3.1.20 \(\int \frac {(a+b \tanh ^{-1}(c+d x))^2}{(c e+d e x)^3} \, dx\) [20]

Optimal. Leaf size=119 \[ -\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {b^2 \log (c+d x)}{d e^3}-\frac {b^2 \log \left (1-(c+d x)^2\right )}{2 d e^3} \]

[Out]

-b*(a+b*arctanh(d*x+c))/d/e^3/(d*x+c)+1/2*(a+b*arctanh(d*x+c))^2/d/e^3-1/2*(a+b*arctanh(d*x+c))^2/d/e^3/(d*x+c
)^2+b^2*ln(d*x+c)/d/e^3-1/2*b^2*ln(1-(d*x+c)^2)/d/e^3

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Rubi [A]
time = 0.13, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {6242, 12, 6037, 6129, 272, 36, 31, 29, 6095} \begin {gather*} -\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}+\frac {b^2 \log (c+d x)}{d e^3}-\frac {b^2 \log \left (1-(c+d x)^2\right )}{2 d e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(a + b*ArcTanh[c + d*x]))/(d*e^3*(c + d*x))) + (a + b*ArcTanh[c + d*x])^2/(2*d*e^3) - (a + b*ArcTanh[c +
d*x])^2/(2*d*e^3*(c + d*x)^2) + (b^2*Log[c + d*x])/(d*e^3) - (b^2*Log[1 - (c + d*x)^2])/(2*d*e^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 6095

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]

Rule 6129

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[(f*x)^(m + 2)*((a + b*ArcTanh[c*x])^p/(d +
e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

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

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Mathematica [A]
time = 0.12, size = 136, normalized size = 1.14 \begin {gather*} \frac {-\frac {a^2}{(c+d x)^2}-\frac {2 a b}{c+d x}-\frac {2 b (a+b (c+d x)) \tanh ^{-1}(c+d x)}{(c+d x)^2}+\frac {b^2 \left (-1+c^2+2 c d x+d^2 x^2\right ) \tanh ^{-1}(c+d x)^2}{(c+d x)^2}-b (a+b) \log (1-c-d x)+2 b^2 \log (c+d x)+(a-b) b \log (1+c+d x)}{2 d e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2/(c + d*x)^2) - (2*a*b)/(c + d*x) - (2*b*(a + b*(c + d*x))*ArcTanh[c + d*x])/(c + d*x)^2 + (b^2*(-1 + c^
2 + 2*c*d*x + d^2*x^2)*ArcTanh[c + d*x]^2)/(c + d*x)^2 - b*(a + b)*Log[1 - c - d*x] + 2*b^2*Log[c + d*x] + (a
- b)*b*Log[1 + c + d*x])/(2*d*e^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(113)=226\).
time = 0.77, size = 324, normalized size = 2.72

