3.119 \(\int \coth ^3(c+d x) (a+b \text{sech}^2(c+d x))^2 \, dx\)

Optimal. Leaf size=55 \[ \frac{\left (a^2-b^2\right ) \log (\sinh (c+d x))}{d}-\frac{(a+b)^2 \text{csch}^2(c+d x)}{2 d}+\frac{b^2 \log (\cosh (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.089335, antiderivative size = 55, 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 = {4138, 446, 88} \[ \frac{\left (a^2-b^2\right ) \log (\sinh (c+d x))}{d}-\frac{(a+b)^2 \text{csch}^2(c+d x)}{2 d}+\frac{b^2 \log (\cosh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.208623, size = 82, normalized size = 1.49 \[ -\frac{2 \left (a \cosh ^2(c+d x)+b\right )^2 \left ((a+b)^2 \text{csch}^2(c+d x)-2 \left (\left (a^2-b^2\right ) \log (\sinh (c+d x))+b^2 \log (\cosh (c+d x))\right )\right )}{d (a \cosh (2 (c+d x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 86, normalized size = 1.6 \begin{align*}{\frac{{a}^{2}\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}}{2\,d}}-{\frac{ab \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{d \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2}}{2\,d \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2}\ln \left ( \tanh \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.79006, size = 278, normalized size = 5.05 \begin{align*} a^{2}{\left (x + \frac{c}{d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac{\log \left (e^{\left (-d x - 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 )} - b^{2}{\left (\frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{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{4 \, a b}{d{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.27925, size = 1553, normalized size = 28.24 \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*sech(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-(a^2*d*x*cosh(d*x + c)^4 + 4*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*d*x*sinh(d*x + c)^4 + a^2*d*x - 2*(a
^2*d*x - a^2 - 2*a*b - b^2)*cosh(d*x + c)^2 + 2*(3*a^2*d*x*cosh(d*x + c)^2 - a^2*d*x + a^2 + 2*a*b + b^2)*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*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - ((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))*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 4*(a^2*d*x*cosh(d*x + c)
^3 - (a^2*d*x - a^2 - 2*a*b - 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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.31819, size = 217, normalized size = 3.95 \begin{align*} -\frac{2 \, a^{2} d x - 2 \, b^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - 2 \,{\left (a^{2} e^{\left (2 \, c\right )} - b^{2} e^{\left (2 \, c\right )}\right )} e^{\left (-2 \, c\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac{3 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 10 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a^{2} - 3 \, b^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^2*d*x - 2*b^2*log(e^(2*d*x + 2*c) + 1) - 2*(a^2*e^(2*c) - b^2*e^(2*c))*e^(-2*c)*log(abs(e^(2*d*x + 2
*c) - 1)) + (3*a^2*e^(4*d*x + 4*c) - 3*b^2*e^(4*d*x + 4*c) - 2*a^2*e^(2*d*x + 2*c) + 8*a*b*e^(2*d*x + 2*c) + 1
0*b^2*e^(2*d*x + 2*c) + 3*a^2 - 3*b^2)/(e^(2*d*x + 2*c) - 1)^2)/d