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

Optimal. Leaf size=104 \[ -\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2}+\frac {2 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2}-\frac {b^2 \text {Li}_2\left (\frac {2}{c+d x+1}-1\right )}{d e^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6107, 12, 5916, 5988, 5932, 2447} \[ -\frac {b^2 \text {PolyLog}\left (2,\frac {2}{c+d x+1}-1\right )}{d e^2}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2}+\frac {2 b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 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 5988

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

Rule 6107

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

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 126, normalized size = 1.21 \[ \frac {a \left (2 b (c+d x) \log \left (\frac {c+d x}{\sqrt {1-(c+d x)^2}}\right )-a\right )+2 b \tanh ^{-1}(c+d x) \left (b (c+d x) \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )-a\right )-b^2 (c+d x) \text {Li}_2\left (e^{-2 \tanh ^{-1}(c+d x)}\right )+b^2 (c+d x-1) \tanh ^{-1}(c+d x)^2}{d e^2 (c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {artanh}\left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 396, normalized size = 3.81 \[ -\frac {a^{2}}{d \,e^{2} \left (d x +c \right )}-\frac {b^{2} \arctanh \left (d x +c \right )^{2}}{d \,e^{2} \left (d x +c \right )}+\frac {2 b^{2} \ln \left (d x +c \right ) \arctanh \left (d x +c \right )}{d \,e^{2}}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{d \,e^{2}}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{d \,e^{2}}-\frac {b^{2} \ln \left (d x +c -1\right )^{2}}{4 d \,e^{2}}+\frac {b^{2} \dilog \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d \,e^{2}}+\frac {b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,e^{2}}+\frac {b^{2} \ln \left (d x +c +1\right )^{2}}{4 d \,e^{2}}-\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{2 d \,e^{2}}+\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,e^{2}}-\frac {b^{2} \dilog \left (d x +c \right )}{d \,e^{2}}-\frac {b^{2} \dilog \left (d x +c +1\right )}{d \,e^{2}}-\frac {b^{2} \ln \left (d x +c \right ) \ln \left (d x +c +1\right )}{d \,e^{2}}-\frac {2 a b \arctanh \left (d x +c \right )}{d \,e^{2} \left (d x +c \right )}+\frac {2 a b \ln \left (d x +c \right )}{d \,e^{2}}-\frac {a b \ln \left (d x +c -1\right )}{d \,e^{2}}-\frac {a b \ln \left (d x +c +1\right )}{d \,e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*(log(d*x + c + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2*e^2) + log(d*x + c - 1)/(d^2*e^2)) + 2*arctanh(d*x + c)/
(d^2*e^2*x + c*d*e^2))*a*b - 1/4*b^2*(log(-d*x - c + 1)^2/(d^2*e^2*x + c*d*e^2) + integrate(-((d*x + c - 1)*lo
g(d*x + c + 1)^2 + 2*(d*x - (d*x + c - 1)*log(d*x + c + 1) + c)*log(-d*x - c + 1))/(d^3*e^2*x^3 + c^3*e^2 - c^
2*e^2 + (3*c*d^2*e^2 - d^2*e^2)*x^2 + (3*c^2*d*e^2 - 2*c*d*e^2)*x), x)) - a^2/(d^2*e^2*x + c*d*e^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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