3.22 \(\int \frac{(a+b \tanh ^{-1}(c+d x))^2}{(c e+d e x)^5} \, dx\)

Optimal. Leaf size=172 \[ -\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac{b^2}{12 d e^5 (c+d x)^2}+\frac{2 b^2 \log (c+d x)}{3 d e^5}-\frac{b^2 \log \left (1-(c+d x)^2\right )}{3 d e^5} \]

[Out]

-b^2/(12*d*e^5*(c + d*x)^2) - (b*(a + b*ArcTanh[c + d*x]))/(6*d*e^5*(c + d*x)^3) - (b*(a + b*ArcTanh[c + d*x])
)/(2*d*e^5*(c + d*x)) + (a + b*ArcTanh[c + d*x])^2/(4*d*e^5) - (a + b*ArcTanh[c + d*x])^2/(4*d*e^5*(c + d*x)^4
) + (2*b^2*Log[c + d*x])/(3*d*e^5) - (b^2*Log[1 - (c + d*x)^2])/(3*d*e^5)

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Rubi [A]  time = 0.262447, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {6107, 12, 5916, 5982, 266, 44, 36, 31, 29, 5948} \[ -\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac{b^2}{12 d e^5 (c+d x)^2}+\frac{2 b^2 \log (c+d x)}{3 d e^5}-\frac{b^2 \log \left (1-(c+d x)^2\right )}{3 d e^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])^2/(c*e + d*e*x)^5,x]

[Out]

-b^2/(12*d*e^5*(c + d*x)^2) - (b*(a + b*ArcTanh[c + d*x]))/(6*d*e^5*(c + d*x)^3) - (b*(a + b*ArcTanh[c + d*x])
)/(2*d*e^5*(c + d*x)) + (a + b*ArcTanh[c + d*x])^2/(4*d*e^5) - (a + b*ArcTanh[c + d*x])^2/(4*d*e^5*(c + d*x)^4
) + (2*b^2*Log[c + d*x])/(3*d*e^5) - (b^2*Log[1 - (c + d*x)^2])/(3*d*e^5)

Rule 6107

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]

Rule 12

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5982

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 266

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 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 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 31

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

Rule 29

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

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+d x)\right )^2}{(c e+d e x)^5} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{e^5 x^5} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{x^5} \, dx,x,c+d x\right )}{d e^5}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{b \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{x^4 \left (1-x^2\right )} \, dx,x,c+d x\right )}{2 d e^5}\\ &=-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{b \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{x^4} \, dx,x,c+d x\right )}{2 d e^5}+\frac{b \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{x^2 \left (1-x^2\right )} \, dx,x,c+d x\right )}{2 d e^5}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{b \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{2 d e^5}+\frac{b \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{1-x^2} \, dx,x,c+d x\right )}{2 d e^5}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x^3 \left (1-x^2\right )} \, dx,x,c+d x\right )}{6 d e^5}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{(1-x) x^2} \, dx,x,(c+d x)^2\right )}{12 d e^5}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{2 d e^5}\\ &=-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{b^2 \operatorname{Subst}\left (\int \left (\frac{1}{1-x}+\frac{1}{x^2}+\frac{1}{x}\right ) \, dx,x,(c+d x)^2\right )}{12 d e^5}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{(1-x) x} \, dx,x,(c+d x)^2\right )}{4 d e^5}\\ &=-\frac{b^2}{12 d e^5 (c+d x)^2}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{b^2 \log (c+d x)}{6 d e^5}-\frac{b^2 \log \left (1-(c+d x)^2\right )}{12 d e^5}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,(c+d x)^2\right )}{4 d e^5}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,(c+d x)^2\right )}{4 d e^5}\\ &=-\frac{b^2}{12 d e^5 (c+d x)^2}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac{2 b^2 \log (c+d x)}{3 d e^5}-\frac{b^2 \log \left (1-(c+d x)^2\right )}{3 d e^5}\\ \end{align*}

