3.8 \(\int \text {csch}^4(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=44 \[ \frac {(a-b) \coth (c+d x)}{d}-\frac {a \coth ^3(c+d x)}{3 d}-\frac {b \tanh (c+d x)}{d} \]

[Out]

(a-b)*coth(d*x+c)/d-1/3*a*coth(d*x+c)^3/d-b*tanh(d*x+c)/d

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Rubi [A]  time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3663, 448} \[ \frac {(a-b) \coth (c+d x)}{d}-\frac {a \coth ^3(c+d x)}{3 d}-\frac {b \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

((a - b)*Coth[c + d*x])/d - (a*Coth[c + d*x]^3)/(3*d) - (b*Tanh[c + d*x])/d

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 61, normalized size = 1.39 \[ \frac {2 a \coth (c+d x)}{3 d}-\frac {a \coth (c+d x) \text {csch}^2(c+d x)}{3 d}-\frac {b \tanh (c+d x)}{d}-\frac {b \coth (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

(2*a*Coth[c + d*x])/(3*d) - (b*Coth[c + d*x])/d - (a*Coth[c + d*x]*Csch[c + d*x]^2)/(3*d) - (b*Tanh[c + d*x])/
d

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fricas [B]  time = 0.72, size = 244, normalized size = 5.55 \[ -\frac {8 \, {\left ({\left (a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + 3 \, b\right )} \sinh \left (d x + c\right )^{2} + a - 3 \, b\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} - 2 \, d \cosh \left (d x + c\right )^{4} + {\left (15 \, d \cosh \left (d x + c\right )^{2} - 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} - 2 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )^{2} + {\left (15 \, d \cosh \left (d x + c\right )^{4} - 12 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} - 4 \, d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

-8/3*((a + 3*b)*cosh(d*x + c)^2 + 4*a*cosh(d*x + c)*sinh(d*x + c) + (a + 3*b)*sinh(d*x + c)^2 + a - 3*b)/(d*co
sh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 - 2*d*cosh(d*x + c)^4 + (15*d*cosh(d*x +
c)^2 - 2*d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 - 2*d*cosh(d*x + c))*sinh(d*x + c)^3 - d*cosh(d*x + c)^2
+ (15*d*cosh(d*x + c)^4 - 12*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^5 - 4*d*cosh(d*x +
c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + 2*d)

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giac [A]  time = 0.15, size = 80, normalized size = 1.82 \[ \frac {2 \, {\left (\frac {3 \, b}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - \frac {3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a + 3 \, b}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.37, size = 55, normalized size = 1.25 \[ \frac {a \left (\frac {2}{3}-\frac {\mathrm {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+b \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}-2 \tanh \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2),x)

[Out]

1/d*(a*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+b*(-1/sinh(d*x+c)/cosh(d*x+c)-2*tanh(d*x+c)))

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maxima [B]  time = 0.36, size = 113, normalized size = 2.57 \[ \frac {4}{3} \, a {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {4 \, b}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.10, size = 173, normalized size = 3.93 \[ \frac {2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {\frac {2\,\left (2\,a-b\right )}{3\,d}+\frac {2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {2\,b}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {\frac {2\,b}{3\,d}+\frac {2\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(c + d*x)^2)/sinh(c + d*x)^4,x)

[Out]

(2*b)/(d*(exp(2*c + 2*d*x) + 1)) - ((2*(2*a - b))/(3*d) + (2*b*exp(2*c + 2*d*x))/(3*d))/(exp(4*c + 4*d*x) - 2*
exp(2*c + 2*d*x) + 1) - (2*b)/(3*d*(exp(2*c + 2*d*x) - 1)) - ((2*b)/(3*d) + (2*b*exp(4*c + 4*d*x))/(3*d) + (4*
exp(2*c + 2*d*x)*(2*a - b))/(3*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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \operatorname {csch}^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**4*(a+b*tanh(d*x+c)**2),x)

[Out]

Integral((a + b*tanh(c + d*x)**2)*csch(c + d*x)**4, x)

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