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]

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

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Rubi [A]  time = 0.08, antiderivative size = 51, 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, 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 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]))))

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 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 ^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*}

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Mathematica [A]  time = 0.26, 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]

-1/4*(b*Coth[c + d*x]^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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fricas [B]  time = 0.43, size = 1099, normalized size = 21.55 \[ \text {result too large to display} \]

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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giac [B]  time = 0.23, size = 117, normalized size = 2.29 \[ -\frac {12 \, a d x - 12 \, a \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac {25 \, a e^{\left (8 \, d x + 8 \, c\right )} - 52 \, a e^{\left (6 \, d x + 6 \, c\right )} + 24 \, b e^{\left (6 \, d x + 6 \, c\right )} + 102 \, a e^{\left (4 \, d x + 4 \, c\right )} - 52 \, a e^{\left (2 \, d x + 2 \, c\right )} + 24 \, b e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4}}}{12 \, d} \]

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

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maple [A]  time = 0.31, size = 78, normalized size = 1.53 \[ \frac {a \ln \left (\sinh \left (d x +c \right )\right )}{d}-\frac {a \left (\coth ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \left (\coth ^{4}\left (d x +c \right )\right )}{4 d}-\frac {b \left (\cosh ^{2}\left (d x +c \right )\right )}{2 d \sinh \left (d x +c \right )^{4}}+\frac {b}{4 d \sinh \left (d x +c \right )^{4}} \]

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*a*coth(d*x+c)^4/d-1/2/d*b/sinh(d*x+c)^4*cosh(d*x+c)^2+1/4/d*b/si
nh(d*x+c)^4

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maxima [B]  time = 0.51, size = 251, normalized size = 4.92 \[ a {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 2 \, b {\left (\frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} + \frac {e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} \]

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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mupad [B]  time = 0.11, size = 179, normalized size = 3.51 \[ \frac {a\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-1\right )}{d}-\frac {8\,\left (a+b\right )}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {2\,\left (2\,a+b\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {4\,\left (a+b\right )}{d\,\left (6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-a\,x-\frac {2\,\left (4\,a+3\,b\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^5*(a + b/cosh(c + d*x)^2),x)

[Out]

(a*log(exp(2*c)*exp(2*d*x) - 1))/d - (8*(a + b))/(d*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x
) - 1)) - (2*(2*a + b))/(d*(exp(2*c + 2*d*x) - 1)) - (4*(a + b))/(d*(6*exp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) -
4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1)) - a*x - (2*(4*a + 3*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 \[ \int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \coth ^{5}{\left (c + d x \right )}\, dx \]

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]

Integral((a + b*sech(c + d*x)**2)*coth(c + d*x)**5, x)

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