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

Optimal. Leaf size=49 \[ -\frac{a \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b \tan ^{-1}(\sinh (c+d x))}{2 d}-\frac{b \tanh (c+d x) \text{sech}(c+d x)}{2 d} \]

[Out]

(b*ArcTan[Sinh[c + d*x]])/(2*d) - (a*ArcTanh[Cosh[c + d*x]])/d - (b*Sech[c + d*x]*Tanh[c + d*x])/(2*d)

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Rubi [A]  time = 0.0743924, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3666, 3770, 2611} \[ -\frac{a \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b \tan ^{-1}(\sinh (c+d x))}{2 d}-\frac{b \tanh (c+d x) \text{sech}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTan[Sinh[c + d*x]])/(2*d) - (a*ArcTanh[Cosh[c + d*x]])/d - (b*Sech[c + d*x]*Tanh[c + d*x])/(2*d)

Rule 3666

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> Int[ExpandTrig[(d*sin[e + f*x])^m*(a + b*(c*tan[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && IGtQ[p, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rubi steps

\begin{align*} \int \text{csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx &=i \int \left (-i a \text{csch}(c+d x)-i b \text{sech}(c+d x) \tanh ^2(c+d x)\right ) \, dx\\ &=a \int \text{csch}(c+d x) \, dx+b \int \text{sech}(c+d x) \tanh ^2(c+d x) \, dx\\ &=-\frac{a \tanh ^{-1}(\cosh (c+d x))}{d}-\frac{b \text{sech}(c+d x) \tanh (c+d x)}{2 d}+\frac{1}{2} b \int \text{sech}(c+d x) \, dx\\ &=\frac{b \tan ^{-1}(\sinh (c+d x))}{2 d}-\frac{a \tanh ^{-1}(\cosh (c+d x))}{d}-\frac{b \text{sech}(c+d x) \tanh (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0311553, size = 75, normalized size = 1.53 \[ \frac{a \log \left (\sinh \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d}-\frac{a \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d}+\frac{b \tan ^{-1}(\sinh (c+d x))}{2 d}-\frac{b \tanh (c+d x) \text{sech}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTan[Sinh[c + d*x]])/(2*d) - (a*Log[Cosh[c/2 + (d*x)/2]])/d + (a*Log[Sinh[c/2 + (d*x)/2]])/d - (b*Sech[c
+ d*x]*Tanh[c + d*x])/(2*d)

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Maple [A]  time = 0.049, size = 65, normalized size = 1.3 \begin{align*} -2\,{\frac{a{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) }{d}}-{\frac{b\sinh \left ( dx+c \right ) }{d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+{\frac{b{\rm sech} \left (dx+c\right )\tanh \left ( dx+c \right ) }{2\,d}}+{\frac{b\arctan \left ({{\rm e}^{dx+c}} \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*a*arctanh(exp(d*x+c))-1/d*b*sinh(d*x+c)/cosh(d*x+c)^2+1/2*b*sech(d*x+c)*tanh(d*x+c)/d+1/d*b*arctan(exp(d*
x+c))

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Maxima [A]  time = 1.51615, size = 112, normalized size = 2.29 \begin{align*} -b{\left (\frac{\arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac{e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac{a \log \left (\tanh \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.63424, size = 1453, normalized size = 29.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*cosh(d*x + c)^3 + 3*b*cosh(d*x + c)*sinh(d*x + c)^2 + b*sinh(d*x + c)^3 - (b*cosh(d*x + c)^4 + 4*b*cosh(d*
x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*b*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + b)*sinh(d*x + c)^2
+ 4*(b*cosh(d*x + c)^3 + b*cosh(d*x + c))*sinh(d*x + c) + b)*arctan(cosh(d*x + c) + sinh(d*x + c)) - b*cosh(d
*x + c) + (a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*a*cosh(d*x + c)^2 + 2
*(3*a*cosh(d*x + c)^2 + a)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) + a)*log(co
sh(d*x + c) + sinh(d*x + c) + 1) - (a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4
+ 2*a*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + a*cosh(d*x + c))*
sinh(d*x + c) + a)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + (3*b*cosh(d*x + c)^2 - b)*sinh(d*x + c))/(d*cosh(d
*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)
^2 + d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right ) \operatorname{csch}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21699, size = 100, normalized size = 2.04 \begin{align*} \frac{b \arctan \left (e^{\left (d x + c\right )}\right ) - a \log \left (e^{\left (d x + c\right )} + 1\right ) + a \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right ) - \frac{b e^{\left (3 \, d x + 3 \, c\right )} - b e^{\left (d x + c\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*arctan(e^(d*x + c)) - a*log(e^(d*x + c) + 1) + a*log(abs(e^(d*x + c) - 1)) - (b*e^(3*d*x + 3*c) - b*e^(d*x
+ c))/(e^(2*d*x + 2*c) + 1)^2)/d