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

Optimal. Leaf size=149 \[ -\frac {a+b \tanh ^{-1}(c x)}{c^3 d^2 (c x+1)}+\frac {2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2}+\frac {a x}{c^2 d^2}-\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{c^3 d^2}-\frac {b}{2 c^3 d^2 (c x+1)}+\frac {b \tanh ^{-1}(c x)}{2 c^3 d^2}+\frac {b x \tanh ^{-1}(c x)}{c^2 d^2}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5940, 5910, 260, 5926, 627, 44, 207, 5918, 2402, 2315} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{c^3 d^2}-\frac {a+b \tanh ^{-1}(c x)}{c^3 d^2 (c x+1)}+\frac {2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^2}+\frac {a x}{c^2 d^2}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^3 d^2}-\frac {b}{2 c^3 d^2 (c x+1)}+\frac {b x \tanh ^{-1}(c x)}{c^2 d^2}+\frac {b \tanh ^{-1}(c x)}{2 c^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

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 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 5926

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

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

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

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Mathematica [A]  time = 0.58, size = 121, normalized size = 0.81 \[ \frac {4 a c x-\frac {4 a}{c x+1}-8 a \log (c x+1)+b \left (2 \log \left (1-c^2 x^2\right )-4 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+\sinh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x) \left (2 c x+4 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+\sinh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (2 \tanh ^{-1}(c x)\right )\right )\right )}{4 c^3 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*a*c*x - (4*a)/(1 + c*x) - 8*a*Log[1 + c*x] + b*(-Cosh[2*ArcTanh[c*x]] + 2*Log[1 - c^2*x^2] - 4*PolyLog[2, -
E^(-2*ArcTanh[c*x])] + Sinh[2*ArcTanh[c*x]] + 2*ArcTanh[c*x]*(2*c*x - Cosh[2*ArcTanh[c*x]] + 4*Log[1 + E^(-2*A
rcTanh[c*x])] + Sinh[2*ArcTanh[c*x]])))/(4*c^3*d^2)

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} \operatorname {artanh}\left (c x\right ) + a x^{2}}{c^{2} d^{2} x^{2} + 2 \, c d^{2} x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 216, normalized size = 1.45 \[ \frac {a x}{c^{2} d^{2}}-\frac {a}{c^{3} d^{2} \left (c x +1\right )}-\frac {2 a \ln \left (c x +1\right )}{c^{3} d^{2}}+\frac {b x \arctanh \left (c x \right )}{c^{2} d^{2}}-\frac {b \arctanh \left (c x \right )}{c^{3} d^{2} \left (c x +1\right )}-\frac {2 b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{c^{3} d^{2}}+\frac {b \ln \left (c x +1\right )^{2}}{2 c^{3} d^{2}}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{c^{3} d^{2}}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{c^{3} d^{2}}+\frac {b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{c^{3} d^{2}}-\frac {b}{2 c^{3} d^{2} \left (c x +1\right )}+\frac {3 b \ln \left (c x +1\right )}{4 c^{3} d^{2}}+\frac {b \ln \left (c x -1\right )}{4 c^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{8} \, {\left (c^{3} {\left (\frac {2}{c^{7} d^{2} x + c^{6} d^{2}} - \frac {4 \, x}{c^{5} d^{2}} + \frac {5 \, \log \left (c x + 1\right )}{c^{6} d^{2}} - \frac {\log \left (c x - 1\right )}{c^{6} d^{2}}\right )} - 4 \, c^{3} \int \frac {x^{3} \log \left (c x + 1\right )}{c^{5} d^{2} x^{3} + c^{4} d^{2} x^{2} - c^{3} d^{2} x - c^{2} d^{2}}\,{d x} - 2 \, c^{2} {\left (\frac {2}{c^{6} d^{2} x + c^{5} d^{2}} + \frac {3 \, \log \left (c x + 1\right )}{c^{5} d^{2}} + \frac {\log \left (c x - 1\right )}{c^{5} d^{2}}\right )} + 12 \, c^{2} \int \frac {x^{2} \log \left (c x + 1\right )}{c^{5} d^{2} x^{3} + c^{4} d^{2} x^{2} - c^{3} d^{2} x - c^{2} d^{2}}\,{d x} + 16 \, c \int \frac {x \log \left (c x + 1\right )}{c^{5} d^{2} x^{3} + c^{4} d^{2} x^{2} - c^{3} d^{2} x - c^{2} d^{2}}\,{d x} + \frac {4 \, {\left (c^{2} x^{2} + c x - 2 \, {\left (c x + 1\right )} \log \left (c x + 1\right ) - 1\right )} \log \left (-c x + 1\right )}{c^{4} d^{2} x + c^{3} d^{2}} + \frac {2}{c^{4} d^{2} x + c^{3} d^{2}} - \frac {\log \left (c x + 1\right )}{c^{3} d^{2}} + \frac {\log \left (c x - 1\right )}{c^{3} d^{2}} + 8 \, \int \frac {\log \left (c x + 1\right )}{c^{5} d^{2} x^{3} + c^{4} d^{2} x^{2} - c^{3} d^{2} x - c^{2} d^{2}}\,{d x}\right )} b - a {\left (\frac {1}{c^{4} d^{2} x + c^{3} d^{2}} - \frac {x}{c^{2} d^{2}} + \frac {2 \, \log \left (c x + 1\right )}{c^{3} d^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c^3*(2/(c^7*d^2*x + c^6*d^2) - 4*x/(c^5*d^2) + 5*log(c*x + 1)/(c^6*d^2) - log(c*x - 1)/(c^6*d^2)) - 4*c^
3*integrate(x^3*log(c*x + 1)/(c^5*d^2*x^3 + c^4*d^2*x^2 - c^3*d^2*x - c^2*d^2), x) - 2*c^2*(2/(c^6*d^2*x + c^5
*d^2) + 3*log(c*x + 1)/(c^5*d^2) + log(c*x - 1)/(c^5*d^2)) + 12*c^2*integrate(x^2*log(c*x + 1)/(c^5*d^2*x^3 +
c^4*d^2*x^2 - c^3*d^2*x - c^2*d^2), x) + 16*c*integrate(x*log(c*x + 1)/(c^5*d^2*x^3 + c^4*d^2*x^2 - c^3*d^2*x
- c^2*d^2), x) + 4*(c^2*x^2 + c*x - 2*(c*x + 1)*log(c*x + 1) - 1)*log(-c*x + 1)/(c^4*d^2*x + c^3*d^2) + 2/(c^4
*d^2*x + c^3*d^2) - log(c*x + 1)/(c^3*d^2) + log(c*x - 1)/(c^3*d^2) + 8*integrate(log(c*x + 1)/(c^5*d^2*x^3 +
c^4*d^2*x^2 - c^3*d^2*x - c^2*d^2), x))*b - a*(1/(c^4*d^2*x + c^3*d^2) - x/(c^2*d^2) + 2*log(c*x + 1)/(c^3*d^2
))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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