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]

-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)+11/144*b^2*arctanh(c*x)/c-1/9*b*(a+b*arctanh(
c*x))/c/(c*x+1)^3-1/12*b*(a+b*arctanh(c*x))/c/(c*x+1)^2-1/12*b*(a+b*arctanh(c*x))/c/(c*x+1)+1/24*(a+b*arctanh(
c*x))^2/c-1/3*(a+b*arctanh(c*x))^2/c/(c*x+1)^3

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Rubi [A]  time = 0.22, antiderivative size = 176, normalized size of antiderivative = 1.00, 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 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 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 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 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 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*}

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Mathematica [A]  time = 0.18, 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]

-1/864*(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])/(c*(1 + c*x)^3)

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fricas [A]  time = 0.62, size = 203, normalized size = 1.15 \[ -\frac {6 \, {\left (12 \, a b + 11 \, b^{2}\right )} c^{2} x^{2} + 54 \, {\left (4 \, a b + 3 \, b^{2}\right )} c x - 9 \, {\left (b^{2} c^{3} x^{3} + 3 \, b^{2} c^{2} x^{2} + 3 \, b^{2} c x - 7 \, b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} + 288 \, a^{2} + 240 \, a b + 112 \, b^{2} - 3 \, {\left ({\left (12 \, a b + 11 \, b^{2}\right )} c^{3} x^{3} + 3 \, {\left (12 \, a b + 7 \, b^{2}\right )} c^{2} x^{2} + 3 \, {\left (12 \, a b - b^{2}\right )} c x - 84 \, a b - 29 \, b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{864 \, {\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \]

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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giac [B]  time = 0.47, size = 333, normalized size = 1.89 \[ \frac {1}{1728} \, c {\left (\frac {18 \, {\left (\frac {3 \, {\left (c x + 1\right )}^{2} b^{2}}{{\left (c x - 1\right )}^{2}} - \frac {3 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + b^{2}\right )} {\left (c x - 1\right )}^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2}}{{\left (c x + 1\right )}^{3} c^{2}} + \frac {6 \, {\left (\frac {36 \, {\left (c x + 1\right )}^{2} a b}{{\left (c x - 1\right )}^{2}} - \frac {36 \, {\left (c x + 1\right )} a b}{c x - 1} + 12 \, a b + \frac {18 \, {\left (c x + 1\right )}^{2} b^{2}}{{\left (c x - 1\right )}^{2}} - \frac {9 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + 2 \, b^{2}\right )} {\left (c x - 1\right )}^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{3} c^{2}} + \frac {{\left (\frac {216 \, {\left (c x + 1\right )}^{2} a^{2}}{{\left (c x - 1\right )}^{2}} - \frac {216 \, {\left (c x + 1\right )} a^{2}}{c x - 1} + 72 \, a^{2} + \frac {216 \, {\left (c x + 1\right )}^{2} a b}{{\left (c x - 1\right )}^{2}} - \frac {108 \, {\left (c x + 1\right )} a b}{c x - 1} + 24 \, a b + \frac {108 \, {\left (c x + 1\right )}^{2} b^{2}}{{\left (c x - 1\right )}^{2}} - \frac {27 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + 4 \, b^{2}\right )} {\left (c x - 1\right )}^{3}}{{\left (c x + 1\right )}^{3} c^{2}}\right )} \]

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]

1/1728*c*(18*(3*(c*x + 1)^2*b^2/(c*x - 1)^2 - 3*(c*x + 1)*b^2/(c*x - 1) + b^2)*(c*x - 1)^3*log(-(c*x + 1)/(c*x
- 1))^2/((c*x + 1)^3*c^2) + 6*(36*(c*x + 1)^2*a*b/(c*x - 1)^2 - 36*(c*x + 1)*a*b/(c*x - 1) + 12*a*b + 18*(c*x
+ 1)^2*b^2/(c*x - 1)^2 - 9*(c*x + 1)*b^2/(c*x - 1) + 2*b^2)*(c*x - 1)^3*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^
3*c^2) + (216*(c*x + 1)^2*a^2/(c*x - 1)^2 - 216*(c*x + 1)*a^2/(c*x - 1) + 72*a^2 + 216*(c*x + 1)^2*a*b/(c*x -
1)^2 - 108*(c*x + 1)*a*b/(c*x - 1) + 24*a*b + 108*(c*x + 1)^2*b^2/(c*x - 1)^2 - 27*(c*x + 1)*b^2/(c*x - 1) + 4
*b^2)*(c*x - 1)^3/((c*x + 1)^3*c^2))

