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

Optimal. Leaf size=145 \[ -\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c d}-\frac {a x}{c^2 d}+\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 c^3 d}-\frac {b \tanh ^{-1}(c x)}{2 c^3 d}+\frac {b x}{2 c^2 d}-\frac {b x \tanh ^{-1}(c x)}{c^2 d}-\frac {b \log \left (1-c^2 x^2\right )}{2 c^3 d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {5930, 5916, 321, 206, 5910, 260, 5918, 2402, 2315} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c^3 d}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c d}-\frac {a x}{c^2 d}-\frac {b \log \left (1-c^2 x^2\right )}{2 c^3 d}+\frac {b x}{2 c^2 d}-\frac {b x \tanh ^{-1}(c x)}{c^2 d}-\frac {b \tanh ^{-1}(c x)}{2 c^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 5930

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

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 97, normalized size = 0.67 \[ \frac {a c^2 x^2-2 a c x+2 a \log (c x+1)-b \log \left (1-c^2 x^2\right )+b \tanh ^{-1}(c x) \left (c^2 x^2-2 c x-2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-1\right )+b \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+b c x}{2 c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} \operatorname {artanh}\left (c x\right ) + a x^{2}}{c d x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 213, normalized size = 1.47 \[ \frac {a \,x^{2}}{2 c d}-\frac {a x}{c^{2} d}+\frac {a \ln \left (c x +1\right )}{c^{3} d}+\frac {b \arctanh \left (c x \right ) x^{2}}{2 c d}-\frac {b x \arctanh \left (c x \right )}{c^{2} d}+\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{c^{3} d}-\frac {b \ln \left (c x +1\right )^{2}}{4 c^{3} d}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 c^{3} d}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{3} d}-\frac {b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{3} d}+\frac {b x}{2 c^{2} d}+\frac {b}{2 c^{3} d}-\frac {3 b \ln \left (c x +1\right )}{4 c^{3} d}-\frac {b \ln \left (c x -1\right )}{4 c^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/c*a/d*x^2-a*x/c^2/d+1/c^3*a/d*ln(c*x+1)+1/2/c*b/d*arctanh(c*x)*x^2-b*x*arctanh(c*x)/c^2/d+1/c^3*b/d*arctan
h(c*x)*ln(c*x+1)-1/4/c^3*b/d*ln(c*x+1)^2+1/2/c^3*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/2/c^3*b/d*ln(-1/2*c*x+1/2)*l
n(1/2+1/2*c*x)-1/2/c^3*b/d*dilog(1/2+1/2*c*x)+1/2*b*x/c^2/d+1/2/c^3*b/d-3/4/c^3*b/d*ln(c*x+1)-1/4/c^3*b/d*ln(c
*x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, {\left (c^{3} {\left (\frac {x^{2}}{c^{4} d} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{6} d}\right )} + 8 \, c^{3} \int \frac {x^{3} \log \left (c x + 1\right )}{2 \, {\left (c^{4} d x^{2} - c^{2} d\right )}}\,{d x} - c^{2} {\left (\frac {2 \, x}{c^{4} d} - \frac {\log \left (c x + 1\right )}{c^{5} d} + \frac {\log \left (c x - 1\right )}{c^{5} d}\right )} - 8 \, c^{2} \int \frac {x^{2} \log \left (c x + 1\right )}{2 \, {\left (c^{4} d x^{2} - c^{2} d\right )}}\,{d x} + 8 \, c \int \frac {x \log \left (c x + 1\right )}{2 \, {\left (c^{4} d x^{2} - c^{2} d\right )}}\,{d x} - \frac {2 \, {\left (c^{2} x^{2} - 2 \, c x + 2 \, \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{c^{3} d} - \frac {2 \, \log \left (2 \, c^{4} d x^{2} - 2 \, c^{2} d\right )}{c^{3} d} + 8 \, \int \frac {\log \left (c x + 1\right )}{2 \, {\left (c^{4} d x^{2} - c^{2} d\right )}}\,{d x}\right )} b + \frac {1}{2} \, a {\left (\frac {c x^{2} - 2 \, x}{c^{2} d} + \frac {2 \, \log \left (c x + 1\right )}{c^{3} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(c^3*(x^2/(c^4*d) + log(c^2*x^2 - 1)/(c^6*d)) + 8*c^3*integrate(1/2*x^3*log(c*x + 1)/(c^4*d*x^2 - c^2*d),
x) - c^2*(2*x/(c^4*d) - log(c*x + 1)/(c^5*d) + log(c*x - 1)/(c^5*d)) - 8*c^2*integrate(1/2*x^2*log(c*x + 1)/(c
^4*d*x^2 - c^2*d), x) + 8*c*integrate(1/2*x*log(c*x + 1)/(c^4*d*x^2 - c^2*d), x) - 2*(c^2*x^2 - 2*c*x + 2*log(
c*x + 1))*log(-c*x + 1)/(c^3*d) - 2*log(2*c^4*d*x^2 - 2*c^2*d)/(c^3*d) + 8*integrate(1/2*log(c*x + 1)/(c^4*d*x
^2 - c^2*d), x))*b + 1/2*a*((c*x^2 - 2*x)/(c^2*d) + 2*log(c*x + 1)/(c^3*d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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