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

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

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5928, 5926, 627, 44, 207, 5948} \[ -\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (c x+1)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (c x+1)^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (c x+1)^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac {3 b^2}{16 c d^3 (c x+1)}-\frac {b^2}{16 c d^3 (c x+1)^2}+\frac {3 b^2 \tanh ^{-1}(c x)}{16 c d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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}{(d+c d x)^3} \, dx &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac {b \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 d^2 (1+c x)^3}+\frac {a+b \tanh ^{-1}(c x)}{4 d^2 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{4 d^2 \left (-1+c^2 x^2\right )}\right ) \, dx}{d}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 d^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 d^3}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c 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 d^3}+\frac {b^2 \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{4 d^3}+\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{4 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}+\frac {b^2 \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 d^3}+\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 d^3}\\ &=-\frac {b^2}{16 c d^3 (1+c x)^2}-\frac {3 b^2}{16 c d^3 (1+c x)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}-\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{16 d^3}-\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{8 d^3}\\ &=-\frac {b^2}{16 c d^3 (1+c x)^2}-\frac {3 b^2}{16 c d^3 (1+c x)}+\frac {3 b^2 \tanh ^{-1}(c x)}{16 c d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c d^3 (1+c x)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{8 c d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c d^3 (1+c x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 183, normalized size = 1.17 \[ \frac {-8 a^2-4 a b-b^2}{16 c d^3 (c x+1)^2}+\frac {\left (-4 a b-3 b^2\right ) \log (1-c x)}{32 c d^3}+\frac {\left (4 a b+3 b^2\right ) \log (c x+1)}{32 c d^3}-\frac {b (4 a+3 b)}{16 c d^3 (c x+1)}-\frac {b \tanh ^{-1}(c x) (4 a+b c x+2 b)}{4 c d^3 (c x+1)^2}+\frac {b^2 \left (c^2 x^2+2 c x-3\right ) \tanh ^{-1}(c x)^2}{8 c d^3 (c x+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.13, size = 156, normalized size = 0.99 \[ -\frac {2 \, {\left (4 \, a b + 3 \, b^{2}\right )} c x - {\left (b^{2} c^{2} x^{2} + 2 \, b^{2} c x - 3 \, b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} + 16 \, a^{2} + 16 \, a b + 8 \, b^{2} - {\left ({\left (4 \, a b + 3 \, b^{2}\right )} c^{2} x^{2} + 2 \, {\left (4 \, a b + b^{2}\right )} c x - 12 \, a b - 5 \, b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{32 \, {\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 232, normalized size = 1.48 \[ \frac {1}{64} \, c {\left (\frac {2 \, {\left (\frac {2 \, {\left (c x + 1\right )} b^{2}}{c x - 1} - b^{2}\right )} {\left (c x - 1\right )}^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2}}{{\left (c x + 1\right )}^{2} c^{2} d^{3}} + \frac {2 \, {\left (\frac {8 \, {\left (c x + 1\right )} a b}{c x - 1} - 4 \, a b + \frac {4 \, {\left (c x + 1\right )} b^{2}}{c x - 1} - b^{2}\right )} {\left (c x - 1\right )}^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{2} c^{2} d^{3}} + \frac {{\left (\frac {16 \, {\left (c x + 1\right )} a^{2}}{c x - 1} - 8 \, a^{2} + \frac {16 \, {\left (c x + 1\right )} a b}{c x - 1} - 4 \, a b + \frac {8 \, {\left (c x + 1\right )} b^{2}}{c x - 1} - b^{2}\right )} {\left (c x - 1\right )}^{2}}{{\left (c x + 1\right )}^{2} c^{2} d^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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^2*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^2*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^2*d^3))

