3.14 \(\int x^4 (a+b \tanh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=162 \[ -\frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{5 c^5}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}-\frac{2 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac{b^2 x^3}{30 c^2}+\frac{3 b^2 x}{10 c^4}-\frac{3 b^2 \tanh ^{-1}(c x)}{10 c^5} \]

[Out]

(3*b^2*x)/(10*c^4) + (b^2*x^3)/(30*c^2) - (3*b^2*ArcTanh[c*x])/(10*c^5) + (b*x^2*(a + b*ArcTanh[c*x]))/(5*c^3)
+ (b*x^4*(a + b*ArcTanh[c*x]))/(10*c) + (a + b*ArcTanh[c*x])^2/(5*c^5) + (x^5*(a + b*ArcTanh[c*x])^2)/5 - (2*
b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(5*c^5) - (b^2*PolyLog[2, 1 - 2/(1 - c*x)])/(5*c^5)

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Rubi [A]  time = 0.302228, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {5916, 5980, 302, 206, 321, 5984, 5918, 2402, 2315} \[ -\frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{5 c^5}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}-\frac{2 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac{b^2 x^3}{30 c^2}+\frac{3 b^2 x}{10 c^4}-\frac{3 b^2 \tanh ^{-1}(c x)}{10 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTanh[c*x])^2,x]

[Out]

(3*b^2*x)/(10*c^4) + (b^2*x^3)/(30*c^2) - (3*b^2*ArcTanh[c*x])/(10*c^5) + (b*x^2*(a + b*ArcTanh[c*x]))/(5*c^3)
+ (b*x^4*(a + b*ArcTanh[c*x]))/(10*c) + (a + b*ArcTanh[c*x])^2/(5*c^5) + (x^5*(a + b*ArcTanh[c*x])^2)/5 - (2*
b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(5*c^5) - (b^2*PolyLog[2, 1 - 2/(1 - c*x)])/(5*c^5)

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 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 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]

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 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 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]

Rubi steps

\begin{align*} \int x^4 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{5} (2 b c) \int \frac{x^5 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(2 b) \int x^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c}-\frac{(2 b) \int \frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c}\\ &=\frac{b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{10} b^2 \int \frac{x^4}{1-c^2 x^2} \, dx+\frac{(2 b) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^3}-\frac{(2 b) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^3}\\ &=\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac{b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{10} b^2 \int \left (-\frac{1}{c^4}-\frac{x^2}{c^2}+\frac{1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx-\frac{(2 b) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{5 c^4}-\frac{b^2 \int \frac{x^2}{1-c^2 x^2} \, dx}{5 c^2}\\ &=\frac{3 b^2 x}{10 c^4}+\frac{b^2 x^3}{30 c^2}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac{b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^5}-\frac{b^2 \int \frac{1}{1-c^2 x^2} \, dx}{10 c^4}-\frac{b^2 \int \frac{1}{1-c^2 x^2} \, dx}{5 c^4}+\frac{\left (2 b^2\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^4}\\ &=\frac{3 b^2 x}{10 c^4}+\frac{b^2 x^3}{30 c^2}-\frac{3 b^2 \tanh ^{-1}(c x)}{10 c^5}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac{b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^5}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{5 c^5}\\ &=\frac{3 b^2 x}{10 c^4}+\frac{b^2 x^3}{30 c^2}-\frac{3 b^2 \tanh ^{-1}(c x)}{10 c^5}+\frac{b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^3}+\frac{b x^4 \left (a+b \tanh ^{-1}(c x)\right )}{10 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{5 c^5}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{5 c^5}\\ \end{align*}

Mathematica [A]  time = 0.451515, size = 161, normalized size = 0.99 \[ \frac{6 b^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+6 a^2 c^5 x^5+3 a b c^4 x^4+6 a b c^2 x^2+6 a b \log \left (c^2 x^2-1\right )+3 b \tanh ^{-1}(c x) \left (4 a c^5 x^5+b \left (c^4 x^4+2 c^2 x^2-3\right )-4 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )-9 a b+b^2 c^3 x^3+6 b^2 \left (c^5 x^5-1\right ) \tanh ^{-1}(c x)^2+9 b^2 c x}{30 c^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4*(a + b*ArcTanh[c*x])^2,x]

[Out]

(-9*a*b + 9*b^2*c*x + 6*a*b*c^2*x^2 + b^2*c^3*x^3 + 3*a*b*c^4*x^4 + 6*a^2*c^5*x^5 + 6*b^2*(-1 + c^5*x^5)*ArcTa
nh[c*x]^2 + 3*b*ArcTanh[c*x]*(4*a*c^5*x^5 + b*(-3 + 2*c^2*x^2 + c^4*x^4) - 4*b*Log[1 + E^(-2*ArcTanh[c*x])]) +
6*a*b*Log[-1 + c^2*x^2] + 6*b^2*PolyLog[2, -E^(-2*ArcTanh[c*x])])/(30*c^5)