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {b^{2} \arctanh \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{2 e^{3}}-\frac {b^{2} \arctanh \left (d x +c \right )}{e^{3} \left (d x +c \right )}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{2 e^{3}}+\frac {b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{4 e^{3}}-\frac {b^{2} \ln \left (d x +c -1\right )^{2}}{8 e^{3}}-\frac {b^{2} \ln \left (d x +c +1\right )}{2 e^{3}}+\frac {b^{2} \ln \left (d x +c \right )}{e^{3}}-\frac {b^{2} \ln \left (d x +c -1\right )}{2 e^{3}}+\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{4 e^{3}}-\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{4 e^{3}}-\frac {b^{2} \ln \left (d x +c +1\right )^{2}}{8 e^{3}}-\frac {a b \arctanh \left (d x +c \right )}{e^{3} \left (d x +c \right )^{2}}+\frac {a b \ln \left (d x +c +1\right )}{2 e^{3}}-\frac {a b}{e^{3} \left (d x +c \right )}-\frac {a b \ln \left (d x +c -1\right )}{2 e^{3}}}{d}\)	\(324\)
default	\(\frac {-\frac {a^{2}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {b^{2} \arctanh \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{2 e^{3}}-\frac {b^{2} \arctanh \left (d x +c \right )}{e^{3} \left (d x +c \right )}-\frac {b^{2} \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{2 e^{3}}+\frac {b^{2} \ln \left (d x +c -1\right ) \ln \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{4 e^{3}}-\frac {b^{2} \ln \left (d x +c -1\right )^{2}}{8 e^{3}}-\frac {b^{2} \ln \left (d x +c +1\right )}{2 e^{3}}+\frac {b^{2} \ln \left (d x +c \right )}{e^{3}}-\frac {b^{2} \ln \left (d x +c -1\right )}{2 e^{3}}+\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{4 e^{3}}-\frac {b^{2} \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {d x}{2}+\frac {c}{2}+\frac {1}{2}\right )}{4 e^{3}}-\frac {b^{2} \ln \left (d x +c +1\right )^{2}}{8 e^{3}}-\frac {a b \arctanh \left (d x +c \right )}{e^{3} \left (d x +c \right )^{2}}+\frac {a b \ln \left (d x +c +1\right )}{2 e^{3}}-\frac {a b}{e^{3} \left (d x +c \right )}-\frac {a b \ln \left (d x +c -1\right )}{2 e^{3}}}{d}\)	\(324\)
risch	\(\frac {b^{2} \left (d^{2} x^{2}+2 c d x +c^{2}-1\right ) \ln \left (d x +c +1\right )^{2}}{8 e^{3} \left (d x +c \right )^{2} d}-\frac {b \left (b \,d^{2} x^{2} \ln \left (-d x -c +1\right )+2 b d x \ln \left (-d x -c +1\right ) c +\ln \left (-d x -c +1\right ) b \,c^{2}+2 b d x +2 b c -b \ln \left (-d x -c +1\right )+2 a \right ) \ln \left (d x +c +1\right )}{4 e^{3} \left (d x +c \right )^{2} d}+\frac {-4 a^{2}-8 a b c -8 a b d x +4 b^{2} d x \ln \left (-d x -c +1\right )-b^{2} \ln \left (-d x -c +1\right )^{2}-8 a b c d x \ln \left (-d x -c +1\right )+b^{2} c^{2} \ln \left (-d x -c +1\right )^{2}-4 a b \,d^{2} x^{2} \ln \left (-d x -c +1\right )+2 b^{2} c d x \ln \left (-d x -c +1\right )^{2}+4 \ln \left (-d x -c -1\right ) a b \,c^{2}+b^{2} d^{2} x^{2} \ln \left (-d x -c +1\right )^{2}+4 \ln \left (-d x -c +1\right ) a b +8 \ln \left (-d x -c -1\right ) a b c d x +4 b^{2} c \ln \left (-d x -c +1\right )-4 \ln \left (-d x -c -1\right ) b^{2} c^{2}-4 \ln \left (-d x -c +1\right ) b^{2} c^{2}+8 \ln \left (d x +c \right ) b^{2} c^{2}+4 \ln \left (-d x -c -1\right ) a b \,d^{2} x^{2}-8 \ln \left (-d x -c -1\right ) b^{2} c d x -8 \ln \left (-d x -c +1\right ) b^{2} c d x +16 \ln \left (d x +c \right ) b^{2} c d x +8 \ln \left (d x +c \right ) b^{2} d^{2} x^{2}-4 \ln \left (-d x -c +1\right ) a b \,c^{2}-4 \ln \left (-d x -c -1\right ) b^{2} d^{2} x^{2}-4 \ln \left (-d x -c +1\right ) b^{2} d^{2} x^{2}}{8 e^{3} \left (d x +c \right )^{2} d}\)	\(568\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (108) = 216\).
