*x)*ln(c*x+1)-1/2/c*a*b/d*ln(c*x+1)^2+1/c*a*b/d*ln(-1/2*c
*x+1/2)*ln(c*x+1)-1/c*a*b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-1/c*a*b/d*dilog(1/2+1/2*c*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b^{2} \log \left (c x + 1\right ) \log \left (-c x + 1\right )^{2}}{4 \, c d} + \frac {a^{2} \log \left (c d x + d\right )}{c d} - \int -\frac {{\left (b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c x - a b\right )} \log \left (c x + 1\right ) - 4 \, {\left (b^{2} c x \log \left (c x + 1\right ) + a b c x - a b\right )} \log \left (-c x + 1\right )}{4 \, {\left (c^{2} d x^{2} - d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*log(c*x + 1)*log(-c*x + 1)^2/(c*d) + a^2*log(c*d*x + d)/(c*d) - integrate(-1/4*((b^2*c*x - b^2)*log(c*
x + 1)^2 + 4*(a*b*c*x - a*b)*log(c*x + 1) - 4*(b^2*c*x*log(c*x + 1) + a*b*c*x - a*b)*log(-c*x + 1))/(c^2*d*x^2
- d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c*x + 1), x) + Integral(b**2*atanh(c*x)**2/(c*x + 1), x) + Integral(2*a*b*atanh(c*x)/(c*x + 1)
, x))/d

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