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

Optimal. Leaf size=176 \[ -\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (c x+1)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (c x+1)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (c x+1)^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (c x+1)^3}-\frac{11 b^2}{144 c (c x+1)}-\frac{5 b^2}{144 c (c x+1)^2}-\frac{b^2}{54 c (c x+1)^3}+\frac{11 b^2 \tanh ^{-1}(c x)}{144 c} \]

[Out]

-b^2/(54*c*(1 + c*x)^3) - (5*b^2)/(144*c*(1 + c*x)^2) - (11*b^2)/(144*c*(1 + c*x)) + (11*b^2*ArcTanh[c*x])/(14
4*c) - (b*(a + b*ArcTanh[c*x]))/(9*c*(1 + c*x)^3) - (b*(a + b*ArcTanh[c*x]))/(12*c*(1 + c*x)^2) - (b*(a + b*Ar
cTanh[c*x]))/(12*c*(1 + c*x)) + (a + b*ArcTanh[c*x])^2/(24*c) - (a + b*ArcTanh[c*x])^2/(3*c*(1 + c*x)^3)

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Rubi [A]  time = 0.217757, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5928, 5926, 627, 44, 207, 5948} \[ -\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (c x+1)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (c x+1)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (c x+1)^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (c x+1)^3}-\frac{11 b^2}{144 c (c x+1)}-\frac{5 b^2}{144 c (c x+1)^2}-\frac{b^2}{54 c (c x+1)^3}+\frac{11 b^2 \tanh ^{-1}(c x)}{144 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/(1 + c*x)^4,x]

[Out]

-b^2/(54*c*(1 + c*x)^3) - (5*b^2)/(144*c*(1 + c*x)^2) - (11*b^2)/(144*c*(1 + c*x)) + (11*b^2*ArcTanh[c*x])/(14
4*c) - (b*(a + b*ArcTanh[c*x]))/(9*c*(1 + c*x)^3) - (b*(a + b*ArcTanh[c*x]))/(12*c*(1 + c*x)^2) - (b*(a + b*Ar
cTanh[c*x]))/(12*c*(1 + c*x)) + (a + b*ArcTanh[c*x])^2/(24*c) - (a + b*ArcTanh[c*x])^2/(3*c*(1 + c*x)^3)

Rule 5928

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

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 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 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 5948

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^4} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac{1}{3} (2 b) \int \left (\frac{a+b \tanh ^{-1}(c x)}{2 (1+c x)^4}+\frac{a+b \tanh ^{-1}(c x)}{4 (1+c x)^3}+\frac{a+b \tanh ^{-1}(c x)}{8 (1+c x)^2}-\frac{a+b \tanh ^{-1}(c x)}{8 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac{1}{12} b \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx-\frac{1}{12} b \int \frac{a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx+\frac{1}{6} b \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx+\frac{1}{3} b \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^4} \, dx\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac{1}{12} b^2 \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx+\frac{1}{12} b^2 \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx+\frac{1}{9} b^2 \int \frac{1}{(1+c x)^3 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac{1}{12} b^2 \int \frac{1}{(1-c x) (1+c x)^3} \, dx+\frac{1}{12} b^2 \int \frac{1}{(1-c x) (1+c x)^2} \, dx+\frac{1}{9} b^2 \int \frac{1}{(1-c x) (1+c x)^4} \, dx\\ &=-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}+\frac{1}{12} b^2 \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{12} b^2 \int \left (\frac{1}{2 (1+c x)^3}+\frac{1}{4 (1+c x)^2}-\frac{1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx+\frac{1}{9} b^2 \int \left (\frac{1}{2 (1+c x)^4}+\frac{1}{4 (1+c x)^3}+\frac{1}{8 (1+c x)^2}-\frac{1}{8 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{b^2}{54 c (1+c x)^3}-\frac{5 b^2}{144 c (1+c x)^2}-\frac{11 b^2}{144 c (1+c x)}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}-\frac{1}{72} b^2 \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{48} b^2 \int \frac{1}{-1+c^2 x^2} \, dx-\frac{1}{24} b^2 \int \frac{1}{-1+c^2 x^2} \, dx\\ &=-\frac{b^2}{54 c (1+c x)^3}-\frac{5 b^2}{144 c (1+c x)^2}-\frac{11 b^2}{144 c (1+c x)}+\frac{11 b^2 \tanh ^{-1}(c x)}{144 c}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{9 c (1+c x)^3}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)^2}-\frac{b \left (a+b \tanh ^{-1}(c x)\right )}{12 c (1+c x)}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{24 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c (1+c x)^3}\\ \end{align*}

