3.62 \(\int \text {csch}^2(c+d x) (a+b \tanh ^3(c+d x))^2 \, dx\)

Optimal. Leaf size=47 \[ -\frac {a^2 \coth (c+d x)}{d}+\frac {a b \tanh ^2(c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

[Out]

-a^2*coth(d*x+c)/d+a*b*tanh(d*x+c)^2/d+1/5*b^2*tanh(d*x+c)^5/d

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Rubi [A]  time = 0.06, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3663, 270} \[ -\frac {a^2 \coth (c+d x)}{d}+\frac {a b \tanh ^2(c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^2*(a + b*Tanh[c + d*x]^3)^2,x]

[Out]

-((a^2*Coth[c + d*x])/d) + (a*b*Tanh[c + d*x]^2)/d + (b^2*Tanh[c + d*x]^5)/(5*d)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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}^2(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^3\right )^2}{x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2}{x^2}+2 a b x+b^2 x^4\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {a^2 \coth (c+d x)}{d}+\frac {a b \tanh ^2(c+d x)}{d}+\frac {b^2 \tanh ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 94, normalized size = 2.00 \[ -\frac {a^2 \coth (c+d x)}{d}-\frac {a b \text {sech}^2(c+d x)}{d}+\frac {b^2 \tanh (c+d x)}{5 d}+\frac {b^2 \tanh (c+d x) \text {sech}^4(c+d x)}{5 d}-\frac {2 b^2 \tanh (c+d x) \text {sech}^2(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^2*(a + b*Tanh[c + d*x]^3)^2,x]

[Out]

-((a^2*Coth[c + d*x])/d) - (a*b*Sech[c + d*x]^2)/d + (b^2*Tanh[c + d*x])/(5*d) - (2*b^2*Sech[c + d*x]^2*Tanh[c
+ d*x])/(5*d) + (b^2*Sech[c + d*x]^4*Tanh[c + d*x])/(5*d)

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fricas [B]  time = 0.57, size = 518, normalized size = 11.02 \[ -\frac {4 \, {\left ({\left (5 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (5 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (5 \, a b + 3 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + {\left (25 \, a^{2} + 5 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (10 \, {\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 15 \, a b - 3 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + {\left (10 \, {\left (5 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (25 \, a^{2} + 5 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{2} - a b\right )} \cosh \left (d x + c\right ) + {\left (5 \, {\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 9 \, {\left (5 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 10 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )\right )}}{5 \, {\left (d \cosh \left (d x + c\right )^{7} + 7 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + d \sinh \left (d x + c\right )^{7} + 3 \, d \cosh \left (d x + c\right )^{5} + {\left (21 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (7 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + d \cosh \left (d x + c\right )^{3} + {\left (35 \, d \cosh \left (d x + c\right )^{4} + 50 \, d \cosh \left (d x + c\right )^{2} + 9 \, d\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (7 \, d \cosh \left (d x + c\right )^{5} + 10 \, d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 5 \, d \cosh \left (d x + c\right ) + {\left (7 \, d \cosh \left (d x + c\right )^{6} + 25 \, d \cosh \left (d x + c\right )^{4} + 27 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

-4/5*((5*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c)^5 + 5*(5*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 + (5*a
*b + 3*b^2)*sinh(d*x + c)^5 + (25*a^2 + 5*a*b - 2*b^2)*cosh(d*x + c)^3 + (10*(5*a*b + 3*b^2)*cosh(d*x + c)^2 +
15*a*b - 3*b^2)*sinh(d*x + c)^3 + (10*(5*a^2 + 5*a*b + 2*b^2)*cosh(d*x + c)^3 + 3*(25*a^2 + 5*a*b - 2*b^2)*co
sh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^2 - a*b)*cosh(d*x + c) + (5*(5*a*b + 3*b^2)*cosh(d*x + c)^4 + 9*(5*a*b
- b^2)*cosh(d*x + c)^2 + 10*a*b + 10*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)*sinh(d*x + c)^
6 + d*sinh(d*x + c)^7 + 3*d*cosh(d*x + c)^5 + (21*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^5 + 5*(7*d*cosh(d*x +
c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^4 + d*cosh(d*x + c)^3 + (35*d*cosh(d*x + c)^4 + 50*d*cosh(d*x + c)^2
+ 9*d)*sinh(d*x + c)^3 + 3*(7*d*cosh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^2 - 5*
d*cosh(d*x + c) + (7*d*cosh(d*x + c)^6 + 25*d*cosh(d*x + c)^4 + 27*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c))

