3.107 \(\int \frac {a+b \coth ^{-1}(c+d x)}{(e+f x)^2} \, dx\)
Optimal. Leaf size=115 \[ -\frac {a+b \coth ^{-1}(c+d x)}{f (e+f x)}-\frac {b d \log (-c-d x+1)}{2 f (-c f+d e+f)}+\frac {b d \log (c+d x+1)}{2 f (-c f+d e-f)}-\frac {b d \log (e+f x)}{(-c f+d e+f) (d e-(c+1) f)} \]
[Out]
(-a-b*arccoth(d*x+c))/f/(f*x+e)-1/2*b*d*ln(-d*x-c+1)/f/(-c*f+d*e+f)+1/2*b*d*ln(d*x+c+1)/f/(-c*f+d*e-f)-b*d*ln(
f*x+e)/(-c*f+d*e-f)/(-c*f+d*e+f)
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Rubi [A] time = 0.17, antiderivative size = 115, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used =
{6110, 1982, 705, 31, 632} \[ -\frac {a+b \coth ^{-1}(c+d x)}{f (e+f x)}-\frac {b d \log (-c-d x+1)}{2 f (-c f+d e+f)}+\frac {b d \log (c+d x+1)}{2 f (-c f+d e-f)}-\frac {b d \log (e+f x)}{(-c f+d e+f) (d e-(c+1) f)} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcCoth[c + d*x])/(e + f*x)^2,x]
[Out]
-((a + b*ArcCoth[c + d*x])/(f*(e + f*x))) - (b*d*Log[1 - c - d*x])/(2*f*(d*e + f - c*f)) + (b*d*Log[1 + c + d*
x])/(2*f*(d*e - f - c*f)) - (b*d*Log[e + f*x])/((d*e + f - c*f)*(d*e - (1 + c)*f))
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 632
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]
Rule 705
Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
Rule 1982
Int[(u_)^(m_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^p, x] /; FreeQ[{m, p}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] && !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])
Rule 6110
Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(
m + 1)*(a + b*ArcCoth[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*A
rcCoth[c + d*x])^(p - 1))/(1 - (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -
1]
Rubi steps
\begin {align*} \int \frac {a+b \coth ^{-1}(c+d x)}{(e+f x)^2} \, dx &=-\frac {a+b \coth ^{-1}(c+d x)}{f (e+f x)}+\frac {(b d) \int \frac {1}{(e+f x) \left (1-(c+d x)^2\right )} \, dx}{f}\\ &=-\frac {a+b \coth ^{-1}(c+d x)}{f (e+f x)}+\frac {(b d) \int \frac {1}{(e+f x) \left (1-c^2-2 c d x-d^2 x^2\right )} \, dx}{f}\\ &=-\frac {a+b \coth ^{-1}(c+d x)}{f (e+f x)}+\frac {(b d) \int \frac {-d^2 e+2 c d f+d^2 f x}{1-c^2-2 c d x-d^2 x^2} \, dx}{f \left (-d^2 e^2+2 c d e f+\left (1-c^2\right ) f^2\right )}+\frac {(b d f) \int \frac {1}{e+f x} \, dx}{-d^2 e^2+2 c d e f+\left (1-c^2\right ) f^2}\\ &=-\frac {a+b \coth ^{-1}(c+d x)}{f (e+f x)}-\frac {b d \log (e+f x)}{(d e-f-c f) (d e+f-c f)}-\frac {\left (b d^3\right ) \int \frac {1}{-d-c d-d^2 x} \, dx}{2 f (d e-f-c f)}+\frac {\left (b d^3\right ) \int \frac {1}{d-c d-d^2 x} \, dx}{2 f (d e+f-c f)}\\ &=-\frac {a+b \coth ^{-1}(c+d x)}{f (e+f x)}-\frac {b d \log (1-c-d x)}{2 f (d e+f-c f)}+\frac {b d \log (1+c+d x)}{2 f (d e-f-c f)}-\frac {b d \log (e+f x)}{(d e-f-c f) (d e+f-c f)}\\ \end {align*}
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Mathematica [A] time = 0.19, size = 125, normalized size = 1.09 \[ \frac {1}{2} \left (-\frac {2 a}{f (e+f x)}-\frac {2 b d \log (e+f x)}{\left (c^2-1\right ) f^2-2 c d e f+d^2 e^2}+\frac {b d \log (-c-d x+1)}{f ((c-1) f-d e)}-\frac {b d \log (c+d x+1)}{f (c f-d e+f)}-\frac {2 b \coth ^{-1}(c+d x)}{f (e+f x)}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcCoth[c + d*x])/(e + f*x)^2,x]
[Out]
((-2*a)/(f*(e + f*x)) - (2*b*ArcCoth[c + d*x])/(f*(e + f*x)) + (b*d*Log[1 - c - d*x])/(f*(-(d*e) + (-1 + c)*f)
) - (b*d*Log[1 + c + d*x])/(f*(-(d*e) + f + c*f)) - (2*b*d*Log[e + f*x])/(d^2*e^2 - 2*c*d*e*f + (-1 + c^2)*f^2
))/2