time = 0.28, size = 309, normalized size = 2.60 \begin {gather*} \frac {1}{2} \, {\left (d {\left (\frac {e^{\left (-3\right )} \log \left (d x + c + 1\right )}{d^{2}} - \frac {e^{\left (-3\right )} \log \left (d x + c - 1\right )}{d^{2}} - \frac {2}{d^{3} x e^{3} + c d^{2} e^{3}}\right )} - \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{d^{3} x^{2} e^{3} + 2 \, c d^{2} x e^{3} + c^{2} d e^{3}}\right )} a b - \frac {1}{8} \, {\left (d^{2} {\left (\frac {{\left (\log \left (d x + c + 1\right )^{2} - 2 \, \log \left (d x + c + 1\right ) \log \left (d x + c - 1\right ) + \log \left (d x + c - 1\right )^{2} + 4 \, \log \left (d x + c - 1\right )\right )} e^{\left (-3\right )}}{d^{3}} + \frac {4 \, e^{\left (-3\right )} \log \left (d x + c + 1\right )}{d^{3}} - \frac {8 \, e^{\left (-3\right )} \log \left (d x + c\right )}{d^{3}}\right )} - 4 \, d {\left (\frac {e^{\left (-3\right )} \log \left (d x + c + 1\right )}{d^{2}} - \frac {e^{\left (-3\right )} \log \left (d x + c - 1\right )}{d^{2}} - \frac {2}{d^{3} x e^{3} + c d^{2} e^{3}}\right )} \operatorname {artanh}\left (d x + c\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (d x + c\right )^{2}}{2 \, {\left (d^{3} x^{2} e^{3} + 2 \, c d^{2} x e^{3} + c^{2} d e^{3}\right )}} - \frac {a^{2}}{2 \, {\left (d^{3} x^{2} e^{3} + 2 \, c d^{2} x e^{3} + c^{2} d e^{3}\right )}} \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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (108) = 216\).
time = 0.38, size = 348, normalized size = 2.92 \begin {gather*} -\frac {8 \, a b d x + 8 \, a b c - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - b^{2}\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )^{2} + 4 \, a^{2} - 4 \, {\left ({\left (a b - b^{2}\right )} d^{2} x^{2} + 2 \, {\left (a b - b^{2}\right )} c d x + {\left (a b - b^{2}\right )} c^{2}\right )} \log \left (d x + c + 1\right ) - 8 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (d x + c\right ) + 4 \, {\left ({\left (a b + b^{2}\right )} d^{2} x^{2} + 2 \, {\left (a b + b^{2}\right )} c d x + {\left (a b + b^{2}\right )} c^{2}\right )} \log \left (d x + c - 1\right ) + 4 \, {\left (b^{2} d x + b^{2} c + a b\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{8 \, {\left ({\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )} \cosh \left (1\right )^{3} + 3 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )} \cosh \left (1\right )^{2} \sinh \left (1\right ) + 3 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )} \cosh \left (1\right ) \sinh \left (1\right )^{2} + {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )} \sinh \left (1\right )^{3}\right )}} \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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1102 vs. \(2 (100) = 200\).
time = 1.59, size = 1102, normalized size = 9.26 \begin {gather*} \begin {cases} - \frac {a^{2}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 a b c^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {4 a b c d x \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 a b c}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 a b d^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 a b d x}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 a b \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} c^{2} \log {\left (\frac {c}{d} + x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b^{2} c^{2} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {b^{2} c^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} c^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {4 b^{2} c d x \log {\left (\frac {c}{d} + x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {4 b^{2} c d x \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} c d x \operatorname {atanh}^{2}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {4 b^{2} c d x \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b^{2} c \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} d^{2} x^{2} \log {\left (\frac {c}{d} + x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b^{2} d^{2} x^{2} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {b^{2} d^{2} x^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b^{2} d^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {2 b^{2} d x \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atanh}{\left (c \right )}\right )^{2}}{c^{3} e^{3}} & \text {otherwise} \end {cases} \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)**3,x)

