3.114 \(\int \frac {x (a+b \tanh ^{-1}(c x))^2}{(d+c d x)^3} \, dx\)

Optimal. Leaf size=157 \[ -\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (c x+1)}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (c x+1)^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (c x+1)^2}-\frac {5 b^2}{16 c^2 d^3 (c x+1)}+\frac {b^2}{16 c^2 d^3 (c x+1)^2}+\frac {5 b^2 \tanh ^{-1}(c x)}{16 c^2 d^3} \]

[Out]

1/16*b^2/c^2/d^3/(c*x+1)^2-5/16*b^2/c^2/d^3/(c*x+1)+5/16*b^2*arctanh(c*x)/c^2/d^3+1/4*b*(a+b*arctanh(c*x))/c^2
/d^3/(c*x+1)^2-3/4*b*(a+b*arctanh(c*x))/c^2/d^3/(c*x+1)-1/8*(a+b*arctanh(c*x))^2/c^2/d^3+1/2*x^2*(a+b*arctanh(
c*x))^2/d^3/(c*x+1)^2

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Rubi [A]  time = 0.21, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {37, 5938, 5926, 627, 44, 207, 5948} \[ -\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (c x+1)}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (c x+1)^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (c x+1)^2}-\frac {5 b^2}{16 c^2 d^3 (c x+1)}+\frac {b^2}{16 c^2 d^3 (c x+1)^2}+\frac {5 b^2 \tanh ^{-1}(c x)}{16 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcTanh[c*x])^2)/(d + c*d*x)^3,x]

[Out]

b^2/(16*c^2*d^3*(1 + c*x)^2) - (5*b^2)/(16*c^2*d^3*(1 + c*x)) + (5*b^2*ArcTanh[c*x])/(16*c^2*d^3) + (b*(a + b*
ArcTanh[c*x]))/(4*c^2*d^3*(1 + c*x)^2) - (3*b*(a + b*ArcTanh[c*x]))/(4*c^2*d^3*(1 + c*x)) - (a + b*ArcTanh[c*x
])^2/(8*c^2*d^3) + (x^2*(a + b*ArcTanh[c*x])^2)/(2*d^3*(1 + c*x)^2)

Rule 37

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

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 5938

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_), x_Symbol] :> With[{
u = IntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[(a + b*ArcTanh[c*x])^p, u, x] - Dist[b*c*p, Int[ExpandIntegrand[(a
+ b*ArcTanh[c*x])^(p - 1), u/(1 - c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && IGtQ[p, 1] && Eq
Q[c^2*d^2 - e^2, 0] && IntegersQ[m, q] && NeQ[m, -1] && NeQ[q, -1] && ILtQ[m + q + 1, 0] && LtQ[m*q, 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 {x \left (a+b \tanh ^{-1}(c x)\right )^2}{(d+c d x)^3} \, dx &=\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-(2 b c) \int \left (\frac {a+b \tanh ^{-1}(c x)}{4 c^2 d^3 (1+c x)^3}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{8 c^2 d^3 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{8 c^2 d^3 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 c d^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 c d^3}+\frac {(3 b) \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 c d^3}\\ &=\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac {b^2 \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{4 c d^3}+\frac {\left (3 b^2\right ) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 c d^3}\\ &=\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{4 c d^3}+\frac {\left (3 b^2\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{4 c d^3}\\ &=\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}-\frac {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}{4 c d^3}+\frac {\left (3 b^2\right ) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c d^3}\\ &=\frac {b^2}{16 c^2 d^3 (1+c x)^2}-\frac {5 b^2}{16 c^2 d^3 (1+c x)}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}+\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{16 c d^3}-\frac {\left (3 b^2\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{8 c d^3}\\ &=\frac {b^2}{16 c^2 d^3 (1+c x)^2}-\frac {5 b^2}{16 c^2 d^3 (1+c x)}+\frac {5 b^2 \tanh ^{-1}(c x)}{16 c^2 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)^2}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^3 (1+c x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 150, normalized size = 0.96 \[ \frac {-2 \left (16 a^2+12 a b+5 b^2\right ) (c x+1)+2 \left (8 a^2+4 a b+b^2\right )-b (12 a+5 b) (c x+1)^2 \log (1-c x)+b (12 a+5 b) (c x+1)^2 \log (c x+1)-8 b \tanh ^{-1}(c x) (a (8 c x+4)+b (3 c x+2))+4 b^2 \left (3 c^2 x^2-2 c x-1\right ) \tanh ^{-1}(c x)^2}{32 c^2 d^3 (c x+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*ArcTanh[c*x])^2)/(d + c*d*x)^3,x]

[Out]