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Maple [B]  time = 0.016, size = 306, normalized size = 1.9 \begin{align*}{\frac{{x}^{5}{a}^{2}}{5}}+{\frac{{x}^{5}{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{5}}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ){x}^{4}}{10\,c}}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ){x}^{2}}{5\,{c}^{3}}}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{5\,{c}^{5}}}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{5\,{c}^{5}}}+{\frac{{b}^{2}{x}^{3}}{30\,{c}^{2}}}+{\frac{3\,{b}^{2}x}{10\,{c}^{4}}}+{\frac{3\,{b}^{2}\ln \left ( cx-1 \right ) }{20\,{c}^{5}}}-{\frac{3\,{b}^{2}\ln \left ( cx+1 \right ) }{20\,{c}^{5}}}+{\frac{{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{20\,{c}^{5}}}-{\frac{{b}^{2}}{5\,{c}^{5}}{\it dilog} \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{b}^{2}\ln \left ( cx-1 \right ) }{10\,{c}^{5}}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{b}^{2}\ln \left ( cx+1 \right ) }{10\,{c}^{5}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }-{\frac{{b}^{2}}{10\,{c}^{5}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{20\,{c}^{5}}}+{\frac{2\,{x}^{5}ab{\it Artanh} \left ( cx \right ) }{5}}+{\frac{{x}^{4}ab}{10\,c}}+{\frac{ab{x}^{2}}{5\,{c}^{3}}}+{\frac{ab\ln \left ( cx-1 \right ) }{5\,{c}^{5}}}+{\frac{ab\ln \left ( cx+1 \right ) }{5\,{c}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c*x))^2,x)

[Out]

1/5*x^5*a^2+1/5*x^5*b^2*arctanh(c*x)^2+1/10/c*b^2*arctanh(c*x)*x^4+1/5/c^3*b^2*arctanh(c*x)*x^2+1/5/c^5*b^2*ar
ctanh(c*x)*ln(c*x-1)+1/5/c^5*b^2*arctanh(c*x)*ln(c*x+1)+1/30*b^2*x^3/c^2+3/10*b^2*x/c^4+3/20/c^5*b^2*ln(c*x-1)
-3/20/c^5*b^2*ln(c*x+1)+1/20/c^5*b^2*ln(c*x-1)^2-1/5/c^5*b^2*dilog(1/2+1/2*c*x)-1/10/c^5*b^2*ln(c*x-1)*ln(1/2+
1/2*c*x)+1/10/c^5*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/10/c^5*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-1/20/c^5*b^2*ln
(c*x+1)^2+2/5*x^5*a*b*arctanh(c*x)+1/10/c*x^4*a*b+1/5*a*b*x^2/c^3+1/5/c^5*a*b*ln(c*x-1)+1/5/c^5*a*b*ln(c*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^2,x, algorithm="maxima")

[Out]

1/5*a^2*x^5 + 1/10*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*a*b - 1/36000*(24
*c^6*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^10 - 15*log(c*x + 1)/c^11 + 15*log(c*x - 1)/c^11) - 45*c^5*((c^2*x^4
+ 2*x^2)/c^8 + 2*log(c^2*x^2 - 1)/c^10) - 1080000*c^5*integrate(1/150*x^5*log(c*x + 1)/(c^6*x^2 - c^4), x) + 5
0*c^4*(2*(c^2*x^3 + 3*x)/c^8 - 3*log(c*x + 1)/c^9 + 3*log(c*x - 1)/c^9) - 300*c^3*(x^2/c^6 + log(c^2*x^2 - 1)/
c^8) + 900*c^2*(2*x/c^6 - log(c*x + 1)/c^7 + log(c*x - 1)/c^7) - 540000*c*integrate(1/150*x*log(c*x + 1)/(c^6*
x^2 - c^4), x) - 60*(30*c^5*x^5*log(c*x + 1)^2 + (12*c^5*x^5 - 15*c^4*x^4 + 20*c^3*x^3 - 30*c^2*x^2 + 60*c*x -
60*(c^5*x^5 + 1)*log(c*x + 1))*log(-c*x + 1))/c^5 - (72*(c*x - 1)^5*(25*log(-c*x + 1)^2 - 10*log(-c*x + 1) +
2) + 1125*(c*x - 1)^4*(8*log(-c*x + 1)^2 - 4*log(-c*x + 1) + 1) + 2000*(c*x - 1)^3*(9*log(-c*x + 1)^2 - 6*log(
-c*x + 1) + 2) + 9000*(c*x - 1)^2*(2*log(-c*x + 1)^2 - 2*log(-c*x + 1) + 1) + 9000*(c*x - 1)*(log(-c*x + 1)^2
- 2*log(-c*x + 1) + 2))/c^5 + 1800*log(150*c^6*x^2 - 150*c^4)/c^5 - 540000*integrate(1/150*log(c*x + 1)/(c^6*x
^2 - c^4), x))*b^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{2} x^{4} \operatorname{artanh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname{artanh}\left (c x\right ) + a^{2} x^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^4*arctanh(c*x)^2 + 2*a*b*x^4*arctanh(c*x) + a^2*x^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atanh(c*x))**2,x)

[Out]

Integral(x**4*(a + b*atanh(c*x))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^2,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2*x^4, x)