\(\int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) [32]

Optimal result

Integrand size = 23, antiderivative size = 70 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\sqrt {b} (a+b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(a+b) \coth (c+d x)}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d} \]

[Out]

(a+b)*coth(d*x+c)/a^2/d-1/3*coth(d*x+c)^3/a/d+(a+b)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))*b^(1/2)/a^(5/2)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3744, 464, 331, 211} \[ \int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\sqrt {b} (a+b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(a+b) \coth (c+d x)}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d} \]

[In]

Int[Csch[c + d*x]^4/(a + b*Tanh[c + d*x]^2),x]

[Out]

(Sqrt[b]*(a + b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a^(5/2)*d) + ((a + b)*Coth[c + d*x])/(a^2*d) - Coth
[c + d*x]^3/(3*a*d)

Rule 211

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

Rule 331

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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 3744

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1-x^2}{x^4 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = -\frac {\coth ^3(c+d x)}{3 a d}-\frac {(a+b) \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{a d} \\ & = \frac {(a+b) \coth (c+d x)}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d}+\frac {(b (a+b)) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{a^2 d} \\ & = \frac {\sqrt {b} (a+b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(a+b) \coth (c+d x)}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {3 \sqrt {b} (a+b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )+\sqrt {a} \coth (c+d x) \left (2 a+3 b-a \text {csch}^2(c+d x)\right )}{3 a^{5/2} d} \]

[In]

Integrate[Csch[c + d*x]^4/(a + b*Tanh[c + d*x]^2),x]

[Out]

(3*Sqrt[b]*(a + b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]] + Sqrt[a]*Coth[c + d*x]*(2*a + 3*b - a*Csch[c + d*x
]^2))/(3*a^(5/2)*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(245\) vs. \(2(60)=120\).

Time = 1.57 (sec) , antiderivative size = 246, normalized size of antiderivative = 3.51

method	result	size
risch	\(-\frac {2 \left (-3 b \,{\mathrm e}^{4 d x +4 c}+6 \,{\mathrm e}^{2 d x +2 c} a +6 b \,{\mathrm e}^{2 d x +2 c}-2 a -3 b \right )}{3 d \,a^{2} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{2 a^{2} d}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right ) b}{2 a^{3} d}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{2 a^{2} d}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right ) b}{2 a^{3} d}\)	\(246\)
derivativedivides	\(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}-3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a -4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2}}-\frac {2 b \left (a +b \right ) \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{a}-\frac {1}{24 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-3 a -4 b}{8 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(247\)
default	\(\frac {-\frac {\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}-3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a -4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2}}-\frac {2 b \left (a +b \right ) \left (\frac {\left (a +\sqrt {\left (a +b \right ) b}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}+a +2 b \right ) a}}-\frac {\left (-a +\sqrt {\left (a +b \right ) b}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{2 a \sqrt {\left (a +b \right ) b}\, \sqrt {\left (2 \sqrt {\left (a +b \right ) b}-a -2 b \right ) a}}\right )}{a}-\frac {1}{24 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-3 a -4 b}{8 a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\)	\(247\)

[In]

int(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(-3*b*exp(4*d*x+4*c)+6*exp(2*d*x+2*c)*a+6*b*exp(2*d*x+2*c)-2*a-3*b)/d/a^2/(exp(2*d*x+2*c)-1)^3+1/2/a^2*(-
a*b)^(1/2)/d*ln(exp(2*d*x+2*c)+(2*(-a*b)^(1/2)+a-b)/(a+b))+1/2/a^3*(-a*b)^(1/2)/d*ln(exp(2*d*x+2*c)+(2*(-a*b)^
(1/2)+a-b)/(a+b))*b-1/2/a^2*(-a*b)^(1/2)/d*ln(exp(2*d*x+2*c)-(2*(-a*b)^(1/2)-a+b)/(a+b))-1/2/a^3*(-a*b)^(1/2)/
d*ln(exp(2*d*x+2*c)-(2*(-a*b)^(1/2)-a+b)/(a+b))*b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 653 vs. \(2 (60) = 120\).

