3.13 \(\int \frac{a+b \tanh ^{-1}(c+d x)}{(c e+d e x)^2} \, dx\)
Optimal. Leaf size=63 \[ -\frac{a+b \tanh ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \log (c+d x)}{d e^2}-\frac{b \log \left (1-(c+d x)^2\right )}{2 d e^2} \]
[Out]
-((a + b*ArcTanh[c + d*x])/(d*e^2*(c + d*x))) + (b*Log[c + d*x])/(d*e^2) - (b*Log[1 - (c + d*x)^2])/(2*d*e^2)
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Rubi [A] time = 0.0585116, antiderivative size = 63, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{6107, 12, 5916, 266, 36, 31, 29} \[ -\frac{a+b \tanh ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \log (c+d x)}{d e^2}-\frac{b \log \left (1-(c+d x)^2\right )}{2 d e^2} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTanh[c + d*x])/(c*e + d*e*x)^2,x]
[Out]
-((a + b*ArcTanh[c + d*x])/(d*e^2*(c + d*x))) + (b*Log[c + d*x])/(d*e^2) - (b*Log[1 - (c + d*x)^2])/(2*d*e^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 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]
Rubi steps
\begin{align*} \int \frac{a+b \tanh ^{-1}(c+d x)}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{a+b \tanh ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac{a+b \tanh ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \operatorname{Subst}\left (\int \frac{1}{(1-x) x} \, dx,x,(c+d x)^2\right )}{2 d e^2}\\ &=-\frac{a+b \tanh ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,(c+d x)^2\right )}{2 d e^2}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^2}\\ &=-\frac{a+b \tanh ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac{b \log (c+d x)}{d e^2}-\frac{b \log \left (1-(c+d x)^2\right )}{2 d e^2}\\ \end{align*}
Mathematica [A] time = 0.0676478, size = 69, normalized size = 1.1 \[ -\frac{\frac{2 a}{c+d x}+b \log \left (-c^2-2 c d x-d^2 x^2+1\right )-2 b \log (c+d x)+\frac{2 b \tanh ^{-1}(c+d x)}{c+d x}}{2 d e^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcTanh[c + d*x])/(c*e + d*e*x)^2,x]
[Out]
-((2*a)/(c + d*x) + (2*b*ArcTanh[c + d*x])/(c + d*x) - 2*b*Log[c + d*x] + b*Log[1 - c^2 - 2*c*d*x - d^2*x^2])/
(2*d*e^2)
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Maple [A] time = 0.036, size = 86, normalized size = 1.4 \begin{align*} -{\frac{a}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{b{\it Artanh} \left ( dx+c \right ) }{d{e}^{2} \left ( dx+c \right ) }}-{\frac{b\ln \left ( dx+c-1 \right ) }{2\,d{e}^{2}}}+{\frac{b\ln \left ( dx+c \right ) }{d{e}^{2}}}-{\frac{b\ln \left ( dx+c+1 \right ) }{2\,d{e}^{2}}} \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)^2,x)
[Out]
-1/d*a/e^2/(d*x+c)-1/d*b/e^2/(d*x+c)*arctanh(d*x+c)-1/2/d*b/e^2*ln(d*x+c-1)+b*ln(d*x+c)/d/e^2-1/2/d*b/e^2*ln(d
*x+c+1)
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Maxima [A] time = 0.971283, size = 128, normalized size = 2.03 \begin{align*} -\frac{1}{2} \,{\left (d{\left (\frac{\log \left (d x + c + 1\right )}{d^{2} e^{2}} - \frac{2 \, \log \left (d x + c\right )}{d^{2} e^{2}} + \frac{\log \left (d x + c - 1\right )}{d^{2} e^{2}}\right )} + \frac{2 \, \operatorname{artanh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} b - \frac{a}{d^{2} e^{2} x + c d e^{2}} \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)^2,x, algorithm="maxima")
[Out]
-1/2*(d*(log(d*x + c + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2*e^2) + log(d*x + c - 1)/(d^2*e^2)) + 2*arctanh(d*x +
c)/(d^2*e^2*x + c*d*e^2))*b - a/(d^2*e^2*x + c*d*e^2)
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Fricas [A] time = 2.33926, size = 205, normalized size = 3.25 \begin{align*} -\frac{{\left (b d x + b c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right ) - 2 \,{\left (b d x + b c\right )} \log \left (d x + c\right ) + b \log \left (-\frac{d x + c + 1}{d x + c - 1}\right ) + 2 \, a}{2 \,{\left (d^{2} e^{2} x + c d e^{2}\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)^2,x, algorithm="fricas")
[Out]
-1/2*((b*d*x + b*c)*log(d^2*x^2 + 2*c*d*x + c^2 - 1) - 2*(b*d*x + b*c)*log(d*x + c) + b*log(-(d*x + c + 1)/(d*
x + c - 1)) + 2*a)/(d^2*e^2*x + c*d*e^2)
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Sympy [A] time = 152.211, size = 2749, normalized size = 43.63 \begin{align*} \text{result too large to display} \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)**2,x)
[Out]
Piecewise((x*(a + b*atanh(c))/(c**2*e**2), Eq(d, 0)), (-3*sqrt(3)*a*d**3*x**3/(9*d**4*e**2*x**3 - 9*sqrt(3)*d*
*3*e**2*x**2 + 2*sqrt(3)*d*e**2) + 6*sqrt(3)*a*d*x/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*
e**2) + 4*a/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) + 9*b*d**3*x**3*log(x - sqrt(3)/(
3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) - 9*b*d**3*x**3*log(x - 1/d - sqrt(3)/(