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giac [B]  time = 0.32, size = 122, normalized size = 2.60 \[ -\frac {2 \, {\left (\frac {5 \, a^{2}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + \frac {10 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 5 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 30 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 30 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, a b e^{\left (2 \, d x + 2 \, c\right )} + b^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}\right )}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="giac")

[Out]

-2/5*(5*a^2/(e^(2*d*x + 2*c) - 1) + (10*a*b*e^(8*d*x + 8*c) + 5*b^2*e^(8*d*x + 8*c) + 30*a*b*e^(6*d*x + 6*c) +
30*a*b*e^(4*d*x + 4*c) + 10*b^2*e^(4*d*x + 4*c) + 10*a*b*e^(2*d*x + 2*c) + b^2)/(e^(2*d*x + 2*c) + 1)^5)/d

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maple [B]  time = 0.46, size = 98, normalized size = 2.09 \[ \frac {-a^{2} \coth \left (d x +c \right )-\frac {a b}{\cosh \left (d x +c \right )^{2}}+b^{2} \left (-\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x)

[Out]

1/d*(-a^2*coth(d*x+c)-a*b/cosh(d*x+c)^2+b^2*(-1/2*sinh(d*x+c)^3/cosh(d*x+c)^5-3/8*sinh(d*x+c)/cosh(d*x+c)^5+3/
8*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c)))

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maxima [B]  time = 0.31, size = 256, normalized size = 5.45 \[ \frac {2}{5} \, b^{2} {\left (\frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {5 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {2 \, a^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} - \frac {4 \, a b}{d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

2/5*b^2*(10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x
- 8*c) + e^(-10*d*x - 10*c) + 1)) + 5*e^(-8*d*x - 8*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6
*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) +
10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 2*a^2/(d*(e^(-2*d*x - 2*c) - 1)) - 4*a
*b/(d*(e^(d*x + c) + e^(-d*x - c))^2)

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mupad [B]  time = 1.26, size = 483, normalized size = 10.28 \[ -\frac {\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+a\,b\right )}{5\,d}-\frac {2\,\left (2\,a\,b-b^2\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a\,b-b^2\right )}{5\,d}+\frac {12\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,b^2}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b-b^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {2\,\left (a\,b-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {2\,\left (b^2+a\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a\,b-b^2\right )}{5\,d}+\frac {6\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {2\,a^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(c + d*x)^3)^2/sinh(c + d*x)^2,x)

[Out]

- ((2*exp(8*c + 8*d*x)*(2*a*b + b^2))/(5*d) - (8*exp(2*c + 2*d*x)*(a*b + b^2))/(5*d) - (2*(2*a*b - b^2))/(5*d)
+ (8*exp(6*c + 6*d*x)*(a*b - b^2))/(5*d) + (12*b^2*exp(4*c + 4*d*x))/(5*d))/(5*exp(2*c + 2*d*x) + 10*exp(4*c
+ 4*d*x) + 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) - ((2*b^2)/(5*d) + (2*exp(4*c +
4*d*x)*(2*a*b + b^2))/(5*d) + (4*exp(2*c + 2*d*x)*(a*b - b^2))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x)
+ exp(6*c + 6*d*x) + 1) - ((2*(a*b - b^2))/(5*d) + (2*exp(2*c + 2*d*x)*(2*a*b + b^2))/(5*d))/(2*exp(2*c + 2*d
*x) + exp(4*c + 4*d*x) + 1) - ((2*exp(6*c + 6*d*x)*(2*a*b + b^2))/(5*d) - (2*(a*b + b^2))/(5*d) + (6*exp(4*c +
4*d*x)*(a*b - b^2))/(5*d) + (6*b^2*exp(2*c + 2*d*x))/(5*d))/(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(
6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - (2*a^2)/(d*(exp(2*c + 2*d*x) - 1)) - (2*(2*a*b + b^2))/(5*d*(exp(2*c +
2*d*x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**2*(a+b*tanh(d*x+c)**3)**2,x)

[Out]

Integral((a + b*tanh(c + d*x)**3)**2*csch(c + d*x)**2, x)

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