3.83 \(\int \frac {1}{(a+b \coth (c+d x))^3} \, dx\)

Optimal. Leaf size=129 \[ \frac {2 a b}{d \left (a^2-b^2\right )^2 (a+b \coth (c+d x))}+\frac {b}{2 d \left (a^2-b^2\right ) (a+b \coth (c+d x))^2}-\frac {b \left (3 a^2+b^2\right ) \log (a \sinh (c+d x)+b \cosh (c+d x))}{d \left (a^2-b^2\right )^3}+\frac {a x \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3} \]

[Out]

a*(a^2+3*b^2)*x/(a^2-b^2)^3+1/2*b/(a^2-b^2)/d/(a+b*coth(d*x+c))^2+2*a*b/(a^2-b^2)^2/d/(a+b*coth(d*x+c))-b*(3*a
^2+b^2)*ln(b*cosh(d*x+c)+a*sinh(d*x+c))/(a^2-b^2)^3/d

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Rubi [A]  time = 0.18, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3483, 3529, 3531, 3530} \[ \frac {2 a b}{d \left (a^2-b^2\right )^2 (a+b \coth (c+d x))}+\frac {b}{2 d \left (a^2-b^2\right ) (a+b \coth (c+d x))^2}-\frac {b \left (3 a^2+b^2\right ) \log (a \sinh (c+d x)+b \cosh (c+d x))}{d \left (a^2-b^2\right )^3}+\frac {a x \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x])^(-3),x]

[Out]

(a*(a^2 + 3*b^2)*x)/(a^2 - b^2)^3 + b/(2*(a^2 - b^2)*d*(a + b*Coth[c + d*x])^2) + (2*a*b)/((a^2 - b^2)^2*d*(a
+ b*Coth[c + d*x])) - (b*(3*a^2 + b^2)*Log[b*Cosh[c + d*x] + a*Sinh[c + d*x]])/((a^2 - b^2)^3*d)

Rule 3483

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

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

Rubi steps

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

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Mathematica [A]  time = 3.62, size = 134, normalized size = 1.04 \[ -\frac {\frac {b \left (2 \left (3 a^2+b^2\right ) \log (a \tanh (c+d x)+b)+\frac {b \left (b^2-a^2\right ) \left (\left (2 a b^2-6 a^3\right ) \tanh (c+d x)-5 a^2 b+b^3\right )}{a^2 (a \tanh (c+d x)+b)^2}\right )}{\left (a^2-b^2\right )^3}+\frac {\log (1-\tanh (c+d x))}{(a+b)^3}-\frac {\log (\tanh (c+d x)+1)}{(a-b)^3}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Coth[c + d*x])^(-3),x]

[Out]

-1/2*(Log[1 - Tanh[c + d*x]]/(a + b)^3 - Log[1 + Tanh[c + d*x]]/(a - b)^3 + (b*(2*(3*a^2 + b^2)*Log[b + a*Tanh
[c + d*x]] + (b*(-a^2 + b^2)*(-5*a^2*b + b^3 + (-6*a^3 + 2*a*b^2)*Tanh[c + d*x]))/(a^2*(b + a*Tanh[c + d*x])^2
)))/(a^2 - b^2)^3)/d

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fricas [B]  time = 0.44, size = 1431, normalized size = 11.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c))^3,x, algorithm="fricas")

[Out]

((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*x*cosh(d*x + c)^4 + 4*(a^5 + 5*a^4*b + 10*a^3*b^2
+ 10*a^2*b^3 + 5*a*b^4 + b^5)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 +
5*a*b^4 + b^5)*d*x*sinh(d*x + c)^4 + 6*a^3*b^2 - 12*a^2*b^3 + 6*a*b^4 + (a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 +
a*b^4 + b^5)*d*x - 2*(3*a^3*b^2 - a^2*b^3 - 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4
- b^5)*d*x)*cosh(d*x + c)^2 - 2*(3*a^3*b^2 - a^2*b^3 - 3*a*b^4 + b^5 - 3*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*
b^3 + 5*a*b^4 + b^5)*d*x*cosh(d*x + c)^2 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*x)*sinh(d
*x + c)^2 - (3*a^4*b - 6*a^3*b^2 + 4*a^2*b^3 - 2*a*b^4 + b^5 + (3*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^
5)*cosh(d*x + c)^4 + 4*(3*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^5)*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a^
4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^5)*sinh(d*x + c)^4 - 2*(3*a^4*b - 2*a^2*b^3 - b^5)*cosh(d*x + c)^2 -
2*(3*a^4*b - 2*a^2*b^3 - b^5 - 3*(3*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^5)*cosh(d*x + c)^2)*sinh(d*x
+ c)^2 + 4*((3*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 + b^5)*cosh(d*x + c)^3 - (3*a^4*b - 2*a^2*b^3 - b^5)*co
sh(d*x + c))*sinh(d*x + c))*log(2*(b*cosh(d*x + c) + a*sinh(d*x + c))/(cosh(d*x + c) - sinh(d*x + c))) + 4*((a
^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*x*cosh(d*x + c)^3 - (3*a^3*b^2 - a^2*b^3 - 3*a*b^4 +
b^5 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^8 + 2*a^
7*b - 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*d*cosh(d*x + c)^4 + 4*(a^8 + 2*a^7*b - 2*
a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^8 + 2*a^7*b
- 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*d*sinh(d*x + c)^4 - 2*(a^8 - 4*a^6*b^2 + 6*a^
4*b^4 - 4*a^2*b^6 + b^8)*d*cosh(d*x + c)^2 + 2*(3*(a^8 + 2*a^7*b - 2*a^6*b^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b
^6 - 2*a*b^7 - b^8)*d*cosh(d*x + c)^2 - (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sinh(d*x + c)^2 + (
a^8 - 2*a^7*b - 2*a^6*b^2 + 6*a^5*b^3 - 6*a^3*b^5 + 2*a^2*b^6 + 2*a*b^7 - b^8)*d + 4*((a^8 + 2*a^7*b - 2*a^6*b
^2 - 6*a^5*b^3 + 6*a^3*b^5 + 2*a^2*b^6 - 2*a*b^7 - b^8)*d*cosh(d*x + c)^3 - (a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a
^2*b^6 + b^8)*d*cosh(d*x + c))*sinh(d*x + c))