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maple [B]  time = 0.07, size = 386, normalized size = 2.19 \[ -\frac {a^{2}}{3 c \left (c x +1\right )^{3}}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{3 c \left (c x +1\right )^{3}}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{24 c}-\frac {b^{2} \arctanh \left (c x \right )}{9 c \left (c x +1\right )^{3}}-\frac {b^{2} \arctanh \left (c x \right )}{12 c \left (c x +1\right )^{2}}-\frac {b^{2} \arctanh \left (c x \right )}{12 c \left (c x +1\right )}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{24 c}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{96 c}+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{48 c}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{96 c}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{48 c}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{48 c}-\frac {11 b^{2} \ln \left (c x -1\right )}{288 c}-\frac {b^{2}}{54 c \left (c x +1\right )^{3}}-\frac {5 b^{2}}{144 c \left (c x +1\right )^{2}}-\frac {11 b^{2}}{144 c \left (c x +1\right )}+\frac {11 b^{2} \ln \left (c x +1\right )}{288 c}-\frac {2 a b \arctanh \left (c x \right )}{3 c \left (c x +1\right )^{3}}-\frac {a b \ln \left (c x -1\right )}{24 c}-\frac {a b}{9 c \left (c x +1\right )^{3}}-\frac {a b}{12 c \left (c x +1\right )^{2}}-\frac {a b}{12 c \left (c x +1\right )}+\frac {a b \ln \left (c x +1\right )}{24 c} \]

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)-1/96/c*b^2*ln(c*x+1)^2+1/48/c*b^2*ln(-1/2*c*x+1
/2)*ln(c*x+1)-1/48/c*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-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 = 0.35, size = 445, normalized size = 2.53 \[ -\frac {1}{72} \, {\left (c {\left (\frac {2 \, {\left (3 \, c^{2} x^{2} + 9 \, c x + 10\right )}}{c^{5} x^{3} + 3 \, c^{4} x^{2} + 3 \, c^{3} x + c^{2}} - \frac {3 \, \log \left (c x + 1\right )}{c^{2}} + \frac {3 \, \log \left (c x - 1\right )}{c^{2}}\right )} + \frac {48 \, \operatorname {artanh}\left (c x\right )}{c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c}\right )} a b - \frac {1}{864} \, {\left (12 \, c {\left (\frac {2 \, {\left (3 \, c^{2} x^{2} + 9 \, c x + 10\right )}}{c^{5} x^{3} + 3 \, c^{4} x^{2} + 3 \, c^{3} x + c^{2}} - \frac {3 \, \log \left (c x + 1\right )}{c^{2}} + \frac {3 \, \log \left (c x - 1\right )}{c^{2}}\right )} \operatorname {artanh}\left (c x\right ) + \frac {{\left (66 \, c^{2} x^{2} + 9 \, {\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x + 1\right )^{2} + 9 \, {\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right )^{2} + 162 \, c x - 3 \, {\left (11 \, c^{3} x^{3} + 33 \, c^{2} x^{2} + 33 \, c x + 6 \, {\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right ) + 11\right )} \log \left (c x + 1\right ) + 33 \, {\left (c^{3} x^{3} + 3 \, c^{2} x^{2} + 3 \, c x + 1\right )} \log \left (c x - 1\right ) + 112\right )} c^{2}}{c^{6} x^{3} + 3 \, c^{5} x^{2} + 3 \, c^{4} x + c^{3}}\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (c x\right )^{2}}{3 \, {\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} - \frac {a^{2}}{3 \, {\left (c^{4} x^{3} + 3 \, c^{3} x^{2} + 3 \, c^{2} x + c\right )}} \]

