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

Optimal. Leaf size=31 \[ \frac{a \log (\sinh (c+d x))}{d}-\frac{(a+b) \text{csch}^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0565121, antiderivative size = 31, 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 = {4138, 444, 43} \[ \frac{a \log (\sinh (c+d x))}{d}-\frac{(a+b) \text{csch}^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b)*Csch[c + d*x]^2)/(2*d) + (a*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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.168375, size = 52, normalized size = 1.68 \[ -\frac{a \left (\coth ^2(c+d x)-2 \log (\tanh (c+d x))-2 \log (\cosh (c+d x))\right )}{2 d}-\frac{b \text{csch}^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 50, normalized size = 1.6 \begin{align*}{\frac{a\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}}-{\frac{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}a}{2\,d}}-{\frac{b \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{2\,d \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}} \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),x)

[Out]

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

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Maxima [B]  time = 1.1838, size = 146, normalized size = 4.71 \begin{align*} a{\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 )} - \frac{2 \, 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),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.1493, size = 990, normalized size = 31.94 \begin{align*} -\frac{a d x \cosh \left (d x + c\right )^{4} + 4 \, a d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a d x \sinh \left (d x + c\right )^{4} + a d x - 2 \,{\left (a d x - a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a d x \cosh \left (d x + c\right )^{2} - a d x + a + b\right )} \sinh \left (d x + c\right )^{2} -{\left (a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} - 2 \, a \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a \cosh \left (d x + c\right )^{2} - a\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (a \cosh \left (d x + c\right )^{3} - a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a\right )} \log \left (\frac{2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \,{\left (a d x \cosh \left (d x + c\right )^{3} -{\left (a d x - a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} - 2 \, d \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (d \cosh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 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),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.2267, size = 109, normalized size = 3.52 \begin{align*} -\frac{2 \, a d x - 2 \, a \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac{3 \, a e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a}{{\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),x, algorithm="giac")

[Out]

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