3.149 \(\int \coth (c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=49 \[ \frac{a^2 \log (\tanh (c+d x))}{d}+\frac{(a+b)^2 \log (\cosh (c+d x))}{d}-\frac{b^2 \tanh ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0769667, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3670, 446, 72} \[ \frac{a^2 \log (\tanh (c+d x))}{d}+\frac{(a+b)^2 \log (\cosh (c+d x))}{d}-\frac{b^2 \tanh ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.125025, size = 48, normalized size = 0.98 \[ \frac{2 \left (a^2 \log (\tanh (c+d x))+(a+b)^2 \log (\cosh (c+d x))\right )-b^2 \tanh ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 60, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{ab\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.52864, size = 140, normalized size = 2.86 \begin{align*} b^{2}{\left (x + \frac{c}{d} + \frac{\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac{2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac{2 \, a b \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}{d} + \frac{a^{2} \log \left (\sinh \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.09311, size = 1715, normalized size = 35. \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)*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.26452, size = 198, normalized size = 4.04 \begin{align*} \frac{a^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right )}{2 \, d} + \frac{{\left (2 \, a b + b^{2}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right )}{2 \, d} - \frac{2 \, a b{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b^{2}{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 4 \, a b - 2 \, b^{2}}{2 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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