3.1.19 \(\int \frac {(a+b \tanh ^{-1}(c+d x))^2}{(c e+d e x)^2} \, dx\) [19]

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

[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 = 0.13, 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 = {6242, 12, 6037, 6135, 6079, 2497} \begin {gather*} -\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} \end {gather*}

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 2497

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 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6079

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 6135

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 6242

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 {\text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {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) \text {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) \text {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 ) \text {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 = 0.17, size = 126, normalized size = 1.21 \begin {gather*} \frac {b^2 (-1+c+d x) \tanh ^{-1}(c+d x)^2+2 b \tanh ^{-1}(c+d x) \left (-a+b (c+d x) \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+a \left (-a+2 b (c+d x) \log \left (\frac {c+d x}{\sqrt {1-(c+d x)^2}}\right )\right )-b^2 (c+d x) \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c+d x)}\right )}{d e^2 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(345\) vs. \(2(104)=208\).
time = 1.79, size = 346, normalized size = 3.33

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arctanh \left (d x +c \right )^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{e^{2}}+\frac {2 b^{2} \ln \left (d x +c \right ) \arctanh \left (d x +c \right )}{e^{2}}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{e^{2}}+\frac {b^{2} \dilog \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{e^{2}}+\frac {b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{2 e^{2}}-\frac {b^{2} \ln \left (d x +c -1\right )^{2}}{4 e^{2}}-\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{2 e^{2}}+\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{2 e^{2}}+\frac {b^{2} \ln \left (d x +c +1\right )^{2}}{4 e^{2}}-\frac {b^{2} \dilog \left (d x +c +1\right )}{e^{2}}-\frac {b^{2} \ln \left (d x +c \right ) \ln \left (d x +c +1\right )}{e^{2}}-\frac {b^{2} \dilog \left (d x +c \right )}{e^{2}}-\frac {2 a b \arctanh \left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {a b \ln \left (d x +c +1\right )}{e^{2}}+\frac {2 a b \ln \left (d x +c \right )}{e^{2}}-\frac {a b \ln \left (d x +c -1\right )}{e^{2}}}{d}\)	\(346\)
default	\(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arctanh \left (d x +c \right )^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{e^{2}}+\frac {2 b^{2} \ln \left (d x +c \right ) \arctanh \left (d x +c \right )}{e^{2}}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{e^{2}}+\frac {b^{2} \dilog \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{e^{2}}+\frac {b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{2 e^{2}}-\frac {b^{2} \ln \left (d x +c -1\right )^{2}}{4 e^{2}}-\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{2 e^{2}}+\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{2 e^{2}}+\frac {b^{2} \ln \left (d x +c +1\right )^{2}}{4 e^{2}}-\frac {b^{2} \dilog \left (d x +c +1\right )}{e^{2}}-\frac {b^{2} \ln \left (d x +c \right ) \ln \left (d x +c +1\right )}{e^{2}}-\frac {b^{2} \dilog \left (d x +c \right )}{e^{2}}-\frac {2 a b \arctanh \left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {a b \ln \left (d x +c +1\right )}{e^{2}}+\frac {2 a b \ln \left (d x +c \right )}{e^{2}}-\frac {a b \ln \left (d x +c -1\right )}{e^{2}}}{d}\)	\(346\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*e^(-2)/(d^2*x^2 + 2*c*d*x + c^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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}} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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