(2*(8*a^2 + 4*a*b + b^2) - 2*(16*a^2 + 12*a*b + 5*b^2)*(1 + c*x) - 8*b*(b*(2 + 3*c*x) + a*(4 + 8*c*x))*ArcTanh
[c*x] + 4*b^2*(-1 - 2*c*x + 3*c^2*x^2)*ArcTanh[c*x]^2 - b*(12*a + 5*b)*(1 + c*x)^2*Log[1 - c*x] + b*(12*a + 5*
b)*(1 + c*x)^2*Log[1 + c*x])/(32*c^2*d^3*(1 + c*x)^2)

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fricas [A]  time = 0.63, size = 164, normalized size = 1.04 \[ -\frac {2 \, {\left (16 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} c x - {\left (3 \, b^{2} c^{2} x^{2} - 2 \, b^{2} c x - b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} + 16 \, a^{2} + 16 \, a b + 8 \, b^{2} - {\left ({\left (12 \, a b + 5 \, b^{2}\right )} c^{2} x^{2} - 2 \, {\left (4 \, a b + b^{2}\right )} c x - 4 \, a b - 3 \, b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{32 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

-1/32*(2*(16*a^2 + 12*a*b + 5*b^2)*c*x - (3*b^2*c^2*x^2 - 2*b^2*c*x - b^2)*log(-(c*x + 1)/(c*x - 1))^2 + 16*a^
2 + 16*a*b + 8*b^2 - ((12*a*b + 5*b^2)*c^2*x^2 - 2*(4*a*b + b^2)*c*x - 4*a*b - 3*b^2)*log(-(c*x + 1)/(c*x - 1)
))/(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3)

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giac [A]  time = 0.32, size = 226, normalized size = 1.44 \[ \frac {1}{64} \, c {\left (\frac {2 \, {\left (\frac {2 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + b^{2}\right )} {\left (c x - 1\right )}^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2}}{{\left (c x + 1\right )}^{2} c^{3} d^{3}} + \frac {2 \, {\left (\frac {8 \, {\left (c x + 1\right )} a b}{c x - 1} + 4 \, a b + \frac {4 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + b^{2}\right )} {\left (c x - 1\right )}^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{2} c^{3} d^{3}} + \frac {{\left (\frac {16 \, {\left (c x + 1\right )} a^{2}}{c x - 1} + 8 \, a^{2} + \frac {16 \, {\left (c x + 1\right )} a b}{c x - 1} + 4 \, a b + \frac {8 \, {\left (c x + 1\right )} b^{2}}{c x - 1} + b^{2}\right )} {\left (c x - 1\right )}^{2}}{{\left (c x + 1\right )}^{2} c^{3} d^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="giac")

[Out]

1/64*c*(2*(2*(c*x + 1)*b^2/(c*x - 1) + b^2)*(c*x - 1)^2*log(-(c*x + 1)/(c*x - 1))^2/((c*x + 1)^2*c^3*d^3) + 2*
(8*(c*x + 1)*a*b/(c*x - 1) + 4*a*b + 4*(c*x + 1)*b^2/(c*x - 1) + b^2)*(c*x - 1)^2*log(-(c*x + 1)/(c*x - 1))/((
c*x + 1)^2*c^3*d^3) + (16*(c*x + 1)*a^2/(c*x - 1) + 8*a^2 + 16*(c*x + 1)*a*b/(c*x - 1) + 4*a*b + 8*(c*x + 1)*b
^2/(c*x - 1) + b^2)*(c*x - 1)^2/((c*x + 1)^2*c^3*d^3))

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maple [B]  time = 0.07, size = 460, normalized size = 2.93 \[ -\frac {a^{2}}{c^{2} d^{3} \left (c x +1\right )}+\frac {a^{2}}{2 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{c^{2} d^{3} \left (c x +1\right )}+\frac {b^{2} \arctanh \left (c x \right )^{2}}{2 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {3 b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{8 c^{2} d^{3}}+\frac {b^{2} \arctanh \left (c x \right )}{4 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {3 b^{2} \arctanh \left (c x \right )}{4 c^{2} d^{3} \left (c x +1\right )}+\frac {3 b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{8 c^{2} d^{3}}-\frac {3 b^{2} \ln \left (c x -1\right )^{2}}{32 c^{2} d^{3}}+\frac {3 b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{16 c^{2} d^{3}}-\frac {3 b^{2} \ln \left (c x +1\right )^{2}}{32 c^{2} d^{3}}+\frac {3 b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{16 c^{2} d^{3}}-\frac {3 b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{16 c^{2} d^{3}}-\frac {5 b^{2} \ln \left (c x -1\right )}{32 c^{2} d^{3}}+\frac {b^{2}}{16 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {5 b^{2}}{16 c^{2} d^{3} \left (c x +1\right )}+\frac {5 b^{2} \ln \left (c x +1\right )}{32 c^{2} d^{3}}-\frac {2 a b \arctanh \left (c x \right )}{c^{2} d^{3} \left (c x +1\right )}+\frac {a b \arctanh \left (c x \right )}{c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {3 a b \ln \left (c x -1\right )}{8 c^{2} d^{3}}+\frac {a b}{4 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {3 a b}{4 c^{2} d^{3} \left (c x +1\right )}+\frac {3 a b \ln \left (c x +1\right )}{8 c^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x)