Mathematica [A]  time = 0.267043, size = 218, normalized size = 1.27 \[ -\frac{\frac{3 a^2}{(c+d x)^4}+\frac{2 b \tanh ^{-1}(c+d x) \left (3 a+b \left (9 c^2 d x+3 c^3+9 c d^2 x^2+c+3 d^3 x^3+d x\right )\right )}{(c+d x)^4}+\frac{6 a b}{c+d x}+\frac{2 a b}{(c+d x)^3}+b (3 a+4 b) \log (-c-d x+1)-b (3 a-4 b) \log (c+d x+1)-\frac{3 b^2 \left (6 c^2 d^2 x^2+4 c^3 d x+c^4+4 c d^3 x^3+d^4 x^4-1\right ) \tanh ^{-1}(c+d x)^2}{(c+d x)^4}+\frac{b^2}{(c+d x)^2}-8 b^2 \log (c+d x)}{12 d e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c + d*x])^2/(c*e + d*e*x)^5,x]

[Out]

-((3*a^2)/(c + d*x)^4 + (2*a*b)/(c + d*x)^3 + b^2/(c + d*x)^2 + (6*a*b)/(c + d*x) + (2*b*(3*a + b*(c + 3*c^3 +
d*x + 9*c^2*d*x + 9*c*d^2*x^2 + 3*d^3*x^3))*ArcTanh[c + d*x])/(c + d*x)^4 - (3*b^2*(-1 + c^4 + 4*c^3*d*x + 6*
c^2*d^2*x^2 + 4*c*d^3*x^3 + d^4*x^4)*ArcTanh[c + d*x]^2)/(c + d*x)^4 + b*(3*a + 4*b)*Log[1 - c - d*x] - 8*b^2*
Log[c + d*x] - (3*a - 4*b)*b*Log[1 + c + d*x])/(12*d*e^5)

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Maple [B]  time = 0.076, size = 431, normalized size = 2.5 \begin{align*} -{\frac{{a}^{2}}{4\,d{e}^{5} \left ( dx+c \right ) ^{4}}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}}{4\,d{e}^{5} \left ( dx+c \right ) ^{4}}}-{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) \ln \left ( dx+c-1 \right ) }{4\,d{e}^{5}}}-{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) }{6\,d{e}^{5} \left ( dx+c \right ) ^{3}}}-{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) }{2\,d{e}^{5} \left ( dx+c \right ) }}+{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) \ln \left ( dx+c+1 \right ) }{4\,d{e}^{5}}}-{\frac{{b}^{2} \left ( \ln \left ( dx+c-1 \right ) \right ) ^{2}}{16\,d{e}^{5}}}+{\frac{{b}^{2}\ln \left ( dx+c-1 \right ) }{8\,d{e}^{5}}\ln \left ({\frac{1}{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{{b}^{2}\ln \left ( dx+c+1 \right ) }{8\,d{e}^{5}}\ln \left ( -{\frac{dx}{2}}-{\frac{c}{2}}+{\frac{1}{2}} \right ) }-{\frac{{b}^{2}}{8\,d{e}^{5}}\ln \left ( -{\frac{dx}{2}}-{\frac{c}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{{b}^{2} \left ( \ln \left ( dx+c+1 \right ) \right ) ^{2}}{16\,d{e}^{5}}}-{\frac{{b}^{2}\ln \left ( dx+c-1 \right ) }{3\,d{e}^{5}}}-{\frac{{b}^{2}}{12\,d{e}^{5} \left ( dx+c \right ) ^{2}}}+{\frac{2\,{b}^{2}\ln \left ( dx+c \right ) }{3\,d{e}^{5}}}-{\frac{{b}^{2}\ln \left ( dx+c+1 \right ) }{3\,d{e}^{5}}}-{\frac{ab{\it Artanh} \left ( dx+c \right ) }{2\,d{e}^{5} \left ( dx+c \right ) ^{4}}}-{\frac{ab\ln \left ( dx+c-1 \right ) }{4\,d{e}^{5}}}-{\frac{ab}{6\,d{e}^{5} \left ( dx+c \right ) ^{3}}}-{\frac{ab}{2\,d{e}^{5} \left ( dx+c \right ) }}+{\frac{ab\ln \left ( dx+c+1 \right ) }{4\,d{e}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^5,x)

[Out]

