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

Optimal. Leaf size=48 \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{a^{3/2} d}-\frac{\coth (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.062818, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3663, 325, 205} \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{a^{3/2} d}-\frac{\coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.115079, size = 48, normalized size = 1. \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{a^{3/2} d}-\frac{\coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.076, size = 413, normalized size = 8.6 \begin{align*} -{\frac{1}{2\,da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{b}{d}{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }-a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{b \left ( a+b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }-a-2\,b \right ) a}}}}-{\frac{b}{da}{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }-a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }-a-2\,b \right ) a}}}}+{\frac{{b}^{2}}{da}{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }-a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{b \left ( a+b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }-a-2\,b \right ) a}}}}+{\frac{b}{d}\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }+a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{b \left ( a+b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }+a+2\,b \right ) a}}}}+{\frac{b}{da}\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }+a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }+a+2\,b \right ) a}}}}+{\frac{{b}^{2}}{da}\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }+a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{b \left ( a+b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{b \left ( a+b \right ) }+a+2\,b \right ) a}}}}-{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.20358, size = 1636, normalized size = 34.08 \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)^2/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/2*((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(-b/a)*log(((a^2 + 2*a*b + b
^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^
4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^
2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*((a^2
+ a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 - a*b)*
sqrt(-b/a))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(
a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 +
(a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) - 4)/(a*d*cosh(d*x + c)^2 + 2*a*d*cosh(d*x + c)*sinh(d*x + c) +
a*d*sinh(d*x + c)^2 - a*d), -((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(b/
a)*arctan(1/2*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a -
b)*sqrt(b/a)/b) + 2)/(a*d*cosh(d*x + c)^2 + 2*a*d*cosh(d*x + c)*sinh(d*x + c) + a*d*sinh(d*x + c)^2 - a*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{2}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.42002, size = 93, normalized size = 1.94 \begin{align*} -\frac{\frac{b \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt{a b}}\right )}{\sqrt{a b} a} + \frac{2}{a{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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