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fricas [B] time = 0.80, size = 262, normalized size = 2.28 \[ -\frac {2 \, a d^{2} e^{2} - 4 \, a c d e f + 2 \, {\left (a c^{2} - a\right )} f^{2} - {\left (b d^{2} e^{2} - {\left (b c - b\right )} d e f + {\left (b d^{2} e f - {\left (b c - b\right )} d f^{2}\right )} x\right )} \log \left (d x + c + 1\right ) + {\left (b d^{2} e^{2} - {\left (b c + b\right )} d e f + {\left (b d^{2} e f - {\left (b c + b\right )} d f^{2}\right )} x\right )} \log \left (d x + c - 1\right ) + 2 \, {\left (b d f^{2} x + b d e f\right )} \log \left (f x + e\right ) + {\left (b d^{2} e^{2} - 2 \, b c d e f + {\left (b c^{2} - b\right )} f^{2}\right )} \log \left (\frac {d x + c + 1}{d x + c - 1}\right )}{2 \, {\left (d^{2} e^{3} f - 2 \, c d e^{2} f^{2} + {\left (c^{2} - 1\right )} e f^{3} + {\left (d^{2} e^{2} f^{2} - 2 \, c d e f^{3} + {\left (c^{2} - 1\right )} f^{4}\right )} x\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arccoth(d*x+c))/(f*x+e)^2,x, algorithm="fricas")
[Out]
-1/2*(2*a*d^2*e^2 - 4*a*c*d*e*f + 2*(a*c^2 - a)*f^2 - (b*d^2*e^2 - (b*c - b)*d*e*f + (b*d^2*e*f - (b*c - b)*d*
f^2)*x)*log(d*x + c + 1) + (b*d^2*e^2 - (b*c + b)*d*e*f + (b*d^2*e*f - (b*c + b)*d*f^2)*x)*log(d*x + c - 1) +
2*(b*d*f^2*x + b*d*e*f)*log(f*x + e) + (b*d^2*e^2 - 2*b*c*d*e*f + (b*c^2 - b)*f^2)*log((d*x + c + 1)/(d*x + c
- 1)))/(d^2*e^3*f - 2*c*d*e^2*f^2 + (c^2 - 1)*e*f^3 + (d^2*e^2*f^2 - 2*c*d*e*f^3 + (c^2 - 1)*f^4)*x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcoth}\left (d x + c\right ) + a}{{\left (f x + e\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arccoth(d*x+c))/(f*x+e)^2,x, algorithm="giac")
[Out]
integrate((b*arccoth(d*x + c) + a)/(f*x + e)^2, x)
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maple [A] time = 0.04, size = 141, normalized size = 1.23 \[ -\frac {d a}{\left (d f x +d e \right ) f}-\frac {d b \,\mathrm {arccoth}\left (d x +c \right )}{\left (d f x +d e \right ) f}-\frac {d b \ln \left (\left (d x +c \right ) f -c f +d e \right )}{\left (c f -d e -f \right ) \left (c f -d e +f \right )}+\frac {d b \ln \left (d x +c -1\right )}{f \left (2 c f -2 d e -2 f \right )}-\frac {d b \ln \left (d x +c +1\right )}{f \left (2 c f -2 d e +2 f \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arccoth(d*x+c))/(f*x+e)^2,x)
[Out]
-d*a/(d*f*x+d*e)/f-d*b/(d*f*x+d*e)/f*arccoth(d*x+c)-d*b/(c*f-d*e-f)/(c*f-d*e+f)*ln((d*x+c)*f-c*f+d*e)+d*b/f/(2
*c*f-2*d*e-2*f)*ln(d*x+c-1)-d*b/f/(2*c*f-2*d*e+2*f)*ln(d*x+c+1)
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maxima [A] time = 0.31, size = 121, normalized size = 1.05 \[ \frac {1}{2} \, {\left (d {\left (\frac {\log \left (d x + c + 1\right )}{d e f - {\left (c + 1\right )} f^{2}} - \frac {\log \left (d x + c - 1\right )}{d e f - {\left (c - 1\right )} f^{2}} - \frac {2 \, \log \left (f x + e\right )}{d^{2} e^{2} - 2 \, c d e f + {\left (c^{2} - 1\right )} f^{2}}\right )} - \frac {2 \, \operatorname {arcoth}\left (d x + c\right )}{f^{2} x + e f}\right )} b - \frac {a}{f^{2} x + e f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arccoth(d*x+c))/(f*x+e)^2,x, algorithm="maxima")
[Out]
1/2*(d*(log(d*x + c + 1)/(d*e*f - (c + 1)*f^2) - log(d*x + c - 1)/(d*e*f - (c - 1)*f^2) - 2*log(f*x + e)/(d^2*
e^2 - 2*c*d*e*f + (c^2 - 1)*f^2)) - 2*arccoth(d*x + c)/(f^2*x + e*f))*b - a/(f^2*x + e*f)
