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]

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

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Rubi [A]  time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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 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]

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*}

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Mathematica [B]  time = 0.75, 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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fricas [B]  time = 0.56, size = 588, normalized size = 11.09 \[ \left [\frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \sqrt {\frac {b}{a + b}} \log \left (\frac {a^{2} \cosh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{2} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + a^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} \cosh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 4 \, {\left ({\left (a^{2} + a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{2} + a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} + a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {\frac {b}{a + b}}}{a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} + 2 \, {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a \cosh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a \cosh \left (d x + c\right )^{3} + {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a}\right ) - 4}{2 \, {\left ({\left (a + b\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} d \sinh \left (d x + c\right )^{2} - {\left (a + b\right )} d\right )}}, \frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \sqrt {-\frac {b}{a + b}} \arctan \left (\frac {{\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sqrt {-\frac {b}{a + b}}}{2 \, b}\right ) - 2}{{\left (a + b\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} d \sinh \left (d x + c\right )^{2} - {\left (a + b\right )} d}\right ] \]

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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giac [A]  time = 0.63, size = 75, normalized size = 1.42 \[ \frac {\frac {b \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} {\left (a + b\right )}} - \frac {2}{{\left (a + b\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}}{d} \]

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

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maple [B]  time = 0.30, size = 147, normalized size = 2.77 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \left (a +b \right )}-\frac {1}{2 d \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\sqrt {b}\, \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{2 d \left (a +b \right )^{\frac {3}{2}}}+\frac {\sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \left (a +b \right )^{\frac {3}{2}}} \]

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/(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)*tan
h(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2*c)-(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*b^(1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))

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maxima [B]  time = 0.43, size = 100, normalized size = 1.89 \[ -\frac {b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} {\left (a + b\right )} d} + \frac {2}{{\left ({\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} - a - b\right )} d} \]

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]

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

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mupad [B]  time = 2.47, size = 847, normalized size = 15.98 \[ -\frac {2}{\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a\,d+b\,d\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {2\,\left (8\,b^{5/2}\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}+8\,a\,b^{3/2}\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}+a^2\,\sqrt {b}\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}\right )\,\left (a^2+8\,a\,b+8\,b^2\right )}{a^5\,d\,{\left (a+b\right )}^3\,\left (a^2+2\,a\,b+b^2\right )\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}}+\frac {4\,\sqrt {b}\,\left (2\,a+4\,b\right )\,\left (4\,d\,a^3\,b+16\,d\,a^2\,b^2+20\,d\,a\,b^3+8\,d\,b^4\right )}{a^5\,\left (a+b\right )\,\sqrt {-d^2\,{\left (a+b\right )}^3}\,\left (a^2+2\,a\,b+b^2\right )\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}}\right )+\frac {2\,\left (2\,a\,b^{3/2}\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}+a^2\,\sqrt {b}\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}\right )\,\left (a^2+8\,a\,b+8\,b^2\right )}{a^5\,d\,{\left (a+b\right )}^3\,\left (a^2+2\,a\,b+b^2\right )\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}}+\frac {4\,\sqrt {b}\,\left (2\,a+4\,b\right )\,\left (2\,d\,a^3\,b+4\,d\,a^2\,b^2+2\,d\,a\,b^3\right )}{a^5\,\left (a+b\right )\,\sqrt {-d^2\,{\left (a+b\right )}^3}\,\left (a^2+2\,a\,b+b^2\right )\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}}\right )\,\left (a^5\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}+3\,a^4\,b\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}+a^2\,b^3\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}+3\,a^3\,b^2\,\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}\right )}{4\,b}\right )}{\sqrt {-a^3\,d^2-3\,a^2\,b\,d^2-3\,a\,b^2\,d^2-b^3\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)^2*(a + b/cosh(c + d*x)^2)),x)

[Out]

- 2/((exp(2*c + 2*d*x) - 1)*(a*d + b*d)) - (b^(1/2)*atan(((exp(2*c)*exp(2*d*x)*((2*(8*b^(5/2)*(- a^3*d^2 - b^3
*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2) + 8*a*b^(3/2)*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2)
+ a^2*b^(1/2)*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2))*(8*a*b + a^2 + 8*b^2))/(a^5*d*(a + b)^3
*(2*a*b + a^2 + b^2)*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2)) + (4*b^(1/2)*(2*a + 4*b)*(8*b^4*
d + 16*a^2*b^2*d + 20*a*b^3*d + 4*a^3*b*d))/(a^5*(a + b)*(-d^2*(a + b)^3)^(1/2)*(2*a*b + a^2 + b^2)*(- a^3*d^2
- b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2))) + (2*(2*a*b^(3/2)*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b
*d^2)^(1/2) + a^2*b^(1/2)*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2))*(8*a*b + a^2 + 8*b^2))/(a^5
*d*(a + b)^3*(2*a*b + a^2 + b^2)*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2)) + (4*b^(1/2)*(2*a +
4*b)*(4*a^2*b^2*d + 2*a*b^3*d + 2*a^3*b*d))/(a^5*(a + b)*(-d^2*(a + b)^3)^(1/2)*(2*a*b + a^2 + b^2)*(- a^3*d^2
- b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2)))*(a^5*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2) +
3*a^4*b*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2) + a^2*b^3*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2
- 3*a^2*b*d^2)^(1/2) + 3*a^3*b^2*(- a^3*d^2 - b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2)))/(4*b)))/(- a^3*d^2
- b^3*d^2 - 3*a*b^2*d^2 - 3*a^2*b*d^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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