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giac [A]  time = 0.18, size = 203, normalized size = 1.57 \[ -\frac {\frac {{\left (3 \, a^{2} b + b^{3}\right )} \log \left ({\left | a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} - a + b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {d x + c}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {2 \, {\left ({\left (3 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} e^{\left (2 \, d x + 2 \, c\right )} - \frac {3 \, {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )}}{a + b}\right )}}{{\left (a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} - a + b\right )}^{2} {\left (a + b\right )}^{2} {\left (a - b\right )}^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c))^3,x, algorithm="giac")

[Out]

-((3*a^2*b + b^3)*log(abs(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) - a + b))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)
- (d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 2*((3*a^2*b^2 - 4*a*b^3 + b^4)*e^(2*d*x + 2*c) - 3*(a^3*b^2 - 2*
a^2*b^3 + a*b^4)/(a + b))/((a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) - a + b)^2*(a + b)^2*(a - b)^3))/d

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maple [A]  time = 0.12, size = 166, normalized size = 1.29 \[ -\frac {\ln \left (\coth \left (d x +c \right )-1\right )}{2 d \left (a +b \right )^{3}}+\frac {\ln \left (\coth \left (d x +c \right )+1\right )}{2 d \left (a -b \right )^{3}}+\frac {b}{2 d \left (a -b \right ) \left (a +b \right ) \left (a +b \coth \left (d x +c \right )\right )^{2}}+\frac {2 a b}{d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \coth \left (d x +c \right )\right )}-\frac {3 b \ln \left (a +b \coth \left (d x +c \right )\right ) a^{2}}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {b^{3} \ln \left (a +b \coth \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*coth(d*x+c))^3,x)

[Out]

-1/2/d/(a+b)^3*ln(coth(d*x+c)-1)+1/2/d/(a-b)^3*ln(coth(d*x+c)+1)+1/2/d*b/(a-b)/(a+b)/(a+b*coth(d*x+c))^2+2/d*a
*b/(a+b)^2/(a-b)^2/(a+b*coth(d*x+c))-3/d*b/(a+b)^3/(a-b)^3*ln(a+b*coth(d*x+c))*a^2-1/d*b^3/(a+b)^3/(a-b)^3*ln(
a+b*coth(d*x+c))

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maxima [B]  time = 0.34, size = 322, normalized size = 2.50 \[ -\frac {{\left (3 \, a^{2} b + b^{3}\right )} \log \left (-{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a + b\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d} - \frac {2 \, {\left (3 \, a^{2} b^{2} + 3 \, a b^{3} - {\left (3 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )}}{{\left (a^{7} + a^{6} b - 3 \, a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6} - b^{7} - 2 \, {\left (a^{7} - a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - 3 \, a^{2} b^{5} - a b^{6} + b^{7}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{7} - 3 \, a^{6} b + a^{5} b^{2} + 5 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c))^3,x, algorithm="maxima")

[Out]

-(3*a^2*b + b^3)*log(-(a - b)*e^(-2*d*x - 2*c) + a + b)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d) - 2*(3*a^2*b^2
+ 3*a*b^3 - (3*a^2*b^2 - 2*a*b^3 - b^4)*e^(-2*d*x - 2*c))/((a^7 + a^6*b - 3*a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 +
3*a^2*b^5 - a*b^6 - b^7 - 2*(a^7 - a^6*b - 3*a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 - 3*a^2*b^5 - a*b^6 + b^7)*e^(-2
*d*x - 2*c) + (a^7 - 3*a^6*b + a^5*b^2 + 5*a^4*b^3 - 5*a^3*b^4 - a^2*b^5 + 3*a*b^6 - b^7)*e^(-4*d*x - 4*c))*d)
+ (d*x + c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d)

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mupad [B]  time = 1.40, size = 195, normalized size = 1.51 \[ \frac {x}{{\left (a-b\right )}^3}-\frac {\ln \left (b-a+a\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (3\,a^2\,b+b^3\right )}{d\,a^6-3\,d\,a^4\,b^2+3\,d\,a^2\,b^4-d\,b^6}+\frac {2\,b^3}{d\,{\left (a+b\right )}^3\,\left (a-b\right )\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^2+{\left (a-b\right )}^2-2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )\,\left (a-b\right )\right )}-\frac {2\,\left (3\,a\,b^2-b^3\right )}{d\,{\left (a+b\right )}^3\,{\left (a-b\right )}^2\,\left (b-a+{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*coth(c + d*x))^3,x)

[Out]

x/(a - b)^3 - (log(b - a + a*exp(2*c)*exp(2*d*x) + b*exp(2*c)*exp(2*d*x))*(3*a^2*b + b^3))/(a^6*d - b^6*d + 3*
a^2*b^4*d - 3*a^4*b^2*d) + (2*b^3)/(d*(a + b)^3*(a - b)*(exp(4*c + 4*d*x)*(a + b)^2 + (a - b)^2 - 2*exp(2*c +
2*d*x)*(a + b)*(a - b))) - (2*(3*a*b^2 - b^3))/(d*(a + b)^3*(a - b)^2*(b - a + exp(2*c + 2*d*x)*(a + b)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c))**3,x)

[Out]

Timed out

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