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

Optimal. Leaf size=93 \[ -\frac {a+b \tanh ^{-1}(c x)}{e (d+e x)}-\frac {b c \log (d+e x)}{c^2 d^2-e^2}-\frac {b c \log (1-c x)}{2 e (c d+e)}+\frac {b c \log (c x+1)}{2 e (c d-e)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5926, 706, 31, 633} \[ -\frac {a+b \tanh ^{-1}(c x)}{e (d+e x)}-\frac {b c \log (d+e x)}{c^2 d^2-e^2}-\frac {b c \log (1-c x)}{2 e (c d+e)}+\frac {b c \log (c x+1)}{2 e (c d-e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 5926

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 102, normalized size = 1.10 \[ -\frac {a}{e (d+e x)}-\frac {b c \log (d+e x)}{c^2 d^2-e^2}-\frac {b c \log (1-c x)}{2 e (c d+e)}-\frac {b c \log (c x+1)}{2 e (e-c d)}-\frac {b \tanh ^{-1}(c x)}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.94, size = 182, normalized size = 1.96 \[ -\frac {2 \, a c^{2} d^{2} - 2 \, a e^{2} - {\left (b c^{2} d^{2} + b c d e + {\left (b c^{2} d e + b c e^{2}\right )} x\right )} \log \left (c x + 1\right ) + {\left (b c^{2} d^{2} - b c d e + {\left (b c^{2} d e - b c e^{2}\right )} x\right )} \log \left (c x - 1\right ) + 2 \, {\left (b c e^{2} x + b c d e\right )} \log \left (e x + d\right ) + {\left (b c^{2} d^{2} - b e^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{2 \, {\left (c^{2} d^{3} e - d e^{3} + {\left (c^{2} d^{2} e^{2} - e^{4}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(e*x+d)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.18, size = 374, normalized size = 4.02 \[ -\frac {{\left (\frac {{\left (c x + 1\right )} b c d \log \left (\frac {{\left (c x + 1\right )} c d}{c x - 1} - c d + \frac {{\left (c x + 1\right )} e}{c x - 1} + e\right )}{c x - 1} - b c d \log \left (\frac {{\left (c x + 1\right )} c d}{c x - 1} - c d + \frac {{\left (c x + 1\right )} e}{c x - 1} + e\right ) - \frac {{\left (c x + 1\right )} b c d \log \left (-\frac {c x + 1}{c x - 1}\right )}{c x - 1} - 2 \, a c d + \frac {{\left (c x + 1\right )} b e \log \left (\frac {{\left (c x + 1\right )} c d}{c x - 1} - c d + \frac {{\left (c x + 1\right )} e}{c x - 1} + e\right )}{c x - 1} + b e \log \left (\frac {{\left (c x + 1\right )} c d}{c x - 1} - c d + \frac {{\left (c x + 1\right )} e}{c x - 1} + e\right ) - \frac {{\left (c x + 1\right )} b e \log \left (-\frac {c x + 1}{c x - 1}\right )}{c x - 1} + 2 \, a e\right )} c}{\frac {{\left (c x + 1\right )} c^{3} d^{3}}{c x - 1} - c^{3} d^{3} + \frac {{\left (c x + 1\right )} c^{2} d^{2} e}{c x - 1} + c^{2} d^{2} e - \frac {{\left (c x + 1\right )} c d e^{2}}{c x - 1} + c d e^{2} - \frac {{\left (c x + 1\right )} e^{3}}{c x - 1} - e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(e*x+d)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 114, normalized size = 1.23 \[ -\frac {c a}{\left (c x e +c d \right ) e}-\frac {c b \arctanh \left (c x \right )}{\left (c x e +c d \right ) e}-\frac {c b \ln \left (c x e +c d \right )}{\left (c d +e \right ) \left (c d -e \right )}-\frac {c b \ln \left (c x -1\right )}{e \left (2 c d +2 e \right )}+\frac {c b \ln \left (c x +1\right )}{e \left (2 c d -2 e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 99, normalized size = 1.06 \[ \frac {1}{2} \, {\left (c {\left (\frac {\log \left (c x + 1\right )}{c d e - e^{2}} - \frac {\log \left (c x - 1\right )}{c d e + e^{2}} - \frac {2 \, \log \left (e x + d\right )}{c^{2} d^{2} - e^{2}}\right )} - \frac {2 \, \operatorname {artanh}\left (c x\right )}{e^{2} x + d e}\right )} b - \frac {a}{e^{2} x + d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(e*x+d)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.60, size = 115, normalized size = 1.24 \[ -\frac {d^2\,\left (\frac {b\,c\,\ln \left (c^2\,x^2-1\right )}{2}-b\,c\,\ln \left (d+e\,x\right )+a\,c^2\,x+b\,c^2\,x\,\mathrm {atanh}\left (c\,x\right )\right )+d\,e\,\left (b\,\mathrm {atanh}\left (c\,x\right )-b\,c\,x\,\ln \left (d+e\,x\right )+\frac {b\,c\,x\,\ln \left (c^2\,x^2-1\right )}{2}\right )-a\,e^2\,x}{d\,\left (e^2-c^2\,d^2\right )\,\left (d+e\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(d + e*x)^2,x)