-1/4/d*a^2/e^5/(d*x+c)^4-1/4/d*b^2/e^5/(d*x+c)^4*arctanh(d*x+c)^2-1/4/d*b^2/e^5*arctanh(d*x+c)*ln(d*x+c-1)-1/6
/d*b^2/e^5*arctanh(d*x+c)/(d*x+c)^3-1/2/d*b^2/e^5*arctanh(d*x+c)/(d*x+c)+1/4/d*b^2/e^5*arctanh(d*x+c)*ln(d*x+c
+1)-1/16/d*b^2/e^5*ln(d*x+c-1)^2+1/8/d*b^2/e^5*ln(d*x+c-1)*ln(1/2+1/2*d*x+1/2*c)+1/8/d*b^2/e^5*ln(-1/2*d*x-1/2
*c+1/2)*ln(d*x+c+1)-1/8/d*b^2/e^5*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2+1/2*d*x+1/2*c)-1/16/d*b^2/e^5*ln(d*x+c+1)^2-1/
3/d*b^2/e^5*ln(d*x+c-1)-1/12*b^2/d/e^5/(d*x+c)^2+2/3*b^2*ln(d*x+c)/d/e^5-1/3/d*b^2/e^5*ln(d*x+c+1)-1/2/d*a*b/e
^5/(d*x+c)^4*arctanh(d*x+c)-1/4/d*a*b/e^5*ln(d*x+c-1)-1/6/d*a*b/e^5/(d*x+c)^3-1/2/d*a*b/e^5/(d*x+c)+1/4/d*a*b/
e^5*ln(d*x+c+1)

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Maxima [B]  time = 1.1316, size = 828, normalized size = 4.81 \begin{align*} -\frac{1}{12} \,{\left (d{\left (\frac{2 \,{\left (3 \, d^{2} x^{2} + 6 \, c d x + 3 \, c^{2} + 1\right )}}{d^{5} e^{5} x^{3} + 3 \, c d^{4} e^{5} x^{2} + 3 \, c^{2} d^{3} e^{5} x + c^{3} d^{2} e^{5}} - \frac{3 \, \log \left (d x + c + 1\right )}{d^{2} e^{5}} + \frac{3 \, \log \left (d x + c - 1\right )}{d^{2} e^{5}}\right )} + \frac{6 \, \operatorname{artanh}\left (d x + c\right )}{d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}}\right )} a b - \frac{1}{48} \,{\left (d^{2}{\left (\frac{3 \,{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \log \left (d x + c + 1\right )^{2} + 3 \,{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \log \left (d x + c - 1\right )^{2} + 2 \,{\left (8 \, d^{2} x^{2} + 16 \, c d x + 8 \, c^{2} - 3 \,{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \log \left (d x + c - 1\right )\right )} \log \left (d x + c + 1\right ) + 16 \,{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \log \left (d x + c - 1\right ) + 4}{d^{5} e^{5} x^{2} + 2 \, c d^{4} e^{5} x + c^{2} d^{3} e^{5}} - \frac{32 \, \log \left (d x + c\right )}{d^{3} e^{5}}\right )} + 4 \, d{\left (\frac{2 \,{\left (3 \, d^{2} x^{2} + 6 \, c d x + 3 \, c^{2} + 1\right )}}{d^{5} e^{5} x^{3} + 3 \, c d^{4} e^{5} x^{2} + 3 \, c^{2} d^{3} e^{5} x + c^{3} d^{2} e^{5}} - \frac{3 \, \log \left (d x + c + 1\right )}{d^{2} e^{5}} + \frac{3 \, \log \left (d x + c - 1\right )}{d^{2} e^{5}}\right )} \operatorname{artanh}\left (d x + c\right )\right )} b^{2} - \frac{b^{2} \operatorname{artanh}\left (d x + c\right )^{2}}{4 \,{\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} - \frac{a^{2}}{4 \,{\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^5,x, algorithm="maxima")

[Out]

