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]

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

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Rubi [A]  time = 0.06, antiderivative size = 31, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.18, 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]

-1/2*(b*Csch[c + d*x]^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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fricas [B]  time = 0.42, size = 378, normalized size = 12.19 \[ -\frac {a d x \cosh \left (d x + c\right )^{4} + 4 \, a d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a d x \sinh \left (d x + c\right )^{4} + a d x - 2 \, {\left (a d x - a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a d x \cosh \left (d x + c\right )^{2} - a d x + a + b\right )} \sinh \left (d x + c\right )^{2} - {\left (a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} - 2 \, a \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a \cosh \left (d x + c\right )^{2} - a\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a \cosh \left (d x + c\right )^{3} - a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \, {\left (a d x \cosh \left (d x + c\right )^{3} - {\left (a d x - a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} - 2 \, d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (d \cosh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + d} \]

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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giac [B]  time = 0.21, size = 81, normalized size = 2.61 \[ -\frac {2 \, a d x - 2 \, a \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac {3 \, a e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \]

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

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maple [A]  time = 0.33, size = 42, normalized size = 1.35 \[ \frac {a \ln \left (\sinh \left (d x +c \right )\right )}{d}-\frac {a \left (\coth ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b}{2 d \sinh \left (d x +c \right )^{2}} \]

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/sinh(d*x+c)^2*b

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maxima [B]  time = 0.33, size = 108, normalized size = 3.48 \[ a {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} - \frac {2 \, b}{d {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2}} \]

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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mupad [B]  time = 1.41, size = 76, normalized size = 2.45 \[ \frac {a\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-1\right )}{d}-a\,x-\frac {2\,\left (a+b\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,\left (a+b\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \coth ^{3}{\left (c + d x \right )}\, dx \]

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]

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

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