3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) - 9*b*d**3*x**3*atanh(d*x - sqrt(3)/3)/
(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) - 9*sqrt(3)*b*d**2*x**2*log(x - sqrt(3)/(3*d)
)/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) + 9*sqrt(3)*b*d**2*x**2*log(x - 1/d - sqrt(
3)/(3*d))/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) - 9*b*d**2*x**2*atanh(d*x - sqrt(3)
/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) + 9*sqrt(3)*b*d**2*x**2*atanh(d*x - sqrt(
3)/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) + 6*sqrt(3)*b*d*x*atanh(d*x - sqrt(3)/3
)/(9*d**4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) + 2*sqrt(3)*b*log(x - sqrt(3)/(3*d))/(9*d**
4*e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) - 2*sqrt(3)*b*log(x - 1/d - sqrt(3)/(3*d))/(9*d**4*
e**2*x**3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) - 2*sqrt(3)*b*atanh(d*x - sqrt(3)/3)/(9*d**4*e**2*x**
3 - 9*sqrt(3)*d**3*e**2*x**2 + 2*sqrt(3)*d*e**2) + 6*b*atanh(d*x - sqrt(3)/3)/(9*d**4*e**2*x**3 - 9*sqrt(3)*d*
*3*e**2*x**2 + 2*sqrt(3)*d*e**2), Eq(c, -sqrt(3)/3)), (3*sqrt(3)*a*d**3*x**3/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**
3*e**2*x**2 - 2*sqrt(3)*d*e**2) - 6*sqrt(3)*a*d*x/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e
**2) + 4*a/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) + 9*b*d**3*x**3*log(x + sqrt(3)/(3
*d))/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) - 9*b*d**3*x**3*log(x - 1/d + sqrt(3)/(3
*d))/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) - 9*b*d**3*x**3*atanh(d*x + sqrt(3)/3)/(
9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) + 9*sqrt(3)*b*d**2*x**2*log(x + sqrt(3)/(3*d))
/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) - 9*sqrt(3)*b*d**2*x**2*log(x - 1/d + sqrt(3
)/(3*d))/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) - 9*sqrt(3)*b*d**2*x**2*atanh(d*x +
sqrt(3)/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) - 9*b*d**2*x**2*atanh(d*x + sqrt(3
)/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) - 6*sqrt(3)*b*d*x*atanh(d*x + sqrt(3)/3)
/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) - 2*sqrt(3)*b*log(x + sqrt(3)/(3*d))/(9*d**4
*e**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) + 2*sqrt(3)*b*log(x - 1/d + sqrt(3)/(3*d))/(9*d**4*e
**2*x**3 + 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) + 2*sqrt(3)*b*atanh(d*x + sqrt(3)/3)/(9*d**4*e**2*x**3
+ 9*sqrt(3)*d**3*e**2*x**2 - 2*sqrt(3)*d*e**2) + 6*b*atanh(d*x + sqrt(3)/3)/(9*d**4*e**2*x**3 + 9*sqrt(3)*d**
3*e**2*x**2 - 2*sqrt(3)*d*e**2), Eq(c, sqrt(3)/3)), (zoo*a*x, Eq(c, -d*x)), (-a*c**2/(3*c**3*d*e**2 + 3*c**2*d
**2*e**2*x - c*d*e**2 - d**2*e**2*x) + 2*a*c*d*x/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x)
+ a/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) + 3*b*c**3*log(c/d + x)/(3*c**3*d*e**2 + 3*
c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) - 3*b*c**3*log(c/d + x + 1/d)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x -
c*d*e**2 - d**2*e**2*x) + 3*b*c**3*atanh(c + d*x)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*
x) + 3*b*c**2*d*x*log(c/d + x)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) - 3*b*c**2*d*x*lo
g(c/d + x + 1/d)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) + 3*b*c**2*d*x*atanh(c + d*x)/(
3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) - 3*b*c**2*atanh(c + d*x)/(3*c**3*d*e**2 + 3*c**2
*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) - b*c*log(c/d + x)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**
2*e**2*x) + b*c*log(c/d + x + 1/d)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) - b*c*atanh(c
+ d*x)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) - b*d*x*log(c/d + x)/(3*c**3*d*e**2 + 3*
c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) + b*d*x*log(c/d + x + 1/d)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*
d*e**2 - d**2*e**2*x) - b*d*x*atanh(c + d*x)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x) + b
*atanh(c + d*x)/(3*c**3*d*e**2 + 3*c**2*d**2*e**2*x - c*d*e**2 - d**2*e**2*x), True))
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Giac [A] time = 1.14218, size = 115, normalized size = 1.83 \begin{align*} -\frac{1}{2} \, b{\left (\frac{e^{\left (-2\right )} \log \left ({\left | \frac{e^{2}}{{\left (d x e + c e\right )}^{2}} - 1 \right |}\right )}{d} + \frac{e^{\left (-1\right )} \log \left (-\frac{d x + c + 1}{d x + c - 1}\right )}{{\left (d x e + c e\right )} d}\right )} - \frac{a e^{\left (-1\right )}}{{\left (d x e + c e\right )} d} \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)^2,x, algorithm="giac")
[Out]
-1/2*b*(e^(-2)*log(abs(e^2/(d*x*e + c*e)^2 - 1))/d + e^(-1)*log(-(d*x + c + 1)/(d*x + c - 1))/((d*x*e + c*e)*d
)) - a*e^(-1)/((d*x*e + c*e)*d)