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

Optimal. Leaf size=63 \[ -\frac{a+b \tanh ^{-1}(c+d x)}{2 d e^3 (c+d x)^2}-\frac{b}{2 d e^3 (c+d x)}+\frac{b \tanh ^{-1}(c+d x)}{2 d e^3} \]

[Out]

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

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Rubi [A]  time = 0.048815, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {6107, 12, 5916, 325, 206} \[ -\frac{a+b \tanh ^{-1}(c+d x)}{2 d e^3 (c+d x)^2}-\frac{b}{2 d e^3 (c+d x)}+\frac{b \tanh ^{-1}(c+d x)}{2 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0557101, size = 100, normalized size = 1.59 \[ -\frac{a}{2 d e^3 (c+d x)^2}-\frac{b}{2 d e^3 (c+d x)}-\frac{b \log (-c-d x+1)}{4 d e^3}+\frac{b \log (c+d x+1)}{4 d e^3}-\frac{b \tanh ^{-1}(c+d x)}{2 d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 88, normalized size = 1.4 \begin{align*} -{\frac{a}{2\,d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{b{\it Artanh} \left ( dx+c \right ) }{2\,d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{b}{2\,d{e}^{3} \left ( dx+c \right ) }}-{\frac{b\ln \left ( dx+c-1 \right ) }{4\,d{e}^{3}}}+{\frac{b\ln \left ( dx+c+1 \right ) }{4\,d{e}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.9821, size = 177, normalized size = 2.81 \begin{align*} -\frac{1}{4} \,{\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 )} b - \frac{a}{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))/(d*e*x+c*e)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.2796, size = 194, normalized size = 3.08 \begin{align*} -\frac{2 \, b d x + 2 \, b c -{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} - b\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 2 \, a}{4 \,{\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))/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.22598, size = 196, normalized size = 3.11 \begin{align*} \frac{b d^{2} x^{2} \log \left (d x + c + 1\right ) - b d^{2} x^{2} \log \left (d x + c - 1\right ) + 2 \, b c d x \log \left (d x + c + 1\right ) - 2 \, b c d x \log \left (d x + c - 1\right ) + b c^{2} \log \left (d x + c + 1\right ) - b c^{2} \log \left (d x + c - 1\right ) - 2 \, b d x - 2 \, b c - b \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) - 2 \, a}{4 \,{\left (d^{3} x^{2} e^{3} + 2 \, c d^{2} x e^{3} + 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))/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

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