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mupad [B] time = 2.08, size = 175, normalized size = 1.52 \[ \ln \left (e+f\,x\right )\,\left (\frac {b\,\left (c-1\right )}{2\,e\,\left (d\,e-f\,\left (c-1\right )\right )}-\frac {b\,\left (c+1\right )}{2\,e\,\left (d\,e-f\,\left (c+1\right )\right )}\right )-\frac {a}{x\,f^2+e\,f}-\frac {b\,\ln \left (\frac {1}{c+d\,x}+1\right )}{2\,f\,\left (e+f\,x\right )}-\frac {b\,d\,\ln \left (c+d\,x-1\right )}{2\,f^2-2\,c\,f^2+2\,d\,e\,f}-\frac {b\,d\,\ln \left (c+d\,x+1\right )}{2\,c\,f^2+2\,f^2-2\,d\,e\,f}+\frac {b\,\ln \left (1-\frac {1}{c+d\,x}\right )}{f\,\left (2\,e+2\,f\,x\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*acoth(c + d*x))/(e + f*x)^2,x)
[Out]
log(e + f*x)*((b*(c - 1))/(2*e*(d*e - f*(c - 1))) - (b*(c + 1))/(2*e*(d*e - f*(c + 1)))) - a/(e*f + f^2*x) - (
b*log(1/(c + d*x) + 1))/(2*f*(e + f*x)) - (b*d*log(c + d*x - 1))/(2*f^2 - 2*c*f^2 + 2*d*e*f) - (b*d*log(c + d*
x + 1))/(2*c*f^2 + 2*f^2 - 2*d*e*f) + (b*log(1 - 1/(c + d*x)))/(f*(2*e + 2*f*x))
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sympy [A] time = 9.48, size = 1658, normalized size = 14.42 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*acoth(d*x+c))/(f*x+e)**2,x)
[Out]
Piecewise(((a*x + b*c*acoth(c + d*x)/d + b*x*acoth(c + d*x) + b*log(c/d + x + 1/d)/d - b*acoth(c + d*x)/d)/e**
2, Eq(f, 0)), (-2*a*f/(2*e*f**2 + 2*f**3*x) + b*d*e*acoth(d*e/f + d*x - 1)/(2*e*f**2 + 2*f**3*x) + b*d*f*x*aco
th(d*e/f + d*x - 1)/(2*e*f**2 + 2*f**3*x) - 2*b*f*acoth(d*e/f + d*x - 1)/(2*e*f**2 + 2*f**3*x) - b*f/(2*e*f**2
+ 2*f**3*x), Eq(c, (d*e - f)/f)), (-2*a*f/(2*e*f**2 + 2*f**3*x) - b*d*e*acoth(d*e/f + d*x + 1)/(2*e*f**2 + 2*
f**3*x) - b*d*f*x*acoth(d*e/f + d*x + 1)/(2*e*f**2 + 2*f**3*x) - 2*b*f*acoth(d*e/f + d*x + 1)/(2*e*f**2 + 2*f*
*3*x) + b*f/(2*e*f**2 + 2*f**3*x), Eq(c, (d*e + f)/f)), (zoo*(a*x + b*c*acoth(c + d*x)/d + b*x*acoth(c + d*x)
+ b*log(c/d + x + 1/d)/d - b*acoth(c + d*x)/d), Eq(e, -f*x)), (-(a + b*acoth(c))/(e*f + f**2*x), Eq(d, 0)), (-
a*c**2*f**2/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e
*f**3 - f**4*x) + 2*a*c*d*e*f/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d*
*2*e**2*f**2*x - e*f**3 - f**4*x) - a*d**2*e**2/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x
+ d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + a*f**2/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*
c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) - b*c**2*f**2*acoth(c + d*x)/(c**2*e*f**3 + c
**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + b*c*d*e*f*
acoth(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x
- e*f**3 - f**4*x) - b*c*d*f**2*x*acoth(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x
+ d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + b*d**2*e*f*x*acoth(c + d*x)/(c**2*e*f**3 + c**2*f**4*x
- 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) - b*d*e*f*log(e/f + x)/
(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4
*x) + b*d*e*f*log(c/d + x + 1/d)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f +
d**2*e**2*f**2*x - e*f**3 - f**4*x) - b*d*e*f*acoth(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2
*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) - b*d*f**2*x*log(e/f + x)/(c**2*e*f**3 + c**
2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + b*d*f**2*x*l
og(c/d + x + 1/d)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2
*x - e*f**3 - f**4*x) - b*d*f**2*x*acoth(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c*d*e**2*f**2 - 2*c*d*e*f**3*
x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x) + b*f**2*acoth(c + d*x)/(c**2*e*f**3 + c**2*f**4*x - 2*c
*d*e**2*f**2 - 2*c*d*e*f**3*x + d**2*e**3*f + d**2*e**2*f**2*x - e*f**3 - f**4*x), True))
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