3.178 \(\int \frac{\coth ^3(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\)

Optimal. Leaf size=85 \[ \frac{b^2 \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^2 d (a+b)}+\frac{(a-b) \log (\tanh (c+d x))}{a^2 d}+\frac{\log (\cosh (c+d x))}{d (a+b)}-\frac{\coth ^2(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.137222, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3670, 446, 72} \[ \frac{b^2 \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^2 d (a+b)}+\frac{(a-b) \log (\tanh (c+d x))}{a^2 d}+\frac{\log (\cosh (c+d x))}{d (a+b)}-\frac{\coth ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^3/(a + b*Tanh[c + d*x]^2),x]

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.163663, size = 60, normalized size = 0.71 \[ -\frac{-\frac{b^2 \log \left (a \coth ^2(c+d x)+b\right )}{a^2 (a+b)}-\frac{2 \log (\sinh (c+d x))}{a+b}+\frac{\coth ^2(c+d x)}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^3/(a + b*Tanh[c + d*x]^2),x]

[Out]

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

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Maple [B]  time = 0.086, size = 180, normalized size = 2.1 \begin{align*} -{\frac{1}{8\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{d \left ( a+b \right ) }\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{{b}^{2}}{2\,d{a}^{2} \left ( a+b \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a \right ) }-{\frac{1}{8\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{b}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{d \left ( a+b \right ) }\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3/(a+b*tanh(d*x+c)^2),x)

[Out]

-1/8/d/a*tanh(1/2*d*x+1/2*c)^2-1/d/(a+b)*ln(tanh(1/2*d*x+1/2*c)+1)+1/2/d*b^2/a^2/(a+b)*ln(tanh(1/2*d*x+1/2*c)^
4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)-1/8/d/a/tanh(1/2*d*x+1/2*c)^2+1/d/a*ln(tanh(1/2*d*x
+1/2*c))-1/d/a^2*b*ln(tanh(1/2*d*x+1/2*c))-1/d/(a+b)*ln(tanh(1/2*d*x+1/2*c)-1)

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Maxima [A]  time = 1.08124, size = 215, normalized size = 2.53 \begin{align*} \frac{b^{2} \log \left (2 \,{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \,{\left (a^{3} + a^{2} b\right )} d} + \frac{d x + c}{{\left (a + b\right )} d} + \frac{2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (2 \, a e^{\left (-2 \, d x - 2 \, c\right )} - a e^{\left (-4 \, d x - 4 \, c\right )} - a\right )} d} + \frac{{\left (a - b\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} + \frac{{\left (a - b\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*tanh(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.48711, size = 1817, normalized size = 21.38 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/2*(2*a^2*d*x*cosh(d*x + c)^4 + 8*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*a^2*d*x*sinh(d*x + c)^4 + 2*a^2*
d*x - 4*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^2 + 4*(3*a^2*d*x*cosh(d*x + c)^2 - a^2*d*x + a^2 + a*b)*sinh(d*x +
c)^2 - (b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 - 2*b^2*cosh(d*x + c)
^2 + 2*(3*b^2*cosh(d*x + c)^2 - b^2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 - b^2*cosh(d*x + c))*sinh(
d*x + c))*log(2*((a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)
*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*((a^2 - b^2)*cosh(d*x + c)^4 + 4*(a^2 - b^2)*cosh(d*x + c)*sinh(d*x + c
)^3 + (a^2 - b^2)*sinh(d*x + c)^4 - 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 - b^2)*cosh(d*x + c)^2 - a^2 + b
^2)*sinh(d*x + c)^2 + a^2 - b^2 + 4*((a^2 - b^2)*cosh(d*x + c)^3 - (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c))*l
og(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*(a^2*d*x*cosh(d*x + c)^3 - (a^2*d*x - a^2 - a*b)*cosh(
d*x + c))*sinh(d*x + c))/((a^3 + a^2*b)*d*cosh(d*x + c)^4 + 4*(a^3 + a^2*b)*d*cosh(d*x + c)*sinh(d*x + c)^3 +
(a^3 + a^2*b)*d*sinh(d*x + c)^4 - 2*(a^3 + a^2*b)*d*cosh(d*x + c)^2 + 2*(3*(a^3 + a^2*b)*d*cosh(d*x + c)^2 - (
a^3 + a^2*b)*d)*sinh(d*x + c)^2 + (a^3 + a^2*b)*d + 4*((a^3 + a^2*b)*d*cosh(d*x + c)^3 - (a^3 + a^2*b)*d*cosh(
d*x + c))*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{3}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**3/(a+b*tanh(d*x+c)**2),x)

[Out]

Integral(coth(c + d*x)**3/(a + b*tanh(c + d*x)**2), x)

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Giac [A]  time = 1.19926, size = 190, normalized size = 2.24 \begin{align*} \frac{b^{2} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{2 \,{\left (a^{3} d + a^{2} b d\right )}} - \frac{d x + c}{a d + b d} + \frac{{\left (a - b\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a^{2} d} - \frac{2 \, e^{\left (2 \, d x + 2 \, c\right )}}{a d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

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