[Out]

-1/c^2*a^2/d^3/(c*x+1)+1/2/c^2*a^2/d^3/(c*x+1)^2-1/c^2*b^2/d^3*arctanh(c*x)^2/(c*x+1)+1/2/c^2*b^2/d^3*arctanh(
c*x)^2/(c*x+1)^2-3/8/c^2*b^2/d^3*arctanh(c*x)*ln(c*x-1)+1/4/c^2*b^2/d^3*arctanh(c*x)/(c*x+1)^2-3/4/c^2*b^2/d^3
*arctanh(c*x)/(c*x+1)+3/8/c^2*b^2/d^3*arctanh(c*x)*ln(c*x+1)-3/32/c^2*b^2/d^3*ln(c*x-1)^2+3/16/c^2*b^2/d^3*ln(
c*x-1)*ln(1/2+1/2*c*x)-3/32/c^2*b^2/d^3*ln(c*x+1)^2+3/16/c^2*b^2/d^3*ln(-1/2*c*x+1/2)*ln(c*x+1)-3/16/c^2*b^2/d
^3*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-5/32/c^2*b^2/d^3*ln(c*x-1)+1/16*b^2/c^2/d^3/(c*x+1)^2-5/16*b^2/c^2/d^3/(c*
x+1)+5/32/c^2*b^2/d^3*ln(c*x+1)-2/c^2*a*b/d^3*arctanh(c*x)/(c*x+1)+1/c^2*a*b/d^3*arctanh(c*x)/(c*x+1)^2-3/8/c^
2*a*b/d^3*ln(c*x-1)+1/4/c^2*a*b/d^3/(c*x+1)^2-3/4/c^2*a*b/d^3/(c*x+1)+3/8/c^2*a*b/d^3*ln(c*x+1)

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maxima [B]  time = 0.34, size = 429, normalized size = 2.73 \[ -\frac {{\left (2 \, c x + 1\right )} b^{2} \operatorname {artanh}\left (c x\right )^{2}}{2 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} - \frac {1}{8} \, {\left (c {\left (\frac {2 \, {\left (3 \, c x + 2\right )}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} - \frac {3 \, \log \left (c x + 1\right )}{c^{3} d^{3}} + \frac {3 \, \log \left (c x - 1\right )}{c^{3} d^{3}}\right )} + \frac {8 \, {\left (2 \, c x + 1\right )} \operatorname {artanh}\left (c x\right )}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}}\right )} a b - \frac {1}{32} \, {\left (4 \, c {\left (\frac {2 \, {\left (3 \, c x + 2\right )}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} - \frac {3 \, \log \left (c x + 1\right )}{c^{3} d^{3}} + \frac {3 \, \log \left (c x - 1\right )}{c^{3} d^{3}}\right )} \operatorname {artanh}\left (c x\right ) + \frac {{\left (3 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x + 1\right )^{2} + 3 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right )^{2} + 10 \, c x - {\left (5 \, c^{2} x^{2} + 10 \, c x + 6 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 5\right )} \log \left (c x + 1\right ) + 5 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 8\right )} c^{2}}{c^{6} d^{3} x^{2} + 2 \, c^{5} d^{3} x + c^{4} d^{3}}\right )} b^{2} - \frac {{\left (2 \, c x + 1\right )} a^{2}}{2 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

-1/2*(2*c*x + 1)*b^2*arctanh(c*x)^2/(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3) - 1/8*(c*(2*(3*c*x + 2)/(c^5*d^3*x^2
+ 2*c^4*d^3*x + c^3*d^3) - 3*log(c*x + 1)/(c^3*d^3) + 3*log(c*x - 1)/(c^3*d^3)) + 8*(2*c*x + 1)*arctanh(c*x)/
(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3))*a*b - 1/32*(4*c*(2*(3*c*x + 2)/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) -
3*log(c*x + 1)/(c^3*d^3) + 3*log(c*x - 1)/(c^3*d^3))*arctanh(c*x) + (3*(c^2*x^2 + 2*c*x + 1)*log(c*x + 1)^2 +
3*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1)^2 + 10*c*x - (5*c^2*x^2 + 10*c*x + 6*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1) +
5)*log(c*x + 1) + 5*(c^2*x^2 + 2*c*x + 1)*log(c*x - 1) + 8)*c^2/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3))*b^2 -
1/2*(2*c*x + 1)*a^2/(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3)

