3.111 \(\int \coth ^5(c+d x) (a+b \text{sech}^2(c+d x)) \, dx\)

Optimal. Leaf size=51 \[ -\frac{(a+b) \text{csch}^4(c+d x)}{4 d}-\frac{(2 a+b) \text{csch}^2(c+d x)}{2 d}+\frac{a \log (\sinh (c+d x))}{d} \]

[Out]

-((2*a + b)*Csch[c + d*x]^2)/(2*d) - ((a + b)*Csch[c + d*x]^4)/(4*d) + (a*Log[Sinh[c + d*x]])/d

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Rubi [A]  time = 0.0781343, antiderivative size = 51, 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, 446, 77} \[ -\frac{(a+b) \text{csch}^4(c+d x)}{4 d}-\frac{(2 a+b) \text{csch}^2(c+d x)}{2 d}+\frac{a \log (\sinh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^5*(a + b*Sech[c + d*x]^2),x]

[Out]

-((2*a + b)*Csch[c + d*x]^2)/(2*d) - ((a + b)*Csch[c + d*x]^4)/(4*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 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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \coth ^5(c+d x) \left (a+b \text{sech}^2(c+d x)\right ) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (b+a x^2\right )}{\left (1-x^2\right )^3} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x (b+a x)}{(1-x)^3} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{-a-b}{(-1+x)^3}+\frac{-2 a-b}{(-1+x)^2}-\frac{a}{-1+x}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{(2 a+b) \text{csch}^2(c+d x)}{2 d}-\frac{(a+b) \text{csch}^4(c+d x)}{4 d}+\frac{a \log (\sinh (c+d x))}{d}\\ \end{align*}

Mathematica [A]  time = 0.262425, size = 62, normalized size = 1.22 \[ -\frac{a \left (\coth ^4(c+d x)+2 \coth ^2(c+d x)-4 \log (\tanh (c+d x))-4 \log (\cosh (c+d x))\right )}{4 d}-\frac{b \coth ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^5*(a + b*Sech[c + d*x]^2),x]

[Out]

-(b*Coth[c + d*x]^4)/(4*d) - (a*(2*Coth[c + d*x]^2 + Coth[c + d*x]^4 - 4*Log[Cosh[c + d*x]] - 4*Log[Tanh[c + d
*x]]))/(4*d)

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Maple [A]  time = 0.041, size = 86, normalized size = 1.7 \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{a \left ({\rm coth} \left (dx+c\right ) \right ) ^{4}}{4\,d}}-{\frac{b \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{4\,d \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}}-{\frac{b \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{4\,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)^5*(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-1/4/d*a*coth(d*x+c)^4-1/4/d*b/sinh(d*x+c)^4*cosh(d*x+c)^2-1/4/d*b*co
sh(d*x+c)^2/sinh(d*x+c)^2

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Maxima [B]  time = 1.2081, size = 339, normalized size = 6.65 \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{4 \,{\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d{\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 2 \, b{\left (\frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} + \frac{e^{\left (-6 \, d x - 6 \, c\right )}}{d{\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^5*(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 + 4*(e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c) + e^(-
6*d*x - 6*c))/(d*(4*e^(-2*d*x - 2*c) - 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) - e^(-8*d*x - 8*c) - 1))) + 2*b
*(e^(-2*d*x - 2*c)/(d*(4*e^(-2*d*x - 2*c) - 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) - e^(-8*d*x - 8*c) - 1)) +
e^(-6*d*x - 6*c)/(d*(4*e^(-2*d*x - 2*c) - 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) - e^(-8*d*x - 8*c) - 1)))

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Fricas [B]  time = 2.24184, size = 2898, normalized size = 56.82 \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)^5*(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

