3.104 \(\int (e+f x) (a+b \coth ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=97 \[ \frac {(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}+\frac {b (-c f+d e+f)^2 \log (-c-d x+1)}{4 d^2 f}-\frac {b (d e-(c+1) f)^2 \log (c+d x+1)}{4 d^2 f}+\frac {b f x}{2 d} \]

[Out]

1/2*b*f*x/d+1/2*(f*x+e)^2*(a+b*arccoth(d*x+c))/f+1/4*b*(-c*f+d*e+f)^2*ln(-d*x-c+1)/d^2/f-1/4*b*(d*e-(1+c)*f)^2
*ln(d*x+c+1)/d^2/f

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Rubi [A]  time = 0.17, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6112, 5927, 702, 633, 31} \[ \frac {(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}+\frac {b (-c f+d e+f)^2 \log (-c-d x+1)}{4 d^2 f}-\frac {b (d e-(c+1) f)^2 \log (c+d x+1)}{4 d^2 f}+\frac {b f x}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)*(a + b*ArcCoth[c + d*x]),x]

[Out]

(b*f*x)/(2*d) + ((e + f*x)^2*(a + b*ArcCoth[c + d*x]))/(2*f) + (b*(d*e + f - c*f)^2*Log[1 - c - d*x])/(4*d^2*f
) - (b*(d*e - (1 + c)*f)^2*Log[1 + c + d*x])/(4*d^2*f)

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 702

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

Rule 5927

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcCoth[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]

Rule 6112

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCoth[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &
& IGtQ[p, 0]

Rubi steps

\begin {align*} \int (e+f x) \left (a+b \coth ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \left (\frac {d e-c f}{d}+\frac {f x}{d}\right ) \left (a+b \coth ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac {b \operatorname {Subst}\left (\int \frac {\left (\frac {d e-c f}{d}+\frac {f x}{d}\right )^2}{1-x^2} \, dx,x,c+d x\right )}{2 f}\\ &=\frac {(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {f^2}{d^2}+\frac {d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x}{d^2 \left (1-x^2\right )}\right ) \, dx,x,c+d x\right )}{2 f}\\ &=\frac {b f x}{2 d}+\frac {(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac {b \operatorname {Subst}\left (\int \frac {d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x}{1-x^2} \, dx,x,c+d x\right )}{2 d^2 f}\\ &=\frac {b f x}{2 d}+\frac {(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac {\left (b (d e+f-c f)^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,c+d x\right )}{4 d^2 f}+\frac {\left (b (d e-(1+c) f)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x} \, dx,x,c+d x\right )}{4 d^2 f}\\ &=\frac {b f x}{2 d}+\frac {(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}+\frac {b (d e+f-c f)^2 \log (1-c-d x)}{4 d^2 f}-\frac {b (d e-(1+c) f)^2 \log (1+c+d x)}{4 d^2 f}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 138, normalized size = 1.42 \[ a e x+\frac {1}{2} a f x^2+\frac {b \left (c^2-2 c+1\right ) f \log (-c-d x+1)}{4 d^2}+\frac {b \left (-c^2-2 c-1\right ) f \log (c+d x+1)}{4 d^2}+\frac {b e ((c+1) \log (c+d x+1)-(c-1) \log (-c-d x+1))}{2 d}+b e x \coth ^{-1}(c+d x)+\frac {1}{2} b f x^2 \coth ^{-1}(c+d x)+\frac {b f x}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)*(a + b*ArcCoth[c + d*x]),x]

[Out]

a*e*x + (b*f*x)/(2*d) + (a*f*x^2)/2 + b*e*x*ArcCoth[c + d*x] + (b*f*x^2*ArcCoth[c + d*x])/2 + (b*(1 - 2*c + c^
2)*f*Log[1 - c - d*x])/(4*d^2) + (b*(-1 - 2*c - c^2)*f*Log[1 + c + d*x])/(4*d^2) + (b*e*(-((-1 + c)*Log[1 - c
- d*x]) + (1 + c)*Log[1 + c + d*x]))/(2*d)

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fricas [A]  time = 0.45, size = 133, normalized size = 1.37 \[ \frac {2 \, a d^{2} f x^{2} + 2 \, {\left (2 \, a d^{2} e + b d f\right )} x + {\left (2 \, {\left (b c + b\right )} d e - {\left (b c^{2} + 2 \, b c + b\right )} f\right )} \log \left (d x + c + 1\right ) - {\left (2 \, {\left (b c - b\right )} d e - {\left (b c^{2} - 2 \, b c + b\right )} f\right )} \log \left (d x + c - 1\right ) + {\left (b d^{2} f x^{2} + 2 \, b d^{2} e x\right )} \log \left (\frac {d x + c + 1}{d x + c - 1}\right )}{4 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(a+b*arccoth(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x + e\right )} {\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(a+b*arccoth(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)*(b*arccoth(d*x + c) + a), x)

