3.1.13 \(\int \frac {a+b \tanh ^{-1}(c+d x)}{(c e+d e x)^2} \, dx\) [13]

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 = 0.04, 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 = {6242, 12, 6037, 272, 36, 31, 29} \begin {gather*} -\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 {gather*}

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 272

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 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6242

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 {\text {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {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 \text {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 \text {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 \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,(c+d x)^2\right )}{2 d e^2}+\frac {b \text {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 = 0.08, size = 69, normalized size = 1.10 \begin {gather*} -\frac {\frac {2 a}{c+d x}+\frac {2 b \tanh ^{-1}(c+d x)}{c+d x}-2 b \log (c+d x)+b \log \left (1-c^2-2 c d x-d^2 x^2\right )}{2 d e^2} \end {gather*}

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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Maple [A]
time = 0.59, size = 75, normalized size = 1.19

method	result	size
derivativedivides	\(\frac {-\frac {a}{e^{2} \left (d x +c \right )}-\frac {b \arctanh \left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {b \ln \left (d x +c +1\right )}{2 e^{2}}+\frac {b \ln \left (d x +c \right )}{e^{2}}-\frac {b \ln \left (d x +c -1\right )}{2 e^{2}}}{d}\)	\(75\)
default	\(\frac {-\frac {a}{e^{2} \left (d x +c \right )}-\frac {b \arctanh \left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {b \ln \left (d x +c +1\right )}{2 e^{2}}+\frac {b \ln \left (d x +c \right )}{e^{2}}-\frac {b \ln \left (d x +c -1\right )}{2 e^{2}}}{d}\)	\(75\)
risch	\(-\frac {b \ln \left (d x +c +1\right )}{2 d \left (d x +c \right ) e^{2}}-\frac {\ln \left (d^{2} x^{2}+2 c d x +c^{2}-1\right ) b d x -2 \ln \left (-d x -c \right ) b d x +\ln \left (d^{2} x^{2}+2 c d x +c^{2}-1\right ) b c -2 \ln \left (-d x -c \right ) b c -b \ln \left (-d x -c +1\right )+2 a}{2 e^{2} \left (d x +c \right ) d}\)	\(127\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 88, normalized size = 1.40 \begin {gather*} -\frac {1}{2} \, {\left (d {\left (\frac {e^{\left (-2\right )} \log \left (d x + c + 1\right )}{d^{2}} - \frac {2 \, e^{\left (-2\right )} \log \left (d x + c\right )}{d^{2}} + \frac {e^{\left (-2\right )} \log \left (d x + c - 1\right )}{d^{2}}\right )} + \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{d^{2} x e^{2} + c d e^{2}}\right )} b - \frac {a}{d^{2} x e^{2} + c d e^{2}} \end {gather*}

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

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Fricas [A]
time = 0.43, size = 114, normalized size = 1.81 \begin {gather*} -\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 ({\left (d^{2} x + c d\right )} \cosh \left (1\right )^{2} + 2 \, {\left (d^{2} x + c d\right )} \cosh \left (1\right ) \sinh \left (1\right ) + {\left (d^{2} x + c d\right )} \sinh \left (1\right )^{2}\right )}} \end {gather*}

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*x + c*d)*cosh(1)^2 + 2*(d^2*x + c*d)*cosh(1)*sinh(1) + (d^2*x + c*d)*sinh(1)^2)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (53) = 106\).
time = 1.20, size = 235, normalized size = 3.73 \begin {gather*} \begin {cases} \frac {\tilde {\infty } a}{e^{2} x} & \text {for}\: c = 0 \wedge d = 0 \\\frac {x \left (a + b \operatorname {atanh}{\left (c \right )}\right )}{c^{2} e^{2}} & \text {for}\: d = 0 \\\tilde {\infty } a x & \text {for}\: c = - d x \\- \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} \end {gather*}

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)), (x*(a + b*atanh(c))/(c**2*e**2), Eq(d, 0)), (zoo*a*x, Eq(c, -
d*x)), (-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*atan
h(c + d*x)/(c*d*e**2 + d**2*e**2*x), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (61) = 122\).
time = 0.41, size = 152, normalized size = 2.41 \begin {gather*} \frac {1}{2} \, {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )} {\left (\frac {b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{\frac {{\left (d x + c + 1\right )} d^{2} e^{2}}{d x + c - 1} + d^{2} e^{2}} + \frac {2 \, a}{\frac {{\left (d x + c + 1\right )} d^{2} e^{2}}{d x + c - 1} + d^{2} e^{2}} + \frac {b \log \left (-\frac {d x + c + 1}{d x + c - 1} - 1\right )}{d^{2} e^{2}} - \frac {b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d^{2} e^{2}}\right )} \end {gather*}

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

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Mupad [B]
time = 1.37, size = 122, normalized size = 1.94 \begin {gather*} \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} \end {gather*}

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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