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

Optimal. Leaf size=53 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{d (a+b)^{3/2}}-\frac{\coth (c+d x)}{d (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.0709458, antiderivative size = 53, 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 = {4132, 325, 208} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{d (a+b)^{3/2}}-\frac{\coth (c+d x)}{d (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/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 208

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

Rubi steps

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

Mathematica [B]  time = 0.70113, size = 179, normalized size = 3.38 \[ \frac{\text{sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\sqrt{a+b} \text{csch}(c) \sinh (d x) \sqrt{b (\cosh (c)-\sinh (c))^4} \text{csch}(c+d x)+b (\cosh (2 c)-\sinh (2 c)) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )\right )}{2 d (a+b)^{3/2} \sqrt{b (\cosh (c)-\sinh (c))^4} \left (a+b \text{sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(b*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh
[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(Cosh[2*c] - Sinh[2*c]) + Sqrt[a +
b]*Csch[c]*Csch[c + d*x]*Sqrt[b*(Cosh[c] - Sinh[c])^4]*Sinh[d*x]))/(2*(a + b)^(3/2)*d*(a + b*Sech[c + d*x]^2)*
Sqrt[b*(Cosh[c] - Sinh[c])^4])

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Maple [B]  time = 0.061, size = 147, normalized size = 2.8 \begin{align*} -{\frac{1}{2\,d \left ( a+b \right ) }\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{1}{2\,d}\sqrt{b}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ) \left ( a+b \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{2\,d}\sqrt{b}\ln \left ( -\sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}-\sqrt{a+b} \right ) \left ( a+b \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{2\,d \left ( a+b \right ) } \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*sech(d*x+c)^2),x)

[Out]

-1/2/d/(a+b)*tanh(1/2*d*x+1/2*c)+1/2/d*b^(1/2)/(a+b)^(3/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x
+1/2*c)*b^(1/2)+(a+b)^(1/2))-1/2/d*b^(1/2)/(a+b)^(3/2)*ln(-(a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/
2*c)*b^(1/2)-(a+b)^(1/2))-1/2/d/(a+b)/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*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.80905, size = 1589, normalized size = 29.98 \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*sech(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 + b))*log((a^2*cosh(d*
x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*
a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a
*b)*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 + 3*a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*s
inh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x
+ c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) - 4)/((a + b)*d*cosh(d*x + c)^2 +
2*(a + b)*d*cosh(d*x + c)*sinh(d*x + c) + (a + b)*d*sinh(d*x + c)^2 - (a + b)*d), ((cosh(d*x + c)^2 + 2*cosh(
d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(-b/(a + b))*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x +
c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b) - 2)/((a + b)*d*cosh(d*x + c)^2 + 2*(a + b
)*d*cosh(d*x + c)*sinh(d*x + c) + (a + b)*d*sinh(d*x + c)^2 - (a + b)*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 \operatorname{sech}^{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*sech(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.2643, size = 107, normalized size = 2.02 \begin{align*} \frac{b \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}}{\left (a d + b d\right )}} - \frac{2}{{\left (a d + b d\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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