Mathematica [A]  time = 0.173419, size = 168, normalized size = 0.95 \[ -\frac{16 \left (18 a^2+6 a b+b^2\right )+24 b \tanh ^{-1}(c x) \left (24 a+b \left (3 c^2 x^2+9 c x+10\right )\right )+6 b (12 a+11 b) (c x+1)^2+6 b (12 a+5 b) (c x+1)+3 b (12 a+11 b) (c x+1)^3 \log (1-c x)-3 b (12 a+11 b) (c x+1)^3 \log (c x+1)-36 b^2 \left (c^3 x^3+3 c^2 x^2+3 c x-7\right ) \tanh ^{-1}(c x)^2}{864 c (c x+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/(1 + c*x)^4,x]

[Out]

-(16*(18*a^2 + 6*a*b + b^2) + 6*b*(12*a + 5*b)*(1 + c*x) + 6*b*(12*a + 11*b)*(1 + c*x)^2 + 24*b*(24*a + b*(10
+ 9*c*x + 3*c^2*x^2))*ArcTanh[c*x] - 36*b^2*(-7 + 3*c*x + 3*c^2*x^2 + c^3*x^3)*ArcTanh[c*x]^2 + 3*b*(12*a + 11
*b)*(1 + c*x)^3*Log[1 - c*x] - 3*b*(12*a + 11*b)*(1 + c*x)^3*Log[1 + c*x])/(864*c*(1 + c*x)^3)

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Maple [B]  time = 0.062, size = 386, normalized size = 2.2 \begin{align*} -{\frac{{a}^{2}}{3\,c \left ( cx+1 \right ) ^{3}}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{3\,c \left ( cx+1 \right ) ^{3}}}-{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{24\,c}}-{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) }{9\,c \left ( cx+1 \right ) ^{3}}}-{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) }{12\,c \left ( cx+1 \right ) ^{2}}}-{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) }{12\,c \left ( cx+1 \right ) }}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{24\,c}}-{\frac{{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{96\,c}}+{\frac{{b}^{2}\ln \left ( cx-1 \right ) }{48\,c}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{b}^{2}}{48\,c}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{b}^{2}\ln \left ( cx+1 \right ) }{48\,c}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }-{\frac{{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{96\,c}}-{\frac{11\,{b}^{2}\ln \left ( cx-1 \right ) }{288\,c}}-{\frac{{b}^{2}}{54\,c \left ( cx+1 \right ) ^{3}}}-{\frac{5\,{b}^{2}}{144\,c \left ( cx+1 \right ) ^{2}}}-{\frac{11\,{b}^{2}}{144\,c \left ( cx+1 \right ) }}+{\frac{11\,{b}^{2}\ln \left ( cx+1 \right ) }{288\,c}}-{\frac{2\,ab{\it Artanh} \left ( cx \right ) }{3\,c \left ( cx+1 \right ) ^{3}}}-{\frac{ab\ln \left ( cx-1 \right ) }{24\,c}}-{\frac{ab}{9\,c \left ( cx+1 \right ) ^{3}}}-{\frac{ab}{12\,c \left ( cx+1 \right ) ^{2}}}-{\frac{ab}{12\,c \left ( cx+1 \right ) }}+{\frac{ab\ln \left ( cx+1 \right ) }{24\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^2/(c*x+1)^4,x)

[Out]

-1/3/c*a^2/(c*x+1)^3-1/3/c*b^2/(c*x+1)^3*arctanh(c*x)^2-1/24/c*b^2*arctanh(c*x)*ln(c*x-1)-1/9/c*b^2*arctanh(c*
x)/(c*x+1)^3-1/12/c*b^2*arctanh(c*x)/(c*x+1)^2-1/12/c*b^2*arctanh(c*x)/(c*x+1)+1/24/c*b^2*arctanh(c*x)*ln(c*x+
1)-1/96/c*b^2*ln(c*x-1)^2+1/48/c*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-1/48/c*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/4
8/c*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/96/c*b^2*ln(c*x+1)^2-11/288/c*b^2*ln(c*x-1)-1/54*b^2/c/(c*x+1)^3-5/144*b^
2/c/(c*x+1)^2-11/144*b^2/c/(c*x+1)+11/288/c*b^2*ln(c*x+1)-2/3/c*a*b*arctanh(c*x)/(c*x+1)^3-1/24/c*a*b*ln(c*x-1
)-1/9/c*a*b/(c*x+1)^3-1/12/c*a*b/(c*x+1)^2-1/12/c*a*b/(c*x+1)+1/24/c*a*b*ln(c*x+1)

