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]

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

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Rubi [A]  time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 12

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

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 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)^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*}

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Mathematica [A]  time = 0.06, 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]

-1/2*a/(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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fricas [A]  time = 0.57, size = 88, normalized size = 1.40 \[ -\frac {2 \, b d x + 2 \, b c - {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} - b\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right ) + 2 \, a}{4 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

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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giac [B]  time = 0.33, size = 134, normalized size = 2.13 \[ \frac {{\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )} {\left (\frac {{\left (d x + c + 1\right )} b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d x + c - 1} + \frac {2 \, {\left (d x + c + 1\right )} a}{d x + c - 1} + \frac {{\left (d x + c + 1\right )} b}{d x + c - 1} + b\right )}}{2 \, {\left (\frac {{\left (d x + c + 1\right )}^{2} d^{2} e^{3}}{{\left (d x + c - 1\right )}^{2}} + \frac {2 \, {\left (d x + c + 1\right )} d^{2} e^{3}}{d x + c - 1} + d^{2} e^{3}\right )}} \]

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]

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

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maple [A]  time = 0.04, size = 88, normalized size = 1.40 \[ -\frac {a}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {b \arctanh \left (d x +c \right )}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {b}{2 d \,e^{3} \left (d x +c \right )}-\frac {b \ln \left (d x +c -1\right )}{4 d \,e^{3}}+\frac {b \ln \left (d x +c +1\right )}{4 d \,e^{3}} \]

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 = 0.32, size = 131, normalized size = 2.08 \[ -\frac {1}{4} \, {\left (d {\left (\frac {2}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac {\log \left (d x + c + 1\right )}{d^{2} e^{3}} + \frac {\log \left (d x + c - 1\right )}{d^{2} e^{3}}\right )} + \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} b - \frac {a}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

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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mupad [B]  time = 1.73, size = 67, normalized size = 1.06 \[ \frac {b\,\mathrm {atanh}\left (c+d\,x\right )}{2\,d\,e^3}-\frac {\frac {a}{2}+\frac {b\,c}{2}+\frac {b\,\ln \left (c+d\,x+1\right )}{4}-\frac {b\,\ln \left (1-d\,x-c\right )}{4}+\frac {b\,d\,x}{2}}{d\,e^3\,{\left (c+d\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.55, size = 313, normalized size = 4.97 \[ \begin {cases} - \frac {a}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {b c^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {2 b c d x \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b c}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} + \frac {b d^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b d x}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} - \frac {b \operatorname {atanh}{\left (c + d x \right )}}{2 c^{2} d e^{3} + 4 c d^{2} e^{3} x + 2 d^{3} e^{3} x^{2}} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atanh}{\relax (c )}\right )}{c^{3} e^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))/(d*e*x+c*e)**3,x)

[Out]

Piecewise((-a/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + b*c**2*atanh(c + d*x)/(2*c**2*d*e**3 + 4*
c*d**2*e**3*x + 2*d**3*e**3*x**2) + 2*b*c*d*x*atanh(c + d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x*
*2) - b*c/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) + b*d**2*x**2*atanh(c + d*x)/(2*c**2*d*e**3 + 4
*c*d**2*e**3*x + 2*d**3*e**3*x**2) - b*d*x/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2) - b*atanh(c +
d*x)/(2*c**2*d*e**3 + 4*c*d**2*e**3*x + 2*d**3*e**3*x**2), Ne(d, 0)), (x*(a + b*atanh(c))/(c**3*e**3), True))

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