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mupad [B]  time = 2.69, size = 405, normalized size = 2.58 \[ -\frac {16\,a\,b+17\,b^2\,\ln \left (c\,x+1\right )-17\,b^2\,\ln \left (1-c\,x\right )+b^2\,{\ln \left (c\,x+1\right )}^2+b^2\,{\ln \left (1-c\,x\right )}^2-28\,b^2\,\mathrm {atanh}\left (c\,x\right )+16\,a^2+8\,b^2+16\,a\,b\,\ln \left (c\,x+1\right )-16\,a\,b\,\ln \left (1-c\,x\right )-2\,b^2\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )-24\,a\,b\,\mathrm {atanh}\left (c\,x\right )+32\,a^2\,c\,x+10\,b^2\,c\,x+30\,b^2\,c\,x\,\ln \left (c\,x+1\right )-30\,b^2\,c\,x\,\ln \left (1-c\,x\right )-3\,b^2\,c^2\,x^2\,{\ln \left (c\,x+1\right )}^2-3\,b^2\,c^2\,x^2\,{\ln \left (1-c\,x\right )}^2-28\,b^2\,c^2\,x^2\,\mathrm {atanh}\left (c\,x\right )+2\,b^2\,c\,x\,{\ln \left (c\,x+1\right )}^2+2\,b^2\,c\,x\,{\ln \left (1-c\,x\right )}^2-56\,b^2\,c\,x\,\mathrm {atanh}\left (c\,x\right )+9\,b^2\,c^2\,x^2\,\ln \left (c\,x+1\right )-9\,b^2\,c^2\,x^2\,\ln \left (1-c\,x\right )+24\,a\,b\,c\,x+32\,a\,b\,c\,x\,\ln \left (c\,x+1\right )-32\,a\,b\,c\,x\,\ln \left (1-c\,x\right )-4\,b^2\,c\,x\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )-24\,a\,b\,c^2\,x^2\,\mathrm {atanh}\left (c\,x\right )-48\,a\,b\,c\,x\,\mathrm {atanh}\left (c\,x\right )+6\,b^2\,c^2\,x^2\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )}{32\,c^2\,d^3\,{\left (c\,x+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*atanh(c*x))^2)/(d + c*d*x)^3,x)

[Out]

-(16*a*b + 17*b^2*log(c*x + 1) - 17*b^2*log(1 - c*x) + b^2*log(c*x + 1)^2 + b^2*log(1 - c*x)^2 - 28*b^2*atanh(
c*x) + 16*a^2 + 8*b^2 + 16*a*b*log(c*x + 1) - 16*a*b*log(1 - c*x) - 2*b^2*log(c*x + 1)*log(1 - c*x) - 24*a*b*a
tanh(c*x) + 32*a^2*c*x + 10*b^2*c*x + 30*b^2*c*x*log(c*x + 1) - 30*b^2*c*x*log(1 - c*x) - 3*b^2*c^2*x^2*log(c*
x + 1)^2 - 3*b^2*c^2*x^2*log(1 - c*x)^2 - 28*b^2*c^2*x^2*atanh(c*x) + 2*b^2*c*x*log(c*x + 1)^2 + 2*b^2*c*x*log
(1 - c*x)^2 - 56*b^2*c*x*atanh(c*x) + 9*b^2*c^2*x^2*log(c*x + 1) - 9*b^2*c^2*x^2*log(1 - c*x) + 24*a*b*c*x + 3
2*a*b*c*x*log(c*x + 1) - 32*a*b*c*x*log(1 - c*x) - 4*b^2*c*x*log(c*x + 1)*log(1 - c*x) - 24*a*b*c^2*x^2*atanh(
c*x) - 48*a*b*c*x*atanh(c*x) + 6*b^2*c^2*x^2*log(c*x + 1)*log(1 - c*x))/(32*c^2*d^3*(c*x + 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b^{2} x \operatorname {atanh}^{2}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {2 a b x \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atanh(c*x))**2/(c*d*x+d)**3,x)

[Out]

(Integral(a**2*x/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b**2*x*atanh(c*x)**2/(c**3*x**3 + 3*c**2
*x**2 + 3*c*x + 1), x) + Integral(2*a*b*x*atanh(c*x)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x))/d**3

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