[Out]

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

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sympy [A]  time = 3.62, size = 690, normalized size = 7.42 \[ \begin {cases} \frac {a x + b x \operatorname {atanh}{\left (c x \right )} + \frac {b \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b \operatorname {atanh}{\left (c x \right )}}{c}}{d^{2}} & \text {for}\: e = 0 \\- \frac {a}{d e + e^{2} x} & \text {for}\: c = 0 \\- \frac {2 a d}{2 d^{2} e + 2 d e^{2} x} + \frac {b d \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{2 d^{2} e + 2 d e^{2} x} + \frac {b d}{2 d^{2} e + 2 d e^{2} x} - \frac {b e x \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{2 d^{2} e + 2 d e^{2} x} & \text {for}\: c = - \frac {e}{d} \\- \frac {2 a d}{2 d^{2} e + 2 d e^{2} x} - \frac {b d \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{2 d^{2} e + 2 d e^{2} x} - \frac {b d}{2 d^{2} e + 2 d e^{2} x} + \frac {b e x \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{2 d^{2} e + 2 d e^{2} x} & \text {for}\: c = \frac {e}{d} \\\tilde {\infty } \left (a x + b x \operatorname {atanh}{\left (c x \right )} + \frac {b \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b \operatorname {atanh}{\left (c x \right )}}{c}\right ) & \text {for}\: d = - e x \\- \frac {a c^{2} d^{2}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} + \frac {a e^{2}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} + \frac {b c^{2} d e x \operatorname {atanh}{\left (c x \right )}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} + \frac {b c d e \log {\left (x - \frac {1}{c} \right )}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} - \frac {b c d e \log {\left (\frac {d}{e} + x \right )}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} + \frac {b c d e \operatorname {atanh}{\left (c x \right )}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} + \frac {b c e^{2} x \log {\left (x - \frac {1}{c} \right )}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} - \frac {b c e^{2} x \log {\left (\frac {d}{e} + x \right )}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} + \frac {b c e^{2} x \operatorname {atanh}{\left (c x \right )}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} + \frac {b e^{2} \operatorname {atanh}{\left (c x \right )}}{c^{2} d^{3} e + c^{2} d^{2} e^{2} x - d e^{3} - e^{4} x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((a*x + b*x*atanh(c*x) + b*log(x - 1/c)/c + b*atanh(c*x)/c)/d**2, Eq(e, 0)), (-a/(d*e + e**2*x), Eq(
c, 0)), (-2*a*d/(2*d**2*e + 2*d*e**2*x) + b*d*atanh(e*x/d)/(2*d**2*e + 2*d*e**2*x) + b*d/(2*d**2*e + 2*d*e**2*
x) - b*e*x*atanh(e*x/d)/(2*d**2*e + 2*d*e**2*x), Eq(c, -e/d)), (-2*a*d/(2*d**2*e + 2*d*e**2*x) - b*d*atanh(e*x
/d)/(2*d**2*e + 2*d*e**2*x) - b*d/(2*d**2*e + 2*d*e**2*x) + b*e*x*atanh(e*x/d)/(2*d**2*e + 2*d*e**2*x), Eq(c,
e/d)), (zoo*(a*x + b*x*atanh(c*x) + b*log(x - 1/c)/c + b*atanh(c*x)/c), Eq(d, -e*x)), (-a*c**2*d**2/(c**2*d**3
*e + c**2*d**2*e**2*x - d*e**3 - e**4*x) + a*e**2/(c**2*d**3*e + c**2*d**2*e**2*x - d*e**3 - e**4*x) + b*c**2*
d*e*x*atanh(c*x)/(c**2*d**3*e + c**2*d**2*e**2*x - d*e**3 - e**4*x) + b*c*d*e*log(x - 1/c)/(c**2*d**3*e + c**2
*d**2*e**2*x - d*e**3 - e**4*x) - b*c*d*e*log(d/e + x)/(c**2*d**3*e + c**2*d**2*e**2*x - d*e**3 - e**4*x) + b*
c*d*e*atanh(c*x)/(c**2*d**3*e + c**2*d**2*e**2*x - d*e**3 - e**4*x) + b*c*e**2*x*log(x - 1/c)/(c**2*d**3*e + c
**2*d**2*e**2*x - d*e**3 - e**4*x) - b*c*e**2*x*log(d/e + x)/(c**2*d**3*e + c**2*d**2*e**2*x - d*e**3 - e**4*x
) + b*c*e**2*x*atanh(c*x)/(c**2*d**3*e + c**2*d**2*e**2*x - d*e**3 - e**4*x) + b*e**2*atanh(c*x)/(c**2*d**3*e
+ c**2*d**2*e**2*x - d*e**3 - e**4*x), True))

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