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

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

[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)

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Rubi [A]  time = 0.174176, antiderivative size = 119, normalized size of antiderivative = 1., 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 = {6107, 12, 5916, 5982, 266, 36, 31, 29, 5948} \[ -\frac{b \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^3}+\frac{b^2 \log (c+d x)}{d e^3}-\frac{b^2 \log \left (1-(c+d x)^2\right )}{2 d e^3} \]

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

Mathematica [A]  time = 0.178121, size = 136, normalized size = 1.14 \[ \frac{-\frac{a^2}{(c+d x)^2}-\frac{2 a b}{c+d x}-b (a+b) \log (-c-d x+1)+b (a-b) \log (c+d x+1)-\frac{2 b \tanh ^{-1}(c+d x) (a+b (c+d x))}{(c+d x)^2}+\frac{b^2 \left (c^2+2 c d x+d^2 x^2-1\right ) \tanh ^{-1}(c+d x)^2}{(c+d x)^2}+2 b^2 \log (c+d x)}{2 d e^3} \]

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]  time = 0.064, size = 371, normalized size = 3.1 \begin{align*} -{\frac{{a}^{2}}{2\,d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}}{2\,d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) }{d{e}^{3} \left ( dx+c \right ) }}-{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) \ln \left ( dx+c-1 \right ) }{2\,d{e}^{3}}}+{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) \ln \left ( dx+c+1 \right ) }{2\,d{e}^{3}}}-{\frac{{b}^{2} \left ( \ln \left ( dx+c-1 \right ) \right ) ^{2}}{8\,d{e}^{3}}}+{\frac{{b}^{2}\ln \left ( dx+c-1 \right ) }{4\,d{e}^{3}}\ln \left ({\frac{1}{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{{b}^{2}\ln \left ( dx+c-1 \right ) }{2\,d{e}^{3}}}+{\frac{{b}^{2}\ln \left ( dx+c \right ) }{d{e}^{3}}}-{\frac{{b}^{2}\ln \left ( dx+c+1 \right ) }{2\,d{e}^{3}}}-{\frac{{b}^{2}}{4\,d{e}^{3}}\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}\ln \left ( dx+c+1 \right ) }{4\,d{e}^{3}}\ln \left ( -{\frac{dx}{2}}-{\frac{c}{2}}+{\frac{1}{2}} \right ) }-{\frac{{b}^{2} \left ( \ln \left ( dx+c+1 \right ) \right ) ^{2}}{8\,d{e}^{3}}}-{\frac{ab{\it Artanh} \left ( dx+c \right ) }{d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{ab}{d{e}^{3} \left ( dx+c \right ) }}-{\frac{ab\ln \left ( dx+c-1 \right ) }{2\,d{e}^{3}}}+{\frac{ab\ln \left ( dx+c+1 \right ) }{2\,d{e}^{3}}} \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)^3,x)

[Out]

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

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Maxima [B]  time = 1.02258, size = 444, normalized size = 3.73 \begin{align*} -\frac{1}{2} \,{\left (d{\left (\frac{2}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac{\log \left (d x + c + 1\right )}{d^{2} e^{3}} + \frac{\log \left (d x + c - 1\right )}{d^{2} e^{3}}\right )} + \frac{2 \, \operatorname{artanh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} a b - \frac{1}{8} \,{\left (d^{2}{\left (\frac{\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 )}{d^{3} e^{3}} + \frac{4 \, \log \left (d x + c + 1\right )}{d^{3} e^{3}} - \frac{8 \, \log \left (d x + c\right )}{d^{3} e^{3}}\right )} + 4 \, d{\left (\frac{2}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac{\log \left (d x + c + 1\right )}{d^{2} e^{3}} + \frac{\log \left (d x + c - 1\right )}{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} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - \frac{a^{2}}{2 \,{\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.23881, size = 598, normalized size = 5.03 \begin{align*} -\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 (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\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)^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*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3
)

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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)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{3}}\,{d x} \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)^3,x, algorithm="giac")

[Out]

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