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maple [B]  time = 0.03, size = 184, normalized size = 1.90 \[ \frac {a \,x^{2} f}{2}-\frac {a \,c^{2} f}{2 d^{2}}+a x e +\frac {a c e}{d}+\frac {b \,\mathrm {arccoth}\left (d x +c \right ) f \,x^{2}}{2}-\frac {b \,\mathrm {arccoth}\left (d x +c \right ) f \,c^{2}}{2 d^{2}}+\mathrm {arccoth}\left (d x +c \right ) x b e +\frac {\mathrm {arccoth}\left (d x +c \right ) b c e}{d}+\frac {b f x}{2 d}+\frac {b c f}{2 d^{2}}-\frac {b \ln \left (d x +c -1\right ) c f}{2 d^{2}}+\frac {b \ln \left (d x +c -1\right ) e}{2 d}+\frac {b \ln \left (d x +c -1\right ) f}{4 d^{2}}-\frac {b \ln \left (d x +c +1\right ) c f}{2 d^{2}}+\frac {b \ln \left (d x +c +1\right ) e}{2 d}-\frac {b \ln \left (d x +c +1\right ) f}{4 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*(a+b*arccoth(d*x+c)),x)

[Out]

1/2*a*x^2*f-1/2/d^2*a*c^2*f+a*x*e+1/d*a*c*e+1/2*b*arccoth(d*x+c)*f*x^2-1/2/d^2*b*arccoth(d*x+c)*f*c^2+arccoth(
d*x+c)*x*b*e+1/d*arccoth(d*x+c)*b*c*e+1/2*b*f*x/d+1/2/d^2*b*c*f-1/2/d^2*b*ln(d*x+c-1)*c*f+1/2/d*b*ln(d*x+c-1)*
e+1/4/d^2*b*ln(d*x+c-1)*f-1/2/d^2*b*ln(d*x+c+1)*c*f+1/2/d*b*ln(d*x+c+1)*e-1/4/d^2*b*ln(d*x+c+1)*f

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maxima [A]  time = 0.30, size = 109, normalized size = 1.12 \[ \frac {1}{2} \, a f x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {arcoth}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} b f + a e x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b e}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(a+b*arccoth(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.40, size = 136, normalized size = 1.40 \[ a\,e\,x+\frac {a\,f\,x^2}{2}+\frac {b\,e\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2-1\right )}{2\,d}-\frac {b\,f\,\mathrm {acoth}\left (c+d\,x\right )}{2\,d^2}+\frac {b\,f\,x^2\,\mathrm {acoth}\left (c+d\,x\right )}{2}+\frac {b\,f\,x}{2\,d}+b\,e\,x\,\mathrm {acoth}\left (c+d\,x\right )-\frac {b\,c^2\,f\,\mathrm {acoth}\left (c+d\,x\right )}{2\,d^2}-\frac {b\,c\,f\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2-1\right )}{2\,d^2}+\frac {b\,c\,e\,\mathrm {acoth}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)*(a + b*acoth(c + d*x)),x)

[Out]

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

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sympy [A]  time = 2.27, size = 173, normalized size = 1.78 \[ \begin {cases} a e x + \frac {a f x^{2}}{2} - \frac {b c^{2} f \operatorname {acoth}{\left (c + d x \right )}}{2 d^{2}} + \frac {b c e \operatorname {acoth}{\left (c + d x \right )}}{d} - \frac {b c f \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{d^{2}} + \frac {b c f \operatorname {acoth}{\left (c + d x \right )}}{d^{2}} + b e x \operatorname {acoth}{\left (c + d x \right )} + \frac {b f x^{2} \operatorname {acoth}{\left (c + d x \right )}}{2} + \frac {b e \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{d} - \frac {b e \operatorname {acoth}{\left (c + d x \right )}}{d} + \frac {b f x}{2 d} - \frac {b f \operatorname {acoth}{\left (c + d x \right )}}{2 d^{2}} & \text {for}\: d \neq 0 \\\left (a + b \operatorname {acoth}{\relax (c )}\right ) \left (e x + \frac {f x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(a+b*acoth(d*x+c)),x)

[Out]

Piecewise((a*e*x + a*f*x**2/2 - b*c**2*f*acoth(c + d*x)/(2*d**2) + b*c*e*acoth(c + d*x)/d - b*c*f*log(c/d + x
+ 1/d)/d**2 + b*c*f*acoth(c + d*x)/d**2 + b*e*x*acoth(c + d*x) + b*f*x**2*acoth(c + d*x)/2 + b*e*log(c/d + x +
1/d)/d - b*e*acoth(c + d*x)/d + b*f*x/(2*d) - b*f*acoth(c + d*x)/(2*d**2), Ne(d, 0)), ((a + b*acoth(c))*(e*x
+ f*x**2/2), True))

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