-(a*d*x*cosh(d*x + c)^8 + 8*a*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + a*d*x*sinh(d*x + c)^8 - 2*(2*a*d*x - 2*a - b
)*cosh(d*x + c)^6 + 2*(14*a*d*x*cosh(d*x + c)^2 - 2*a*d*x + 2*a + b)*sinh(d*x + c)^6 + 4*(14*a*d*x*cosh(d*x +
c)^3 - 3*(2*a*d*x - 2*a - b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*a*d*x - 2*a)*cosh(d*x + c)^4 + 2*(35*a*d*x*
cosh(d*x + c)^4 + 3*a*d*x - 15*(2*a*d*x - 2*a - b)*cosh(d*x + c)^2 - 2*a)*sinh(d*x + c)^4 + 8*(7*a*d*x*cosh(d*
x + c)^5 - 5*(2*a*d*x - 2*a - b)*cosh(d*x + c)^3 + (3*a*d*x - 2*a)*cosh(d*x + c))*sinh(d*x + c)^3 + a*d*x - 2*
(2*a*d*x - 2*a - b)*cosh(d*x + c)^2 + 2*(14*a*d*x*cosh(d*x + c)^6 - 15*(2*a*d*x - 2*a - b)*cosh(d*x + c)^4 - 2
*a*d*x + 6*(3*a*d*x - 2*a)*cosh(d*x + c)^2 + 2*a + b)*sinh(d*x + c)^2 - (a*cosh(d*x + c)^8 + 8*a*cosh(d*x + c)
*sinh(d*x + c)^7 + a*sinh(d*x + c)^8 - 4*a*cosh(d*x + c)^6 + 4*(7*a*cosh(d*x + c)^2 - a)*sinh(d*x + c)^6 + 8*(
7*a*cosh(d*x + c)^3 - 3*a*cosh(d*x + c))*sinh(d*x + c)^5 + 6*a*cosh(d*x + c)^4 + 2*(35*a*cosh(d*x + c)^4 - 30*
a*cosh(d*x + c)^2 + 3*a)*sinh(d*x + c)^4 + 8*(7*a*cosh(d*x + c)^5 - 10*a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c))*
sinh(d*x + c)^3 - 4*a*cosh(d*x + c)^2 + 4*(7*a*cosh(d*x + c)^6 - 15*a*cosh(d*x + c)^4 + 9*a*cosh(d*x + c)^2 -
a)*sinh(d*x + c)^2 + 8*(a*cosh(d*x + c)^7 - 3*a*cosh(d*x + c)^5 + 3*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) - sinh(d*x + c))) + 4*(2*a*d*x*cosh(d*x + c)^7 - 3*(2*a*d*x -
2*a - b)*cosh(d*x + c)^5 + 2*(3*a*d*x - 2*a)*cosh(d*x + c)^3 - (2*a*d*x - 2*a - b)*cosh(d*x + c))*sinh(d*x +
c))/(d*cosh(d*x + c)^8 + 8*d*cosh(d*x + c)*sinh(d*x + c)^7 + d*sinh(d*x + c)^8 - 4*d*cosh(d*x + c)^6 + 4*(7*d*
cosh(d*x + c)^2 - d)*sinh(d*x + c)^6 + 8*(7*d*cosh(d*x + c)^3 - 3*d*cosh(d*x + c))*sinh(d*x + c)^5 + 6*d*cosh(
d*x + c)^4 + 2*(35*d*cosh(d*x + c)^4 - 30*d*cosh(d*x + c)^2 + 3*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c)^5 -
10*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^3 - 4*d*cosh(d*x + c)^2 + 4*(7*d*cosh(d*x + c)^6 - 15*
d*cosh(d*x + c)^4 + 9*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 8*(d*cosh(d*x + c)^7 - 3*d*cosh(d*x + c)^5 + 3*
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)**5*(a+b*sech(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.27999, size = 158, normalized size = 3.1 \begin{align*} -\frac{12 \, a d x - 12 \, a \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac{25 \, a e^{\left (8 \, d x + 8 \, c\right )} - 52 \, a e^{\left (6 \, d x + 6 \, c\right )} + 24 \, b e^{\left (6 \, d x + 6 \, c\right )} + 102 \, a e^{\left (4 \, d x + 4 \, c\right )} - 52 \, a e^{\left (2 \, d x + 2 \, c\right )} + 24 \, b e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^5*(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

-1/12*(12*a*d*x - 12*a*log(abs(e^(2*d*x + 2*c) - 1)) + (25*a*e^(8*d*x + 8*c) - 52*a*e^(6*d*x + 6*c) + 24*b*e^(
6*d*x + 6*c) + 102*a*e^(4*d*x + 4*c) - 52*a*e^(2*d*x + 2*c) + 24*b*e^(2*d*x + 2*c) + 25*a)/(e^(2*d*x + 2*c) -
1)^4)/d