-1/12*(d*(2*(3*d^2*x^2 + 6*c*d*x + 3*c^2 + 1)/(d^5*e^5*x^3 + 3*c*d^4*e^5*x^2 + 3*c^2*d^3*e^5*x + c^3*d^2*e^5)
- 3*log(d*x + c + 1)/(d^2*e^5) + 3*log(d*x + c - 1)/(d^2*e^5)) + 6*arctanh(d*x + c)/(d^5*e^5*x^4 + 4*c*d^4*e^5
*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5))*a*b - 1/48*(d^2*((3*(d^2*x^2 + 2*c*d*x + c^2)*log(d*x
+ c + 1)^2 + 3*(d^2*x^2 + 2*c*d*x + c^2)*log(d*x + c - 1)^2 + 2*(8*d^2*x^2 + 16*c*d*x + 8*c^2 - 3*(d^2*x^2 +
2*c*d*x + c^2)*log(d*x + c - 1))*log(d*x + c + 1) + 16*(d^2*x^2 + 2*c*d*x + c^2)*log(d*x + c - 1) + 4)/(d^5*e^
5*x^2 + 2*c*d^4*e^5*x + c^2*d^3*e^5) - 32*log(d*x + c)/(d^3*e^5)) + 4*d*(2*(3*d^2*x^2 + 6*c*d*x + 3*c^2 + 1)/(
d^5*e^5*x^3 + 3*c*d^4*e^5*x^2 + 3*c^2*d^3*e^5*x + c^3*d^2*e^5) - 3*log(d*x + c + 1)/(d^2*e^5) + 3*log(d*x + c
- 1)/(d^2*e^5))*arctanh(d*x + c))*b^2 - 1/4*b^2*arctanh(d*x + c)^2/(d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*
e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5) - 1/4*a^2/(d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^
2*e^5*x + c^4*d*e^5)

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Fricas [B]  time = 2.41814, size = 1181, normalized size = 6.87 \begin{align*} -\frac{24 \, a b d^{3} x^{3} + 24 \, a b c^{3} + 4 \,{\left (18 \, a b c + b^{2}\right )} d^{2} x^{2} + 4 \, b^{2} c^{2} + 8 \, a b c + 8 \,{\left (9 \, a b c^{2} + b^{2} c + a b\right )} d x - 3 \,{\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4} - b^{2}\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )^{2} + 12 \, a^{2} - 4 \,{\left ({\left (3 \, a b - 4 \, b^{2}\right )} d^{4} x^{4} + 4 \,{\left (3 \, a b - 4 \, b^{2}\right )} c d^{3} x^{3} + 6 \,{\left (3 \, a b - 4 \, b^{2}\right )} c^{2} d^{2} x^{2} + 4 \,{\left (3 \, a b - 4 \, b^{2}\right )} c^{3} d x +{\left (3 \, a b - 4 \, b^{2}\right )} c^{4}\right )} \log \left (d x + c + 1\right ) - 32 \,{\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (d x + c\right ) + 4 \,{\left ({\left (3 \, a b + 4 \, b^{2}\right )} d^{4} x^{4} + 4 \,{\left (3 \, a b + 4 \, b^{2}\right )} c d^{3} x^{3} + 6 \,{\left (3 \, a b + 4 \, b^{2}\right )} c^{2} d^{2} x^{2} + 4 \,{\left (3 \, a b + 4 \, b^{2}\right )} c^{3} d x +{\left (3 \, a b + 4 \, b^{2}\right )} c^{4}\right )} \log \left (d x + c - 1\right ) + 4 \,{\left (3 \, b^{2} d^{3} x^{3} + 9 \, b^{2} c d^{2} x^{2} + 3 \, b^{2} c^{3} + b^{2} c +{\left (9 \, b^{2} c^{2} + b^{2}\right )} d x + 3 \, a b\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )}{48 \,{\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^5,x, algorithm="fricas")

[Out]

