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

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Rubi [A]  time = 0.06, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.07, size = 69, normalized size = 1.10 \[ -\frac {\frac {2 a}{c+d x}+b \log \left (-c^2-2 c d x-d^2 x^2+1\right )-2 b \log (c+d x)+\frac {2 b \tanh ^{-1}(c+d x)}{c+d x}}{2 d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((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])/(d*e^2)

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fricas [A]  time = 0.82, size = 85, normalized size = 1.35 \[ -\frac {{\left (b d x + b c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right ) - 2 \, {\left (b d x + b c\right )} \log \left (d x + c\right ) + b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right ) + 2 \, a}{2 \, {\left (d^{2} e^{2} x + c d e^{2}\right )}} \]

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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giac [B]  time = 0.18, size = 137, normalized size = 2.17 \[ \frac {{\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )} {\left (\frac {{\left (d x + c + 1\right )} b \log \left (-\frac {d x + c + 1}{d x + c - 1} - 1\right )}{d x + c - 1} + b \log \left (-\frac {d x + c + 1}{d x + c - 1} - 1\right ) - \frac {{\left (d x + c + 1\right )} b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} + 2 \, a\right )}}{2 \, {\left (\frac {{\left (d x + c + 1\right )} d^{2} e^{2}}{d x + c - 1} + d^{2} e^{2}\right )}} \]

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

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maple [A]  time = 0.04, size = 86, normalized size = 1.37 \[ -\frac {a}{d \,e^{2} \left (d x +c \right )}-\frac {b \arctanh \left (d x +c \right )}{d \,e^{2} \left (d x +c \right )}+\frac {b \ln \left (d x +c \right )}{d \,e^{2}}-\frac {b \ln \left (d x +c -1\right )}{2 d \,e^{2}}-\frac {b \ln \left (d x +c +1\right )}{2 d \,e^{2}} \]

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

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maxima [A]  time = 0.32, size = 95, normalized size = 1.51 \[ -\frac {1}{2} \, {\left (d {\left (\frac {\log \left (d x + c + 1\right )}{d^{2} e^{2}} - \frac {2 \, \log \left (d x + c\right )}{d^{2} e^{2}} + \frac {\log \left (d x + c - 1\right )}{d^{2} e^{2}}\right )} + \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} b - \frac {a}{d^{2} e^{2} x + c d e^{2}} \]

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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mupad [B]  time = 1.37, size = 122, normalized size = 1.94 \[ \frac {b\,\ln \left (1-d\,x-c\right )}{2\,x\,d^2\,e^2+2\,c\,d\,e^2}-\frac {b\,\ln \left (c+d\,x+1\right )}{2\,\left (x\,d^2\,e^2+c\,d\,e^2\right )}-\frac {a}{x\,d^2\,e^2+c\,d\,e^2}-\frac {b\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2-1\right )}{2\,d\,e^2}+\frac {b\,\ln \left (c+d\,x\right )}{d\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.83, size = 270, normalized size = 4.29 \[ \begin {cases} \frac {\tilde {\infty } a}{e^{2} x} & \text {for}\: c = 0 \wedge d = 0 \\\frac {- \frac {a}{x} + b d \log {\relax (x )} - b d \log {\left (x - \frac {1}{d} \right )} - b d \operatorname {atanh}{\left (d x \right )} - \frac {b \operatorname {atanh}{\left (d x \right )}}{x}}{d^{2} e^{2}} & \text {for}\: c = 0 \\\frac {x \left (a + b \operatorname {atanh}{\relax (c )}\right )}{c^{2} e^{2}} & \text {for}\: d = 0 \\- \frac {a}{c d e^{2} + d^{2} e^{2} x} + \frac {b c \log {\left (\frac {c}{d} + x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b c \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {b c \operatorname {atanh}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {b d x \log {\left (\frac {c}{d} + x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b d x \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {b d x \operatorname {atanh}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b \operatorname {atanh}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} & \text {otherwise} \end {cases} \]

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((zoo*a/(e**2*x), Eq(c, 0) & Eq(d, 0)), ((-a/x + b*d*log(x) - b*d*log(x - 1/d) - b*d*atanh(d*x) - b*a
tanh(d*x)/x)/(d**2*e**2), Eq(c, 0)), (x*(a + b*atanh(c))/(c**2*e**2), Eq(d, 0)), (-a/(c*d*e**2 + d**2*e**2*x)
+ b*c*log(c/d + x)/(c*d*e**2 + d**2*e**2*x) - b*c*log(c/d + x + 1/d)/(c*d*e**2 + d**2*e**2*x) + b*c*atanh(c +
d*x)/(c*d*e**2 + d**2*e**2*x) + b*d*x*log(c/d + x)/(c*d*e**2 + d**2*e**2*x) - b*d*x*log(c/d + x + 1/d)/(c*d*e*
*2 + d**2*e**2*x) + b*d*x*atanh(c + d*x)/(c*d*e**2 + d**2*e**2*x) - b*atanh(c + d*x)/(c*d*e**2 + d**2*e**2*x),
True))

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