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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mupad [B]  time = 2.30, size = 498, normalized size = 2.83 \[ \ln \left (1-c\,x\right )\,\left (\ln \left (c\,x+1\right )\,\left (\frac {b^2}{3\,c\,\left (2\,c^3\,x^3+6\,c^2\,x^2+6\,c\,x+2\right )}-\frac {b^2\,\left (c^3\,x^3+3\,c^2\,x^2+3\,c\,x+1\right )}{24\,c\,\left (2\,c^3\,x^3+6\,c^2\,x^2+6\,c\,x+2\right )}\right )+\frac {b^2}{3\,c\,\left (6\,c^3\,x^3+18\,c^2\,x^2+18\,c\,x+6\right )}+\frac {b\,\left (6\,a-b\right )}{3\,c\,\left (6\,c^3\,x^3+18\,c^2\,x^2+18\,c\,x+6\right )}+\frac {b^2\,\left (11\,c^3\,x^3+45\,c^2\,x^2+69\,c\,x+51\right )}{48\,c\,\left (6\,c^3\,x^3+18\,c^2\,x^2+18\,c\,x+6\right )}\right )-\frac {x\,\left (27\,b^2+36\,a\,b\right )+x^2\,\left (11\,c\,b^2+12\,a\,c\,b\right )+\frac {8\,\left (18\,a^2+15\,a\,b+7\,b^2\right )}{3\,c}}{144\,c^3\,x^3+432\,c^2\,x^2+432\,c\,x+144}+{\ln \left (c\,x+1\right )}^2\,\left (\frac {b^2}{96\,c}-\frac {b^2}{12\,c^2\,\left (3\,x+3\,c\,x^2+\frac {1}{c}+c^2\,x^3\right )}\right )+{\ln \left (1-c\,x\right )}^2\,\left (\frac {b^2}{96\,c}-\frac {b^2}{3\,c\,\left (4\,c^3\,x^3+12\,c^2\,x^2+12\,c\,x+4\right )}\right )-\frac {\ln \left (c\,x+1\right )\,\left (\frac {7\,b^2}{96\,c^2}+\frac {5\,b^2\,x^2}{32}+\frac {23\,b^2\,x}{96\,c}+\frac {11\,b^2\,c\,x^3}{288}+\frac {b\,\left (16\,a+5\,b\right )}{48\,c^2}\right )}{3\,x+3\,c\,x^2+\frac {1}{c}+c^2\,x^3}-\frac {b\,\mathrm {atan}\left (c\,x\,1{}\mathrm {i}\right )\,\left (6\,a+11\,b\right )\,1{}\mathrm {i}}{72\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1 - c*x)*(log(c*x + 1)*(b^2/(3*c*(6*c*x + 6*c^2*x^2 + 2*c^3*x^3 + 2)) - (b^2*(3*c*x + 3*c^2*x^2 + c^3*x^3
+ 1))/(24*c*(6*c*x + 6*c^2*x^2 + 2*c^3*x^3 + 2))) + b^2/(3*c*(18*c*x + 18*c^2*x^2 + 6*c^3*x^3 + 6)) + (b*(6*a
- b))/(3*c*(18*c*x + 18*c^2*x^2 + 6*c^3*x^3 + 6)) + (b^2*(69*c*x + 45*c^2*x^2 + 11*c^3*x^3 + 51))/(48*c*(18*c*
x + 18*c^2*x^2 + 6*c^3*x^3 + 6))) - (x*(36*a*b + 27*b^2) + x^2*(11*b^2*c + 12*a*b*c) + (8*(15*a*b + 18*a^2 + 7
*b^2))/(3*c))/(432*c*x + 432*c^2*x^2 + 144*c^3*x^3 + 144) + log(c*x + 1)^2*(b^2/(96*c) - b^2/(12*c^2*(3*x + 3*
c*x^2 + 1/c + c^2*x^3))) + log(1 - c*x)^2*(b^2/(96*c) - b^2/(3*c*(12*c*x + 12*c^2*x^2 + 4*c^3*x^3 + 4))) - (lo
g(c*x + 1)*((7*b^2)/(96*c^2) + (5*b^2*x^2)/32 + (23*b^2*x)/(96*c) + (11*b^2*c*x^3)/288 + (b*(16*a + 5*b))/(48*
c^2)))/(3*x + 3*c*x^2 + 1/c + c^2*x^3) - (b*atan(c*x*1i)*(6*a + 11*b)*1i)/(72*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}}{\left (c x + 1\right )^{4}}\, dx \]

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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