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maple [B]  time = 0.07, size = 398, normalized size = 2.54 \[ -\frac {a^{2}}{2 c \,d^{3} \left (c x +1\right )^{2}}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{2 c \,d^{3} \left (c x +1\right )^{2}}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{8 c \,d^{3}}-\frac {b^{2} \arctanh \left (c x \right )}{4 c \,d^{3} \left (c x +1\right )^{2}}-\frac {b^{2} \arctanh \left (c x \right )}{4 c \,d^{3} \left (c x +1\right )}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{8 c \,d^{3}}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{32 c \,d^{3}}+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{16 c \,d^{3}}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{32 c \,d^{3}}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{16 c \,d^{3}}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{16 c \,d^{3}}-\frac {3 b^{2} \ln \left (c x -1\right )}{32 c \,d^{3}}-\frac {b^{2}}{16 c \,d^{3} \left (c x +1\right )^{2}}-\frac {3 b^{2}}{16 c \,d^{3} \left (c x +1\right )}+\frac {3 b^{2} \ln \left (c x +1\right )}{32 c \,d^{3}}-\frac {a b \arctanh \left (c x \right )}{c \,d^{3} \left (c x +1\right )^{2}}-\frac {a b \ln \left (c x -1\right )}{8 c \,d^{3}}-\frac {a b}{4 c \,d^{3} \left (c x +1\right )^{2}}-\frac {a b}{4 c \,d^{3} \left (c x +1\right )}+\frac {a b \ln \left (c x +1\right )}{8 c \,d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 399, normalized size = 2.54 \[ -\frac {1}{8} \, {\left (c {\left (\frac {2 \, {\left (c x + 2\right )}}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}} - \frac {\log \left (c x + 1\right )}{c^{2} d^{3}} + \frac {\log \left (c x - 1\right )}{c^{2} d^{3}}\right )} + \frac {8 \, \operatorname {artanh}\left (c x\right )}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}}\right )} a b - \frac {1}{32} \, {\left (4 \, c {\left (\frac {2 \, {\left (c x + 2\right )}}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}} - \frac {\log \left (c x + 1\right )}{c^{2} d^{3}} + \frac {\log \left (c x - 1\right )}{c^{2} d^{3}}\right )} \operatorname {artanh}\left (c x\right ) + \frac {{\left ({\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x + 1\right )^{2} + {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right )^{2} + 6 \, c x - {\left (3 \, c^{2} x^{2} + 6 \, c x + 2 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 3\right )} \log \left (c x + 1\right ) + 3 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 8\right )} c^{2}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}}\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (c x\right )^{2}}{2 \, {\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} - \frac {a^{2}}{2 \, {\left (c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.16, size = 373, normalized size = 2.38 \[ \frac {11\,b^2\,\ln \left (1-c\,x\right )-11\,b^2\,\ln \left (c\,x+1\right )-16\,a\,b-3\,b^2\,{\ln \left (c\,x+1\right )}^2-3\,b^2\,{\ln \left (1-c\,x\right )}^2+12\,b^2\,\mathrm {atanh}\left (c\,x\right )-16\,a^2-8\,b^2-16\,a\,b\,\ln \left (c\,x+1\right )+16\,a\,b\,\ln \left (1-c\,x\right )+6\,b^2\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )+8\,a\,b\,\mathrm {atanh}\left (c\,x\right )-6\,b^2\,c\,x-10\,b^2\,c\,x\,\ln \left (c\,x+1\right )+10\,b^2\,c\,x\,\ln \left (1-c\,x\right )+b^2\,c^2\,x^2\,{\ln \left (c\,x+1\right )}^2+b^2\,c^2\,x^2\,{\ln \left (1-c\,x\right )}^2+12\,b^2\,c^2\,x^2\,\mathrm {atanh}\left (c\,x\right )+2\,b^2\,c\,x\,{\ln \left (c\,x+1\right )}^2+2\,b^2\,c\,x\,{\ln \left (1-c\,x\right )}^2+24\,b^2\,c\,x\,\mathrm {atanh}\left (c\,x\right )-3\,b^2\,c^2\,x^2\,\ln \left (c\,x+1\right )+3\,b^2\,c^2\,x^2\,\ln \left (1-c\,x\right )-8\,a\,b\,c\,x-4\,b^2\,c\,x\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )+8\,a\,b\,c^2\,x^2\,\mathrm {atanh}\left (c\,x\right )+16\,a\,b\,c\,x\,\mathrm {atanh}\left (c\,x\right )-2\,b^2\,c^2\,x^2\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )}{32\,c\,d^3\,{\left (c\,x+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*b^2*log(1 - c*x) - 11*b^2*log(c*x + 1) - 16*a*b - 3*b^2*log(c*x + 1)^2 - 3*b^2*log(1 - c*x)^2 + 12*b^2*ata
nh(c*x) - 16*a^2 - 8*b^2 - 16*a*b*log(c*x + 1) + 16*a*b*log(1 - c*x) + 6*b^2*log(c*x + 1)*log(1 - c*x) + 8*a*b
*atanh(c*x) - 6*b^2*c*x - 10*b^2*c*x*log(c*x + 1) + 10*b^2*c*x*log(1 - c*x) + b^2*c^2*x^2*log(c*x + 1)^2 + b^2
*c^2*x^2*log(1 - c*x)^2 + 12*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 + 24
*b^2*c*x*atanh(c*x) - 3*b^2*c^2*x^2*log(c*x + 1) + 3*b^2*c^2*x^2*log(1 - c*x) - 8*a*b*c*x - 4*b^2*c*x*log(c*x
+ 1)*log(1 - c*x) + 8*a*b*c^2*x^2*atanh(c*x) + 16*a*b*c*x*atanh(c*x) - 2*b^2*c^2*x^2*log(c*x + 1)*log(1 - c*x)
)/(32*c*d^3*(c*x + 1)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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