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

Optimal. Leaf size=52 \[ \frac {a^2 \log (\sinh (c+d x))}{d}-\frac {(a+b)^2 \text {csch}^4(c+d x)}{4 d}-\frac {a (a+b) \text {csch}^2(c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4138, 444, 43} \[ \frac {a^2 \log (\sinh (c+d x))}{d}-\frac {(a+b)^2 \text {csch}^4(c+d x)}{4 d}-\frac {a (a+b) \text {csch}^2(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.23, size = 77, normalized size = 1.48 \[ -\frac {\left (a \cosh ^2(c+d x)+b\right )^2 \left (-4 a^2 \log (\sinh (c+d x))+(a+b)^2 \text {csch}^4(c+d x)+4 a (a+b) \text {csch}^2(c+d x)\right )}{d (a \cosh (2 (c+d x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 1252, normalized size = 24.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*d*x*cosh(d*x + c)^8 + 8*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*d*x*sinh(d*x + c)^8 - 4*(a^2*d*x - a
^2 - a*b)*cosh(d*x + c)^6 + 4*(7*a^2*d*x*cosh(d*x + c)^2 - a^2*d*x + a^2 + a*b)*sinh(d*x + c)^6 + 8*(7*a^2*d*x
*cosh(d*x + c)^3 - 3*(a^2*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*a^2*d*x - 2*a^2 + 2*b^2)*cosh
(d*x + c)^4 + 2*(35*a^2*d*x*cosh(d*x + c)^4 + 3*a^2*d*x - 30*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^2 - 2*a^2 + 2
*b^2)*sinh(d*x + c)^4 + a^2*d*x + 8*(7*a^2*d*x*cosh(d*x + c)^5 - 10*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^3 + (3
*a^2*d*x - 2*a^2 + 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 - 4*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^2 + 4*(7*a^2*
d*x*cosh(d*x + c)^6 - 15*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^4 - a^2*d*x + 3*(3*a^2*d*x - 2*a^2 + 2*b^2)*cosh(
d*x + c)^2 + a^2 + a*b)*sinh(d*x + c)^2 - (a^2*cosh(d*x + c)^8 + 8*a^2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sin
h(d*x + c)^8 - 4*a^2*cosh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 - a^2)*sinh(d*x + c)^6 + 6*a^2*cosh(d*x + c)^4
+ 8*(7*a^2*cosh(d*x + c)^3 - 3*a^2*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*a^2*cosh(d*x + c)^4 - 30*a^2*cosh(d
*x + c)^2 + 3*a^2)*sinh(d*x + c)^4 - 4*a^2*cosh(d*x + c)^2 + 8*(7*a^2*cosh(d*x + c)^5 - 10*a^2*cosh(d*x + c)^3
+ 3*a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(7*a^2*cosh(d*x + c)^6 - 15*a^2*cosh(d*x + c)^4 + 9*a^2*cosh(d*x +
c)^2 - a^2)*sinh(d*x + c)^2 + a^2 + 8*(a^2*cosh(d*x + c)^7 - 3*a^2*cosh(d*x + c)^5 + 3*a^2*cosh(d*x + c)^3 -
a^2*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*(a^2*d*x*cosh(d*x +
c)^7 - 3*(a^2*d*x - a^2 - a*b)*cosh(d*x + c)^5 + (3*a^2*d*x - 2*a^2 + 2*b^2)*cosh(d*x + c)^3 - (a^2*d*x - a^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 = 0.34, size = 147, normalized size = 2.83 \[ -\frac {12 \, a^{2} d x - 12 \, a^{2} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + \frac {25 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} - 52 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 48 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 102 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 52 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 48 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*a^2*d*x - 12*a^2*log(abs(e^(2*d*x + 2*c) - 1)) + (25*a^2*e^(8*d*x + 8*c) - 52*a^2*e^(6*d*x + 6*c) +
48*a*b*e^(6*d*x + 6*c) + 102*a^2*e^(4*d*x + 4*c) + 48*b^2*e^(4*d*x + 4*c) - 52*a^2*e^(2*d*x + 2*c) + 48*a*b*e^
(2*d*x + 2*c) + 25*a^2)/(e^(2*d*x + 2*c) - 1)^4)/d

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maple [B]  time = 0.32, size = 102, normalized size = 1.96 \[ \frac {a^{2} \ln \left (\sinh \left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\coth ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} \left (\coth ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a b \left (\cosh ^{2}\left (d x +c \right )\right )}{d \sinh \left (d x +c \right )^{4}}+\frac {a b}{2 d \sinh \left (d x +c \right )^{4}}-\frac {b^{2}}{4 d \sinh \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 282, normalized size = 5.42 \[ a^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 4 \, a b {\left (\frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} + \frac {e^{\left (-6 \, d x - 6 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} - \frac {4 \, b^{2}}{d {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.47, size = 207, normalized size = 3.98 \[ \frac {a^2\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-1\right )}{d}-\frac {4\,\left (a^2+2\,a\,b+b^2\right )}{d\,\left (6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {4\,\left (2\,a^2+3\,a\,b+b^2\right )}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-a^2\,x-\frac {4\,\left (a^2+b\,a\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {8\,\left (a^2+2\,a\,b+b^2\right )}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*log(exp(2*c)*exp(2*d*x) - 1))/d - (4*(2*a*b + a^2 + b^2))/(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)) - (4*(3*a*b + 2*a^2 + b^2))/(d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*
x) + 1)) - a^2*x - (4*(a*b + a^2))/(d*(exp(2*c + 2*d*x) - 1)) - (8*(2*a*b + a^2 + b^2))/(d*(3*exp(2*c + 2*d*x)
- 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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