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Maxima [B]  time = 1.04532, size = 601, normalized size = 3.41 \begin{align*} -\frac{1}{72} \,{\left (c{\left (\frac{2 \,{\left (3 \, c^{2} x^{2} + 9 \, c x + 10\right )}}{c^{5} x^{3} + 3 \, c^{4} x^{2} + 3 \, c^{3} x + c^{2}} - \frac{3 \, \log \left (c x + 1\right )}{c^{2}} + \frac{3 \, \log \left (c x - 1\right )}{c^{2}}\right )} + \frac{48 \, \operatorname{artanh}\left (c x\right )}{c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c}\right )} a b - \frac{1}{864} \,{\left (12 \, c{\left (\frac{2 \,{\left (3 \, c^{2} x^{2} + 9 \, c x + 10\right )}}{c^{5} x^{3} + 3 \, c^{4} x^{2} + 3 \, c^{3} x + c^{2}} - \frac{3 \, \log \left (c x + 1\right )}{c^{2}} + \frac{3 \, \log \left (c x - 1\right )}{c^{2}}\right )} \operatorname{artanh}\left (c x\right ) + \frac{{\left (66 \, c^{2} x^{2} + 9 \,{\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x + 1\right )^{2} + 9 \,{\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right )^{2} + 162 \, c x - 3 \,{\left (11 \, c^{3} x^{3} + 33 \, c^{2} x^{2} + 33 \, c x + 6 \,{\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right ) + 11\right )} \log \left (c x + 1\right ) + 33 \,{\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right ) + 112\right )} c^{2}}{c^{6} x^{3} + 3 \, c^{5} x^{2} + 3 \, c^{4} x + c^{3}}\right )} b^{2} - \frac{b^{2} \operatorname{artanh}\left (c x\right )^{2}}{3 \,{\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} - \frac{a^{2}}{3 \,{\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/(c*x+1)^4,x, algorithm="maxima")

[Out]

-1/72*(c*(2*(3*c^2*x^2 + 9*c*x + 10)/(c^5*x^3 + 3*c^4*x^2 + 3*c^3*x + c^2) - 3*log(c*x + 1)/c^2 + 3*log(c*x -
1)/c^2) + 48*arctanh(c*x)/(c^4*x^3 + 3*c^3*x^2 + 3*c^2*x + c))*a*b - 1/864*(12*c*(2*(3*c^2*x^2 + 9*c*x + 10)/(
c^5*x^3 + 3*c^4*x^2 + 3*c^3*x + c^2) - 3*log(c*x + 1)/c^2 + 3*log(c*x - 1)/c^2)*arctanh(c*x) + (66*c^2*x^2 + 9
*(c^3*x^3 + 3*c^2*x^2 + 3*c*x + 1)*log(c*x + 1)^2 + 9*(c^3*x^3 + 3*c^2*x^2 + 3*c*x + 1)*log(c*x - 1)^2 + 162*c
*x - 3*(11*c^3*x^3 + 33*c^2*x^2 + 33*c*x + 6*(c^3*x^3 + 3*c^2*x^2 + 3*c*x + 1)*log(c*x - 1) + 11)*log(c*x + 1)
+ 33*(c^3*x^3 + 3*c^2*x^2 + 3*c*x + 1)*log(c*x - 1) + 112)*c^2/(c^6*x^3 + 3*c^5*x^2 + 3*c^4*x + c^3))*b^2 - 1
/3*b^2*arctanh(c*x)^2/(c^4*x^3 + 3*c^3*x^2 + 3*c^2*x + c) - 1/3*a^2/(c^4*x^3 + 3*c^3*x^2 + 3*c^2*x + c)

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Fricas [A]  time = 1.96812, size = 458, normalized size = 2.6 \begin{align*} -\frac{6 \,{\left (12 \, a b + 11 \, b^{2}\right )} c^{2} x^{2} + 54 \,{\left (4 \, a b + 3 \, b^{2}\right )} c x - 9 \,{\left (b^{2} c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + 3 \, b^{2} c x - 7 \, b^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} + 288 \, a^{2} + 240 \, a b + 112 \, b^{2} - 3 \,{\left ({\left (12 \, a b + 11 \, b^{2}\right )} c^{3} x^{3} + 3 \,{\left (12 \, a b + 7 \, b^{2}\right )} c^{2} x^{2} + 3 \,{\left (12 \, a b - b^{2}\right )} c x - 84 \, a b - 29 \, b^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{864 \,{\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/(c*x+1)^4,x, algorithm="fricas")

[Out]

-1/864*(6*(12*a*b + 11*b^2)*c^2*x^2 + 54*(4*a*b + 3*b^2)*c*x - 9*(b^2*c^3*x^3 + 3*b^2*c^2*x^2 + 3*b^2*c*x - 7*
b^2)*log(-(c*x + 1)/(c*x - 1))^2 + 288*a^2 + 240*a*b + 112*b^2 - 3*((12*a*b + 11*b^2)*c^3*x^3 + 3*(12*a*b + 7*
b^2)*c^2*x^2 + 3*(12*a*b - b^2)*c*x - 84*a*b - 29*b^2)*log(-(c*x + 1)/(c*x - 1)))/(c^4*x^3 + 3*c^3*x^2 + 3*c^2
*x + c)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{2}}{\left (c x + 1\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**2/(c*x+1)**4,x)

[Out]

Integral((a + b*atanh(c*x))**2/(c*x + 1)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{{\left (c x + 1\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/(c*x+1)^4,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2/(c*x + 1)^4, x)