Time = 0.28 (sec) , antiderivative size = 1628, normalized size of antiderivative = 23.26 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/6*(12*b*cosh(d*x + c)^4 + 48*b*cosh(d*x + c)*sinh(d*x + c)^3 + 12*b*sinh(d*x + c)^4 - 24*(a + b)*cosh(d*x +
c)^2 + 24*(3*b*cosh(d*x + c)^2 - a - b)*sinh(d*x + c)^2 + 3*((a + b)*cosh(d*x + c)^6 + 6*(a + b)*cosh(d*x + c
)*sinh(d*x + c)^5 + (a + b)*sinh(d*x + c)^6 - 3*(a + b)*cosh(d*x + c)^4 + 3*(5*(a + b)*cosh(d*x + c)^2 - a - b
)*sinh(d*x + c)^4 + 4*(5*(a + b)*cosh(d*x + c)^3 - 3*(a + b)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(a + b)*cosh(d
*x + c)^2 + 3*(5*(a + b)*cosh(d*x + c)^4 - 6*(a + b)*cosh(d*x + c)^2 + a + b)*sinh(d*x + c)^2 + 6*((a + b)*cos
h(d*x + c)^5 - 2*(a + b)*cosh(d*x + c)^3 + (a + b)*cosh(d*x + c))*sinh(d*x + c) - a - b)*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)*sin
h(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)) + 48*(b*cosh(d*x + c)^3 - (a + b)*cosh(d*x + c))*sin
h(d*x + c) + 8*a + 12*b)/(a^2*d*cosh(d*x + c)^6 + 6*a^2*d*cosh(d*x + c)*sinh(d*x + c)^5 + a^2*d*sinh(d*x + c)^
6 - 3*a^2*d*cosh(d*x + c)^4 + 3*a^2*d*cosh(d*x + c)^2 + 3*(5*a^2*d*cosh(d*x + c)^2 - a^2*d)*sinh(d*x + c)^4 +
4*(5*a^2*d*cosh(d*x + c)^3 - 3*a^2*d*cosh(d*x + c))*sinh(d*x + c)^3 - a^2*d + 3*(5*a^2*d*cosh(d*x + c)^4 - 6*a
^2*d*cosh(d*x + c)^2 + a^2*d)*sinh(d*x + c)^2 + 6*(a^2*d*cosh(d*x + c)^5 - 2*a^2*d*cosh(d*x + c)^3 + a^2*d*cos
h(d*x + c))*sinh(d*x + c)), 1/3*(6*b*cosh(d*x + c)^4 + 24*b*cosh(d*x + c)*sinh(d*x + c)^3 + 6*b*sinh(d*x + c)^
4 - 12*(a + b)*cosh(d*x + c)^2 + 12*(3*b*cosh(d*x + c)^2 - a - b)*sinh(d*x + c)^2 + 3*((a + b)*cosh(d*x + c)^6
+ 6*(a + b)*cosh(d*x + c)*sinh(d*x + c)^5 + (a + b)*sinh(d*x + c)^6 - 3*(a + b)*cosh(d*x + c)^4 + 3*(5*(a + b
)*cosh(d*x + c)^2 - a - b)*sinh(d*x + c)^4 + 4*(5*(a + b)*cosh(d*x + c)^3 - 3*(a + b)*cosh(d*x + c))*sinh(d*x
+ c)^3 + 3*(a + b)*cosh(d*x + c)^2 + 3*(5*(a + b)*cosh(d*x + c)^4 - 6*(a + b)*cosh(d*x + c)^2 + a + b)*sinh(d*
x + c)^2 + 6*((a + b)*cosh(d*x + c)^5 - 2*(a + b)*cosh(d*x + c)^3 + (a + b)*cosh(d*x + c))*sinh(d*x + c) - a -
b)*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) + 24*(b*cosh(d*x + c)^3 - (a + b)*cosh(d*x + c))*sinh(d*x + c) + 4*a + 6*b)/(a^2*d
*cosh(d*x + c)^6 + 6*a^2*d*cosh(d*x + c)*sinh(d*x + c)^5 + a^2*d*sinh(d*x + c)^6 - 3*a^2*d*cosh(d*x + c)^4 + 3
*a^2*d*cosh(d*x + c)^2 + 3*(5*a^2*d*cosh(d*x + c)^2 - a^2*d)*sinh(d*x + c)^4 + 4*(5*a^2*d*cosh(d*x + c)^3 - 3*
a^2*d*cosh(d*x + c))*sinh(d*x + c)^3 - a^2*d + 3*(5*a^2*d*cosh(d*x + c)^4 - 6*a^2*d*cosh(d*x + c)^2 + a^2*d)*s
inh(d*x + c)^2 + 6*(a^2*d*cosh(d*x + c)^5 - 2*a^2*d*cosh(d*x + c)^3 + a^2*d*cosh(d*x + c))*sinh(d*x + c))]

Sympy [F]

\[ \int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int \frac {\operatorname {csch}^{4}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \]

[In]

integrate(csch(d*x+c)**4/(a+b*tanh(d*x+c)**2),x)

[Out]

Integral(csch(c + d*x)**4/(a + b*tanh(c + d*x)**2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (60) = 120\).

Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.91 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {2 \, {\left (6 \, {\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, b e^{\left (-4 \, d x - 4 \, c\right )} - 2 \, a - 3 \, b\right )}}{3 \, {\left (3 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, a^{2} e^{\left (-4 \, d x - 4 \, c\right )} + a^{2} e^{\left (-6 \, d x - 6 \, c\right )} - a^{2}\right )} d} - \frac {{\left (a b + b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{\sqrt {a b} a^{2} d} \]

[In]

integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2),x, algorithm="maxima")

[Out]

2/3*(6*(a + b)*e^(-2*d*x - 2*c) - 3*b*e^(-4*d*x - 4*c) - 2*a - 3*b)/((3*a^2*e^(-2*d*x - 2*c) - 3*a^2*e^(-4*d*x
- 4*c) + a^2*e^(-6*d*x - 6*c) - a^2)*d) - (a*b + b^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b)
)/(sqrt(a*b)*a^2*d)

Giac [F]

\[ \int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int { \frac {\operatorname {csch}\left (d x + c\right )^{4}}{b \tanh \left (d x + c\right )^{2} + a} \,d x } \]

[In]

integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 2.28 (sec) , antiderivative size = 254, normalized size of antiderivative = 3.63 \[ \int \frac {\text {csch}^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {2\,b}{a^2\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {8}{3\,a\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {4}{a\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {\sqrt {-b}\,\ln \left (-\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{a^2}-\frac {2\,\sqrt {-b}\,\left (a\,d+b\,d+a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}-b\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^{5/2}\,d}\right )\,\left (a+b\right )}{2\,a^{5/2}\,d}-\frac {\sqrt {-b}\,\ln \left (\frac {2\,\sqrt {-b}\,\left (a\,d+b\,d+a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}-b\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^{5/2}\,d}-\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{a^2}\right )\,\left (a+b\right )}{2\,a^{5/2}\,d} \]

[In]

int(1/(sinh(c + d*x)^4*(a + b*tanh(c + d*x)^2)),x)

[Out]

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