-1/48*(24*a*b*d^3*x^3 + 24*a*b*c^3 + 4*(18*a*b*c + b^2)*d^2*x^2 + 4*b^2*c^2 + 8*a*b*c + 8*(9*a*b*c^2 + b^2*c +
a*b)*d*x - 3*(b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4 - b^2)*log(-(d*x +
c + 1)/(d*x + c - 1))^2 + 12*a^2 - 4*((3*a*b - 4*b^2)*d^4*x^4 + 4*(3*a*b - 4*b^2)*c*d^3*x^3 + 6*(3*a*b - 4*b^2
)*c^2*d^2*x^2 + 4*(3*a*b - 4*b^2)*c^3*d*x + (3*a*b - 4*b^2)*c^4)*log(d*x + c + 1) - 32*(b^2*d^4*x^4 + 4*b^2*c*
d^3*x^3 + 6*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4)*log(d*x + c) + 4*((3*a*b + 4*b^2)*d^4*x^4 + 4*(3*a*b +
4*b^2)*c*d^3*x^3 + 6*(3*a*b + 4*b^2)*c^2*d^2*x^2 + 4*(3*a*b + 4*b^2)*c^3*d*x + (3*a*b + 4*b^2)*c^4)*log(d*x +
c - 1) + 4*(3*b^2*d^3*x^3 + 9*b^2*c*d^2*x^2 + 3*b^2*c^3 + b^2*c + (9*b^2*c^2 + b^2)*d*x + 3*a*b)*log(-(d*x + c
+ 1)/(d*x + c - 1)))/(d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))**2/(d*e*x+c*e)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.64732, size = 608, normalized size = 3.53 \begin{align*} \frac{{\left (3 \, b^{2} \log \left (\frac{\frac{e}{d x e + c e} + 1}{\frac{e}{d x e + c e} - 1}\right )^{2} - \frac{12 \, b^{2} e \log \left (\frac{\frac{e}{d x e + c e} + 1}{\frac{e}{d x e + c e} - 1}\right )}{d x e + c e} - 12 \, a b \log \left (-\frac{e}{d x e + c e} + 1\right ) - 16 \, b^{2} \log \left (-\frac{e}{d x e + c e} + 1\right ) + 12 \, a b \log \left (-\frac{e}{d x e + c e} - 1\right ) - 16 \, b^{2} \log \left (-\frac{e}{d x e + c e} - 1\right ) - \frac{24 \, a b e}{d x e + c e} - \frac{4 \, b^{2} e^{2}}{{\left (d x e + c e\right )}^{2}} - \frac{4 \, b^{2} e^{3} \log \left (\frac{\frac{e}{d x e + c e} + 1}{\frac{e}{d x e + c e} - 1}\right )}{{\left (d x e + c e\right )}^{3}} - \frac{3 \, b^{2} e^{4} \log \left (\frac{\frac{e}{d x e + c e} + 1}{\frac{e}{d x e + c e} - 1}\right )^{2}}{{\left (d x e + c e\right )}^{4}} - \frac{8 \, a b e^{3}}{{\left (d x e + c e\right )}^{3}} - \frac{12 \, a b e^{4} \log \left (\frac{\frac{e}{d x e + c e} + 1}{\frac{e}{d x e + c e} - 1}\right )}{{\left (d x e + c e\right )}^{4}} - \frac{12 \, a^{2} e^{4}}{{\left (d x e + c e\right )}^{4}}\right )} e^{\left (-5\right )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^5,x, algorithm="giac")

[Out]

1/48*(3*b^2*log((e/(d*x*e + c*e) + 1)/(e/(d*x*e + c*e) - 1))^2 - 12*b^2*e*log((e/(d*x*e + c*e) + 1)/(e/(d*x*e
+ c*e) - 1))/(d*x*e + c*e) - 12*a*b*log(-e/(d*x*e + c*e) + 1) - 16*b^2*log(-e/(d*x*e + c*e) + 1) + 12*a*b*log(
-e/(d*x*e + c*e) - 1) - 16*b^2*log(-e/(d*x*e + c*e) - 1) - 24*a*b*e/(d*x*e + c*e) - 4*b^2*e^2/(d*x*e + c*e)^2
- 4*b^2*e^3*log((e/(d*x*e + c*e) + 1)/(e/(d*x*e + c*e) - 1))/(d*x*e + c*e)^3 - 3*b^2*e^4*log((e/(d*x*e + c*e)
+ 1)/(e/(d*x*e + c*e) - 1))^2/(d*x*e + c*e)^4 - 8*a*b*e^3/(d*x*e + c*e)^3 - 12*a*b*e^4*log((e/(d*x*e + c*e) +
1)/(e/(d*x*e + c*e) - 1))/(d*x*e + c*e)^4 - 12*a^2*e^4/(d*x*e + c*e)^4)*e^(-5)/d