[Out]

Piecewise((-a**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 2*a*b*c**2*atanh(c + d*x)/(2*c**2*d*e*
*3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 4*a*b*c*d*x*atanh(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**
3*e**3*x**2) - 2*a*b*c/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 2*a*b*d**2*x**2*atanh(c + d*x)/(
2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*a*b*d*x/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3
*x**2) - 2*a*b*atanh(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 2*b**2*c**2*log(c/d + x)/
(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*b**2*c**2*log(c/d + x + 1/d)/(2*c**2*d*e**3 + 4*c*d**
2*e**3*x + 2*d**3*e**3*x**2) + b**2*c**2*atanh(c + d*x)**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2
) + 2*b**2*c**2*atanh(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 4*b**2*c*d*x*log(c/d + x
)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 4*b**2*c*d*x*log(c/d + x + 1/d)/(2*c**2*d*e**3 + 4*c*
d**2*e**3*x + 2*d**3*e**3*x**2) + 2*b**2*c*d*x*atanh(c + d*x)**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**
3*x**2) + 4*b**2*c*d*x*atanh(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*b**2*c*atanh(c
+ d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 2*b**2*d**2*x**2*log(c/d + x)/(2*c**2*d*e**3 + 4
*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*b**2*d**2*x**2*log(c/d + x + 1/d)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*
d**3*e**3*x**2) + b**2*d**2*x**2*atanh(c + d*x)**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + 2*b*
*2*d**2*x**2*atanh(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - 2*b**2*d*x*atanh(c + d*x)/(
2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - b**2*atanh(c + d*x)**2/(2*c**2*d*e**3 + 4*c*d**2*e**3*x
+ 2*d**3*e**3*x**2), Ne(d, 0)), (x*(a + b*atanh(c))**2/(c**3*e**3), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (113) = 226\).
time = 0.42, size = 375, normalized size = 3.15 \begin {gather*} \frac {1}{4} \, {\left (\frac {{\left (d x + c + 1\right )} b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )^{2}}{{\left (\frac {{\left (d x + c + 1\right )}^{2} d^{2} e^{3}}{{\left (d x + c - 1\right )}^{2}} + \frac {2 \, {\left (d x + c + 1\right )} d^{2} e^{3}}{d x + c - 1} + d^{2} e^{3}\right )} {\left (d x + c - 1\right )}} + \frac {2 \, {\left (\frac {2 \, {\left (d x + c + 1\right )} a b}{d x + c - 1} + \frac {{\left (d x + c + 1\right )} b^{2}}{d x + c - 1} + b^{2}\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{\frac {{\left (d x + c + 1\right )}^{2} d^{2} e^{3}}{{\left (d x + c - 1\right )}^{2}} + \frac {2 \, {\left (d x + c + 1\right )} d^{2} e^{3}}{d x + c - 1} + d^{2} e^{3}} + \frac {4 \, {\left (\frac {{\left (d x + c + 1\right )} a^{2}}{d x + c - 1} + \frac {{\left (d x + c + 1\right )} a b}{d x + c - 1} + a b\right )}}{\frac {{\left (d x + c + 1\right )}^{2} d^{2} e^{3}}{{\left (d x + c - 1\right )}^{2}} + \frac {2 \, {\left (d x + c + 1\right )} d^{2} e^{3}}{d x + c - 1} + d^{2} e^{3}} + \frac {2 \, b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} - 1\right )}{d^{2} e^{3}} - \frac {2 \, b^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d^{2} e^{3}}\right )} {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )} \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)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.42, size = 776, normalized size = 6.52 \begin {gather*} {\ln \left (1-d\,x-c\right )}^2\,\left (\frac {b^2}{8\,d\,e^3}-\frac {b^2}{2\,d\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}\right )+{\ln \left (c+d\,x+1\right )}^2\,\left (\frac {b^2}{8\,d\,e^3}-\frac {b^2}{8\,d^2\,e^3\,\left (2\,c\,x+d\,x^2+\frac {c^2}{d}\right )}\right )+\ln \left (1-d\,x-c\right )\,\left (\ln \left (c+d\,x+1\right )\,\left (\frac {b^2}{2\,d\,\left (2\,c^2\,e^3+4\,c\,d\,e^3\,x+2\,d^2\,e^3\,x^2\right )}-\frac {b^2\,\left (c^2+2\,c\,d\,x+d^2\,x^2\right )}{2\,d\,\left (2\,c^2\,e^3+4\,c\,d\,e^3\,x+2\,d^2\,e^3\,x^2\right )}\right )+\frac {b^2}{2\,d\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}+\frac {b\,\left (4\,a-b\right )}{2\,d\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}-\frac {b^2\,\left (x\,\left (4\,c\,d-d+d\,\left (2\,c-1\right )\right )-c+c^2+c\,\left (2\,c-1\right )+3\,d^2\,x^2+1\right )}{2\,d\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}+\frac {b^2\,\left (x\,\left (2\,d\,e^3+d\,\left (4\,c\,e^3+2\,e^3\right )+8\,c\,d\,e^3\right )+2\,c\,e^3+2\,e^3+c\,\left (4\,c\,e^3+2\,e^3\right )+2\,c^2\,e^3+6\,d^2\,e^3\,x^2\right )}{4\,d\,e^3\,\left (4\,c^2\,e^3+8\,c\,d\,e^3\,x+4\,d^2\,e^3\,x^2\right )}\right )-\frac {\frac {a^2+2\,b\,c\,a}{2\,d}+a\,b\,x}{c^2\,e^3+2\,c\,d\,e^3\,x+d^2\,e^3\,x^2}-\frac {\ln \left (c+d\,x+1\right )\,\left (x\,\left (\frac {2\,b^2\,c+b^2}{4\,d\,e^3}+\frac {b^2\,c}{4\,d\,e^3}-\frac {b^2\,\left (3\,c-1\right )}{4\,d\,e^3}\right )+\frac {b^2\,c^2+b^2\,c+b^2+4\,a\,b}{8\,d^2\,e^3}-\frac {b^2\,\left (\frac {c^2-c+1}{2\,d}+\frac {c\,\left (2\,c-1\right )}{2\,d}\right )}{4\,d\,e^3}+\frac {c\,\left (2\,b^2\,c+b^2\right )}{8\,d^2\,e^3}\right )}{2\,c\,x+d\,x^2+\frac {c^2}{d}}+\frac {b^2\,\ln \left (c+d\,x\right )}{d\,e^3}-\frac {\ln \left (c+d\,x-1\right )\,\left (b^2+a\,b\right )}{2\,d\,e^3}+\frac {\ln \left (c+d\,x+1\right )\,\left (a\,b-b^2\right )}{2\,d\,e^3} \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)^3,x)

[Out]

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

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